Class 12 Chapter 1

Test your Knowledge

question 1 of 5

What are the key properties of electric charges?

1.Conservation of electric Charge

2.Charge can be transferred

3.Charge is always associated with mass

4.All of the above are correct

Must-Revise Concepts

Forces Between Multiple Charges

Electric Potential And Capacitance

Relation Between Electric Field And Electric Potential

1.0Electric Charges And Fields: Key Concepts

Electric Charges and Fields is a key chapter in electromagnetism that explains the basics of electric charge—how objects can be positively or negatively charged, and how like charges repel while opposites attract. It covers important properties like charge conservation and quantization. The chapter introduces Coulomb’s Law to measure the force between charges, and explains electric fields—invisible regions around a charge where other charges feel a force. It also discusses electric field lines, dipole behavior in fields, and Gauss’s Law, which helps calculate electric fields in cases with symmetry.

2.0Introduction to Electric Charges

Electric charge: It’s a fundamental property of matter responsible for electric and magnetic interactions.

Types of Charges

Positive charge – Occurs when there is a loss of electrons.
Negative charge – Occurs when there is a gain of electrons.

Units of Charge

SI - Coulomb(C)
CGS-Stat coulomb (stat C)
Dimensional Formula -[AT]

3.0Properties of Electric Charges

Additivity of electric Charge.
Conservation of electric Charge
Quantization of Electric Charge
Charge can be transferred
Charge is invariant.
Charge is a scalar quantity
Charge is always associated with mass

Class 12 Physics Chapters List:- JEE & NEET Revision Notes

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: Materials, Devices and Simple Circuits

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Class 12 Physics Chapter 1 Electric Charges and Fields: Download Revision Notes

Master Electric Charges and Fields with expert study material and key concepts. Explore past-year weightage, JEE and NEET exam trends, and prepare smarter with focused resources.

Electric Charges and Fields smart study material curated by ALLEN expert faculty for effective learning, quick revision, and exam preparation. This chapter covers key concepts of electrostatics, including electric charges, Coulomb’s Law, electric field and field lines, electric dipole, electric flux, and Gauss’s Law. Physics Class 12 Chapter 1 revision notes brings together concise theory, important formulas, solved examples, previous-year question trends, and exam-focused tips in a structured format.

1.0Chapter Snapshot

Parameter	Details
Difficulty Level	Moderate to High
JEE Main Weightage	High (Generally 1–2 Questions)
JEE Advanced Weightage	High (2–4 Integrated Concepts)
Key Concepts Covered	Electric Charge, Coulomb's Law, Electric Field, Electric Flux, Electric Dipole, Gauss's Law

2.0Chapter 1 Smart Study Material by ALLEN Experts: Download PDFs

Download below study resources by ALLEN experts, created to help students strengthen their concepts and prepare effectively for competitive and school exams.

NCERT Solutions – Chapter 1
JEE Main PYQs with Solutions
JEE Advanced PYQs with Solutions
JEE Important Questions & Solutions
NEET Important Questions & Solutions

These study materials reinforce conceptual understanding, improve problem-solving speed, and enhance overall exam readiness to help students perform with confidence and score higher.

3.0Learning Outcomes

After completing these Electric Charges and Fields Revision Notes, you will be able to:

Understand the properties of electric charge, including conservation, quantization, additivity, and the transfer of charge.
Apply Coulomb's Law to calculate the electrostatic force between two or more point charges in different mediums.
Use the Superposition Principle to determine the net electric force and electric field due to multiple charges.
Calculate electric field intensity for point charges, continuous charge distributions, and electric dipoles.
Interpret electric field lines and understand their significance in representing the direction and strength of electric fields.
Evaluate electric flux through various surfaces and apply the concept to different electrostatic situations.
Apply Gauss's Law to solve problems involving symmetric charge distributions such as spheres, cylinders, and infinite sheets.
Analyze the behaviour of electric dipoles in uniform electric fields, including torque and potential energy.

Test your Knowledge

question 1 of 5

What are the key properties of electric charges?

1.Conservation of electric Charge

2.Charge can be transferred

3.Charge is always associated with mass

4.All of the above are correct

Must-Revise Concepts

Forces Between Multiple Charges

Electric Potential And Capacitance

Relation Between Electric Field And Electric Potential

1.0Electric Charges And Fields: Key Concepts

Electric Charges and Fields is a key chapter in electromagnetism that explains the basics of electric charge—how objects can be positively or negatively charged, and how like charges repel while opposites attract. It covers important properties like charge conservation and quantization. The chapter introduces Coulomb’s Law to measure the force between charges, and explains electric fields—invisible regions around a charge where other charges feel a force. It also discusses electric field lines, dipole behavior in fields, and Gauss’s Law, which helps calculate electric fields in cases with symmetry.

2.0Introduction to Electric Charges

Electric charge: It’s a fundamental property of matter responsible for electric and magnetic interactions.

Types of Charges

Positive charge – Occurs when there is a loss of electrons.
Negative charge – Occurs when there is a gain of electrons.

Units of Charge

SI - Coulomb(C)
CGS-Stat coulomb (stat C)
Dimensional Formula -[AT]

3.0Properties of Electric Charges

Additivity of electric Charge.
Conservation of electric Charge
Quantization of Electric Charge
Charge can be transferred
Charge is invariant.
Charge is a scalar quantity
Charge is always associated with mass

1.0Coulomb’s Law

The electrostatic force between two point charges is directly proportional to their charges and inversely proportional to the square of the distance between them, always acting along the line joining them.

