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Class 12 Physics Assignments

We have provided below free printable Class 12 Physics Assignments for Download in PDF. The Assignments have been designed based on the latest NCERT Book for Class 12 Physics . These Assignments for Grade 12 Physics cover all important topics which can come in your standard 12 tests and examinations. Free printable Assignments for CBSE Class 12 Physics , school and class assignments, and practice test papers have been designed by our highly experienced class 12 faculty. You can free download CBSE NCERT printable Assignments for Physics Class 12 with solutions and answers. All Assignments and test sheets have been prepared by expert teachers as per the latest Syllabus in Physics Class 12. Students can click on the links below and download all Pdf Assignments for Physics class 12 for free. All latest Kendriya Vidyalaya Class 12 Physics Assignments with Answers and test papers are given below.

Physics Class 12 Assignments Pdf Download

We have provided below the biggest collection of free CBSE NCERT KVS Assignments for Class 12 Physics . Students and teachers can download and save all free Physics assignments in Pdf for grade 12th. Our expert faculty have covered Class 12 important questions and answers for Physics as per the latest syllabus for the current academic year. All test papers and question banks for Class 12 Physics and CBSE Assignments for Physics Class 12 will be really helpful for standard 12th students to prepare for the class tests and school examinations. Class 12th students can easily free download in Pdf all printable practice worksheets given below.

Topicwise Assignments for Class 12 Physics Download in Pdf

Advantages of Class 12 Physics Assignments

• As we have the best and largest collection of Physics assignments for Grade 12, you will be able to easily get full list of solved important questions which can come in your examinations.
• Students will be able to go through all important and critical topics given in your CBSE Physics textbooks for Class 12 .
• All Physics assignments for Class 12 have been designed with answers. Students should solve them yourself and then compare with the solutions provided by us.
• Class 12 Students studying in per CBSE, NCERT and KVS schools will be able to free download all Physics chapter wise worksheets and assignments for free in Pdf
• Class 12 Physics question bank will help to improve subject understanding which will help to get better rank in exams

Frequently Asked Questions by Class 12 Physics students

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We provide here Standard 12 Physics chapter-wise assignments which can be easily downloaded in Pdf format for free.

You can click on the links above and get assignments for Physics in Grade 12, all topic-wise question banks with solutions have been provided here. You can click on the links to download in Pdf.

We have provided here topic-wise Physics Grade 12 question banks, revision notes and questions for all difficult topics, and other study material.

We have provided the best collection of question bank and practice tests for Class 12 for all subjects. You can download them all and use them offline without the internet.

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NCERT Solutions for Class 12 Physics

NCERT Solutions for Class 12 Physics in English and Hindi Medium PDF file format is for free download updated for new session 2024-25 based on latest NCERT Books. As per the new textbooks issued for academic year 2024-25, the additional exercises are not included in curriculum for exams. Class 12 Physics Solutions in English Medium Chapter 1. Electric Charges and Fields Chapter 2. Electrostatic Potential and Capacitance Chapter 3. Current Electricity Chapter 4. Moving Charges and Magnetism Chapter 5. Magnetism and Matter Chapter 6. Electromagnetic Induction Chapter 7. Alternating Current Chapter 8. Electromagnetic Waves Chapter 9. Ray Optics and Optical Instruments Chapter 10. Wave Optics Chapter 11. Dual Nature of Radiation and Matter Chapter 12. Atoms Chapter 13. Nuclei Chapter 14. Semiconductor Electronics Class 12 Physics Solutions in Hindi Medium

UP Board Students for Class 12 Physics can also download from here. Visit to discussion forum to ask your doubts. CBSE Solutions Apps as well as NCERT Solutions and their answers, solutions of additional exercises, intext questions, back exercises questions with assignments from popular books like S L Arora, Concepts of Physics by H C Verma, Pradeep’s fundamental physics, A B C Physics, Arihant publications books, Full Marks question bank etc.

NCERT Solutions for Class 12 Physics in PDF format is given below to free download for new academic session 2024-25 based on latest CBSE Syllabus. For the regular preparation for CBSE, IIT – JEE Mains and Advance, NEET, BITSAT, GGSIPU use latest NCERT books available in the market. A few questions related to these books are given below. Ask your doubts through Discussion Forum about CBSE or NIOS boards.

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Useful Resources in Physics

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• Special Constants Values

Chapter 1: Electric Charges and Fields 1. Does the charge given to a metallic sphere depend on whether it is hollow or solid? Give reason for your answer. [Delhi 2017] Chapter 2: Electrostatic Potential and Capacitance 1. A 12 pF capacitor is connected to a 50 v battery. How much electrostatic energy is stored in the capacitor? If another capacitor of 6 pf is connected in series with it with the same battery connected across the combination, find the charge stored and potential difference across each capacitor. [Delhi 2017] 2. Derive the expression for the electric potential due to an electric dipole at a point on its axial line. Depict the equipotential surfaces due to an electric dipole. [Delhi 2017]

Chapter 3: Current Electricity 1. Derive an expression for drift velocity of electrons in a conductor. Hence deduce Ohm’s law. A wire whose cross sectional area is increasing linearly from its one end to the other, is connected across a battery of V volts. Which of the following quantities remain constant in the wire? (a) drift velocity (b) current density (c) electric current (d) electric field. Justify your answer. [Delhi 2017] 2. State the two Kirchhoff’s laws. Explain briefly how these rules are justified. The current is drawn from a cell of emf E and internal resistance r connected to the network of resistors each of resistance r as shown in the figure. Obtain the expression for (i) the current draw from the cell and (ii) the power consumed in the network. [Delhi 2017]

3. A resistance of R draws current from a potentiometer. The potentiometer wire, AB, has a total resistance of R0. A voltage V is supplied to the potentiometer. Derive an expression for the voltage across R when the sliding contact is in the middle of potentiometer wire. [Delhi 2017]

Chapter 4: Moving Charges and Magnetism 1. Describe the working principle of a moving coil galvanometer. Why is it necessary to use (i) a radial magnetic field and (ii) a cylindrical soft iron core in a galvanometer? Write the expression for current sensitivity of the galvanometer. Can a galvanometer as such be used for measuring the current? Explain. [Delhi 2017] 2. An electron of mass m revolves around a nucleus of charge +Ze. Show that it behaves like a tiny magnetic dipole. Hence prove that the magnetic moment associated with it is expressed as µ = (e/2m)L, where L is the orbital angular momentum of the electron. Give the significance of negative sign. [Delhi 2017]

CHAPTER 5: MAGNETISM AND MATTER 1. At a place, the horizontal component of Earth’s magnetic field is B and angle of dip is 60. What is the value of horizontal component of the Earth’s magnetic field at equator? [Delhi 2017] Chapter 6: Electromagnetic Induction 1. A long straight current carrying wire passed normally through the centre of circular loop. If the current through the wire increases, will there be an increase induced emf in the loop? Justify. [Delhi 2017] 2. Define the term ‘Self Inductance’ and write its S.I. unit. Obtain the expression for the mutual inductance of two long co-axial solenoids S1 and S2 wound one over the other, each of length L and radii r1 and r2 and n1 and n2 number of terns per unit length, when a current I is set up in the outer solenoid S2. [Delhi 2017] 3. Draw a labelled diagram of AC generator. Derive the expression for the instantaneous value of the emf induced in the coil. A circular coil of cross-sectional area 200 sq. cm and 20 terns is rotated about the vertical diameter with angular speed of 50 rad/s in a uniform magnetic field of magnitude 3.0 × 10^-2 T. Calculate the maximum value of the current in the coil. [Delhi 2017]

Chapter 7: Alternating Current 1. Draw a labelled diagram of a step-up transformer. Obtain the ratio of secondary to primary voltage in terms of number of turns and currents in the two coils. A power transmission line feeds input power at 2200 V to a step-down transformer with its primary windings having 3000 turns. Find the number of turns in the secondary to get the power output at 220 V. [Delhi 2017] 2. Find the value of the phase difference between the current and the voltage in the series LCR circuit shown below. Which one leads in phase: current or voltage? Without making any other change, find the value of the additional capacitor C1, to be connected in parallel with the capacitor C, in order to make the power factor of the circuit unity. [Delhi 2017]

Chapter 8: Electromagnetic Waves 1. How is the speed of em-waves in vacuum determined by the electric and magnetic fields? [Delhi 2017] 2. How does Ampere-Maxwell law explain the flow of current through a capacitor when it is being charged by a battery? Write the expression for the displacement current in terms of the rate of change of electric flux. [Delhi 2017] Chapter 9: Ray Optics and Optical Instruments 1. A ray of light incident on face AB of an equilateral glass prism, shows minimum deviation of 30. Calculate the speed of light through the prism. Find the angle of incident at face AB so that the emergent ray grazes along the face AC. [Delhi 2017]

2. Mrs. Rashmi Singh broke her reading glasses. When she went to the shopkeeper to order new spects, he suggested that she should get spectacles with plastic lenses instead of glass lenses. On getting the new spectacles, she found that the new ones could not offer satisfactory explanation for this. At home, Mrs. Singh raised the same question to her daughter Anuja who explained why plastic lenses were thicker. (a) Write the two qualities displayed each by Anuja and her mother. (b) How do you explain this fact using lens maker’s formula? [Delhi 2017]

Chapter 10: Wave Optics 1. Why should the objective of a telescope have large focal length and large aperture? Justify your answer. [Delhi 2017] 2. Distinguish between unpolarised light and linearly polarised light. How does one get linearly polarised light with the help of a Polaroid? A narrow beam of unpolarised light of intensity I0 is incident on a Polaroid P1. The light transmitted by it is then incident on a second Polaroid P2 with its pass axis making angle of 60 relative to the pass axis of P1. Find the intensity of the light transmitted by P2. [Delhi 2017] 3. Explain two features to distinguish between the interference patterns in Young’s double slit experiment with the diffraction pattern obtained due to a single slit. A monochromatic light of wavelength 500 nm is incident normally on a single slit of width 0.2 mm to produce a diffraction pattern. Find the angular width of the central maximum obtained on the screen. Estimate the number of fringes obtained in Young’s double slit experiment with fringe width 0.5 mm, which can be accommodated within the region of total angular spread of the central maximum due to single slit. [Delhi 2017] Chapter 11: Dual Nature of Radiation and Matter 1. In the study of photoelectric effect the graph between the stopping potential V and frequency v of the incident radiation on two different metals P and Q is shown below:

(i) Which one of the two metals has higher threshold frequency? (ii) Determine the work function of the metal which has greater value. (iii) Find the maximum kinetic energy of electron emitted by light of frequency 8 × 10^14 Hz for this metal. [Delhi 2017]

Chapter 12: Atoms 1. Find the wavelength of the electron orbiting in the first exited state in hydrogen atom. [Delhi 2017] 2. Define the distance of closest approach. An α-particle of kinetic energy ‘K’ is bombarded on a thin gold foil. The distance of the closest approach is ‘r’. What will be the distance of closest approach for an α-particle of double the kinetic energy? [Delhi 2017] 3. Write two important limitations of Rutherford nuclear model of the atom. [Delhi 2017] Chapter 13: Nuclei 1. A radioactive nucleus ‘A’ undergoes a series of decays as given below:

The mass number and atomic number of A2 are 176 and 71 respectively. Determine the mass and atomic number of A4 and A. Write the basic nuclear processes underlying β+ and β- decays. [Delhi 2017]

Chapter 14: Semiconductor Electronics: Materials, Devises Simple Circuits 1. Name the junction diode whose I-V characteristics are drawn below:

For a CE-transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 2V. Given the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1 kΩ. [Delhi 2017] 2. Zener diode is fabricated by heavily doping both p- and n- sides of the junction. Explain, why? Briefly explain the use of Zener diode as a dc voltage regulator with the help of a circuit diagram. [Delhi 2017]

Chapter 15: Communication System 1. Distinguish between a transducer and a repeater. [Delhi 2017] Define the term ‘amplitude modulation’. Explain any two factors which justify the need for modulating a low frequency base-band signal. [Delhi 2017]

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Physics Class 12

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Do you want to continue your journey of learning physics and explore the advanced topics and applications of physics? Do you want to understand the concepts and phenomena of electromagnetism, optics, modern physics and more? Do you want to prepare yourself for the board exams, competitive exams and higher studies in physics and related fields? If you answered yes to any of these questions, then this course is for you! 🙌

In this course, you will learn the details and techniques of physics for class 12. You will learn how to:

• Use the concepts of  electromagnetic induction ,  alternating current ,  electromagnetic waves  and  ray optics  to understand the generation, transmission and manipulation of electromagnetic radiation 🌈
• Apply the concepts of  wave optics ,  dual nature of radiation and matter ,  atoms ,  nuclei  and  semiconductors  to explore the phenomena and devices of modern physics 🌟
• Solve numerical problems and derivations using the formulas and principles of physics 📝
• Perform experiments and observations using the apparatus and methods of physics 🛠️
• Analyze and interpret the data and results obtained from the experiments and problems 📊
• Prepare for the board exams, competitive exams and higher studies in physics and related fields 🚀

By the end of this course, you will have a comprehensive understanding of the advanced topics and applications of physics. You will also be able to apply your physics knowledge and skills to various domains such as engineering, medicine, astronomy, nanotechnology and more. 😍

This course is designed for students who have completed class 11 physics or equivalent, but want to learn more about physics and its applications. It is also suitable for anyone who is interested in learning more about physics and its wonders. 🚀

Solutions to CBSE Sample Paper - Physics Class 12

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CBSE Class 12 Physics 2022-23 Chapter-Wise Questions With Solutions

Cbse 12th physics 2023 chapter-wise question bank with solution.

This Post Contains CBSE Class 12 Physics Chapter-Wise Solved Questions. In this post all of you can find Most Likely CBSE Question Bank for Class 12 Physics For Board Exam 2022-23 . This Post Consists of Chapter-Wise and also Category-wise Questions and Answers as per the Latest CBSE Syllabus issued for session 2022-23.

Here we provide Chapter-wise Question Bank in-depth knowledge of different concept of physics questions with solutions and their weightage also to prepare for Class 12th CBSE Board Exam 2022-23 .

Key features of study materials for CBSE Class 12 are as

• Focussed on New CBSE Paper Pattern Question and Answers
• Chapter Summary for each chapters for Easy and Quick Revision for students
• Each chapter Includes MCQs, Case-based, Assertion & Reasoning based, Very Short/Short/Long Answer Type Questions with solutions
• Self Assessment Test for practicing for students

All Students need to create vision boards to establish best study schedules, and maintain daily study logs to measure their progress on monthly basis.

Importance Of NCERT Class 12 Physics Revision Notes and Study Material

These questions with solutions are very helpful to the students for the good preparation for their  Class 12 Physics   Board examinations . So, every student must try to study in very good manner with the help of these study materials and also try to solve given questions with the help of given Solutions and Answers.

These CBSE Class 12 Physics  study materials and revision notes are prepared by our expert team which are on the basis of latest  CBSE  exam pattern for session 2022-23. These notes are very helpful for the preparing final Board examinations 2022-23.

Benefit Of CBSE Class 12 Physics Good Study Material and Quick Revision Notes

There are many important benefits of good study materials and quick revision notes for  CBSE Class 12 Physics  which are listed below

* Student can easily understand the pattern of  CBSE   Class 12 Physics question papers.

