What is Coulomb's Law in electrostatics?

Coulomb's Law states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Mathematically: F = k(q₁q₂)/r² where k = 9×10⁹ Nm²/C².

What is the difference between electric field and electric potential?

Electric Field (E) is a vector quantity representing force per unit charge (F/q), measured in N/C or V/m. Electric Potential (V) is a scalar quantity representing work done per unit charge to bring it from infinity, measured in volts (V). They are related as E = -dV/dr.

What is Gauss Law and its applications?

Gauss Law states that the total electric flux through a closed surface is equal to 1/ε₀ times the charge enclosed: ∮E·dA = Q/ε₀. It is used to find electric fields of symmetric charge distributions like infinite plane sheets, spheres, and cylinders.

What is the formula for capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor is C = ε₀A/d, where ε₀ is permittivity of free space (8.85×10⁻¹² F/m), A is the area of plates, and d is the separation between plates. With dielectric: C = Kε₀A/d.

How is energy stored in a capacitor calculated?

Energy stored in a capacitor can be calculated using three equivalent formulas: U = ½CV², U = ½Q²/C, or U = ½QV, where C is capacitance, V is voltage, and Q is charge stored.

What is electric dipole moment?

Electric dipole moment (p) is a vector quantity equal to the product of charge magnitude and separation distance: p = q×2a, where q is charge and 2a is the distance between +q and -q. Its direction is from negative to positive charge. Unit: C·m.

Electrostatics for JEE 2025-26 | Complete Notes with Formulas, Laws & Solved Examples

Electrostatics JEE notes, Formulas, PYQs — Electrostatics JEE Notes, Formulas, PYQs

Electric Charge

Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electromagnetic field. It is the basis of all electrical phenomena and is conserved in nature.

1.1 Properties of Electric Charge

Basic Properties

Quantization: q = ±ne (n = 1,2,3...)
Conservation: Total charge is constant
Additive: Total charge = Σqᵢ
Two types: Positive (+) and Negative (-)

Important Constants

Elementary charge (e)1.6 × 10⁻¹⁹ C

Proton charge+e

Electron charge-e

SI UnitCoulomb (C)

1.2 Methods of Charging

Method	Process	Final Charge	Example
Friction	Transfer of electrons by rubbing	Both objects charged	Glass rod with silk
Conduction	Direct contact with charged body	Equal charge distribution	Touching metal sphere
Induction	Redistribution without contact	Opposite to inducing charge	Electroscope charging

💡 JEE Quick Facts

Like charges repel, unlike charges attract
Charge is invariant - same in all reference frames
Conductors: Free electrons, allow charge flow
Insulators: No free electrons, don't allow charge flow
Charging by induction gives opposite charge without contact

📝 Solved Example 1

Question: How many electrons are removed from a body to give it a positive charge of 3.2 × 10⁻¹⁵ C?

Solution:

Given: q = 3.2 × 10⁻¹⁵ C

Elementary charge e = 1.6 × 10⁻¹⁹ C

Using quantization of charge:

\[q = ne\]

n = \frac{q}{e} = \frac{3.2 \times 10^{-15}}{1.6 \times 10^{-19}}

n = 2 \times 10^4 = 20,000

\text{Answer: } 20,000 \text{ electrons}

Coulomb's Law

Coulomb's Law is the fundamental law of electrostatics that describes the force between two point charges. It is analogous to Newton's law of gravitation but much stronger.

2.1 Mathematical Formulation

Coulomb's Law Formula

\vec{F} = k\frac{q_1q_2}{r^2}\hat{r} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\hat{r}

Where:

F = Electrostatic force (N)
q₁, q₂ = Point charges (C)
r = Distance between charges (m)
k = Coulomb's constant
ε₀ = Permittivity of free space

Constants:

k = 9 × 10⁹ Nm²/C²
ε₀ = 8.85 × 10⁻¹² C²/Nm²
k = 1/(4πε₀)

2.2 Key Features of Coulomb's Law

Vector Form

\vec{F}_{12} = k\frac{q_1q_2}{r_{12}^2}\hat{r}_{12}

Force on q₁ due to q₂

\vec{F}_{21} = -\vec{F}_{12}

Newton's third law satisfied

Superposition Principle

\vec{F} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 + ...