$F = K \frac{q _{1} q _{2}}{r ^{2}}$

$k = \frac{1}{4 π ϵ _{0}} = 9 \times 1 0^{9} N m^{2} C^{- 2}$ =Electrostatic constant or Coulomb’s Constant

Coulomb's Law in Vector Form

$F_{21} = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2} ( r _{2} - r _{1} )}{∣ r _{2} - r _{1} ∣ ^{3}} or F_{12} = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2} ( r _{1} - r _{2} )}{∣ r _{1} - r _{2} ∣ ^{3}}$

2.0Superposition Principle

When multiple charges are present, the net force on any charge is the vector sum of forces from all the other charges.

Superposition Principle in electrostatics

$F = F_{01} + F_{02} + F_{03} + \dots + F_{0 n}$

$F = \frac{k q _{0} q _{1}}{r _{1}^{2}} \overset{r}{^}_{1} + \frac{k q _{0} q _{2}}{r _{2}^{2}} \overset{r}{^}_{2} + \dots + \frac{k q _{0} q _{n}}{r _{n}^{2}} \overset{r}{^}_{n}$

Relative Permittivity( $ϵ_{r}$ ) or Dielectric Constant (K)

$Dielectric Constant (K) = \frac{Permittivity of a Medium ( ε )}{Permittivity of Vacuum ( ε _{0} )}$

$K = \frac{ε _{r}}{ε _{0}}$

$ε = ε_{0} ε_{r}$

The value of $ε_{r}$ or K ≥1

3.0Electric Field

An electric field is the region around a charge where it exerts a force on other charges.
Mathematically, $E = \frac{F}{q _{0}}$ SI unit : N/C or V/m
$E = q_{0} \to 0 lim \frac{F}{q _{0}}$
Dimensional Formula $[M^{1} L^{1} T^{- 3} A^{- 1}]$
It is a vector quantity.

4.0Electric Field Lines

Electric field lines are imaginary lines, straight or curved, that represent the direction and strength of the electric field around a charged object. At any point on these lines, the tangent indicates the direction of the electric field at that location.
Two electric field lines can never intersect each other.

Electrostatic field lines can never form closed loops.

Electric field lines due to positive or negative charges

Electric line of force due to an electric dipole

5.0Continuous Charge Distribution

A group of closely spaced electric charges forms a continuous charge distribution.

Linear Charge Density()

Surface Charge Density ()

Volume Charge Density()

$λ = \frac{Q}{l}$

SI Unit = $\frac{C}{m}$

$σ = \frac{Q}{S}$

SI Unit = $\frac{C}{m ^{2}}$

$ρ = \frac{Q}{V}$

SI Unit = $\frac{C}{m ^{3}}$

6.0Electric Dipole

It is a pair of equal and opposite charges separated by a small distance.

The dipole moment is the product of the magnitude of either charge and the distance between them. $p = q (2 l)$
S.I. unit- Cm

Electric Field Due to a Dipole

At Axial / End on position

$E_{A x ia l} = \frac{1}{4 π ε _{0}} \frac{2 p}{r ^{3}}$

At Equator/Broadside on position

Electric Dipole at Equator/Broadside on position

$E_{Eq u i t or ia l} = \frac{1}{4 π ε _{0}} \frac{p}{r ^{3}}$

At general position

Electric field due to a dipole at general position

$E = \frac{p}{4 π ε _{0} r ^{3}} 3 co s^{2} θ + 1$

$T an α = \frac{1}{2} T an θ$

Electric Field Intensity due to a charged wire

Special Cases:

For Infinite Wire,( both ends goes to infinite)

$EX = E ⊥= \frac{2 K λ}{r}$

E_Y=E॥=0

For Semi-infinite wire

For Semi-infinite wire - Electric Field Intensity due to a charged wire

EX=E丄= $\frac{K λ}{r}$

$E_{Y} = E_{∣∣} = \frac{K λ}{r}$

$E_{R es u lt an t} = 2 \frac{K λ}{r}$

Electric field due to finite wire at symmetric point:

$E_{X} = E_{⊥} = \frac{2 K λ}{r} s in θ$

$E_{Y} = E_{॥} = 0$

Electric Field due to a uniformly charged Arc

Electric Field due to a uniformly charged Arc

$E = \frac{2 K λ}{R} S in (\frac{θ}{2})$

Electric field at centre of uniformly charged Ring

$E = \frac{2 K λ}{R} S in (\frac{θ}{2}), θ \to$ angle of arc

$θ$ =360⁰

$E = \frac{2 K λ}{R} S in (\frac{360}{2})$

E =0

Electric Field due to a Uniformly charged Ring at its Axis

Electric Field due to a Uniformly charged Ring at its Axis

$E = \frac{k Q x}{( R ^{2} + x ^{2} ) ^{3/2}}$

Special cases:

Electric field on the axis for small values of x : $E = \frac{k Q}{R ^{3}} \cdot x$
Electric field at the centre of the ring is zero because x=0
Electric field at the axis for larger values of x: $E \approx \frac{k Q}{x ^{2}}$
Maximum value of electric field

$\frac{d E}{d x} = 0$

$x = \pm \frac{R}{2}$

$E_{max} = \frac{2 k Q}{3 3 R ^{2}}$

Variation of electric field with distance

7.0Dipole placed in an electric field, torque acts on it

て=p E Sin

(∴ $θ$ is the angle between dipole moment (p)and electric field (E))

$τ = p \times E$

Special Cases:

If =0⁰ then stable Equilibrium.
If =180⁰ then て=o, unstable equilibrium.