* Student can prepared for their final board examination 2022-23 in very short time.

* Good study materials helps to the student for the better preparation for their board examination 2022-23.

* Study material helps to the student to enhance their knowledge for Board Exam 2022-23.

* With the help of a variety of questions and important quick revision notes student can clear their doubt and enhance their knowledge for final Board Exam 2022-23.

* Quick Revision notes are very helpful in the last time revision for their board examination 2022-23 in very short time.

If you are preparing for any other Board examinations or any competitive examinations like  JEE ,  NEET ,  GATE  and  NTSE  etc, then this website is very helpful to the student. 

Maths And Physics With Pandey Sir  is a very dedicated website for the students which always provide a good class notes, revision notes, important questions with solution/answer, case based questions, sample papers etc for their students, so that they can prepared for their Board examinations or any other competitive examination.

Given Below Are The CBSE Class 12 Physics Chapters Name With Their Respective Download Links Containing Study Materials.

NCERT Class 12 Physics Chapter-Wise Notes

NCERT Class 12 Physics Chapter-1 (Electric Charges And Fields) Note s

NCERT Class 12 Physics Chapter-2 (Electrostatic Potential And Capacitance) Note s

NCERT Class 12 Physics Chapter-3 (Curren t Electricity ) Notes

NCERT Class 12 Physics Chapter-4 (Moving Charges And Magnetism) Notes

NCERT Class 12 Physics Chapter-5 (Magnetis m and Matter ) Notes

NCERT Class 12 Physics Chapter-6 (Electromagnetic Induction ) Notes

NCERT Class 12 Physics Chapter-7 (Alternat ing Current ) Notes

NCERT Class 12 Physics Chapter-8 (Electromagnetic Waves) Notes

NCERT Class 12 Physics Chapter-9 (Ray Optic s and Optical Instruments ) Notes

NCERT Class 12 Physics Chapter-10 (Wave Optics) Notes

NCERT Class 12 Physics Chapter-11 (Dual Natur e of Radiation ) Notes

NCERT Class 12 Physics Chapter-12 (Atoms) Notes

NCERT Class 12 Physics Chapter-13 (Nuclei) Notes

NCERT Class 12 Physics Chapter-14 (Semiconductor and Electronic Devices) Notes

NCERT Class 12 Physics Book PDF Free Download

SL Arora Class 12 Physics Book PDF Free Download

All In One Arihant Class 12 Physics Book PDF Free Download

NCERT Solutions for Physics Class 12

Cbse, karnataka board puc class 12 physics solutions guide.

Shaalaa.com provides the CBSE, Karnataka Board PUC Class 12 Physics Solutions Digest. Shaalaa is undoubtedly a site that most of your classmates are using to perform well in exams.

You can solve the Class 12 Physics Book Solutions CBSE, Karnataka Board PUC textbook questions by using Shaalaa.com to verify your answers, which will help you practise better and become more confident.

CBSE, Karnataka Board PUC Class 12 Physics Textbook Solutions

Questions and answers for the Class 12 Physics Textbook are on this page. NCERT Solutions for Class 12 Physics Digest CBSE, Karnataka Board PUC will help students understand the concepts better.

NCERT Solutions for Class 12 Physics Chapterwise List | Class 12 Physics Digest

The answers to the NCERT books are the best study material for students. Listed below are the chapter-wise NCERT Physics Class 12 Solutions CBSE, Karnataka Board PUC.

•  • Chapter 1: Electric Charge and Fields
•  • Chapter 2: Electrostatic Potential and Capacitance
•  • Chapter 3: Current Electricity
•  • Chapter 4: Moving Charges and Magnetism
•  • Chapter 5: Magnetism and Matter
•  • Chapter 6: Electromagnetic Induction
•  • Chapter 7: Alternating Current
•  • Chapter 8: Electromagnetic Waves
•  • Chapter 9: Ray Optics and Optical Instruments
•  • Chapter 10: Wave Optics
•  • Chapter 11: Dual Nature of Radiation and Matter
•  • Chapter 12: Atoms
•  • Chapter 13: Nuclei
•  • Chapter 14: Semiconductor Electronics: Materials, Devices and Simple Circuits
•  • Chapter 15: Communication Systems
• Science (English Medium) Class 12 CBSE
• PUC Science 2nd PUC Class 12 Karnataka Board PUC

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NCERT Class 12 solutions for other subjects

• NCERT solutions for Accountancy - Company Accounts and Analysis of Financial Statements Class 12 CBSE [लेखाशास्त्र - कंपनी खाते एवं वित्तीय विवरणों का विश्लेषण १२ वीं कक्षा]
• NCERT solutions for Accountancy - Not-for-Profit Organisation and Partnership Accounts Class 12 CBSE [लेखाशास्त्र - अलाभकारी संस्थाएँ एवं साझेदारी खाते १२ वीं कक्षा]
• NCERT solutions for Biology Class 12
• NCERT solutions for Biology Class 12 CBSE [जीव विज्ञान १२ वीं कक्षा]
• NCERT solutions for Chemistry Part 1 and 2 Class 12
• NCERT solutions for Chemistry Class 12 CBSE [रसायन विज्ञान १२ वीं कक्षा]
• NCERT solutions for Class 12 Accountancy - Company Accounts and Analysis of Financial Statements
• NCERT solutions for Class 12 Accountancy - Not-for-Profit Organisation and Partnership Accounts
• NCERT solutions for Business Studies Part - 1 Principles and Functions of Management for Class 12
• NCERT solutions for Business Studies Part - 2 Business Finance and Marketing for Class 12
• NCERT solutions for Computer Science Class 12
• NCERT solutions for Economics - Indian Economic Development Class 12
• NCERT solutions for Class 12 Economics - Introductory Macroeconomics
• NCERT solutions for Flamingo - Textbook in English (Core Course) for Class 12
• NCERT solutions for Vistas (Core Course) - Supplementary Reader in English for Class 12
• NCERT solutions for Geography - Fundamentals of Human Geography Class 12 CBSE
• NCERT solutions for Geography - Fundamentals of Human Geography Class 12 CBSE [भूगोल - मानव भूगोल के मूल सिद्धांत १२ वीं कक्षा]
• NCERT solutions for Geography - India: People and Economy Class 12 CBSE
• NCERT solutions for Geography - India: People and Economy Class 12 CBSE [भूगोल - भारत : लोग और अर्थव्यवस्था १२ वीं कक्षा]
• NCERT solutions for Geography - Practical Work in Geography Class 12 CBSE
• NCERT solutions for Geography - Practical Work in Geography Class 12 CBSE [भूगोल - भूगोल में प्रयोगात्मक कार्य १२ वीं कक्षा]
• NCERT solutions for Hindi - Aaroh Class 12 CBSE [हिंदी - आरोह १२ वीं कक्षा]
• NCERT solutions for Hindi - Antaraal Class 12 CBSE [हिंदी - अंतराल १२ वीं कक्षा]
• NCERT solutions for Hindi - Antara Class 12 CBSE [हिंदी - अंतरा १२ वीं कक्षा]
• NCERT solutions for Hindi - Vitaan Class 12 CBSE [हिंदी - वितान १२ वीं कक्षा]
• NCERT solutions for Class 12 History - Themes in Indian History
• NCERT solutions for Class 12 Political Science - Contemporary World Politics
• NCERT solutions for Class 12 Political Science: Politics in India Since Independence
• NCERT solutions for Class 12 Psychology Textbook
• NCERT solutions for Class 12 Sociology - Indian Society
• NCERT solutions for Class 12 Sociology - Social Change and Development in India
• NCERT solutions for Economics Class 12 [अर्थशास्त्र - भारतीय अर्थव्यवस्था का विकास १२ वीं कक्षा]
• NCERT solutions for Economics - Introductory Macroeconomics Class 12 CBSE [अर्थशास्त्र - समष्टि अर्थशास्त्र एक परिचय १२ वीं कक्षा]
• NCERT solutions for English - Kaleidoscope Class 12 CBSE
• NCERT solutions for History Class 12 [भारतीय इतिहास के कुछ विषय १२ वीं कक्षा]
• NCERT solutions for Mathematics Part 1 and 2 Class 12
• NCERT solutions for Mathematics Part 1 and 2 Class 12 [गणित भाग १ व २]
• NCERT solutions for Physics Part 1 and 2 Class 12
• NCERT solutions for Physics Part 1 and 2 Class 12 [भौतिकी भाग १ व २ कक्षा १२ वीं]
• NCERT solutions for Political Science Class 12 [राजनीति विज्ञान - समकालीन विश्व राजनीति १२ वीं कक्षा]
• NCERT solutions for Political Science Class 12 [राजनीति विज्ञान - स्वतंत्र भारत में राजनीति १२ वीं कक्षा]
• NCERT solutions for Psychology Class 12 [मनोविज्ञान १२ वीं कक्षा]
• NCERT solutions for Sanskrit - Bhaswati Class 12 [संस्कृत - भास्वती कक्षा १२]
• NCERT solutions for Sanskrit - Sahitya Parichay Class 11 and 12 [संस्कृत - साहित्य परिचय कक्षा ११ एवं १२]
• NCERT solutions for Sanskrit - Shashwati Class 12 [संस्कृत - शाश्वती कक्षा १२]
• NCERT solutions for Sociology Class 12 [समाजशास्त्र - भारत में सामाजिक परिवर्तन एवं विकास १२ वीं कक्षा]
• NCERT solutions for Sociology Class 12 [समाजशास्त्र - भारतीय समाज १२ वीं कक्षा]

Chapters covered in NCERT Solutions for Physics Class 12

Ncert solutions for class 12 physics chapter 1: electric charge and fields, ncert class 12 physics chapter 1: electric charge and fields exercises.

NCERT Solutions for Class 12 Physics Chapter 2: Electrostatic Potential and Capacitance

Ncert class 12 physics chapter 2: electrostatic potential and capacitance exercises.

NCERT Solutions for Class 12 Physics Chapter 3: Current Electricity

Ncert class 12 physics chapter 3: current electricity exercises.

NCERT Solutions for Class 12 Physics Chapter 4: Moving Charges and Magnetism

Ncert class 12 physics chapter 4: moving charges and magnetism exercises.

NCERT Solutions for Class 12 Physics Chapter 5: Magnetism and Matter

Ncert class 12 physics chapter 5: magnetism and matter exercises.

NCERT Solutions for Class 12 Physics Chapter 6: Electromagnetic Induction

Ncert class 12 physics chapter 6: electromagnetic induction exercises.

NCERT Solutions for Class 12 Physics Chapter 7: Alternating Current

Ncert class 12 physics chapter 7: alternating current exercises.

NCERT Solutions for Class 12 Physics Chapter 8: Electromagnetic Waves

Ncert class 12 physics chapter 8: electromagnetic waves exercises.

NCERT Solutions for Class 12 Physics Chapter 9: Ray Optics and Optical Instruments

Ncert class 12 physics chapter 9: ray optics and optical instruments exercises.

NCERT Solutions for Class 12 Physics Chapter 10: Wave Optics

Ncert class 12 physics chapter 10: wave optics exercises.

NCERT Solutions for Class 12 Physics Chapter 11: Dual Nature of Radiation and Matter

Ncert class 12 physics chapter 11: dual nature of radiation and matter exercises.

NCERT Solutions for Class 12 Physics Chapter 12: Atoms

Ncert class 12 physics chapter 12: atoms exercises.

NCERT Solutions for Class 12 Physics Chapter 13: Nuclei

Ncert class 12 physics chapter 13: nuclei exercises.

NCERT Solutions for Class 12 Physics Chapter 14: Semiconductor Electronics: Materials, Devices and Simple Circuits

Ncert class 12 physics chapter 14: semiconductor electronics: materials, devices and simple circuits exercises.

NCERT Solutions for Class 12 Physics Chapter 15: Communication Systems

Ncert class 12 physics chapter 15: communication systems exercises.

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ASSIGNMENTS FOR CLASS 12 PHYSICS

Download free PDF assignment for CBSE NCERT Class 12 Physics with chapter-wise and topic-wise important Numerical Questions and MCQs based on the latest syllabus and textbooks of CBSE NCERT. These assignments are designed for helping students to understand various topics, practice skills and improve their subject knowledge which in turn helps students to improve their academic performance.

Assignment for Class 12 Free PDF Download

We have proved large number of Assignments for Numerical Questions and MCQs of every chapter for class 12 Physics. By Practicing these assignments will help you in concept clarification and enhance your problem solving skill. These question will help in subjective and well as objective approach of solving questions as latest pattern of CBSE includes Theory based as well as MCQ based questions.

Class XII PHYSICS

Chapter-wise and topic-wise, download free pdf, chapter 1: electric charges and fields.

• Electric Charges and Fields                                   MCQ Set A
• Electric Charges and Fields                                   MCQ Set B
• Electric Charges and Fields                                   MCQ Set C

Electric Charges and Fields                                   MCQ Set D

Assignment Class 8 Science

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Assignment Class 10 Science

Assignment Class 11 Physics

Assignment Class 11 Chemistry

Assignment Class 11 Biology

Assignment Class 12 Physics

Assignment Class 12 Chemistry

Assignment Class 12 Biology

Mathematics Assignments

Assignment Class 8 Math

Assignment Class 9 Math

Assignment Class 10 Math

Assignment Class 11 Math

Assignment Class 12 Math

Chapter 2: ELECTRIC POTENTIAL AND CAPACITACE

• Electric Potential and Capacitance                     MCQ Set A
• Electric Potential and Capacitance                     MCQ Set B
• Electric Potential and Capacitance                     MCQ Set C
• Electric Potential and Capacitance                     MCQ Set D

Chapter 3: CURRENT ELECTRICTIY

• Current Electricity                                                    MCQ Set A
• Current Electricity                                                    MCQ Set B

Chapter 4: MOVING CHARGES AND MAGNETISM

• Moving Charges and Magnetism                          MCQ Set A

Moving Charges and Magnetism                          MCQ Set B

Chapter 5: MAGNETISM AND MATTER

Magnetism and Matter                                           MCQ Set A

Magnetism and Matter                                           MCQ Set B

Chapter 6: ELECTROMAGNETIC INDUCTION

Electromagnetic Induction                                    MCQ Set A

Electromagnetic Induction                                    MCQ Set B

Chapter 7: ALTERNATING CURRENT

Alternating Current                                                 MCQ Set A

• Alternating Current                                                 MCQ Set B

Chapter 8: ELECTROMAGNETIC WAVES

Electromagnetic Waves                                          MCQ Set A

Electromagnetic Waves                                          MCQ Set B

Chapter 9: RAY OPTICS AND OPTICAL INSTRUMENTS

Ray Optics and Optical Instruments                    MCQ Set A

Ray Optics and Optical Instruments                    MCQ Set B

Chapter 10: WAVE OPTICS

Wave Optics                                                                 MCQ Set A

Wave Optics                                                                 MCQ Set B

Chapter 11: DUAL NATURE OF MATTER AND RADIATION

Dual Nature of Matter and Radiation                   MCQ Set A

Dual Nature of Matter and Radiation                   MCQ Set B

 Short, Long Answer Type Questions and Numerical Questions

( as per cbse ncert pattern ), 1. electric charges and fields.