Net force = Vector sum of individual forces

\vec{F} = \sum_{i=1}^{n} \vec{F}_i

2.3 Comparison: Coulomb Force vs Gravitational Force

Aspect	Coulomb Force	Gravitational Force
Formula	F = kq₁q₂/r²	F = Gm₁m₂/r²
Nature	Attractive or Repulsive	Always Attractive
Relative Strength	10³⁶ times stronger	Much weaker
Constant	k = 9 × 10⁹ Nm²/C²	G = 6.67 × 10⁻¹¹ Nm²/kg²
Depends on	Electric charge	Mass

⚠️ Common JEE Mistakes

Sign confusion: Use magnitude formula |F| = k|q₁q₂|/r² and determine direction separately
Forgetting vector nature: Always use vector addition for multiple charges
Using wrong distance: r is center-to-center distance, not surface-to-surface
Medium effect: In medium with dielectric constant K, F reduces to F/K

📝 Solved Example 2

Question: Three charges +2μC, -3μC, and +4μC are placed at the vertices of an equilateral triangle of side 10 cm. Find the net force on the +2μC charge.

Solution:

Given: q₁ = +2μC, q₂ = -3μC, q₃ = +4μC, a = 10 cm = 0.1 m

Step 1: Force due to q₂ on q₁

F_{12} = k\frac{|q_1q_2|}{a^2} = 9 \times 10^9 \times \frac{2 \times 10^{-6} \times 3 \times 10^{-6}}{(0.1)^2}

F_{12} = 5.4 \text{ N (attractive, towards } q_2)

Step 2: Force due to q₃ on q₁

F_{13} = k\frac{q_1q_3}{a^2} = 9 \times 10^9 \times \frac{2 \times 10^{-6} \times 4 \times 10^{-6}}{(0.1)^2}

F_{13} = 7.2 \text{ N (repulsive, away from } q_3)

Step 3: Vector addition (angle = 60°)

F_{net} = \sqrt{F_{12}^2 + F_{13}^2 + 2F_{12}F_{13}\cos60°}

F_{net} = \sqrt{5.4^2 + 7.2^2 + 2(5.4)(7.2)(0.5)}

\text{Answer: } F_{net} = 10.96 \text{ N}

Electric Field

Electric field is a region around a charged particle where another charged particle experiences a force. It is a vector quantity representing force per unit positive charge.

3.1 Electric Field Definition & Formula

Electric Field Intensity

\vec{E} = \frac{\vec{F}}{q_0} = k\frac{Q}{r^2}\hat{r}

Definition:

E = Electric field intensity
F = Force on test charge
q₀ = Test charge (positive)
Q = Source charge
r = Distance from source

Units & Dimensions:

Unit: N/C or V/m
Dimension: [MLT⁻³A⁻¹]
Vector quantity
Direction: Force on +ve charge

3.2 Electric Field of Different Charge Distributions

Charge Distribution	Electric Field Formula	Direction
Point Charge Q	E = kQ/r²	Radially outward (Q>0)
Infinite Line Charge (λ)	E = λ/(2πε₀r)	Perpendicular to line
Infinite Plane Sheet (σ)	E = σ/(2ε₀)	Perpendicular to sheet
Ring (on axis at distance x)	E = kQx/(R²+x²)^(3/2)	Along axis
Uniformly Charged Sphere (r>R)	E = kQ/r²	Radially outward
Uniformly Charged Sphere (r	E = kQr/R³	Radially outward

3.3 Electric Field Lines

Properties of Field Lines

Originate from positive charge
Terminate at negative charge
Never cross each other
Tangent gives direction of E
Density ∝ field strength
Always perpendicular to conductor surface

Field Line Patterns

Single +ve charge: Radial outward
Single -ve charge: Radial inward
Two +ve charges: Repel, no lines between
+ve and -ve: Attract, curved lines
Parallel plates: Uniform field, parallel lines

💡 JEE Important Points

Electric field is independent of test charge q₀
Superposition principle: E_net = E₁ + E₂ + E₃ + ... (vector sum)
Inside a conductor: E = 0 (electrostatic equilibrium)
At the surface: E is perpendicular to surface
Uniform field: E = V/d (parallel plates)

📝 Solved Example 3

Question: Find the electric field at the center of a uniformly charged ring of radius R and total charge Q.

Solution:

Consider a small element dq on the ring

Key Concept:

Due to symmetry, for every element dq at position θ, there's another element at position (180° - θ)

The horizontal components cancel out

The vertical components also cancel due to symmetry

\text{Answer: } E = 0 \text{ (at center)}

Note: At a distance x on the axis: E = kQx/(R²+x²)^(3/2)

Electric Dipole

An electric dipole consists of two equal and opposite charges separated by a small distance. It is one of the most important configurations in electrostatics with applications in molecular physics.