8.0Work done in rotating a dipole in a uniform electric field

When an electric dipole with dipole moment is oriented at an angle to an electric field , torque is exerted on the dipole, causing it to rotate. The resulting work done can be expressed as

$W = pE (C os θ_{1} - C os θ_{2})$

9.0Electric Flux

This physical quantity is used to measure strength of electric field and it is defined as the total number of electric field lines passing through an area.
Electric flux is a scalar quantity.
Unit of Electric Flux:Nm2/C or V m
Dimensional Formula-[M¹L³T^-3A^-1]
Electric flux through a large surface $ϕ = \int d ϕ = \int E \cdot d A$
If Electric Field is uniform= $\Rightarrow E$ =Constant(same everywhere)

$ϕ = E \cdot \int d A (∵ \int d A = total area vector of a plane surface)$

$\Rightarrow ϕ = E \cdot A \Rightarrow ϕ = E A cos θ$

Different cases

If $θ$ =0⁰, $ϕ = E A cos 0^{\circ} = E A (positive flux means outgoing or leaving)$

If $θ$ is 90⁰, $ϕ = E A cos 90° = 0.$
If $θ$ is 180⁰ , $ϕ = E A C os 180° = - E A$ (negative flux means incoming or entering)

10.0Gauss’s Law

According to this law the total electric flux ( $ϕ$ ) through any closed surface (S) in free space is equal to $\frac{1}{ε _{0}}$ times the total electric charge (q) enclosed by the surface.

$ϕ = \oint E . d S = \frac{Q _{e n c l ose d}}{ε _{0}}$

11.0Applications of Gauss Law

Electric field Intensity due to infinitely long wire

Electric field Intensity due to infinitely long wire

Variation of electric field intensity with distance

$E = \frac{λ}{2 π ε _{0} r} = \frac{2 K λ}{r}$

Electric field due to uniformly charged long cylindrical pipe/cylindrical shell

Case 1.Electric field at any point outside the cylinder(r>R): $E = \frac{σ R}{ϵ _{0} r}$
Case 2.For the point lying on the surface(r≈R): $E = \frac{σ}{ε _{0}}$
Case 3.For the point inside the surface(r<R): $E_{in s i d e} = 0$

Electric Field due to Uniformly Charged Infinite Sheet

Non Conducting Sheet: $E = \frac{σ}{2 ε _{0}}$
Conducting sheet or Metal Plate: $E = \frac{σ}{ε _{0}}$

Electric Field due to the charged conducting sphere or charged thin shell

Electric Field at any point outside the sphere (r>R)

Electric Field at any point outside the sphere

$E = \frac{q}{4 π ε _{0} r ^{2}} = \frac{k q}{r ^{2}}$

For any point lying on the surface of sphere (r=R)

$E_{s} = \frac{k q}{R ^{2}} = \frac{q}{4 π ε _{0} R ^{2}} = \frac{σ}{ε _{0}}$ $(∵ σ = \frac{q}{4 π R ^{2}})$

Electric Field at any point Inside the sphere(r<R)

In this case charge enclosed by the gaussian surface is zero, i.e., E_inside=0

Electric Field at any point Inside the sphere

Variation of E with r

variation of eletric field intensity with distance

Electric Field due to uniformly charged non conducting sphere(solid sphere)

Electric field at any point outside the sphere (r>R)

$E = \frac{k q}{r ^{2}}$

Electric field at any point lying on the surface of sphere(r=R)

$E_{S} = \frac{ρR}{3 ε _{0}}$ (∴ $ρ$ is volume charge density)

Electric field at any point inside the sphere(r<R)

$E_{in s i d e} = \frac{1}{4 π ε _{0}} \frac{q}{R ^{3}} r = \frac{r h o}{3 ε _{0}} (r)$

Variation of E with r

variation of electric field intensity with distance for a solid sphere

12.0Sample Questions on Electric Charges And Fields

Q-1. $1 0^{10}$ alpha particles are ejected per second from a body, then after how much time, the body will acquire a charge of 8 μC ?

Solution: Charge appear per second on body, $Q = (\frac{N}{t}) q_{α}$

$Q = 1 0^{10} \times (2 \times 1.6 \times 1 0^{- 19}) = 3.2 \times 1 0^{- 9} C / sec$

Let t be the time taken to acquire charge of 8 μC then

$t = \frac{8 \times 1 0 ^{- 6}}{3.2 \times 1 0 ^{- 9}} = 2.5 \times 1 0^{3} sec$

Q-2.Find final charges on the spheres when switch S is closed.