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NCERT Solutions For Class 12 Physics Chapter 1 Electric Charges and Fields

NCERT Solutions for class 12 Physics Chapter 1 Electric Charges and Fields is prepared by our senior and renowned teachers of Physics Wallah. Read now.

February 9, 2024

Table of Contents

NCERT Solutions for class-12 Physics  Chapter 1 Electric Charges and Fields is prepared by our senior and renowned teachers of Physics Wallah primary focus while solving these questions of class-12 in NCERT textbook, also do read theory of this Chapter 1 Electric Charges and Fields while going before solving the NCERT questions. You can download and share  NCERT Solutions  of Class 12 Physics from Physics Wallah.

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NCERT Solutions Class 12 Science Chapter 1 Overview

Developing a solid understanding of each chapter is crucial for students. Chapter 1 of NCERT Solutions for Class 12 Science covers significant topics. To fully comprehend the concepts presented in this chapter and effectively utilize the provided solutions, it’s recommended that students meticulously study each topic.

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Physics Wallah’s instructors have specifically created these solutions to aid in understanding the concepts within this chapter. The goal is to equip students to confidently tackle tests after reviewing and practicing these solutions.

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NCERT Solutions For Class 12 Physics Chapter 1

Answer The Following Question Answers of class 12 physics Chapter 1 – Electric Charges and Fields:

Question 1. What is the force between two small charged spheres having charges of 2 × 10−7 C and 3 × 10−7 C placed 30 cm apart in air?

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Question 2. The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge − 0.8 μC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first?

Solution : (a) Electrostatic force on the first sphere, F = 0.2 N

Charge on this sphere, q1 = 0.4 μC = 0.4 × 10−6 C

Charge on the second sphere, q2 = − 0.8 μC = − 0.8 × 10−6 C

Electrostatic force between the spheres is given by the relation,

Where, ∈0 = Permittivity of free space

The distance between the two spheres is 0.12 m.

(b) Both the spheres attract each other with the same force. Therefore, the force on the second sphere due to the first is 0.2 N.

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Question 3. Check that the ratio ke2/G memp is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

G = Gravitational constant

Its unit is N m2 kg−2.

me and mp = Masses of electron and proton.

Their unit is kg.

e = Electric charge.

Its unit is C.

∈0 = Permittivity of free space

Its unit is N m2 C−2.

Hence, the given ratio is dimensionless.

e = 1.6 × 10−19 C

G = 6.67 × 10−11 N m2 kg-2

me= 9.1 × 10−31 kg

mp = 1.66 × 10−27 kg

Hence, the numerical value of the given ratio is

This is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant.

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Question 4. (a) Explain the meaning of the statement ‘electric charge of a body is quantised’.

(b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?

Solution : (a) Electric charge of a body is quantized. This means that only integral (1, 2, …., n) number of electrons can be transferred from one body to the other. Charges are not transferred in fraction. Hence, a body possesses total charge only in integral multiples of electric charge.

(b) In macroscopic or large scale charges, the charges used are huge as compared to the magnitude of electric charge. Hence, quantization of electric charge is of no use on macroscopic scale. Therefore, it is ignored and it is considered that electric charge is continuous.

Question 5. When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.

Solution : Rubbing produces charges of equal magnitude but of opposite nature on the two bodies because charges are created in pairs. This phenomenon of charging is called charging by friction. The net charge on the system of two rubbed bodies is zero. This is because equal amount of opposite charges annihilate each other. When a glass rod is rubbed with a silk cloth, opposite natured charges appear on both the bodies. This phenomenon is in consistence with the law of conservation of energy. A similar phenomenon is observed with many other pairs of bodies.

Question 6. Four point charges qA = 2 μC, qB = −5 μC, qC = 2 μC, and qD = −5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?

Solution : The given figure shows a square of side 10 cm with four charges placed at its corners. O is the centre of the square.

(Sides) AB = BC = CD = AD = 10 cm

A charge of amount 1μC is placed at point O.

Force of repulsion between charges placed at corner A and centre O is equal in magnitude but opposite in direction relative to the force of repulsion between the charges placed at corner C and centre O. Hence, they will cancel each other. Similarly, force of attraction between charges placed at corner B and centre O is equal in magnitude but opposite in direction relative to the force of attraction between the charges placed at corner D and centre O. Hence, they will also cancel each other. Therefore, net force caused by the four charges placed at the corner of the square on 1 μC charge at centre O is zero.

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Question 7. (a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?

(b) Explain why two field lines never cross each other at any point?

Solution : (a) An electrostatic field line is a continuous curve because a charge experiences a continuous force when traced in an electrostatic field. The field line cannot have sudden breaks because the charge moves continuously and does not jump from one point to the other.

(b) If two field lines cross each other at a point, then electric field intensity will show two directions at that point. This is not possible. Hence, two field lines never cross each other.

Question 8. Two point charges qA = 3 μC and qB = −3 μC are located 20 cm apart in vacuum.

(a) What is the electric field at the midpoint O of the line AB joining the two charges?

(b) If a negative test charge of magnitude 1.5 × 10−9 C is placed at this point, what is the force experienced by the test charge?

Solution : (a) The situation is represented in the given figure. O is the mid-point of line AB.

Distance between the two charges, AB = 20 cm

∴AO = OB = 10 cm

Net electric field at point O = E

Electric field at point O caused by +3μC charge,

Magnitude of electric field at point O caused by −3μC charge,

= 5.4 × 106 N/C along OB

Therefore, the electric field at mid-point O is 5.4 × 106 N C−1 along OB.

(b) A test charge of amount 1.5 × 10−9 C is placed at mid-point O.

q = 1.5 × 10−9 C

Force experienced by the test charge = F

= 1.5 × 10−9 × 5.4 × 106

= 8.1 × 10−3 N

The force is directed along line OA. This is because the negative test charge is repelled by the charge placed at point B but attracted towards point A.

Therefore, the force experienced by the test charge is 8.1 × 10−3 N along OA.

Question 9. A system has two charges qA = 2.5 × 10−7 C and qB = −2.5 × 10−7 C located at points A: (0, 0, − 15 cm) and B: (0, 0, + 15 cm), respectively. What are the total charge and electric dipole moment of the system?

Solution : Both the charges can be located in a coordinate frame of reference as shown in the given figure.

At A, amount of charge, qA = 2.5 × 10−7C

At B, amount of charge, qB = −2.5 × 10−7 C

Total charge of the system,

q = qA + qB

= 2.5 × 10−7 C − 2.5 × 10−7 C

Distance between two charges at points A and B,

d = 15 + 15 = 30 cm = 0.3 m

Electric dipole moment of the system is given by,

p = qA × d = qB × d

= 2.5 × 10−7 × 0.3

= 7.5 × 10−8 C m along positive z-axis

Therefore, the electric dipole moment of the system is 7.5 × 10−8 C m along positive z−axis.

Question 10. An electric dipole with dipole moment 4 × 10−9 C m is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 104 N C−1. Calculate the magnitude of the torque acting on the dipole.

Solution : Electric dipole moment, p = 4 × 10−9 C m

Angle made by p with a uniform electric field, θ = 30°

Electric field, E = 5 × 104 N C−1

Torque acting on the dipole is given by the relation,

τ = pE sinθ

Therefore, the magnitude of the torque acting on the dipole is 10−4 N m.

Question 11. A polythene piece rubbed with wool is found to have a negative charge of 3 × 10−7 C.

(a) Estimate the number of electrons transferred (from which to which?)

(b) Is there a transfer of mass from wool to polythene?

Solution : (a) When polythene is rubbed against wool, a number of electrons get transferred from wool to polythene. Hence, wool becomes positively charged and polythene becomes negatively charged.

Amount of charge on the polythene piece, q = −3 × 10−7 C

Amount of charge on an electron, e = −1.6 × 10−19 C

Number of electrons transferred from wool to polythene = n

n can be calculated using the relation,

= 1.87 × 1012

Therefore, the number of electrons transferred from wool to polythene is 1.87 × 1012.

There is a transfer of mass taking place. This is because an electron has mass,

me = 9.1 × 10−3 kg

Total mass transferred to polythene from wool,

= 9.1 × 10−31 × 1.85 × 1012

= 1.706 × 10−18 kg

Hence, a negligible amount of mass is transferred from wool to polythene.

Question 12. (a) Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10−7 C? The radii of A and B are negligible compared to the distance of separation.

(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

Solution : (a) Charge on sphere A, qA = Charge on sphere B, qB = 6.5 × 10−7 C

Distance between the spheres, r = 50 cm = 0.5 m

Force of repulsion between the two spheres,

∈0 = Free space permittivity

= 1.52 × 10−2 N

Therefore, the force between the two spheres is 1.52 × 10−2 N.

(b) After doubling the charge, charge on sphere A, qA = Charge on sphere B, qB = 2 × 6.5 × 10−7 C = 1.3 × 10−6 C

The distance between the spheres is halved.

= 16 × 1.52 × 10−2

Therefore, the force between the two spheres is 0.243 N.

Question 13. Suppose the spheres A and B in Exercise 1.12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?

Note: If a conducting uncharged material is brought in contact with a charged surface then the charges are shared uniformly between the two bodies.

Given: Charge on sphere 1, q 1  = 6.5× 10 -7  C

Charge on sphere 2, q 2  = 6.5× 10 -7  C

Charge on sphere 3, q 3  = 0

Step 1: The uncharged sphere is brought in contact with sphere 1. Since sphere 1 has charge ‘q’, it gets distributed among sphere 1 and sphere 3.

Now, charge on sphere 1 = q/2

Charge on sphere 2 = q/2

At this point the sphere 3 which was initially uncharged has a charge “q/2”.

Step 2: Now sphere 3 is brought in contact with sphere 2 due to which 1/4 × q will flow from sphere 2 to sphere 3. Now sphere 2 and sphere 3 have “3/4 × q” charge.

Now, q 1  = ½ × 6.5 × 10 -7  C

⇒ 3.25× 10 -7  C

q 2  = ¾ × 6.5× 10 -7  C

⇒ 4.87 × 10 -7  C

q 3  = ¾ × 6.5× 10 -7  C

Question 14. Figure 1.33 shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

Opposite charges attract each other and same charges repel each other. It can be observed that particles 1 and 2 both move towards the positively charged plate and repel away from the negatively charged plate. Hence, these two particles are negatively charged. It can also be observed that particle 3 moves towards the negatively charged plate and repels away from the positively charged plate. Hence, particle 3 is positively charged.

The charge to mass ratio (emf) is directly proportional to the displacement or amount of deflection for a given velocity. Since the deflection of particle 3 is the maximum, it has the highest charge to mass ratio.

Question 15. Consider a uniform electric field E = 3 × 103 îN/C. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?

Solution : Given:

Electric field E = 3× 10 3   N/C

Side of square, s = 10 cm

a)  Flux of field through square whose plane is parallel to yz plane.

We understand that the normal to the plane is parallel to the direction of field.

Ø = E × A× cos(θ)            …(1)

Where, E = Electric field

A = Area through which we have to calculate flux

θ = Angle between normal to surface and the Electric field

A = .01 m 2

Plugging values, of E, A and θ in equation (1)

Ø = 3× 10 3  NC -1 × 0.01m 2 × cos0°

Ø = 30 Nm 2 C -1

b)  If normal to its (square’s) plane makes 60° with the X axis.

Ø = E × A× cos(θ)

Ø = 3× 10 3  NC -1 × 0.01m 2  × cos60°

Ø = 15 Nm 2 C -1

Question 16. What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?

Solution :  All the faces of a cube are parallel to the coordinate axes. Therefore, the number of field lines entering the cube is equal to the number of field lines piercing out of the cube. As a result, net flux through the cube is zero.

Question 17. Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 × 103 N m2/C. (a) What is the net charge inside the box? (b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not?

a)           Ø = 8.0× 10 3  Nm 2 C -1

Let net charge inside the box = q

We know that,

Flux, Ø = q/ε 0                 ..(1)

Where, q = net charged enclosed

ε 0  = permittivity of free space

ε 0  = 8.85× 10 -12  N -1  m -2 C 2

Plugging values of Ø and ε 0  in equation (1) we get,

q = Ø × ε 0

⇒ q = 8.0 × 10 3  Nm 2 C -1 × 8.85× 10 -12 N -1 C 2 m -2

⇒ q = 7.08 × 10 -8  C

Hence, the net charge inside the box is 0.07 µC.

(b) No, we cannot conclude that the body doesn’t have any charge. The flux is due to the Net charge of the body. There may still be equal amount of positive and negative charges. So, it is not necessary that if flux is zero then there will be no charges.

Question 18. A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)

Assume the charge to be enclosed by a cube, where the square is one of its sides.

Now, let us find the total flux through the imaginary cube.

Flux, Ø = q/ε            …(1)

Now plugging the values of q and ε 0  in equation (2)

⇒ Ø = 11.28 × 10 5  Nm -2 C -1

We understand that flux through all the faces of cube will be equal;

Let flux through the square = Ø a

Explanation: The net flux will be distributed equally among all 6 faces of the cube. Hence, the square will have one sixth of the total flux.

Ø a  = 1.88 Nm -2 C -1

Question 19. A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?

Total charge inside the cube, q = 2.0 µC

Edge length of Cube, a = 9.0 cm

Question 20. A point charge causes an electric flux of −1.0 × 103 Nm2/C to pass through a spherical Gaussian surface of 10.0 cm radius centered on the charge. (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? (b) What is the value of the point charge?

Ø = -1.0× 10 3  Nm 2 C -1

r­ 1  = 10.0 cm.

a)  Flux if the radius of the Gaussian surface is doubled.

If the radius is doubled then the flux would remain same i.e. -1.0× 10 3  Nm 2 C -1 .

The geometry of the Gaussian surface doesn’t affect the total flux through it. The net charge enclosed by Gaussian surface determines the net flux.

b)  Let value of point charge enclosed by Gaussian surface = q

Flux, Ø = q/ε 0            …(1)

Now plugging, the values Ø and ε 0  in equation (1).

⇒ q = -1.0× 10 3 Nm 2 C -1  × 8.85× 10 -12 N -1 m -2 C 2

⇒ q = -8.85 × 10 -9  C

The charge enclosed by the surface is -8.85 × 10 -9  C.

Question 21. A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 103 N/C and points radially inward, what is the net charge on the sphere?

Radius of charged sphere, r = 10 cm

Electric field, 20 cm away from centre of sphere, E = 1.5× 10 3  NC -1

We know that, electric field intensity at a point P, located at a distance R, due to net charge q is given by,

Now plugging the values of q and R in equation (1)

q = 4× π× ε 0 × R 2 × E

⇒ q = 4× 3.14× 8.85× 10 -12 N -1 m -2 C -2 × (0.2m) 2  × 1.5× 10 3 C

⇒ q = 6.67× 10 -9 C

The net charge on the sphere is -6.67× 10 -9 C. Since the electric field points radially inwards, we can infer that charges on sphere are negative.

Question 22. A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μC/m2. (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere?

Solution : Given: radius of sphere, r = 1.2 m

Surface charge density, σ = 80.0 μC/m 2

a)  Let charge on sphere = q

We understand that,

Total charge, Q = Surface charge density× surface area.