4.1 Electric Dipole Moment

Dipole Moment Formula

\vec{p} = q \times 2\vec{a}

Where:

p = Dipole moment
q = Magnitude of charge
2a = Separation between charges
Direction: -q to +q

Units & Properties:

Unit: C·m (Coulomb-meter)
Also: Debye (D)
1 D = 3.33 × 10⁻³⁰ C·m
Vector quantity

4.2 Electric Field Due to Dipole

Position	Electric Field	Direction	Condition
Axial Line	E = 2kp/r³	Along dipole axis	r >> 2a
Equatorial Line	E = kp/r³	Opposite to p	r >> 2a
General Point	E = (kp/r³)√(3cos²θ + 1)	At angle θ	r >> 2a

Important Relations

Ratio of Fields:

\frac{E_{axial}}{E_{equatorial}} = \frac{2kp/r^3}{kp/r^3} = 2:1

Potential at Axial Point:

V = \frac{kp}{r^2}

Potential at Equatorial Point:

\[V = 0\]

4.3 Dipole in Uniform Electric Field

Torque on Dipole

\vec{\tau} = \vec{p} \times \vec{E}

Magnitude:

\tau = pE\sin\theta

Maximum when θ = 90°: τ_max = pE
Minimum when θ = 0° or 180°: τ = 0
Tends to align dipole with field

Potential Energy

U = -\vec{p} \cdot \vec{E}

Magnitude:

U = -pE\cos\theta

Minimum when θ = 0°: U_min = -pE (stable)
Maximum when θ = 180°: U_max = +pE (unstable)
Zero when θ = 90°

⚠️ JEE Critical Points

Net force on dipole in uniform field = 0 (only torque exists)
In non-uniform field: Both force and torque exist
Work done to rotate: W = pE(cosθ₁ - cosθ₂)
Oscillation: If displaced, dipole oscillates with period T = 2π√(I/pE)

📝 Solved Example 4

Question: An electric dipole of dipole moment 4 × 10⁻⁹ C·m is aligned at 30° with an electric field of 5 × 10⁴ N/C. Find (a) torque (b) potential energy.

Solution:

Given: p = 4 × 10⁻⁹ C·m, E = 5 × 10⁴ N/C, θ = 30°

(a) Torque:

\tau = pE\sin\theta

\tau = (4 \times 10^{-9})(5 \times 10^4)\sin30°

\tau = 2 \times 10^{-4} \times 0.5 = 1 \times 10^{-4} \text{ N·m}

(b) Potential Energy:

U = -pE\cos\theta

U = -(4 \times 10^{-9})(5 \times 10^4)\cos30°

U = -2 \times 10^{-4} \times 0.866 = -1.732 \times 10^{-4} \text{ J}

\text{Answer: (a) } \tau = 10^{-4} \text{ N·m, (b) } U = -1.732 \times 10^{-4} \text{ J}

Gauss Law

Gauss Law is one of the four Maxwell's equations and provides an elegant way to calculate electric fields for symmetric charge distributions. It relates electric flux through a closed surface to the charge enclosed.

5.1 Electric Flux

Electric Flux Definition

\phi = \int \vec{E} \cdot d\vec{A} = \int E \cdot dA \cos\theta

Definition:

φ = Electric flux
E = Electric field
dA = Area element
θ = Angle between E and dA

Units & Properties:

Unit: N·m²/C or V·m
Scalar quantity
Can be +ve, -ve or zero
Dimension: [ML³T⁻³A⁻¹]

5.2 Gauss Law Statement & Formula

Gauss Law (Integral Form)

\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\epsilon_0}

"The total electric flux through a closed surface is equal to 1/ε₀ times the charge enclosed by that surface."