Sample Questions on Electric Charges And Fields

Solution: Total charge of the system = $Q = 60 μ C + 0 = 60 μ C$

Then after conduction

$\frac{Q _{1}^{'}}{Q _{2}^{'}} = \frac{R _{1}}{R _{2}} \Rightarrow \frac{2}{3}$

$Q 1^{'} = (\frac{2}{2 + 3}) 60 μ C = 24 μ C$

$Q 2^{'} = (\frac{3}{2 + 3}) 60 μ C = 36 μ C$

Q-3.Why can we not use Coulomb’s law for large size bodies ?

Solution:

When large size charged conducting spheres brought close to each other, there charges moves away due to repulsion hence effective distance between their centers increases r'>r

$F_{a c t u a l} = F_{c a l c u l a t e d} = k \frac{q _{1} q _{2}}{r ^{2}}$

Q-4.Two identical conducting spheres A and B carry equal charges and are placed at a distance r apart in vacuum. The electrostatic force between them is F. A third identical uncharged sphere C is first brought into contact with sphere A, then with sphere B, and is finally removed. What will be the new electrostatic force between spheres A and B?

Solution:

$F = k \frac{q ^{2}}{r ^{2}}$ ……….(1)

Now C (uncharged sphere) is touched with A first

Then C is touched with B

New force between A & B now is

$F^{'} = \frac{k ( \frac{Q}{2} ) ( \frac{3 Q}{4} )}{r ^{2}} = \frac{3 k Q ^{2}}{8 r ^{2}} = \frac{3 F}{8}$

Q-5.Two equally charged spheres are placed far apart. If the gravitational force equals the electrostatic force between them, find the ratio of specific charge $\frac{q}{m}$ .

Solution: Equating the gravitational and electrostatics force

$\frac{k q ^{2}}{r ^{2}} = \frac{G m ^{2}}{r ^{2}} \Rightarrow \frac{q}{m} = \frac{G}{k}$ $= \frac{6.67 \times 1 0 ^{- 11}}{9 \times 1 0 ^{9}} = \frac{0.74}{1 0 ^{20}} = \frac{0.86}{1 0 ^{10}} = 0.86 \times 1 0^{- 10}$

1.0High Weightage Topics

The following topics have consistently appeared in JEE Main and JEE Advanced examinations and should be given higher priority during revision.

Topic	Focus Area
Coulomb's Law	Force between charges, distance variation
Superposition Principle	Net force and electric field due to multiple charges
Electric Field Intensity	Point charge, line charge, ring, and symmetry
Electric Flux	Surface orientation and flux calculations
Gauss's Law	Symmetric charge distributions
Electric Dipole	Dipole moment, torque, energy
Dipole in Uniform Electric Field	Torque and equilibrium conditions
Continuous Charge Distribution	Ring, rod, and disc-based numericals

2.0Formula Handbook

The following formulas are the most important for solving JEE Main and JEE Advanced questions from Electrostatics.

3.0Common Mistakes & JEE Tips

Electric Charges and Fields is a concept-driven chapter where even well-prepared students lose marks due to errors in sign conventions, vector addition, formula selection, and understanding electric field concepts. Avoiding these common mistakes can significantly improve your accuracy and confidence in JEE Main and JEE Advanced.

Common Mistake	Correct Approach
Confusing Electric Force with Electric Field	Electric Force depends on both the source charge and the test charge, whereas the Electric Field depends only on the source charge.
Using algebraic addition instead of vector addition	Electric force and electric field are vector quantities. Always use vector addition while solving multiple-charge problems.
Applying Coulomb's Law incorrectly	Ensure the charges are treated as point charges and use the correct distance between them.
Applying Gauss's Law to asymmetric charge distributions	Gauss's Law gives simple results only for highly symmetric charge distributions such as spherical, cylindrical, or planar symmetry.
Measuring the wrong angle in Electric Flux	The angle should always be measured between the electric field vector and the outward surface normal.
Forgetting the direction of Electric Field Lines	Electric field lines originate from positive charges and terminate on negative charges.
Confusing axial and equatorial dipole formulas	The electric field expressions for axial and equatorial positions are different and should be used carefully.
Incorrect SI unit conversion	Always use SI units such as Coulomb (C), metre (m), Newton (N), and Volt (V) while solving numerical problems.

ALLEN Faculty Tips

Draw a neat diagram before solving Electric Charges and Fields problems to visualize the charge configuration and field directions.
Identify whether the problem involves Coulomb's Law, Electric Field, Electric Flux, Electric Dipole, or Gauss's Law before selecting the appropriate formula.
Use vector resolution carefully while solving problems involving multiple charges.
Apply Gauss's Law only when sufficient symmetry exists.
Memorize standard results for electric fields due to point charges, infinite line charges, conducting spheres, infinite sheets, and electric dipoles, as they are frequently used in JEE questions.
Revise important formulas and practice previous year questions regularly to improve speed and accuracy.

4.0PYQ Trend Analysis

Electric Charges and Fields is one of the highest-weightage chapters in Class 12 Physics for both JEE Main and JEE Advanced. Questions are primarily numerical and concept-based, with a strong emphasis on vector analysis, electric field calculations, Gauss's Law, and electric dipoles.

Note: The trend below is based on the analysis of previous years' JEE Main and JEE Advanced question papers. The exact number of questions may vary each year.