Surface area of sphere, S = 4πr 2

S = 18.08 m 2

Q = 80× 10 -6 Cm -2 × 18.08m -2

⇒ Q = 1.447 × 10 -3 C

b)  Let total electric flux leaving the surface of sphere =

Where, Q = net charged.

By, plugging the values of q and ε 0  in equation (1), we get,(b)

Question 23. An infinite line charge produces a field of 9 × 104 N/C at a distance of 2 cm. Calculate the linear charge density.

E = 9× 10 4  NC -1

Let the linear charge density = λ Coulomb/metre

Where, λ = linear charge density.

d = distance

From equation (1) we have,

⇒ λ = E× 2× π× ε 0 × d

⇒ λ = 9× 10 4  NC -1 × 2× π×8.85× 10 -12  N -1  m -2 C 2 × 0.02m

⇒ λ = 10 µ Cm -1

The linear charge density is 10 µ Cm -1 .

Question 24. Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 × 10−22 C/m2. What is E: (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates?

Surface charge density on plate A, σ A  = -17.0 × 10 -22  Cm -2

Surface charge density on plate B, σ B  = 17.0 × 10 -22  Cm -2

The arrangement of plates are as shonw:

Let electric field in region 1 = E 1

Region 2 = E 2

Region 3 = E 3

The electric field in region I and region III is zero because no charge is present in these regions.

E 3  = σ/ε 0    …(1)

Where, σ Surface charge density

Now, plugging the Values in equation (1).

⇒ E 3  = 1.92 × 10 -10  NC -1

The electric field in the region enclosed by the plates is found to be 1.92 × 10 -10  NC -1 .

The electric field in region III is 1.92 × 10 -10  NC -1 .

ADDITIONAL EXERCISES

Question 25. An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 104 N C−1 in Millikan’s oil drop experiment. The density of the oil is 1.26 g cm−3. Estimate the radius of the drop. (g = 9.81 m s−2; e = 1.60 × 10−19 C).

Solution : If the oil drop is stationary, the net force on it must be Zero or the resultant of all the forces on oil drop is zero. There are two forces acting on the oil drop, its weight or force due to Earth’s gravity which is pulling it vertically downwards, and Electrostatic force which is acting in vertically upward direction.Both forces should be equal in magnitude and opposite direction so that they cancel each other.

The arrangement is shown in the figure

Note: Electric Field is in vertically downward direction, because force on an negatively charged body in opposite direction of the field, so force on the drop is in vertically upward direction, which balances weight of the drop acting in vertically downward direction.

There are nine excess electrons which make the drop negatively charged because the electron is negatively charged, the net magnitude charge on a body is given by

n is the excess of electrons or protons on the body,

the charge of an electron is denoted by e

e = 1.60 × 10 –19  C

here since there are twelve excess electrons so

i.e. q = 12 × 1.60 × 10 –19  C

= 1.92 × 10 -18  C

Now the magnitude electrostatic force on a charged particle held in an electric field is given by

where, F force is acting on a particle having charge q held in an electric field of magnitude E

here, charge on the oil drop is

q = 1.92 × 10 -18  C

magnitude of Electric field is

E = 2.55 × 10 4  NC –1

So Electrostatic force on the oil drop is

F = 1.92 × 10 -18  C × 2.55 × 10 4  NC –1

= 4.896 ×10 -14  N

This force is acting in vertically upward direction, so Weight should have same magnitude of Force and is acting in vertically downward direction.

Let us assume oil drop to be Spherical in shape, so the volume of drop will be

Both the forces should be equal in magnitude so equating them

i.e. putting F = W

4.896 ×10 -14  N = (4/3 πr 3 ) ×1.26 × 10 3  Kgm -3  × 9.81ms -2

Question 26. Which among the curves shown in Fig. 1.35 cannot possibly represent electrostatic field lines?

(a) The field lines showed in (a) do not represent electrostatic field lines because field lines must be normal to the surface of the conductor.

(b) The field lines showed in (b) do not represent electrostatic field lines because the field lines cannot emerge from a negative charge and cannot terminate at a positive charge.

(c) The field lines showed in (c) represent electrostatic field lines. This is because the field lines emerge from the positive charges and repel each other.

(d) The field lines showed in (d) do not represent electrostatic field lines because the field lines should not intersect each other.

(e) The field lines showed in (e) do not represent electrostatic field lines because closed loops are not formed in the area between the field lines.

Question 27. In a certain region of space, electric field is along the z-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of 105 NC−1 per metre. What are the force and torque experienced by a system having a total dipole moment equal to 10−7 Cm in the negative z-direction?

Solution : A dipole is a system consisting of two charges equal in magnitude and opposite in nature i.e. a positive charge + q and a negative charge –q separated by some distance d

Electric dipole moment of a dipole is given by,

Where, ‘q’ is magnitude of either of charge (in column)

And ‘d’ is separation between the pair of charges (in metres)

Dipole moment is a vector quantity and its direction is from negative charge to positive charge

We are given dipole moment of the system,

P = 10 -7  Cm , along negative Z axis

Here Electric Field is varying at the rate of 10 5  NC –1  per metre in positive Z direction

i.e. dE/dl = 10 5  NC –1 m -1  along Z direction

so the electric field at any point on z axis at distance of a metres

Force on a charged particle in an electric field is given by

Where q is the magnitude of charge and E is the magnitude of Electric Field and the force is same as the direction of electric field in case of a positively charged particle and opposite to the direction of field in case of negatively charged particle

so Force on the + q charge located at a distance l from origin will be

So whatever be the magnitude of  P  and  E  putting  in the equation  = PESin

We get the torque  τ  = 0 Nm

So the torque of the system is zero

Question 28. (a) A conductor A with a cavity as shown in Fig. 1.36(a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor. (b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q [Fig. 1.36(b)]. (c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.

In Electrostatic we deal with charges at rest so there is no current inside or on the surface of the conductor i.e. in the static situation, the electric field is zero everywhere inside the conductor. This is the basic property of a conductor. A conductor has free electrons. As long as electric field is not zero, the free charge carriers would experience force and drift and re distribute themselves in such a way so that net electric field inside a conductor is zero.

We will use the above stated property to solve the problems

(a)Here in this situation a conductor with a cavity having no charge inside is given a charge Q on outer surface, we have to show that all charge resides on the external surface only and there is no charge on the inner surface.

From Gauss Theorem we know, net electric flux through an Gaussian surface enclosing a charge is equal to net charge enclosed by surface divided by permittivity of free space

we know that net electric field everywhere inside a conductor is zero

so the electric flux through this Gaussian surface would also be zero, this means no charge is enclosed by the Gaussian surface, since cavity does not contain any charge so no charge should be induced on the cavity’s metal surface as well so whole of the charge Q is distributed on the outer surface only.

(b) Again using Gauss Theorem we know,

If electric field at all the points on a Gaussian surface is zero then the net charge enclosed by the Gaussian surface is also zero, or there is no charge inside the Gaussian surface

Now if we consider a Gaussian surface enclosing the cavity inside the conductor  as shown in the figure

so the electric flux through this Gaussian surface would also be zero, this means no charge is enclosed by the Gaussian surface, but the cavity has a charge q , this means a charge –q must be induced on the inner metal surface because net charge enclosed by Gaussian surface must be zero (q + (-q) = 0) , but the metal surface has only been given charge Q so net charge of system must only be q + Q , i.e. a charge q is induced on outer surface of the conductor as which makes the total charge on the conductor equal to Q only Q + (-q) + q = Q

and total charge of the system as

Q + (-q) + q + q = Q + q

This arrangement is also in accordance with the law of conservation of charge

note: Only Q charge has been given on the conductor’s outer surface , the charge appearing (Q + q) on it is only due to induction due to charge inside the cavity.

(c) To shield the sensitive instrument from strong electrostatic fields in its environment. We need to enclose it inside a metal piece’s Cavity, this process is also known as electrostatic shielding.

we consider a metal Block with a cavity having no charge inside and Instrument inside it  as shown in the figure

any cavity inside a metal surface the electric field is zero no matter in which environment it is placed. So in order to save the Instrument from External Electrostatic fiels we need to keep it inside a metal Cavity.

Question 30. Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss’s law. [Hint: Use Coulomb’s law directly and evaluate the necessary integral.]

Solution : Let AB be a long thin wire of uniform linear charge density λ.Let us consider the electric field intensity due to AB at point P at a distance h from it as shown in the figure.

Question 31. It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so called ‘up’ quark (denoted by u) of charge (+2/3) e, and the ‘down’ quark (denoted by d) of charge (−1/3) e, together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.

Question 32. (a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = 0) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.

(b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.

Solution : (a) Let the equilibrium of the test charge be stable. If a test charge is in equilibrium and displaced from its position in any direction, then it experiences a restoring force towards a null point, where the electric field is zero. All the field lines near the null point are directed inwards towards the null point. There is a net inward flux of electric field through a closed surface around the null point. According to Gauss’s law, the flux of electric field through a surface, which is not enclosing any charge, is zero. Hence, the equilibrium of the test charge can be stable.

(b) Two charges of same magnitude and same sign are placed at a certain distance. The mid-point of the joining line of the charges is the null point. When a test charged is displaced along the line, it experiences a restoring force. If it is displaced normal to the joining line, then the net force takes it away from the null point. Hence, the charge is unstable because stability of equilibrium requires restoring force in all directions.

Question 33. A particle of mass m and charge (−q) enters the region between the two charged plates initially moving along x-axis with speed vx (like particle 1 in Fig. 1.33). The length of plate is L and an uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is qEL2/ (2mV x 2 ).

Compare this motion with motion of a projectile in gravitational field discussed in Section 4.10 of Class XI Textbook of Physics.

Solution : Given,

Mass of the particle = m

Charge of the particle = -q

Velocity of the particle = v x

Length of the plate = L

Electric field between the plates = E

From Newton’s second law of motion,

This is the vertical displacement of the particle at the far edge of the plate.

This motion is very similar to the motion of a projectile in a gravitational field. In a gravitational field, the force acting on the particle is mg and in the given case it is qE. The trajectory followed by the object will be similar in both the cases.

NOTE: A projectile is any object thrown into space by the exertion of a force. The path followed by a projectile is known as its trajectory.

Question 34. Suppose that the particle in Exercise in 1.33 is an electron projected with velocity vx= 2.0 × 106 m s−1. If E between the plates separated by 0.5 cm is 9.1 × 102 N/C, where will the electron strike the upper plate? (| e | =1.6 × 10−19 C, me = 9.1 × 10−31 kg.)

Velocity of the electron, v x  = 2.0 × 10 6  m s –1

Separation of plates, l = 0.5cm = 0.5 × 10 -2  m

Electric field between the plates, E = 9.1 × 10 2  N C -1

Charge on the electron, q = |e| = 1.6 × 10 -19  C

Mass of the electron, m e  = 9.1 × 10 -31  kg

The vertical deflection of the particle (s) is given by

NCERT Solutions For Class 12 Physics FAQs

Electric charge is a fundamental property of matter that determines its interactions with electric fields. It can be positive or negative and is quantized in nature, meaning it exists as discrete units.

An electric field is a region around a charged object where another charged object experiences a force. It is a vector quantity and is defined as the force experienced per unit positive charge placed at a point in space. Electric field lines represent the direction and strength of the electric field.

The principle of superposition states that the net electric force on a charged particle due to multiple charges is the vector sum of the forces exerted by individual charges acting independently. In other words, the total force on a charge in the presence of multiple charges is the sum of forces exerted by each individual charge.

An electric dipole consists of two equal and opposite charges separated by a small distance. It is characterized by a dipole moment, which is the product of one of the charges and the separation between the charges. An electric dipole experiences a torque in an electric field.

The SI unit of electric charge is the Coulomb (C).

NCERT Solutions For Class 12 Physics Chapter 8 Electromagnetic Waves

NCERT Solutions For Class 12 Physics Chapter 3 Current Electricity

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NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity

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Class 12 Physics NCERT Solutions Current Electricity Important Questions

Question 3.1:

The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?

Answer 3.1:

In the given question,

The EMF of the battery is given as E = 12 V

The internal resistance of the battery is given as R = 0.4 Ω

The amount of maximum current drawn from the battery is I

According to Ohm’s law,

Rearranging, we get

Substituting values in the above equation, we get

Therefore, the maximum current drawn from the given battery is 30 A.

Question 3.2:

A battery of EMF 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

Answer 3.2 :

The EMF of the battery ( E = 10 V)

The internal resistance of the battery ( R = 3 Ω)

The current in the circuit ( I = 0.5 A)

Consider the resistance of the resistor to be R .

The current in the circuit can be found out using Ohm’s Law as,

Rewriting the above equation, we get

Consider the Terminal voltage of the resistor to be V.

Then, according to Ohm’s law,

Substituting values in the equation, we get

V = 0.5 × 17

Therefore, the resistance of the resistor is 17 Ω, and the terminal voltage is 8.5 V.

Question 3.3:

a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of the combination?

b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.

Answer 3.3:

a) We know that resistors r 1 = 1 Ω , r 2 = 2 Ω and r 3 = 3 Ω are combined in series.

The total resistance of the above series combination can be calculated by the algebraic sum of individual resistances as follows:

Total resistance = 1 Ω + 2 Ω + 3 Ω = 6 Ω

Thus calculated Total Resistance = 6 Ω

b) Let us consider I to be the current flowing the given circuit

The emf of the battery is E = 12 V

The total resistance of the circuit ( calculated above ) = R = 6 Ω

Using Ohm’s law, the relation for current can be obtained as

Therefore, the current calculated is 2 A

Let the Potential drop across 1 Ω resistor = V 1

The value of V 1 can be obtained from Ohm’s law as :

V 1 = 2 x 1 = 2 V

Let the Potential drop across 2 Ω resistor = V 2

The value of V 2 can be obtained from Ohm’s law as:

V 2 = 2 x 2 = 4 V

Let the Potential drop across 3 Ω resistor = V 3

The value of V 3 can be obtained from Ohm’s law as:

V 3 = 2 x 3 = 6 V

Therefore, the potential drops across the given resistors r 1 = 1 Ω , r 2 = 2 Ω and r 3 = 3 Ω are calculated to be

V 2 = 2 x 1 = 4 V

V 3 = 2 x 1 = 6 V

Question 3.4:

a) Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is the total resistance of the combination?

b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.

Answer 3.4:

A ) Resistors r 1 = 2 Ω , r 2 = 4 Ω and r 3 = 5 Ω are combined in parallel

Hence the total resistance of the above circuit can be calculated  by the following formula:

Therefore, the total resistance of the parallel combination given above is given by:

B ) Given that emf of the battery, E = 20 V

Let the current flowing through resistor R 1 be I 1

I 1 is given by :

Let the current flowing through resistor R 2 be I 2

I 2 is given by :

Let the current flowing through resistor R 3 be I 3

I 3 is given by :

Therefore, the total current can be found by the following formula :

I = I 1 + I 2 + I 3 = 10 + 5 + 4 = 19 A

therefore the current flowing through each resistor is calculated to be :

Therefore, the total current is I = 19 A

Question 3.5:

At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10 –4 °C –1 .