Key Points:

Closed surface is called Gaussian surface
Shape of Gaussian surface is arbitrary
Only charge inside the surface contributes to flux
Outside charges don't contribute (but they affect E)

5.3 Applications of Gauss Law

System	Gaussian Surface	Electric Field	Region
Infinite Line Charge (λ)	Cylindrical	E = λ/(2πε₀r)	Outside
Infinite Plane Sheet (σ)	Cylindrical pillbox	E = σ/(2ε₀)	Both sides
Uniformly Charged Sphere	Concentric sphere	E = kQ/r²	r > R
Uniformly Charged Sphere	Concentric sphere	E = kQr/R³	r < R
Spherical Shell	Concentric sphere	E = 0	r < R (inside)
Spherical Shell	Concentric sphere	E = kQ/r²	r > R (outside)

5.4 Electric Field Due to Parallel Charged Sheets

Two Parallel Infinite Sheets

Same Sign (+σ, +σ):

Left of both: E = 0
Between sheets: E = σ/ε₀
Right of both: E = 0

Opposite Sign (+σ, -σ):

Left of both: E = 0
Between sheets: E = σ/ε₀
Right of both: E = 0

Parallel Plate Capacitor:

Two oppositely charged plates create uniform field E = σ/ε₀ between them

💡 Gauss Law Strategy for JEE

Identify symmetry: Spherical, cylindrical, or planar
Choose Gaussian surface: Must match symmetry
Calculate flux: ∮E·dA (often E is constant on surface)
Find enclosed charge: Q_enc (use charge density)
Apply Gauss Law: Solve for E

📝 Solved Example 5

Question: A uniformly charged sphere of radius R has volume charge density ρ. Find electric field at a distance r from center (a) when r > R (b) when r < R.

Solution:

(a) When r > R (Outside the sphere):

Choose Gaussian surface: Sphere of radius r

\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}

E \times 4\pi r^2 = \frac{\rho \times \frac{4}{3}\pi R^3}{\epsilon_0}

E = \frac{\rho R^3}{3\epsilon_0 r^2}

(b) When r < R (Inside the sphere):

Choose Gaussian surface: Sphere of radius r (r < R)

E \times 4\pi r^2 = \frac{\rho \times \frac{4}{3}\pi r^3}{\epsilon_0}

E = \frac{\rho r}{3\epsilon_0}

\text{Answer: (a) } E = \frac{\rho R^3}{3\epsilon_0 r^2} \text{ (r > R)}\] \[\text{(b) } E = \frac{\rho r}{3\epsilon_0} \text{ (r < R)}

Note: E ∝ r inside, E ∝ 1/r² outside

Electric Potential & Potential Energy

Electric potential is a scalar quantity that represents the work done per unit charge to bring a test charge from infinity to a point in an electric field. It simplifies many calculations compared to vector field methods.

6.1 Electric Potential Definition

Electric Potential Formula

V = \frac{W}{q_0} = -\int_{\infty}^{r} \vec{E} \cdot d\vec{r}

Definition:

V = Electric potential
W = Work done
q₀ = Test charge
Reference: V(∞) = 0

Units & Properties:

Unit: Volt (V) = J/C
Scalar quantity
Dimension: [ML²T⁻³A⁻¹]
Can be +ve or -ve

6.2 Potential Due to Different Charge Distributions

System	Electric Potential Formula	Special Cases
Point Charge Q	V = kQ/r	V → 0 as r → ∞
Multiple Point Charges	V = Σ(kqᵢ/rᵢ)	Algebraic sum (scalar)
Electric Dipole (axial)	V = kp cosθ/r²	r >> 2a
Electric Dipole (equatorial)	V = 0	θ = 90°
Uniformly Charged Sphere (r > R)	V = kQ/r	Like point charge
Uniformly Charged Sphere (r = R)	V = kQ/R	At surface
Uniformly Charged Sphere (r < R)	V = (kQ/2R³)(3R² - r²)	Inside sphere
Charged Spherical Shell (r < R)	V = kQ/R	Constant inside

6.3 Relation Between E and V

E-V Relationship

General Form (Vector):

\vec{E} = -\nabla V = -\left(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k}\right)

One Dimension:

E = -\frac{dV}{dr}

Electric field is negative gradient of potential

Potential Difference:

V_B - V_A = -\int_{A}^{B} \vec{E} \cdot d\vec{r}

6.4 Electrostatic Potential Energy

Two Point Charges

U = k\frac{q_1q_2}{r}

U > 0 for like charges (repulsion)
U < 0 for unlike charges (attraction)
U = 0 when r → ∞
Work to assemble = U

System of Charges

U = \frac{1}{2}\sum_{i=1}^{n} q_iV_i

Or sum of all pair interactions:

\[U = \sum_{i

💡 Important JEE Concepts

Equipotential surfaces: No work done moving charge along them, always ⊥ to E
Conductor in equilibrium: V = constant throughout, E = 0 inside
Potential is continuous: Even where E is discontinuous
Zero potential ≠ Zero field: E can exist where V = 0
Potential energy of a charge: U = qV

📝 Solved Example 6

Question: Four charges +q, +q, +q, and -q are placed at the corners of a square of side a. Find the potential energy of the system.