Concept	Questions Asked (Last 5 Years)	Average Difficulty
Electric Charge & Coulomb's Law	7–8	Easy–Moderate
Superposition Principle	4–5	Moderate
Electric Field Intensity	8–10	Moderate
Electric Flux	4–5	Moderate
Gauss's Law	6–7	Moderate–Difficult
Electric Dipole	5–6	Moderate

Most Common Question Types

Numerical problems based on Coulomb's Law.
Net electric field due to multiple point charges.
Applications of the Superposition Principle.
Electric field due to line charges, rings, and charged spheres.
Electric flux through open and closed surfaces.
Applications of Gauss's Law to symmetric charge distributions.
Electric dipole in a uniform electric field.
Assertion–Reason and concept-based MCQs.

Smart Revision Strategy

If you're short on time, revise the chapter in the following order:

Electric Charge & Properties
Coulomb's Law
Superposition Principle
Electric Field Intensity
Electric Flux
Gauss's Law
Electric Dipole
Continuous Charge Distribution

Next Chapter -> Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance

Class 12 Physics Chapters List:- JEE & NEET Revision Notes

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: Materials, Devices and Simple Circuits

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Class 12 Physics Chapter 1 Electric Charges and Fields: Download Revision Notes

Master Electric Charges and Fields with expert study material and key concepts. Explore past-year weightage, JEE and NEET exam trends, and prepare smarter with focused resources.

Electric Charges and Fields smart study material curated by ALLEN expert faculty for effective learning, quick revision, and exam preparation. This chapter covers key concepts of electrostatics, including electric charges, Coulomb’s Law, electric field and field lines, electric dipole, electric flux, and Gauss’s Law. Physics Class 12 Chapter 1 revision notes brings together concise theory, important formulas, solved examples, previous-year question trends, and exam-focused tips in a structured format.

1.0Chapter Snapshot

Parameter	Details
Difficulty Level	Moderate to High
JEE Main Weightage	High (Generally 1–2 Questions)
JEE Advanced Weightage	High (2–4 Integrated Concepts)
Key Concepts Covered	Electric Charge, Coulomb's Law, Electric Field, Electric Flux, Electric Dipole, Gauss's Law

2.0Chapter 1 Smart Study Material by ALLEN Experts: Download PDFs

Download below study resources by ALLEN experts, created to help students strengthen their concepts and prepare effectively for competitive and school exams.

NCERT Solutions – Chapter 1
JEE Main PYQs with Solutions
JEE Advanced PYQs with Solutions
JEE Important Questions & Solutions
NEET Important Questions & Solutions

These study materials reinforce conceptual understanding, improve problem-solving speed, and enhance overall exam readiness to help students perform with confidence and score higher.

3.0Learning Outcomes

After completing these Electric Charges and Fields Revision Notes, you will be able to:

Understand the properties of electric charge, including conservation, quantization, additivity, and the transfer of charge.
Apply Coulomb's Law to calculate the electrostatic force between two or more point charges in different mediums.
Use the Superposition Principle to determine the net electric force and electric field due to multiple charges.
Calculate electric field intensity for point charges, continuous charge distributions, and electric dipoles.
Interpret electric field lines and understand their significance in representing the direction and strength of electric fields.
Evaluate electric flux through various surfaces and apply the concept to different electrostatic situations.
Apply Gauss's Law to solve problems involving symmetric charge distributions such as spheres, cylinders, and infinite sheets.
Analyze the behaviour of electric dipoles in uniform electric fields, including torque and potential energy.

1.0Coulomb’s Law

The electrostatic force between two point charges is directly proportional to their charges and inversely proportional to the square of the distance between them, always acting along the line joining them.

$F = K \frac{q _{1} q _{2}}{r ^{2}}$

$k = \frac{1}{4 π ϵ _{0}} = 9 \times 1 0^{9} N m^{2} C^{- 2}$ =Electrostatic constant or Coulomb’s Constant

Coulomb's Law in Vector Form

$F_{21} = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2} ( r _{2} - r _{1} )}{∣ r _{2} - r _{1} ∣ ^{3}} or F_{12} = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2} ( r _{1} - r _{2} )}{∣ r _{1} - r _{2} ∣ ^{3}}$

2.0Superposition Principle

When multiple charges are present, the net force on any charge is the vector sum of forces from all the other charges.

$F = F_{01} + F_{02} + F_{03} + \dots + F_{0 n}$

$F = \frac{k q _{0} q _{1}}{r _{1}^{2}} \overset{r}{^}_{1} + \frac{k q _{0} q _{2}}{r _{2}^{2}} \overset{r}{^}_{2} + \dots + \frac{k q _{0} q _{n}}{r _{n}^{2}} \overset{r}{^}_{n}$

Relative Permittivity( $ϵ_{r}$ ) or Dielectric Constant (K)

$Dielectric Constant (K) = \frac{Permittivity of a Medium ( ε )}{Permittivity of Vacuum ( ε _{0} )}$

$K = \frac{ε _{r}}{ε _{0}}$

$ε = ε_{0} ε_{r}$

The value of $ε_{r}$ or K ≥1

3.0Electric Field

An electric field is the region around a charge where it exerts a force on other charges.
Mathematically, $E = \frac{F}{q _{0}}$ SI unit : N/C or V/m
$E = q_{0} \to 0 lim \frac{F}{q _{0}}$
Dimensional Formula $[M^{1} L^{1} T^{- 3} A^{- 1}]$
It is a vector quantity.