Answer 3.5:

Given that the room temperature, T = 27 ° C

The heating element has a resistance of R = 100 Ω

Let the increased temperature of the filament be T 1

At T 1 , the resistance of the heating element is  R 1 = 117 Ω

Temperature coefficient of the material used for the element is 1.70 x 10 – 4 C – 1

α = 1.70 x 10 -4 C – 1

α is given by the relation ,

T 1 = 1027 ° C

Therefore, the resistance of the element is 117 Ω at T 1 = 1027 ° C

Question 3.6:

A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10 –7 m 2 , and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?

Answer 3.6:

Given that the length of the wire, L = 15 m

The area of cross-section is given as , a = 6.0 x 10 – 7 m 2

Let the resistance of the material of the wire be, R, i.e., R = 5.0 Ω

The resistivity of the material is given as ρ

Question 3.7:

A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.

Answer 3.7:

Given that temperature T 1 = 27.5 ° C

Resistance R 1 at temperature T 1 is given as :

R 1 =  2.1 Ω ( at T 1 )

Given that temperature T 2 = 100 ° C

Resistance R 2 at temperature T 2 is given as :

R 2 =  2.7 Ω ( at T 2 )

Temperature coefficient of resistivity of silver = α

Question 3.8:

A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10 –4 °C –1 .

Answer 3.8:

In the given problem,

The supply voltage, V = 230 V

The initial current drawn, I 1 = 3.2 A

Consider the initial resistance to be R 1 , which can be found by the following relation:

Substituting the values, we get

Value of current at steady state, I 2 = 2.8 A

Value of resistance at steady state = R 2

R 2 can be calculated by the following equation:

The temperature coefficient of nichrome averaged over the temperature range involved is 1.70 x 10 – 4 ° C – 1

Value of initial temperature of nichrome , T 1 = 27.0 ° C

Value of steady-state temperature reached by nichrome = T 2

This temperature T 2 can be obtained by the following formula:

T 2 = 840.5 + 27 = 867.5 ° C

Hence, the steady temperature of the heating element is 867.5 ° C

Question 3.9 :

Determine the current in each branch of the network shown in the figure:

Answer 3.9 :

The current flowing through various branches of the network is shown in the figure given below:

Let I 1 be the current flowing through the outer circuit

Let I 2 be the current flowing through AB branch

Let I 3 be the current flowing through AD branch

Let I 2 – I 4 be the current flowing through branch BC

Let I 3 + I 4 be the current flowing through branch DC

Let us take closed-circuit ABDA into consideration, we know that potential is zero.

i.e ,  10 I 2 + 5 I 4 – 5 I 3 = 0

2 I 2 +  I 4 –  I 3 = 0

I 3 = 2 I 2 +  I 4                                                                                               . . . . . . . . . . . . . . . . . . . . . . eq ( 1 )

Let us take closed circuit BCDB into consideration, we know that potential is zero.

i.e.,  5 ( I 2 – I 4 ) – 10 ( I 3 + I 4 ) – 5 I 4 = 0

5 I 2 – 5 I 4 – 10 I 3 – 10 I 4 –  5 I 4 = 0

5 I 2 – 10 I 3 –  20 I 4 = 0

I 2 = 2 I 3 – 4 I 4                                                                                           . . . . . . . . . . . . . . . . . . . . . . eq ( 2 )

Let us take closed-circuit ABCFEA into consideration, we know that potential is zero.

i.e ,  – 10 + 10 ( I 1 ) + 10 ( I 2 ) + 5 ( I 2 –  I 4 ) = 0

10 = 15 I 2 + 10 I 1 – 5 I 4

3 I 2 + 2 I 2 – I 4 = 2                                                                                    . . . . . . . . . . . . . . . . . . . . . . eq ( 3 )

From equation ( 1 )  and ( 2 ), we have :

I 3 = 2 ( 2 I 3 + 4 I 4 ) + I 4

I 3 = 4 I 3 + 8 I 4 + I 4

– 3 I 3 = 9 I 4

– 3 I 4 = + I 3                                                                                             . . . . . . . . . . . . . . . . . . . . . . eq ( 4 )

Putting equation  ( 4 ) in equation ( 1 ) , we have  :

I 3 = 2 I 2 + I 4

– 4 I 4 = 2 I 2

I 2 = – 2 I 4                                                                                                . . . . . . . . . . . . . . . . . . . . . . eq ( 5 )

From the above equation , we infer that :

I 1 = I 3 + I 2                                                                                                    . . . . . . . . . . . . . . . . . . . . . . eq ( 6 )

Putting equation ( 4 ) in equation ( 1 ), we obtain

3 I 2 + 2 ( I 3 + I 2 ) – I 4 = 2

5 I 2 + 2 I 3 – I 4 = 2                                                                                  . . . . . . . . . . . . . . . . . . . . . . eq ( 7 )

Putting equations ( 4 ) and ( 5 ) in equation ( 7 ), we obtain

5 ( – 2 I 4 ) + 2 ( – 3 I 4 ) – I 4 = 2

– 10 I 4 –  6 I 4 – I 4 = 2

17 I 4 = – 2

Equation ( 4 ) reduces to

I 3 = – 3 ( I 4 )

I 2 = – 2 ( I 4 )

Therefore, current in each branch is given as :

Question 3. 10:

A ) In a meter bridge given below, the balance point is found to be at 39.5 cm from the end A, when the resistor S is of 12.5 Ω. Determine the resistance of R. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?

B ) Determine the balance point of the bridge above if R and S are interchanged.

C ) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?

Answer 3. 10:

( a ) Let L 1 be the balance point from end A,

Given that , L 1 = 39.5 cm

Given that resistance of the resistor S = 12.5 Ω

We know that condition for the balance is given by the equation:

(b) If R and S are interchanged, then the lengths will also be interchanged.

Hence, the length modifies to

l = 100 – 39.5 = 60.5 cm.

(c) If the galvanometer and the cell are interchanged, the position of the balance point remains unchanged. Therefore, the galvanometer will show no current.

Question 3.11:

A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

Answer 3.11:

The EMF of the given storage battery is E = 8.0 V

The Internal resistance of the battery is given by r = 0.5 Ω

The given DC supply voltage is V = 120 V

The resistance of the resistor is R = 15.5 Ω

Effective voltage in the circuit = V 1

R is connected to the storage battery in series.

Hence, it can be written as

V 1 = V – E

V 1 = 120 – 8 = 112 V

Current flowing in the circuit = I, which is given by the relation,

We know that Voltage across a resistor R given by the product,

I x R = 7 × 15.5 = 108.5 V

We know that,

DC supply voltage = Terminal voltage + voltage drop across R

Terminal voltage of battery = 120 – 108.5 = 11.5 V

A series resistor, when connected to a charging circuit, limits the current drawn from the external source.

The current will become extremely high in its absence. This is extremely dangerous.

Question 3.12 :

In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?

Answer 3.12:

Emf of the cell, E 1 = 1.25 V

The balance point of the potentiometer, l 1 = 35 cm

The cell is replaced by another cell of emf E 2 .

The new balance point of the potentiometer, l 2 = 63 cm

The balance condition is given by the relation,

Question 3.13 :

The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 10 28 m –3 . How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 × 10 –6 m 2 and it is carrying a current of 3.0 A.

  Answer 3.13 :

Given that Number density of free electrons in a copper conductor, n =  8.5 x 10 28 m – 3

Let the Length of the copper wire be l

Given , l = 3.0 m

Let the area of the cross-section of the wire be A = 2.0 x 10 – 6 m 2

The value of the current carried by the wire, I = 3.0 A, which is given  by the equation,

I = n A e V d

e = electric charge = 1.6 x 10 – 19 C

Question 3.14:

The earth’s surface has a negative surface charge density of 10 –9 C  m –2 . The potential difference of 400 kV between the top of the  atmosphere and the surface results (due to the low conductivity of  the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the  earth’s surface? (This never happens in practice because there is a  mechanism to replenish electric charges, namely the continual  thunderstorms and lightning in different parts of the globe). (Radius  of earth = 6.37 × 10 6  m.)

Surface charge density of the earth, σ=10 −9 cm −2 Potential difference between the top of the atmosphere and the surface, V= 400 kV Current over the entire globe, I=1800 A Radius of the earth, r=6.37×10 6 m Surface area of the earth, A=4πr 2 

= 4 x 3.14 x (6.37×10 6  ) 2 =5.09×10 14 m 2

Charge on the earth’s surface, q=σA= 10 −9  x 5.09×10 14 

= 5.09×10 5 C

Time taken to neutralize the earth’s surface, t = q/I

⟹t=5.09×10 5 /1800=283 s

Question 3.15:

(a) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. What is the current drawn from the supply and its terminal voltage? (b) A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 Ω. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?

(a) Emf of the secondary cells, ε = 2.0 V

Number of secondary cells, n = 6

Total EMF, E= nε= 6 x 2 = 12 V

The internal resistance of the secondary cells, r= 0.015 Ω

Resistance to which the secondary cells are connected, R = 8.5 Ω Total resistance in circuit R total = nr+ R = 6×0.015+8.5=8.59Ω

Current drawn from the supply, I=E/R total =12/8.59 = 1.4 A Terminal voltage, V= IR = 1.4×8.5=11.9 V

(b) Emf of the secondary cell, ε = 1.9 V

Internal resistance, r = 380 Ω Maximum current drawn from the cell,I =ε/r =  1.9/380=0.005A.

The current required to start a motor is 100 Amp. Here, the current produced is 0.005 A, so the starting motor of the car cannot be started with this current.

Question 3.16:

Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables. (ρ Al = 2.63 × 10 –8 Ω m, ρ Cu = 1.72 × 10 –8 Ω m, Relative density of Al = 2.7, of Cu = 8.9.)

Length of aluminium = l 1

Resistance of aluminium = R

Resistivity of aluminium, ρ AI =ρ 1 = 2.63×10 −8  Ωm

Relative density of aluminium, d 1 =2.7

Area of the cross-section of the aluminium wire =A 1

Length of copper = l 2

Resistance of copper = R 2

Resistivity of copper, ρ cu  =ρ 2 =1.72 × 10 –8 Ω m

Relative density of copper, d 2 = 8.9

Area of the cross-section of the copper wire =A 2

Therefore, \(\begin{array}{l}R_{1}=\rho _{1}\frac{l_{1}}{A_{1}}\end{array} \) ——(1)

It is given that R 1 = R 2 Therefore,

Given l 1 = l 2

=  (2.63×10 −8 )/(1.72 × 10 –8 ) ​= 1.52

Mass of aluminium, m 1 = Volume x density

= A 1 l 1 x d 1

Mass of copper = m 2  = Volume x density

= A 2 l 2 x d 2

m 1 /m 2 =(A 1 l 1 x d 1 /  A 2 l 2 x d 2 ) Since l 1 = l 2

m 1 /m 2 =(A 1  d 1 /  A 2 d 2 )

m 1 /m 2 = (1.52) x (2.7/8.9)

= (1.52) x (0.303)

m 1 /m 2  = 0.46

The mass ratio of aluminium to copper is 0.46. Since aluminium is lighter, it is preferred for long suspensions of cables

​ Question 3.17:

What conclusion can you draw from the following observations on a resistor made of alloy manganin?

Ohm’s law is valid with high accuracy. This means that the resistivity of the alloy manganin is nearly independent of temperature.

Question 3.18:

Answer the following questions: (a) A steady current flows in a metallic conductor of the non-uniform cross-section. Which of these quantities is constant along the conductor: current, current density, electric field, drift speed? (b) Is Ohm’s law universally applicable for all conducting elements? If not, give examples of elements that do not obey Ohm’s law. (c) A low voltage supply from which one needs high currents must have very low internal resistance. Why? (d) A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why?

(a) Current is given to be steady. Therefore, it is a constant.  The current density, electric field, drift speed depends on the area of the cross-section inversely. (b) No, examples of non-ohmic elements are vacuum diodes, semiconductor diodes, etc. (c) Because the maximum current drawn from a source = ε/r. (d) If the circuit is shorted (accidentally), the current drawn will exceed safety limits if the internal resistance is not large.

Question 3. 19:

Choose the correct alternative: (a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals. (b) Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals. (c) The resistivity of the alloy manganin is nearly independent of/ increases rapidly with increase of temperature. (d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of (10 22 /10 23 ).

(a) greater

(c) nearly independent of

Question 3.20:

(a) Given n resistors each of resistance R, how will you combine  them to get the (i) maximum (ii) minimum effective resistance?  What is the ratio of the maximum to minimum resistance? (b) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent resistance of (i) (11/3) Ω (ii) (11/5) Ω, (iii) 6 Ω, (iv) (6/11) Ω? (c) Determine the equivalent resistance of networks shown in figure

(a) Total number of resistors = n

Resistance of each resistor = R

(i) The maximum effective resistance is got when the resistors are connected in series. The effective resistance R 1 = nR

(ii) The minimum effective resistance is got when the resistors are connected in parallel, the effective resistance R 2 = R/n

The ratio of maximum resistance to minimum resistance = nR/(R/n) = n 2

(b) The resistance given are 1 Ω, 2 Ω, 3 Ω

Consider the following combination

R’ = 2/3 Ω

These resistors are connected in series with 3 Ω. Therefore the effective resistance R = R’ + 3 = (2/3) +3 = 11/3 Ω

(ii)  11/5 Ω

R’ = 6/5 Ω

These resistors are connected in series with 3 Ω. Therefore the effective resistance R = R’ + 3 = (6/5) +1 = 11/5 Ω

When the resistors with resistance 1 Ω, 2 Ω, 3 Ω are connected in series the effective resistance is given by

1 Ω +2 Ω + 3 Ω = 6 Ω

(iv) (6/11) Ω

R’ = (6/11) Ω

(c) (a) In all the loops, two resistors of resistance 1Ω are connected in series. Therefore, the effective resistance is (1+1) = 2 Ω

Similarly, In all the loops, two resistors of resistance 2Ω are connected in series. Therefore, the effective resistance is (2+2) = 4 Ω

R’ = 4/3 Ω

All four resistors are connected in series. Hence the equivalent resistance R each is (4/3) x 4 = 16/3 Ω

(c) (b) The resistors are connected in series. Therefore, the effective resistance is R + R+R+R+R = 5R

Question 3.21:

Determine the current drawn from a 12V supply with internal  resistance 0.5Ω by the infinite network shown in the figure. Each  resistor has 1Ω resistance.

Let the effective resistance of the infinite network be X. Since it is an infinite network, adding three resistors of 1 Ω resistance will not change the total resistance, i.e., it will remain X.  The circuit will look like this if three resistors are added.

The equivalent resistance of this network is R’ = R + (equivalent resistance when X and R are parallel) + R

= R + [XR/(X+R)] + R

As said above, since it is an infinite network, adding three resistors of 1 Ω resistance will not change the total resistance.

R’ = X

⇒  2R +  [XR/(X+R)] = X

Since R = 1Ω we get

2 x 1 +  [X x 1/(X+1)] = X

X 2 – 2X -2 = 0

The value of resistance cannot be negative. Therefore, X= 1+√3 = 2.732 Ω

Given E = 12 V ; r = 0.5 Ω

If I is the current drawn by the network, then

I = E/(X+r) = 12/(2.732+ 0.5) = 3.713 A

Question 3. 22:

Figure shows a potentiometer with a cell of 2.0 V and internal  resistance 0.40 Ω maintaining a potential drop across the resistor  wire AB. A standard cell which maintains a constant emf of 1.02 V  (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

(a) What is the value ε? 