Solution:

Total number of pairs = C(4,2) = 6

Pairs at distance a (4 pairs):

• (+q, +q): U₁ = kq²/a

• (+q, +q): U₂ = kq²/a

• (+q, -q): U₃ = -kq²/a

• (+q, -q): U₄ = -kq²/a

Pairs at distance a√2 (2 pairs - diagonals):

• (+q, +q): U₅ = kq²/(a√2)

• (+q, -q): U₆ = -kq²/(a√2)

Total potential energy:

U_{total} = U_1 + U_2 + U_3 + U_4 + U_5 + U_6

U = \frac{kq^2}{a} + \frac{kq^2}{a} - \frac{kq^2}{a} - \frac{kq^2}{a} + \frac{kq^2}{a\sqrt{2}} - \frac{kq^2}{a\sqrt{2}}

\[U = 0\]

\text{Answer: } U = 0

Capacitors and Capacitance

A capacitor is a device that stores electrical energy in an electric field. It consists of two conductors separated by an insulator (dielectric). Capacitance is the ability to store charge per unit potential difference.

7.1 Capacitance Definition

Capacitance Formula

C = \frac{Q}{V}

Definition:

C = Capacitance
Q = Charge stored
V = Potential difference
Property of conductor/system

Units & Properties:

Unit: Farad (F) = C/V
1 F = 1 Coulomb/Volt
Practical: μF, nF, pF
Scalar quantity

7.2 Capacitance of Different Configurations

Capacitor Type	Capacitance Formula	With Dielectric (K)
Parallel Plate	C = ε₀A/d	C = Kε₀A/d
Spherical (Radius R)	C = 4πε₀R	C = 4πKε₀R
Spherical (radii a, b)	C = 4πε₀ab/(b-a)	C = 4πKε₀ab/(b-a)
Cylindrical (length L)	C = 2πε₀L/ln(b/a)	C = 2πKε₀L/ln(b/a)

7.3 Combination of Capacitors

Series Combination

\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + ...

For two capacitors:

C_{eq} = \frac{C_1C_2}{C_1 + C_2}

Same charge Q on all
Different voltages V₁, V₂, V₃...
V_total = V₁ + V₂ + V₃...
C_eq < smallest C

Parallel Combination

C_{eq} = C_1 + C_2 + C_3 + ...

Simple algebraic sum

Same voltage V on all
Different charges Q₁, Q₂, Q₃...
Q_total = Q₁ + Q₂ + Q₃...
C_eq > largest C

7.4 Energy Stored in Capacitor

Energy Formulas (All Equivalent)

U = \frac{1}{2}CV^2

Using C and V

U = \frac{1}{2}\frac{Q^2}{C}

Using Q and C

U = \frac{1}{2}QV

Using Q and V

Energy Density (per unit volume):

u = \frac{1}{2}\epsilon_0E^2

For parallel plate: u = ½ε₀(V/d)²

7.5 Effect of Dielectric

Parameter	Without Dielectric	With Dielectric (K)	Change Factor
Capacitance	C₀	C = KC₀	Increases K times
Electric Field	E₀	E = E₀/K	Decreases K times
Potential (battery disconnected)	V₀	V = V₀/K	Decreases K times
Charge (battery connected)	Q₀	Q = KQ₀	Increases K times
Energy (battery disconnected)	U₀	U = U₀/K	Decreases K times

⚠️ Common JEE Mistakes

Battery connected vs disconnected: Very different behavior with dielectric
Series combination: Use reciprocal formula, not direct sum
Energy formula: Choose correct one based on given quantities
Parallel plate C = ε₀A/d: A is area of ONE plate, d is separation

📝 Solved Example 7

Question: A parallel plate capacitor with plate area 100 cm² and separation 2 mm is charged to 100V. After disconnecting battery, a dielectric slab (K=5) is inserted. Find: (a) Initial capacitance (b) Final capacitance (c) Initial and final energy.