4.0Electric Field Lines

Electric field lines are imaginary lines, straight or curved, that represent the direction and strength of the electric field around a charged object. At any point on these lines, the tangent indicates the direction of the electric field at that location.
Two electric field lines can never intersect each other.

Electrostatic field lines can never form closed loops.

Electric field lines due to positive or negative charges

Electric line of force due to an electric dipole

5.0Continuous Charge Distribution

A group of closely spaced electric charges forms a continuous charge distribution.

Linear Charge Density()

Surface Charge Density ()

Volume Charge Density()

$λ = \frac{Q}{l}$

SI Unit = $\frac{C}{m}$

$σ = \frac{Q}{S}$

SI Unit = $\frac{C}{m ^{2}}$

$ρ = \frac{Q}{V}$

SI Unit = $\frac{C}{m ^{3}}$

6.0Electric Dipole

It is a pair of equal and opposite charges separated by a small distance.

The dipole moment is the product of the magnitude of either charge and the distance between them. $p = q (2 l)$
S.I. unit- Cm

Electric Field Due to a Dipole

At Axial / End on position

$E_{A x ia l} = \frac{1}{4 π ε _{0}} \frac{2 p}{r ^{3}}$

At Equator/Broadside on position

$E_{Eq u i t or ia l} = \frac{1}{4 π ε _{0}} \frac{p}{r ^{3}}$

At general position

$E = \frac{p}{4 π ε _{0} r ^{3}} 3 co s^{2} θ + 1$

$T an α = \frac{1}{2} T an θ$

Electric Field Intensity due to a charged wire

Special Cases:

For Infinite Wire,( both ends goes to infinite)

$EX = E ⊥= \frac{2 K λ}{r}$

E_Y=E॥=0

For Semi-infinite wire

EX=E丄= $\frac{K λ}{r}$

$E_{Y} = E_{∣∣} = \frac{K λ}{r}$

$E_{R es u lt an t} = 2 \frac{K λ}{r}$

Electric field due to finite wire at symmetric point:

$E_{X} = E_{⊥} = \frac{2 K λ}{r} s in θ$

$E_{Y} = E_{॥} = 0$

Electric Field due to a uniformly charged Arc

$E = \frac{2 K λ}{R} S in (\frac{θ}{2})$

Electric field at centre of uniformly charged Ring

$E = \frac{2 K λ}{R} S in (\frac{θ}{2}), θ \to$ angle of arc

$θ$ =360⁰

$E = \frac{2 K λ}{R} S in (\frac{360}{2})$

E =0

Electric Field due to a Uniformly charged Ring at its Axis

$E = \frac{k Q x}{( R ^{2} + x ^{2} ) ^{3/2}}$

Special cases:

Electric field on the axis for small values of x : $E = \frac{k Q}{R ^{3}} \cdot x$
Electric field at the centre of the ring is zero because x=0
Electric field at the axis for larger values of x: $E \approx \frac{k Q}{x ^{2}}$
Maximum value of electric field

$\frac{d E}{d x} = 0$

$x = \pm \frac{R}{2}$

$E_{max} = \frac{2 k Q}{3 3 R ^{2}}$

7.0Dipole placed in an electric field, torque acts on it

て=p E Sin

(∴ $θ$ is the angle between dipole moment (p)and electric field (E))

$τ = p \times E$

Special Cases:

If =0⁰ then stable Equilibrium.
If =180⁰ then て=o, unstable equilibrium.

8.0Work done in rotating a dipole in a uniform electric field

When an electric dipole with dipole moment is oriented at an angle to an electric field , torque is exerted on the dipole, causing it to rotate. The resulting work done can be expressed as

$W = pE (C os θ_{1} - C os θ_{2})$

9.0Electric Flux

This physical quantity is used to measure strength of electric field and it is defined as the total number of electric field lines passing through an area.
Electric flux is a scalar quantity.
Unit of Electric Flux:Nm2/C or V m
Dimensional Formula-[M¹L³T^-3A^-1]
Electric flux through a large surface $ϕ = \int d ϕ = \int E \cdot d A$
If Electric Field is uniform= $\Rightarrow E$ =Constant(same everywhere)

$ϕ = E \cdot \int d A (∵ \int d A = total area vector of a plane surface)$

$\Rightarrow ϕ = E \cdot A \Rightarrow ϕ = E A cos θ$

Different cases

If $θ$ =0⁰, $ϕ = E A cos 0^{\circ} = E A (positive flux means outgoing or leaving)$

If $θ$ is 90⁰, $ϕ = E A cos 90° = 0.$
If $θ$ is 180⁰ , $ϕ = E A C os 180° = - E A$ (negative flux means incoming or entering)

10.0Gauss’s Law

According to this law the total electric flux ( $ϕ$ ) through any closed surface (S) in free space is equal to $\frac{1}{ε _{0}}$ times the total electric charge (q) enclosed by the surface.