(b) What purpose does the high resistance of 600 kΩ have?

(c) Is the balance point affected by this high resistance 

(d) Would the method work in the above situation if the driver cell  of the potentiometer had an emf of 1.0V instead of 2.0V?

(e) Would the circuit work well for determining an extremely small  emf, say of the order of a few mV (such as the typical emf of a  thermo-couple)? If not, how will you modify the circuit?

(a) Constant emf of the standard cell, E 1 = 1.02 V

The balance point on the wire, l 1 = 67.3 cm

The standard cell is then replaced by a cell of unknown emf ε and the balance point changes to l =  82.3 cm

The relation between Emf and the balancing point

(E 1 /l 1 ) = (ε/l)

ε = (l x E 1 /l 1 ) = (82.3  x 1.02)/67.3  = 1.247 V

(b) The purpose of using high resistance of 600 kΩ is to reduce current through the galvanometer when the movable contact is far from the balance point.

(d) No. If ε is greater than the emf of the driver cell of the potentiometer, there will be no balance point on the wire AB.

(e) The circuit will not be suitable, because the balance point (for ε of the order of a few mV) will be very close to the end A and the percentage of error in the measurement will be very large. The circuit can be modified by putting a suitable resistor R in series with the wire AB so that the potential drop across AB is only slightly greater than the emf to be measured. Then, the balance point will be at a larger length of the wire and the percentage error will be much smaller.

Question 3. 23:

Figure below shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.

Internal resistance of the cell, r = 1.5 V cell

Balance point of the cell in open circuit, l = 76.3 cm

External resistance, R = 9.5 Ω

New balance point, l 1 = 64. 8 cm

The expression for internal resistance is given as

Class 12 Physics NCERT Solutions for Current Electricity

Chapter 3 Current Electricity of Class 12 Physics is framed as per the current CBSE Syllabus 2023-24. The important questions in Chapter 3 of Physics NCERT Solutions for Class 12 that are frequently asked are definitions of resistance, conductance, drift velocity, cell EMF, resistivity, conductance and internal resistance. The graphing of resistivity for metals, semiconductors and alloys must be practised thoroughly. Students must be well-versed in the conversion of complex circuits into simple series and parallel circuits. Students should practise a wide range of problems to get better at this. Questions from potentiometer or meter bridge are regularly asked. The resistivity chart for different colour codes can be asked in the board exam.

Topics covered in NCERT Class 12 Chapter 3 of Physics Current Electricity

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Dropped Topics – 

3.7 Resistivity of Various Materials (delete Tables 3.1 and 3.2 and Carbon resistors, Colour code for carbon resistor) 3.10 Combinations of Resistors – Series and Parallel Example 3.5 3.15 Meter Bridge 3.16 Potentiometer Exercises 3.3, 3.4, 3.10, 3.12, 3.14–3.23

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NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields

• NCERT Solutions
• Chapter 1 Electric Charges And Fields

NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields - FREE PDF Download

NCERT Electric Charges And Fields Class 12 Physics Chapter 1 Solutions by Vedantu, helps to understanding the principles of electrostatics. Electrostatics is the branch of physics that deals with the study of forces, fields, and potentials arising from static electric charges. Understanding these concepts is essential for more advanced topics in electromagnetism. In simple terms, this chapter gives you the foundational knowledge of electric charges. Through these solutions, students can learn the behaviour of electric charges, the laws governing their interactions, and the resultant electric fields. By practising with these Class 12 Physics NCERT Solutions , students can build confidence in their understanding and excel in their studies.

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Glance on Physics Chapter 1 Class 12 - Electric Charges and Fields

Chapter 1 of Class 12 Physics equips you with the foundational knowledge of electric charges and the invisible forces they exert through electric fields. 

Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electromagnetic field.

The electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.

F = K$\frac{q_{1}q_{2}}{r2}$

Gauss’s law: The flux of the electric field through any closed surface S is $1/\varepsilon _0$ times the total charge enclosed by S. The law is especially useful in determining electric field E when the source distribution has simple symmetry.

Electric flux is the measure of the field lines crossing a surface. It is a scalar quantity. “The number of field lines passing through the perpendicular unit area will be proportional to the magnitude of Electric Field there” 

An electric dipole is a pair of equal and opposite charges q and –q separated by some distance 2a. Its dipole moment vector p has magnitude 2qa and is in the direction of the dipole axis from –q to q.

This article contains chapter notes, important questions, exemplar solutions, exercises, and video links for Chapter 12 - Electric Charges and Fields, which you can download as PDFs.

There are 23 fully solved questions in physics class 12 chapter 1 Electric Charges and Fields.

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Access NCERT Solutions for Class 12 Physics Chapter 1 – Electric Charges and Fields

$2\times {{10}^{-7}}C$ and $3\times {{10}^{-7}}C$ placed $30cm$ apart in air?

Ans: We are given the following information:

Repulsive force of magnitude,$F=6\times {{10}^{-3}}N$

Charge on the first sphere, ${{q}_{1}}=2\times {{10}^{-7}}C$

Charge on the second sphere, ${{q}_{2}}=3\times {{10}^{-7}}C$

Distance between the two spheres, $r=30cm=0.3m$

Electrostatic force between the two spheres is given by Coulomb’s law as,

$F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$

Where, ${{\varepsilon }_{0}}$is the permittivity of free space and, 

$\frac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}$

Now on substituting the given values, Coulomb’s law becomes, 

\[F = \frac{9\times 10^{9}\times 2\times 10^{-7}\times 3\times 10^{-7}}{(0.3)^{2}}\]

Therefore, we found the electrostatic force between the given charged spheres to be $F=6\times {{10}^{-3}}N$. Since the charges are of the same nature, we could say that the force is repulsive.

2. The electrostatic force on a small sphere of charge $0.4\mu C$ due to another small sphere of charge $-0.8\mu C$ in air is 0.2N.

a) What is the distance between the two spheres?

Ans: Electrostatic force on the first sphere is given to be, $F=0.2N$

Charge of the first sphere is, ${{q}_{1}}=0.4\mu C=0.4\times {{10}^{-6}}C$

Charge of the second sphere is, ${{q}_{2}}=-0.8\mu C=-0.8\times {{10}^{-6}}C$

We have the electrostatic force given by Coulomb’s law as,

$\Rightarrow r=\sqrt{\frac{{{q}_{1}}{{q}_{2}}}{4\pi {{\varepsilon }_{0}}F}}$

Substituting the given values in the above equation, we get, 

$\Rightarrow r=\sqrt{\frac{0.4\times {{10}^{-6}}\times 8\times {{10}^{-6}}\times 9\times {{10}^{9}}}{0.2}}$

$\Rightarrow r=\sqrt{144\times {{10}^{-4}}}$

$\therefore r=0.12m$

Therefore, we found the distance between charged spheres to be $r=0.12m$.

b) What is the force on the second sphere due to the first?

Ans: From Newton’s third law of motion, we know that every action has an equal and opposite reaction. 

Thus, we could say that the given two spheres would attract each other with the same force. 

So, the force on the second sphere due to the first sphere will be $0.2N$. 

3. Check whether the ratio $\frac{k{{e}^{2}}}{G{{m}_{e}}{{m}_{p}}}$is dimensionless. Look up a table of physical constants and hence determine the value of the given ratio. What does the ratio signify?

Ans: We are given the ratio, $\frac{k{{e}^{2}}}{G{{m}_{e}}{{m}_{p}}}$.

Here, G is the gravitational constant which has its unit $N{{m}^{2}}k{{g}^{-2}}$;

${{m}_{e}}$and ${{m}_{p}}$ are the masses of electron and proton in $kg$ respectively;

$e$ is the electric charge in $C$; 

$k$ is a constant given by $k=\frac{1}{4\pi {{\varepsilon }_{0}}}$

In the expression for k, ${{\varepsilon }_{0}}$ is the permittivity of free space which has its unit $N{{m}^{2}}{{C}^{-2}}$. 

Now, we could find the dimension of the given ratio by considering their units as follows: 

$\frac{k{{e}^{2}}}{G{{m}_{e}}{{m}_{p}}}=\frac{\left[ N{{m}^{2}}{{C}^{-2}} \right]{{\left[ C \right]}^{2}}}{\left[ N{{m}^{2}}k{{g}^{-2}} \right]\left[ kg \right]\left[ kg \right]}={{M}^{0}}{{L}^{0}}{{T}^{0}}$

Clearly, it is understood that the given ratio is dimensionless. 

Now, we know the values for the given physical quantities as, 

$e=1.6\times {{10}^{-19}}C$

$G=6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}$

${{m}_{e}}=9.1\times {{10}^{-31}}kg$

${{m}_{p}}=1.66\times {{10}^{-27}}kg$

Substituting these values into the required ratio, we get, 

$\frac{k{{e}^{2}}}{G{{m}_{e}}{{m}_{p}}}=\frac{9\times {{10}^{9}}\times {{\left( 1.6\times {{10}^{-19}} \right)}^{2}}}{6.67\times {{10}^{-11}}\times 9.1\times {{10}^{-3}}\times 1.67\times {{10}^{-22}}}$

$\Rightarrow \frac{k{{e}^{2}}}{G{{m}_{e}}{{m}_{p}}}\approx 2.3\times {{10}^{39}}$

We could infer that the given ratio is the ratio of electrical force to the gravitational force between a proton and an electron when the distance between them is kept constant. 

a) Explain the meaning of the statements ‘electric charge of a body is quantized’. 

Ans: The given statement ‘Electric charge of a body is quantized’ means that only the integral number $(1,2,3,...,n)$ of electrons can be transferred from one body to another. 

That is, charges cannot be transferred from one body to another in fraction.

b) Why can one ignore quantization of electric charge when dealing with macroscopic i.e., large scale charges?

Ans: On a macroscopic scale or large-scale, the number of charges is as large as the magnitude of an electric charge. 

So, quantization is considered insignificant at a macroscopic scale for an electric charge and electric charges are considered continuous.

5. When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge. 

Ans: Rubbing two objects would produce charges that are equal in magnitude and opposite in nature on the two bodies. 

This happens due to the reason that charges are created in pairs. This phenomenon is called charging by friction. 

The net charge of the system however remains zero as the opposite charges equal in magnitude annihilate each other. 

So, rubbing a glass rod with a silk cloth creates opposite charges of equal magnitude on both of them and this observation is found to be consistent with the law of conservation of charge. 

6. Four point charges ${{q}_{A}}=2\mu C$, ${{q}_{B}}=-5\mu C$, ${{q}_{C}}=2\mu C$and ${{q}_{D}}=-5\mu C$ are located at the corners of a square ABCD with side 10cm. What is the force on the $1\mu C$ charge placed at the centre of this square?

Ans: Consider the square of side length $10cm$ given below with four charges at its corners and let O be its centre.

From the figure we find the diagonals to be, 

$AC=BD=10\sqrt{2}cm$

$\Rightarrow AO=OC=DO=OB=5\sqrt{2}cm$

Now the repulsive force at O due to charge at A, 

${{F}_{AO}}=k\frac{{{q}_{A}}{{q}_{O}}}{O{{A}^{2}}}=k\frac{\left( +2\mu C \right)\left( 1\mu C \right)}{{{\left( 5\sqrt{2} \right)}^{2}}}$…………………………………………… (1)

And the repulsive force at O due to charge at D, 

${{F}_{DO}}=k\frac{{{q}_{D}}{{q}_{O}}}{O{{D}^{2}}}=k\frac{\left( +2\mu C \right)\left( 1\mu C \right)}{{{\left( 5\sqrt{2} \right)}^{2}}}$………………………………………….. (2) 

And the attractive force at O due to charge at B, 

${{F}_{BO}}=k\frac{{{q}_{B}}{{q}_{O}}}{O{{B}^{2}}}=k\frac{\left( -5\mu C \right)\left( 1\mu C \right)}{{{\left( 5\sqrt{2} \right)}^{2}}}$……………………………………………. (3)

And the attractive force at O due to charge at C, 

${{F}_{CO}}=k\frac{{{q}_{C}}{{q}_{O}}}{O{{C}^{2}}}=k\frac{\left( -5\mu C \right)\left( 1\mu C \right)}{{{\left( 5\sqrt{2} \right)}^{2}}}$……………………………………………… (4)

We find that (1) and (2) are of same magnitude but they act in the opposite direction and hence they cancel out each other. 

Similarly, (3) and (4) are of the same magnitude but in the opposite direction and hence they cancel out each other too. 

Hence, the net force on charge at centre O is found to be zero. 

a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?  

Ans: An electrostatic field line is a continuous curve as the charge experiences a continuous force on being placed in an electric field. 

As the charge doesn’t jump from one point to the other, field lines will not have sudden breaks. 

b) Explain why two field lines never cross each other at any point?

Ans: If two field lines are seen to cross each other at a point, it would imply that the electric field intensity has two different directions at that point, as two different tangents (representing the direction of electric field intensity at that point) can be drawn at the point of intersection. 

This is however impossible and thus, two field lines never cross each other.

8. Two point charges ${{q}_{A}}=3\mu C$and ${{q}_{B}}=-3\mu C$are located 20cm apart in vacuum. 

a) What is the electric field at the midpoint O of the line AB joining the two charges?

Ans: The situation could be represented in the following figure. Let O be the midpoint of line AB.

We are given:

\[AB=20cm\]

\[AO=OB=10cm\]

Take E to be the electric field at point O, then, 

The electric field at point O due to charge $+3\mu C$would be, 

${{E}_{1}}=\frac{3\times {{10}^{-6}}}{4\pi {{\varepsilon }_{0}}{{\left( AO \right)}^{2}}}=\frac{3\times {{10}^{-6}}}{4\pi {{\varepsilon }_{0}}{{\left( 10\times {{10}^{-2}} \right)}^{2}}}N{{C}^{-1}}$along OB

The electric field at point O due to charge $-3\mu C$would be, 

${{E}_{2}}=\left| \frac{3\times {{10}^{-6}}}{4\pi {{\varepsilon }_{0}}{{\left( OB \right)}^{2}}} \right|=\frac{3\times {{10}^{-6}}}{4\pi {{\varepsilon }_{0}}{{\left( 10\times {{10}^{-2}} \right)}^{2}}}N{{C}^{-1}}$along OB 

The net electric field,

$\Rightarrow E={{E}_{1}}+{{E}_{2}}$

$\Rightarrow E=2\times \frac{9\times {{10}^{9}}\times 3\times {{10}^{-6}}}{{{\left( 10\times {{10}^{-2}} \right)}^{2}}}$

$\Rightarrow E=5.4\times {{10}^{6}}N{{C}^{-1}}$

Therefore, the electric field at mid-point O is $E=5.4\times {{10}^{6}}N{{C}^{-1}}$ along OB. 

b) If a negative test charge of magnitude $1.5\times {{10}^{-19}}C$ is placed at this point, what is the force experienced by the test charge?