Solution:

Given: A = 100 cm² = 10⁻² m², d = 2 mm = 2×10⁻³ m, V₀ = 100V, K = 5

(a) Initial capacitance:

C_0 = \frac{\epsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 10^{-2}}{2 \times 10^{-3}}

C_0 = 44.25 \times 10^{-12} \text{ F} = 44.25 \text{ pF}

(b) Final capacitance:

C = KC_0 = 5 \times 44.25 = 221.25 \text{ pF}

(c) Initial energy:

U_0 = \frac{1}{2}C_0V_0^2 = \frac{1}{2} \times 44.25 \times 10^{-12} \times (100)^2

U_0 = 2.21 \times 10^{-7} \text{ J}

Final energy (Q constant, battery disconnected):

U = \frac{U_0}{K} = \frac{2.21 \times 10^{-7}}{5} = 4.42 \times 10^{-8} \text{ J}

\text{Answer: (a) } C_0 = 44.25 \text{ pF}\] \[\text{(b) } C = 221.25 \text{ pF}\] \[\text{(c) } U_0 = 2.21 \times 10^{-7} \text{ J, } U = 4.42 \times 10^{-8} \text{ J}

Dielectrics

Dielectrics are insulating materials that can be polarized by an applied electric field. When inserted in a capacitor, they increase the capacitance by reducing the electric field through polarization.

8.1 Polarization of Dielectrics

Polar Molecules

Have permanent dipole moment
Randomly oriented without field
Align with external field
Examples: H₂O, HCl, NH₃

Non-Polar Molecules

No permanent dipole moment
Develop induced dipole in field
Charge separation occurs
Examples: N₂, O₂, CO₂

8.2 Dielectric Constant & Permittivity

Key Definitions

Dielectric Constant (K or εᵣ):

K = \frac{C}{C_0} = \frac{E_0}{E} = \frac{\epsilon}{\epsilon_0}

Ratio of capacitance with/without dielectric

K ≥ 1 (dimensionless)

Permittivity (ε):

\epsilon = K\epsilon_0

where ε₀ = 8.85 × 10⁻¹² F/m (permittivity of free space)

8.3 Dielectric Constants of Common Materials

Material	Dielectric Constant (K)	Application
Vacuum	1	Reference
Air	1.00059 ≈ 1	Practical reference
Paper	3.7	Paper capacitors
Glass	5-10	Glass capacitors
Mica	6	High frequency circuits
Water	80	Universal solvent
Barium Titanate	1200-10000	High-K capacitors

💡 Important Formulas Summary

With dielectric, capacitance increases:

C = KC_0 = K\frac{\epsilon_0 A}{d}

Electric field decreases:

E = \frac{E_0}{K}

If battery disconnected (Q constant):

V = \frac{V_0}{K}, \quad U = \frac{U_0}{K}

If battery connected (V constant):

Q = KQ_0, \quad U = KU_0

📝 Previous Year Questions Analysis

JEE Main (Last 5 Years)

✓ Capacitors: 35%
✓ Gauss Law Applications: 25%
✓ Electric Field & Potential: 20%
✓ Electric Dipole: 15%
✓ Coulomb's Law: 5%

JEE Advanced (Last 5 Years)

✓ Complex Capacitor Networks: 30%
✓ Gauss Law (all cases): 25%
✓ Potential Energy Systems: 20%
✓ Dipole in Non-uniform Field: 15%
✓ Multi-concept Integration: 10%

Weightage Analysis

JEE Main: 12-16 marks (3-4 questions)

JEE Advanced: 18-24 marks (4-6 questions)

Difficulty Level: Medium to Hard

Time Required to Master: 15-20 days

Download Complete PYQ PDF (2015-2024)

🎯 Practice Problem Set

Level 1: Basic (JEE Main Standard)

Two charges +3μC and -3μC are placed 10 cm apart. Find force between them.
Find electric field at a distance of 20 cm from a point charge of 5μC.
Calculate the capacitance of a parallel plate capacitor with area 50 cm² and separation 1 mm.
Three capacitors 2μF, 3μF, and 6μF are connected in series. Find equivalent capacitance.
An electric dipole of moment 4×10⁻⁹ C·m is placed in a field of 10⁵ N/C. Find maximum torque.
Find electric potential at a distance of 30 cm from a charge of +6μC.
Calculate energy stored in a 100μF capacitor charged to 50V.
Using Gauss law, prove that electric field inside a uniformly charged sphere at distance r from center is proportional to r.