$ϕ = \oint E . d S = \frac{Q _{e n c l ose d}}{ε _{0}}$

11.0Applications of Gauss Law

Electric field Intensity due to infinitely long wire

$E = \frac{λ}{2 π ε _{0} r} = \frac{2 K λ}{r}$

Electric field due to uniformly charged long cylindrical pipe/cylindrical shell

Case 1.Electric field at any point outside the cylinder(r>R): $E = \frac{σ R}{ϵ _{0} r}$
Case 2.For the point lying on the surface(r≈R): $E = \frac{σ}{ε _{0}}$
Case 3.For the point inside the surface(r<R): $E_{in s i d e} = 0$

Electric Field due to Uniformly Charged Infinite Sheet

Non Conducting Sheet: $E = \frac{σ}{2 ε _{0}}$
Conducting sheet or Metal Plate: $E = \frac{σ}{ε _{0}}$

Electric Field due to the charged conducting sphere or charged thin shell

Electric Field at any point outside the sphere (r>R)

$E = \frac{q}{4 π ε _{0} r ^{2}} = \frac{k q}{r ^{2}}$

For any point lying on the surface of sphere (r=R)

$E_{s} = \frac{k q}{R ^{2}} = \frac{q}{4 π ε _{0} R ^{2}} = \frac{σ}{ε _{0}}$ $(∵ σ = \frac{q}{4 π R ^{2}})$

Electric Field at any point Inside the sphere(r<R)

In this case charge enclosed by the gaussian surface is zero, i.e., E_inside=0

Variation of E with r

Electric Field due to uniformly charged non conducting sphere(solid sphere)

Electric field at any point outside the sphere (r>R)

$E = \frac{k q}{r ^{2}}$

Electric field at any point lying on the surface of sphere(r=R)

$E_{S} = \frac{ρR}{3 ε _{0}}$ (∴ $ρ$ is volume charge density)

Electric field at any point inside the sphere(r<R)

$E_{in s i d e} = \frac{1}{4 π ε _{0}} \frac{q}{R ^{3}} r = \frac{r h o}{3 ε _{0}} (r)$

Variation of E with r

12.0Sample Questions on Electric Charges And Fields

Q-1. $1 0^{10}$ alpha particles are ejected per second from a body, then after how much time, the body will acquire a charge of 8 μC ?

Solution: Charge appear per second on body, $Q = (\frac{N}{t}) q_{α}$

$Q = 1 0^{10} \times (2 \times 1.6 \times 1 0^{- 19}) = 3.2 \times 1 0^{- 9} C / sec$

Let t be the time taken to acquire charge of 8 μC then

$t = \frac{8 \times 1 0 ^{- 6}}{3.2 \times 1 0 ^{- 9}} = 2.5 \times 1 0^{3} sec$

Q-2.Find final charges on the spheres when switch S is closed.

Solution: Total charge of the system = $Q = 60 μ C + 0 = 60 μ C$

Then after conduction

$\frac{Q _{1}^{'}}{Q _{2}^{'}} = \frac{R _{1}}{R _{2}} \Rightarrow \frac{2}{3}$

$Q 1^{'} = (\frac{2}{2 + 3}) 60 μ C = 24 μ C$

$Q 2^{'} = (\frac{3}{2 + 3}) 60 μ C = 36 μ C$

Q-3.Why can we not use Coulomb’s law for large size bodies ?

Solution:

When large size charged conducting spheres brought close to each other, there charges moves away due to repulsion hence effective distance between their centers increases r'>r

$F_{a c t u a l} = F_{c a l c u l a t e d} = k \frac{q _{1} q _{2}}{r ^{2}}$

Q-4.Two identical conducting spheres A and B carry equal charges and are placed at a distance r apart in vacuum. The electrostatic force between them is F. A third identical uncharged sphere C is first brought into contact with sphere A, then with sphere B, and is finally removed. What will be the new electrostatic force between spheres A and B?

Solution:

$F = k \frac{q ^{2}}{r ^{2}}$ ……….(1)

Now C (uncharged sphere) is touched with A first

Then C is touched with B

New force between A & B now is

$F^{'} = \frac{k ( \frac{Q}{2} ) ( \frac{3 Q}{4} )}{r ^{2}} = \frac{3 k Q ^{2}}{8 r ^{2}} = \frac{3 F}{8}$

Q-5.Two equally charged spheres are placed far apart. If the gravitational force equals the electrostatic force between them, find the ratio of specific charge $\frac{q}{m}$ .

Solution: Equating the gravitational and electrostatics force

$\frac{k q ^{2}}{r ^{2}} = \frac{G m ^{2}}{r ^{2}} \Rightarrow \frac{q}{m} = \frac{G}{k}$ $= \frac{6.67 \times 1 0 ^{- 11}}{9 \times 1 0 ^{9}} = \frac{0.74}{1 0 ^{20}} = \frac{0.86}{1 0 ^{10}} = 0.86 \times 1 0^{- 10}$

1.0High Weightage Topics

The following topics have consistently appeared in JEE Main and JEE Advanced examinations and should be given higher priority during revision.

Topic	Focus Area
Coulomb's Law	Force between charges, distance variation
Superposition Principle	Net force and electric field due to multiple charges
Electric Field Intensity	Point charge, line charge, ring, and symmetry
Electric Flux	Surface orientation and flux calculations
Gauss's Law	Symmetric charge distributions
Electric Dipole	Dipole moment, torque, energy
Dipole in Uniform Electric Field	Torque and equilibrium conditions
Continuous Charge Distribution	Ring, rod, and disc-based numericals

2.0Formula Handbook

The following formulas are the most important for solving JEE Main and JEE Advanced questions from Electrostatics.