Ans: We have a test charge of magnitude $1.5\times {{10}^{-9}}C$ placed at mid-point O and we found the electric field at this point to be $E=5.4\times {{10}^{6}}N{{C}^{-1}}$.

So, the force experienced by the test charge would be F, 

$\Rightarrow F=qE$

$\Rightarrow F=1.5\times {{10}^{-9}}\times 5.4\times {{10}^{6}}$

$\Rightarrow F=8.1\times {{10}^{-3}}N$

This force will be directed along OA since like charges repel and unlike charges attract.

9. A system has two charges ${{q}_{A}}=2.5\times {{10}^{-7}}C$and ${{q}_{B}}=-2.5\times {{10}^{-7}}C$located at points $A:\left( 0,0,-15cm \right)$ and $B:\left( 0,0,+15cm \right)$ respectively. What are the total charge and electric dipole moment of the system?

Ans: The figure given below represents the system mentioned in the question:

The charge at point A, ${{q}_{A}}=2.5\times {{10}^{-7}}C$

The charge at point B, ${{q}_{B}}=-2.5\times {{10}^{-7}}C$

Then, the net charge would be, $q={{q}_{A}}+{{q}_{B}}=2.5\times {{10}^{-7}}C-2.5\times {{10}^{-7}}C=0$

The distance between two charges at A and B would be,

\[d=15+15=30cm\]

The electric dipole moment of the system could be given by,

$P=\mathop{q}_{A}\times d=\mathop{q}_{B}\times d$

$\Rightarrow P=2.5\times {{10}^{-7}}\times 0.3$

$\therefore P=7.5\times {{10}^{-8}}Cm$ along the \[+z\]axis.

Therefore, the electric dipole moment of the system is found to be $7.5\times {{10}^{-8}}Cm$ and it is directed along the positive \[z\]-axis.

10. An electric dipole with dipole moment $4\times {{10}^{-9}}Cm$ is aligned at $30{}^\circ $ with direction of a uniform electric field of magnitude $5\times {{10}^{4}}N{{C}^{-1}}$. Calculate the magnitude of the torque acting on the dipole.

Ans: We are given the following:

Electric dipole moment, $\overrightarrow{p}=4\times {{10}^{-9}}Cm$ 

Angle made by $\overrightarrow{p}$ with uniform electric field, $\theta =30{}^\circ $ 

Electric field, $\overrightarrow{E}=5\times {{10}^{4}}N{{C}^{-1}}$ 

Torque acting on the dipole is given by

$\tau =pE\sin \theta $ 

Substituting the given values we get, 

$\Rightarrow \tau =4\times {{10}^{-9}}\times 5\times {{10}^{4}}\times \sin 30{}^\circ $

$\Rightarrow \tau =20\times {{10}^{-5}}\times \frac{1}{2}$

$\therefore \tau ={{10}^{-4}}Nm$

Thus, the magnitude of the torque acting on the dipole is found to be ${{10}^{-4}}Nm$.

11. A polythene piece rubbed with wool is found to have a negative charge of $3\times {{10}^{-7}}C$ 

a) Estimate the number of electrons transferred (from which to which?)

Ans: When polythene is rubbed against wool, a certain number of electrons get transferred from wool to polythene. 

As a result of which wool becomes positively charged on losing electrons and polythene becomes negatively charged on gaining them.

Charge on the polythene piece, $q=-3\times {{10}^{-7}}C$ 

Charge of an electron, $e=-1.6\times {{10}^{-19}}C$ 

Let n be the number of electrons transferred from wool to polythene, then, from the property of quantization we have, 

$q=ne$ 

$\Rightarrow n=\frac{q}{e}$

Now, on substituting the given values, we get, 

$\Rightarrow n=\frac{-3\times {{10}^{-7}}}{-1.6\times {{10}^{-19}}}$

$\therefore n=1.87\times {{10}^{12}}$

Therefore, the number of electrons transferred from wool to polythene would be$1.87\times {{10}^{12}}$.

b) Is there a transfer of mass from wool to polythene?

Ans: Yes, during the transfer of electrons from wool to polythene, along with charge, mass is transferred too. 

Let $m$ be the mass being transferred in the given case and ${{m}_{e}}$ be the mass of the electron, then,

$m={{m}_{e}}\times n$ 

$\Rightarrow m=9.1\times {{10}^{-31}}\times 1.85\times {{10}^{12}}$

$\therefore m=1.706\times {{10}^{-18}}kg$

Thus, we found that a negligible amount of mass does get transferred from wool to polythene.

a) Two insulated charged copper spheres $A$ and $B$ have their centres separated by a distance of $50cm$. What is the mutual force of electrostatic repulsion if the charge on each is $6.5\times {{10}^{-7}}C$? The radii of A and B are negligible compared to the distance of separation. 

Ans: We are given:

Charges on spheres $A$ and $B$ are equal,

${{q}_{A}}={{q}_{B}}=6.5\times {{10}^{-7}}C$

Distance between the centres of the spheres is given as,

$r=50cm=0.5m$

It is known that the force of repulsion between the two spheres would be given by Coulomb’s law as,

$F=\frac{{{q}_{A}}{{q}_{B}}}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}$

Where, ${{\varepsilon }_{o}}$ is the permittivity of the free space

Substituting the known values into the above expression, we get,

$F=\frac{9\times {{10}^{9}}\times {{(6.5\times {{10}^{-7}})}^{2}}}{{{(0.5)}^{2}}}=1.52\times {{10}^{-2}}N$

Thus, the mutual force of electrostatic repulsion between the two spheres is found to be$F=1.52\times {{10}^{-2}}N$.

b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

Ans: It is told that the charges on both the spheres are doubled and the distance between the centres of the spheres is halved. That is,

${{q}_{A}}'={{q}_{B}}'=2\times 6.5\times {{10}^{-7}}=13\times {{10}^{-7}}C$

$r'=\frac{1}{2}(0.5)=0.25m$

Now, we could substitute these values in Coulomb’s law to get,

$F'=\frac{{{q}_{A}}'{{q}_{B}}'}{4\pi {{\varepsilon }_{0}}r{{'}^{2}}}$

$\Rightarrow F=\frac{9\times {{10}^{9}}\times {{(13\times {{10}^{-7}})}^{2}}}{{{(0.25)}^{2}}}$

$\Rightarrow F=0.243N$

The new mutual force of electrostatic repulsion between the two spheres is found to be $0.243N$.

13. Figure below shows tracks taken by three charged particles in a uniform electrostatic field. Give the signs of the three charges and also mention which particle has the highest charge to mass ratio?

Ans: From the known properties of charges, we know that the unlike charges attract and like charges repel each other. 

So, the particles 1 and 2 that move towards the positively charged plate while repelling away from the negatively charged plate would be negatively charged and the particle 3 that moves towards the negatively charged plate while repelling away from the positively charged plate would be positively charged.

Now, we know that the charge to mass ratio (which is generally known as emf) is directly proportional to the displacement or the amount of deflection for a given velocity. 

Since the deflection of particle 3 is found to be maximum among the three, it would have the highest charge to mass ratio.

14. Consider a uniform electric field $E=3\times {{10}^{3}}\hat{i}N/C$. 

a) Find the flux of this field through a square of side $10cm$whose plane is parallel to the y-z plane. 

Electric field intensity, $\overrightarrow{E}=3\times {{10}^{3}}\hat{i}N/C$

Magnitude of electric field intensity, $\left| \overrightarrow{E} \right|=3\times {{10}^{3}}N/C$

Side of the square, $a=10cm=0.1m$ 

Area of the square, $A={{a}^{2}}=0.01{{m}^{2}}$ 

Since the plane of the square is parallel to the y-z plane, the normal to its plane would be directed in the x direction. So, angle between normal to the plane and the electric field would be, $\theta =0{}^\circ $ 

We know that the flux through a surface is given by the relation, 

$\phi =\left| E \right|\left| A \right|\cos \theta $

Substituting the given values, we get, 

$\Rightarrow \phi =3\times {{10}^{3}}\times 0.01\times \cos 0{}^\circ $

$\therefore \phi =30N{{m}^{2}}/C$

Thus, we found the net flux through the given surface to be $\phi =30N{{m}^{2}}/C$.

b) What would be the flux through the same square if the normal to its plane makes $60{}^\circ $ angle with the x-axis? 

Ans: When the plane makes an angle of $60{}^\circ $ with the x-axis, the flux through the given surface would be,

$\Rightarrow \phi =3\times {{10}^{3}}\times 0.01\times \cos 60{}^\circ $

$\Rightarrow \phi =30\times \frac{1}{2}$

$\Rightarrow \phi =15N{{m}^{2}}/C$

So, we found the flux in this case to be, $\phi =15N{{m}^{2}}/C$.

15. What is the net flux of the uniform electric field of exercise $1.15$ through a cube of side $20cm$ oriented so that its faces are parallel to the coordinate planes?

Ans: We are given that all the faces of the cube are parallel to the coordinate planes. 

Clearly, the number of field lines entering the cube is equal to the number of field lines entering out of the cube. As a result, the net flux through the cube would be zero.

16. Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is $8.0\times {{10}^{3}}N{{m}^{2}}/C$ . 

a) What is the net charge inside the box? 

Ans: We are given that:

Net outward flux through surface of the box,

$\phi =8.0\times {{10}^{3}}N{{m}^{2}}/C$ 

For a body containing of net charge $q$, flux could be given by,

$\phi =\frac{q}{{{\varepsilon }_{0}}}$ 

Where, ${{\varepsilon }_{0}}=8.854\times {{10}^{-12}}{{N}^{-1}}{{C}^{2}}{{m}^{-2}}=$ Permittivity of free space 

Therefore, the charge $q$ is given by

$q=\phi {{\varepsilon }_{0}}$

$\Rightarrow q=8.854\times {{10}^{-12}}\times 8.0\times {{10}^{3}}$

$\Rightarrow q=7.08\times {{10}^{-8}}$

$\Rightarrow q=0.07\mu C$

Therefore, the net charge inside the box is found to be $0.07\mu C$.

b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or why not?

Ans: No, the net flux entering out through a body depends on the net charge contained within the body according to Gauss’s law. 

So, if the net flux is given to be zero, then it can be inferred that the net charge inside the body is zero. 

However, the net charge of the body being zero only implies that the body has equal amount of positive and negative charges and thus, we cannot conclude that there were no charges inside the box.

17. A point charge $+10\mu C$ is a distance $5cm$ directly above the centre of a square of side $10cm$, as shown in Figure below. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge $10cm$) 

Ans: Consider the square as one face of a cube of edge length $10cm$ with a charge $q$ at its centre, according to Gauss's theorem for a cube, total electric flux is through all its six faces.

${{\phi }_{total}}=\frac{q}{{{\varepsilon }_{0}}}$

The electric flux through one face of the cube could be now given by,  \[\phi =\frac{{{\phi }_{total}}}{6}\].

$\phi =\frac{1}{6}\frac{q}{{{\varepsilon }_{0}}}$

${{\varepsilon }_{0}}=8.854\times {{10}^{-12}}{{N}^{-1}}{{C}^{2}}{{m}^{-2}}=$ Permittivity of free space

The net charge enclosed would be, $q=10\mu C=10\times {{10}^{-6}}C$

Substituting the values given in the question, we get, 

$\phi =\frac{1}{6}\times \frac{10\times {{10}^{-6}}}{8.854\times {{10}^{-12}}}$

$\therefore \phi =1.88\times {{10}^{5}}N{{m}^{2}}{{C}^{-1}}$

Therefore, electric flux through the square is found to be $1.88\times {{10}^{5}}N{{m}^{2}}{{C}^{-1}}$.

18. A point charge of $2.0\mu C$ is kept at the centre of a cubic Gaussian surface of edge length $9cm$. What is the net electric flux through this surface? 

Ans: Let us consider one of the faces of the cubical Gaussian surface considered (square).

Since a cube has six such square faces in total, we could say that the flux through one surface would be one-sixth the total flux through the gaussian surface considered. 

The net flux through the cubical Gaussian surface by Gauss’s law could be given by, 

So, the electric flux through one face of the cube would be, \[\phi =\frac{{{\phi }_{total}}}{6}\]

$\Rightarrow \phi =\frac{1}{6}\frac{q}{{{\varepsilon }_{0}}}$……………………………….. (1)

But we have, 

Charge enclosed, $q=10\mu C=10\times {{10}^{-6}}C$

Substituting the given values in (1) we get, 

$\Rightarrow \phi =1.88\times {{10}^{5}}N{{m}^{2}}{{C}^{-1}}$

Therefore, electric flux through the square surface is $1.88\times {{10}^{5}}N{{m}^{2}}{{C}^{-1}}$.

19. A point charge causes an electric flux of $-1.0\times {{10}^{3}}N{{m}^{2}}/C$ to pass through a spherical Gaussian surface of $10cm$ radius centred on the charge. 

a) If the radius of the Gaussian surface were doubled, how much flux could pass through the surface? 

Electric flux due to the given point charge, $\phi =-1.0\times {{10}^{3}}N{{m}^{2}}/C$ 

Radius of the Gaussian surface enclosing the point charge,$r=10.0cm$ 

Electric flux piercing out through a surface depends on the net charge enclosed by the surface according to Gauss’s law and is independent of the dimensions of the arbitrary surface assumed to enclose this charge. 

Hence, if the radius of the Gaussian surface is doubled, then the flux passing through the surface remains the same i.e., $-{{10}^{3}}N{{m}^{2}}/C$ .

b) What is the magnitude of the point charge?

Ans: Electric flux could be given by the relation,

Where,$q=$ net charge enclosed by the spherical surface

$\Rightarrow q=\phi {{\varepsilon }_{0}}$

Substituting the given values,

$\Rightarrow q=-1.0\times {{10}^{3}}\times 8.854\times {{10}^{-12}}=-8.854\times {{10}^{-9}}C$

$\Rightarrow q=-8.854nC$

Thus, the value of the point charge is found to be $-8.854nC$.

20. A conducting sphere of radius $10cm$ has an unknown charge. If the electric field at a point $20cm$ from the centre of the sphere of magnitude $1.5\times {{10}^{3}}N/C$ is directed radially inward, what is the net charge on the sphere?

Ans: We have the relation for electric field intensity $E$ at a distance \[\left( d \right)\] from the centre of a sphere containing net charge $q$ is given by,

$E=\frac{q}{4\pi {{\varepsilon }_{0}}{{d}^{2}}}$ ……………………………………………… (1)

Where, 

Net charge, $q=1.5\times {{10}^{3}}N/C$

Distance from the centre, $d=20cm=0.2m$ 

From (1), the unknown charge would be, 

$q=E\left( 4\pi {{\varepsilon }_{0}} \right){{d}^{2}}$

$\Rightarrow q=\frac{1.5\times {{10}^{3}}\times {{\left( 0.2 \right)}^{2}}}{9\times {{10}^{9}}}=6.67\times {{10}^{-9}}C$

$\Rightarrow q=6.67nC$

Therefore, the net charge on the sphere is found to be$6.67nC$.

21. A uniformly charged conducting sphere of $2.4m$ diameter has a surface charge density of $80.0\mu C/{{m}^{2}}$ .

a) Find the charge on the sphere. 

Ans: Given that,

Diameter of the sphere, $d=2.4m$. 