Level 2: Intermediate (JEE Main/Advanced)

Four charges +q, +q, -q, -q are placed at corners of a square. Find electric field at center.
A capacitor of 5μF charged to 100V is connected to another uncharged capacitor of 3μF. Find common potential and energy loss.
An infinite line charge has linear charge density 2μC/m. Find electric field at 10 cm distance.
Two parallel plates have surface charge densities +σ and -2σ. Find electric field in three regions.
A uniformly charged ring of radius R has total charge Q. Find electric field on axis at distance x from center.
Three capacitors are connected: 2μF in series with parallel combination of 3μF and 6μF. Find equivalent capacitance.
Calculate work done to bring a charge of 5μC from infinity to a point where potential is 100V.
A parallel plate capacitor with dielectric (K=5) has capacitance 100pF. Find capacitance without dielectric.

Level 3: Advanced (JEE Advanced/Olympiad)

A non-conducting sphere of radius R has volume charge density ρ = kr where k is constant. Find E(r) inside and outside.
Two capacitors C₁ and C₂ are charged to potentials V₁ and V₂. They are connected with opposite polarity. Find final charge distribution and energy loss.
An electric dipole is placed at angle θ with a non-uniform electric field E = ax (a = constant). Find force and torque.
A variable capacitor has plates that can overlap area changed from A to A/2. Find change in energy if: (a) battery connected (b) battery disconnected.
A charge Q is distributed uniformly on a ring of radius R. Another charge q is placed at center and released. Find its speed when it reaches distance x on axis.
Three concentric spherical shells of radii R, 2R, 3R have charges Q, -2Q, 3Q. Find potential at all regions.
A capacitor has three dielectrics with constants K₁, K₂, K₃ arranged in series each of thickness d/3. Find equivalent capacitance.
Prove that force per unit area between plates of parallel plate capacitor is ε₀E²/2 where E is electric field.

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Year	JEE Main	JEE Advanced	Most Asked Topic
2024	3 Questions (12 marks)	5 Questions (18 marks)	Capacitor with dielectric, Gauss Law
2023	4 Questions (16 marks)	4 Questions (15 marks)	Energy in capacitors, Electric field
2022	3 Questions (12 marks)	6 Questions (21 marks)	Potential energy, Capacitor combinations
2021	4 Questions (16 marks)	5 Questions (18 marks)	Gauss Law applications, Dipole

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Electrostatics JEE Main & Advanced 2025-26

Electric Charge

1.1 Properties of Electric Charge

Basic Properties

Important Constants

1.2 Methods of Charging

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📝 Solved Example 1

Coulomb's Law

2.1 Mathematical Formulation

Coulomb's Law Formula

2.2 Key Features of Coulomb's Law

Vector Form

Superposition Principle

2.3 Comparison: Coulomb Force vs Gravitational Force

⚠️ Common JEE Mistakes

📝 Solved Example 2

Electric Field

3.1 Electric Field Definition & Formula

Electric Field Intensity

3.2 Electric Field of Different Charge Distributions

3.3 Electric Field Lines

Properties of Field Lines

Field Line Patterns

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📝 Solved Example 3

Electric Dipole

4.1 Electric Dipole Moment

Dipole Moment Formula

4.2 Electric Field Due to Dipole

Important Relations

4.3 Dipole in Uniform Electric Field

Torque on Dipole

Potential Energy

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📝 Solved Example 4

Gauss Law

5.1 Electric Flux

Electric Flux Definition

5.2 Gauss Law Statement & Formula

Gauss Law (Integral Form)

5.3 Applications of Gauss Law

5.4 Electric Field Due to Parallel Charged Sheets

Two Parallel Infinite Sheets

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📝 Solved Example 5

Electric Potential & Potential Energy

6.1 Electric Potential Definition

Electric Potential Formula

6.2 Potential Due to Different Charge Distributions

6.3 Relation Between E and V

E-V Relationship

6.4 Electrostatic Potential Energy

Two Point Charges

System of Charges

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Capacitors and Capacitance

7.1 Capacitance Definition

Capacitance Formula

7.2 Capacitance of Different Configurations

7.3 Combination of Capacitors

Series Combination

Parallel Combination

7.4 Energy Stored in Capacitor

Energy Formulas (All Equivalent)

7.5 Effect of Dielectric

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📝 Solved Example 7

Dielectrics

8.1 Polarization of Dielectrics

Polar Molecules

Non-Polar Molecules

8.2 Dielectric Constant & Permittivity

Key Definitions

8.3 Dielectric Constants of Common Materials

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