3.0Common Mistakes & JEE Tips

Electric Charges and Fields is a concept-driven chapter where even well-prepared students lose marks due to errors in sign conventions, vector addition, formula selection, and understanding electric field concepts. Avoiding these common mistakes can significantly improve your accuracy and confidence in JEE Main and JEE Advanced.

Common Mistake	Correct Approach
Confusing Electric Force with Electric Field	Electric Force depends on both the source charge and the test charge, whereas the Electric Field depends only on the source charge.
Using algebraic addition instead of vector addition	Electric force and electric field are vector quantities. Always use vector addition while solving multiple-charge problems.
Applying Coulomb's Law incorrectly	Ensure the charges are treated as point charges and use the correct distance between them.
Applying Gauss's Law to asymmetric charge distributions	Gauss's Law gives simple results only for highly symmetric charge distributions such as spherical, cylindrical, or planar symmetry.
Measuring the wrong angle in Electric Flux	The angle should always be measured between the electric field vector and the outward surface normal.
Forgetting the direction of Electric Field Lines	Electric field lines originate from positive charges and terminate on negative charges.
Confusing axial and equatorial dipole formulas	The electric field expressions for axial and equatorial positions are different and should be used carefully.
Incorrect SI unit conversion	Always use SI units such as Coulomb (C), metre (m), Newton (N), and Volt (V) while solving numerical problems.

ALLEN Faculty Tips

Draw a neat diagram before solving Electric Charges and Fields problems to visualize the charge configuration and field directions.
Identify whether the problem involves Coulomb's Law, Electric Field, Electric Flux, Electric Dipole, or Gauss's Law before selecting the appropriate formula.
Use vector resolution carefully while solving problems involving multiple charges.
Apply Gauss's Law only when sufficient symmetry exists.
Memorize standard results for electric fields due to point charges, infinite line charges, conducting spheres, infinite sheets, and electric dipoles, as they are frequently used in JEE questions.
Revise important formulas and practice previous year questions regularly to improve speed and accuracy.

4.0PYQ Trend Analysis

Electric Charges and Fields is one of the highest-weightage chapters in Class 12 Physics for both JEE Main and JEE Advanced. Questions are primarily numerical and concept-based, with a strong emphasis on vector analysis, electric field calculations, Gauss's Law, and electric dipoles.

Note: The trend below is based on the analysis of previous years' JEE Main and JEE Advanced question papers. The exact number of questions may vary each year.

Concept	Questions Asked (Last 5 Years)	Average Difficulty
Electric Charge & Coulomb's Law	7–8	Easy–Moderate
Superposition Principle	4–5	Moderate
Electric Field Intensity	8–10	Moderate
Electric Flux	4–5	Moderate
Gauss's Law	6–7	Moderate–Difficult
Electric Dipole	5–6	Moderate

Most Common Question Types

Numerical problems based on Coulomb's Law.
Net electric field due to multiple point charges.
Applications of the Superposition Principle.
Electric field due to line charges, rings, and charged spheres.
Electric flux through open and closed surfaces.
Applications of Gauss's Law to symmetric charge distributions.
Electric dipole in a uniform electric field.
Assertion–Reason and concept-based MCQs.

Smart Revision Strategy

If you're short on time, revise the chapter in the following order:

Electric Charge & Properties
Coulomb's Law
Superposition Principle
Electric Field Intensity
Electric Flux
Gauss's Law
Electric Dipole
Continuous Charge Distribution

Next Chapter -> Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance

Test your Knowledge

question 1 of 5

What are the key properties of electric charges?

1.0Electric Charges And Fields: Key Concepts

2.0Introduction to Electric Charges

Types of Charges

3.0Properties of Electric Charges

Frequently Asked Questions

What are the important topics in Electric Charges and Fields for JEE?

Is Electric Charges and Fields important for JEE Main?

What formulas should I revise from Electric Charges and Fields?

How long does it take to revise this chapter?

Are these revision notes useful for CBSE and JEE?

Join ALLEN!

Class 12 Physics Chapter 1 Electric Charges and Fields: Download Revision Notes

1.0Chapter Snapshot

2.0Chapter 1 Smart Study Material by ALLEN Experts: Download PDFs

3.0Learning Outcomes

Test your Knowledge

question 1 of 5

What are the key properties of electric charges?

1.0Electric Charges And Fields: Key Concepts

2.0Introduction to Electric Charges

Types of Charges

3.0Properties of Electric Charges

1.0High Weightage Topics

2.0Formula Handbook

3.0Common Mistakes & JEE Tips

ALLEN Faculty Tips

4.0PYQ Trend Analysis

Smart Revision Strategy

Frequently Asked Questions

What are the important topics in Electric Charges and Fields for JEE?

Is Electric Charges and Fields important for JEE Main?

What formulas should I revise from Electric Charges and Fields?

How long does it take to revise this chapter?

Are these revision notes useful for CBSE and JEE?

Join ALLEN!

Class 12 Physics Chapter 1 Electric Charges and Fields: Download Revision Notes

1.0Chapter Snapshot

2.0Chapter 1 Smart Study Material by ALLEN Experts: Download PDFs

3.0Learning Outcomes

1.0High Weightage Topics

2.0Formula Handbook

3.0Common Mistakes & JEE Tips

ALLEN Faculty Tips

4.0PYQ Trend Analysis

Smart Revision Strategy