Radius of the sphere, $r=1.2m$. 

Surface charge density, 

\[\sigma =80.0\mu C/{{m}^{2}}=80\times {{10}^{-6}}C/{{m}^{2}}\] 

Total charge on the surface of the sphere,

$Q=\text{Charge density }\times \text{ Surface area}$ 

$\Rightarrow \text{Q}=\sigma \times \text{4}\pi {{\text{r}}^{2}}=80\times {{10}^{-6}}\times 4\times 3.14\times {{\left( 1.2 \right)}^{2}}$

$\Rightarrow Q=1.447\times {{10}^{-3}}C$

Therefore, the charge on the sphere is found to be $1.447\times {{10}^{-3}}C$.

b) What is the total electric flux leaving the surface of the sphere?

Ans: Total electric flux $\left( {{\phi }_{total}} \right)$ leaving out the surface containing net charge $Q$ is given by Gauss’s law as, 

${{\phi }_{total}}=\frac{Q}{{{\varepsilon }_{0}}}$…………………………………………………. (1)

Where, permittivity of free space,

${{\varepsilon }_{0}}=8.854\times {{10}^{-12}}{{N}^{-1}}{{C}^{2}}{{m}^{-2}}$

We found the charge on the sphere to be, 

$Q=1.447\times {{10}^{-3}}C$

Substituting these in (1), we get, 

${{\phi }_{total}}=\frac{1.447\times {{10}^{-3}}}{8.854\times {{10}^{-12}}}$

$\Rightarrow {{\phi }_{total}}=1.63\times {{10}^{-8}}N{{C}^{-1}}{{m}^{2}}$

Therefore, the total electric flux leaving the surface of the sphere is found to be $1.63\times {{10}^{-8}}N{{C}^{-1}}{{m}^{2}}$.

22. An infinite line charge produces a field of magnitude $9\times {{10}^{4}}N/C$ at a distance of $2cm$. Calculate the linear charge density.

Ans: Electric field produced by the given infinite line charge at a distance $d$having linear charge density$\lambda $ could be given by the relation,

$E=\frac{\lambda }{2\pi {{\varepsilon }_{0}}d}$ 

$\Rightarrow \lambda =2\pi {{\varepsilon }_{0}}Ed$…………………………………….. (1)

$d=2cm=0.02m$  

$E=9\times {{10}^{4}}N/C$ 

Permittivity of free space,

${{\varepsilon }_{0}}=8.854\times {{10}^{-12}}{{N}^{-1}}{{C}^{2}}{{m}^{-2}}$ 

Substituting these values in (1) we get, 

$\Rightarrow \lambda =2\pi \left( 8.854\times {{10}^{-12}} \right)\left( 9\times {{10}^{4}} \right)\left( 0.02 \right)$

$\Rightarrow \lambda =10\times {{10}^{-8}}C/m$

Therefore, we found the linear charge density to be $10\times {{10}^{-8}}C/m$.

23. Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude $17.0\times {{10}^{-22}}C{{m}^{-2}}$. What is $E$ in the outer region of the first plate? What is $E$ in the outer region of the second plate? What is E between the plates?

Ans: The given nature of metal plates is represented in the figure below: 

Here, A and B are two parallel plates kept close to each other. The outer region of plate A is denoted as $I$, outer region of plate B is denoted as $III$, and the region between the plates, A and B, is denoted as $II$.

It is given that:

Charge density of plate A, $\sigma =17.0\times {{10}^{-22}}C/{{m}^{2}}$

Charge density of plate B, $\sigma =-17.0\times {{10}^{-22}}C/{{m}^{2}}$

In the regions $I$and$III$, electric field E is zero. This is because the charge is not enclosed within the respective plates.

Now, the electric field $E$ in the region $II$ is given by

$E=\frac{|\sigma |}{{{\varepsilon }_{0}}}$ 

Permittivity of free space ${{\varepsilon }_{0}}=8.854\times {{10}^{-12}}{{N}^{-1}}{{C}^{2}}{{m}^{-2}}$ 

$E=\frac{17.0\times {{10}^{-22}}}{8.854\times {{10}^{-12}}}$

$\Rightarrow E=1.92\times {{10}^{-10}}N/C$

Thus, the electric field between the plates is $1.92\times {{10}^{-10}}N/C$.

List of Important Formulas in Class 12 Physics Chapter 1 Electric Charges And Fields

While preparing a chapter, it is important for students to memorize the formula of a particular topic. This will help them to boost their score. Some important formulas are listed below.

Coulomb’s Law

The electric force between the two-point charges are directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

F represents the electric force.

K represents Coulomb’s constant.

$q_{1}q_{2}$ represents the two charges.

$\lambda$ represents the distance between two charges.  

Electric field intensity 

Electric field intensity is the vector quantity.             

E $\frac{F}{q_{1}}$

Here, 

F is the force experienced by the test charge.

$q_1$ is the test charge.

Electric Flux 

dФ = E.da cos𝛉 

Chapter Summary of Electric Charges & Fields NCERT Solutions

From simple experiments on frictional electricity, one can infer that there are two types of charges in nature; and that like charges repel and unlike charges attract. By convention, the charge on a glass rod rubbed with silk is positive; that on a plastic rod rubbed with fur is then negative.

Charge is not only a scalar (or invariant) under rotation; it is also invariant for frames of reference in relative motion. This is not always true for every scalar. For Example-, kinetic energy is a scalar under rotation, but is not invariant for frames of reference in relative motion.

Conductors allow movement of electric charge through them, insulators do not. In metals, the mobile charges are electrons; in electrolytes both positive and negative ions are mobile.

Coulomb’s Law: The mutual electrostatic force between two point charges q 1 and q 2 is proportional to the product q 1 q 2 and inversely proportional to the square of the distance r 21 separating them. 

$F_{21}=\text{force}\;q_{2} \;\text{du to}\;q_{1}\;=\frac{k(q_{1}q_{2})}{r^2_{21}}\;r_{21}$

Where $r_{21}$ is a unit vector in the direction from q 1 to q 2 and $k=\frac{1}{4\pi \varepsilon _{0}}$ is the constant of proportionality.

In SI units, the unit of charge is coulomb. The experimental value of the constant $\varepsilon _{0}$ is 

$\varepsilon _{0}=8.854\times 10^{-12}C^{2}N^{-}M^{-2}$

The approximate value of k is- 9\times $10^{9}Nm^2C^{-2}$

Superposition Principle: The principle is based on the property that the forces with which two charges attract or repel each other are not affected by the presence of a third (or more) additional charge(s). For an assembly of charges Q 1 , Q 2 , Q 3 ..., the force on any charge, say Q 1 , is the vector sum of the force on  Q 1 due to Q 2 , the force on Q 1 due to Q 3 , and so on. For each pair, the force is given by Coulomb's law for two charges stated earlier.

An electric field line is a curve drawn in such a way that the tangent at each point on the curve gives the direction of the electric field at that point. The relative closeness of field lines indicates the relative strength of electric field at different points; they crowd near each other in regions of strong electric field and are far apart where the electric field is weak. In regions of constant electric field, the field lines are uniformly spaced parallel straight lines.

Some of the important properties of field lines are: (i) Field lines are continuous curves without any breaks. (ii) Two field lines cannot cross each other. (iii) Electrostatic field lines start at positive charges and end at negative charges (iv) They cannot form closed loops.

Electric flux is the measure of the field lines crossing a surface. It is scalar quantity, with SI unit $\frac{N}{C}$-m 2 or V-m. “The number of field lines passing through perpendicular unit area will be proportional to the magnitude of Electric Field there” 

Gauss’s law: The flux of electric field through any closed surface S is $1/\varepsilon _0$ times the total charge enclosed by S. The law is especially useful in determining electric field E, when the source distribution has simple symmetry.

Field of an electric dipole in its equatorial plane (i.e., the plane perpendicular to its axis and passing through its centre) at a distance r from the centre:

$E=\frac{-P}{4\pi \varepsilon _0}\frac{1}{(\alpha^2+r^2)^{3/2}}\cong \frac{-P}{4\pi \varepsilon _0r^3}.\,\text{for}\,r\gg \alpha $

Dipole electric field on the axis at a distance r from the centre:

$E=\frac{2pr}{4\pi\varepsilon_0(r^2-\alpha^2)^3}\cong \frac{2{p}}{4\pi\varepsilon_0{r}^3}\,\,\text{for}\,\,r\gg\alpha$

The 1/r 3 dependence of dipole electric fields should be noted in contrast to the 1/r 2 dependence of electric field due to a point charge.

In a uniform electric field E, a dipole experiences a torque  τ given by: τ = p × E

but experiences no net force.

Electric Field Due to Various Uniform Charge Distribution

Overview of Deleted Syllabus for CBSE Class 12 Physics Electric Charges and Fields

NCERT Solutions for Class 12 Physics Chapter 1 - Electric Charges And Fields focuses on mastering fundamental concepts like Coulomb's law, electric field, Gauss's law, and electric dipole. It's important to understand these principles and their applications thoroughly. Key topics include electric charge properties, force interactions between multiple charges, electric flux, and the behaviour of electric fields in various configurations. Previous year question papers often include 17 marks (5-6 questions) worth of questions from this chapter, emphasizing its importance in the exam. Focus on practising both exercise and additional questions provided in the NCERT solutions to solidify your understanding and improve exam performance.

Other Study Material for CBSE Class 12 Physics Chapter 1

Chapter-Specific NCERT Solutions for Class 12 Physics

Given below are the chapter-wise NCERT Solutions for Class 12 Physics. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

FAQs on NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields

1. What are the Benefits of Class 12 Physics Chapter 1 NCERT Solutions?

NCERT Solutions of the chapter of Electric Charges and Fields class 12 come with an elaborate discussion of the topic and subtopics. Moreover, the solutions are explained in a simple manner for a better understanding of the concepts.

Students who want to ace in the CBSE 12th board examination as well as in competitive exams like JEE can seek the guidance of NCERT solutions. The solutions are created in light of the latest CBSE guidelines that help students to secure better marks. Students can completely rely on the solutions provided by Vedantu, as these are all solved by the subject matter expert of Physics.

2. How can I get the NCERT Class 12 Physics Chapter 1 PDF?

The NCERT solutions of Physics class 12 chapter 1 PDF can be downloaded from the Vedantu website and mobile application. Therefore, students who are facing problems in solving the very first chapter of class 12 Physics can seek guidance in the NCERT solutions. The way every solution is described has been considered helpful. Interestingly, students can download this PDF for absolutely free.

3. What are the Subtopics of the Chapter Electric Charges and Fields?

The first chapter of Physics Class 12 Electric Charges and Fields consists of several sub-topics under it, and these are as follows. 

Introduction

Electric Charge

Conductors and Insulators

Charging by Induction

Basic properties of Electric Charge

Forces between multiple charges

Electric fields

Electric field lines

Electric flux

Electric Dipole

Dipole in a uniform external field

Continuous charge distribution

Gauss’s Law

Application of Gauss’s Law

Students who want to get an in-depth knowledge of these topics can consult the class 12 Physics chapter 1 NCERT solutions provided by Vedantu.

4. Are the NCERT Solutions for Class 12 Physics Chapter 1 sufficient for the exam preparation?

Yes, the NCERT Solutions for Class 12 Physics Chapter 1 is sufficient for the exam preparation. The solutions are well detailed and include every important point that will help students to score high. Each numerical problem is explained in a simple manner so that students can clear their basic concepts and learn the approach to solve questions in Physics. They can refer to  Vedantu’s revision notes  that contain all the important formulas and key points. The solution PDFs and other study materials such as important questions and revision notes can also be downloaded from the Vedantu  app as well for free of cost.

5. How do you solve electric charges and fields?

Electric charges and fields are the first chapters of Class 12 Physics. Electric charges and fields are the basics to understand the concepts in further chapters. Solve every problem from your NCERT Physics book. Refer to  Vedantu’s Solutions for Class 12 Physics  to understand every concept. Every minute detail is included in this material, with an emphasis on strengthening your concept. You can download the Solutions for free from Vedantu’s website. 

6. What is the first chapter of Class 12 Physics?

The first chapter of Class 12 Physics is Electric charges and Fields. It introduces the concepts of charges to the student and how it affects other materials. Students will learn about insulators, conductors, their working procedures, etc. Coulomb’s law and its applications are covered in detail. Vedantu’s NCERT Solution PDF for Class 12 Physics Chapter 1 consists of 34 questions to help students gain command over this topic. It is the best guide to understand the concepts. 

7. What is electric charge and fields?

Electric Charge is the property of a material due to which it experiences a force in an electric, magnetic, or electromagnetic field. There are two different types of charges, positive charge and negative charge. The region around an electrically charged particle up to which it can exert an electrical force on another electrically charged particle is termed an electric field. It can cause either the force of attraction or repulsion, depending on the nature of the charges. 

8. What are some tips to utilise NCERT books effectively for Class 12 exams?

Following are the basic and helpful tips to utilise NCERT books effectively for Class 12 exams:

Make a timetable and stick to it. Your timetable should be unique to you and should be according to your needs and capabilities.

Do not skip topics. Cover the entire syllabus. If you get stuck at any point, you should immediately get your doubts cleared. Refer to Vedantu’s Solutions for Class 12 to clear all your doubts and strengthen your concepts.

Analyse the solutions after you have successfully solved a question. It will help to develop problem-solving capabilities. 

9. What is the focus of Class 12 Physics Chapter 1 NCERT Solutions?

Class 12 Physics Chapter 1 NCERT Solutions focuses on understanding the fundamental principles of electrostatics, which include the nature of electric charges, Coulomb's law, electric fields, and Gauss's law. This chapter lays the groundwork for more advanced topics in electromagnetism.

10. What are electric charges and their properties mentioned in Chapter 1 Physics Class 12?

Electric charges are fundamental properties of matter that cause it to experience a force in an electric field. There are two types of electric charges: positive and negative. Key properties include

Quantization: Charge exists in discrete amounts, typically in multiples of the elementary charge (e.g., the charge of a proton or electron).

Conservation: The total charge in an isolated system remains constant.

11. How are electric field lines used to represent electric fields in Class 12 Physics Ch 1 NCERT Solutions?

In Electric Charges and Fields Class 12 NCERT Solutions, Electric field lines are imaginary lines that represent the direction and strength of an electric field. They start with positive charges and end with negative charges. The density of these lines indicates the strength of the field: closer lines represent stronger fields.

12. What types of questions are included in the Electric Charges and Fields NCERT Solutions?

The NCERT solutions Class 12 Physics Chapter 1 include a variety of questions such as:

Conceptual questions to test understanding of electric charges and fields.

Numerical problems involving Coulomb's Law and electric field calculations.

Application-based questions using Gauss's Law and electric dipoles.

13. Why is it important to master the concepts in Class 12 Physics Chapter 1 Exercise Solutions?

Mastering the concepts in Chapter 1 Physics Class 12 is crucial because these principles are foundational for advanced topics in electromagnetism. They are also widely applicable in various technological and engineering fields, including the design of electronic components and the study of molecular interactions.

NCERT Solutions Class 12 Physics

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NCERT Solutions For Class 12 Physics Chapter 1 Electric Charges And Fields

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