1. Chapter at a Glance
Electrostatics studies charges at rest: the forces between them, the electric field they create, and the electric potential and energy linked to that field. Gauss's law makes fields of symmetric charge distributions easy to find. This is one of the highest-weightage units in JEE Main Physics (together with capacitance, usually two or three questions).
- Coulomb's law and superposition; point where the net field is zero.
- Electric dipole: field, potential, torque and energy in a uniform field.
- Gauss's law: flux, fields of line charges, sheets and spheres.
- Potential, potential energy of a system of charges, work done.
2. Electric Charge and Coulomb's Law
- Charge is quantised (q = ne, e = 1.6 × 10⁻¹⁹ C), conserved and additive. Like charges repel, unlike attract.
- Charging by friction, conduction and induction. Induction does not reduce the charge on the inducing body.
- Coulomb's law: F = kq₁q₂/r², k = 1/(4πε₀) = 9 × 10⁹ N m² C⁻², ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻².
- In a medium of dielectric constant K, the force becomes F/K.
- Forces from several charges add as vectors (superposition).
3. Electric Field and Field Lines
- E = F/q₀; for a point charge E = kq/r², radially outward for positive charge.
- Field lines start on positive and end on negative charges, never cross, and are closer where the field is stronger. They meet a conductor's surface at right angles.
- Inside a conductor in electrostatic equilibrium E = 0, and any excess charge lies on the outer surface.
4. Electric Dipole
- Dipole moment p = q × 2a, directed from −q to +q.
- Axial point (far): E = 2kp/r³; equatorial point (far): E = kp/r³ (opposite to p).
- Potential: axial kp/r², equatorial 0.
- In a uniform field: net force zero, torque τ = p × E (|τ| = pE sinθ), potential energy U = −p·E.
- Stable equilibrium at θ = 0, unstable at θ = 180°. Work to rotate from θ₁ to θ₂: pE(cosθ₁ − cosθ₂).
5. Gauss's Law
| Charge distribution | Electric field |
|---|---|
| Infinite line (λ) | E = λ/(2πε₀r) = 2kλ/r |
| Infinite plane sheet (σ) | E = σ/2ε₀ (independent of distance) |
| Near a charged conductor | E = σ/ε₀ |
| Conducting sphere / shell (Q) | outside kQ/r²; inside 0 |
| Uniformly charged solid sphere | outside kQ/r²; inside kQr/R³ |
A charge at the centre of a cube sends flux q/6ε₀ through each face. Charges outside a closed surface give zero net flux through it.
6. Electric Potential and Energy
- Potential V = W/q₀ = kq/r (scalar — add algebraically). E = −dV/dr.
- Equipotential surfaces are perpendicular to field lines; no work is done moving a charge along one.
- Potential of a conducting sphere: kQ/R everywhere inside and on it; kQ/r outside.
- Potential energy of two charges U = kq₁q₂/r; for a system, sum over all pairs.
- Work done by an external agent = qΔV. An electron accelerated through V volts gains eV of energy.
7. Formula Master Table
| Quantity | Formula |
|---|---|
| Coulomb's law | F = kq₁q₂/r² |
| Point-charge field / potential | E = kq/r²; V = kq/r |
| Dipole (axial, equatorial) | 2kp/r³; kp/r³ |
| Dipole in field | τ = pE sinθ; U = −pE cosθ |
| Gauss's law | Φ = q/ε₀ |
| Line / sheet | λ/2πε₀r; σ/2ε₀ |
| Field–potential | E = −dV/dr |
| Energy of a pair | U = kq₁q₂/r |
8. Mind Map
- Quantisation, conservation
- Coulomb's law
- Superposition
- E = kq/r²
- Field lines
- Dipole
- Flux
- Line, sheet, sphere
- V = kq/r
- Equipotentials
- Potential energy
9. Tips and Tricks
- For two like charges, the null point lies between them, nearer the smaller charge; for unlike charges it lies outside, nearer the smaller charge.
- Field is a vector — watch the directions; potential is a scalar — just add with signs.
- Use symmetry: in the centre of a ring or square of equal charges, E = 0 but V ≠ 0.
- Energy of n charges: count n(n − 1)/2 pairs.
- Adding field magnitudes without directions.
- Using σ/2ε₀ near a conductor (it is σ/ε₀).
- Thinking E = 0 implies V = 0 (inside a charged shell E = 0 but V = kQ/R).
10. Important One-liners
- Electric field inside a conductor in equilibrium is zero.
- Excess charge on a conductor resides on its outer surface.
- Field lines never intersect.
- A dipole in a uniform field feels a torque but no net force.
- No work is done in moving a charge on an equipotential surface.
- Electric flux through a closed surface depends only on the enclosed charge.
11. Solved Numericals: Average to Advanced
Five average-level and five advanced-level problems. Click to reveal each answer and its full working. Take k = 9 × 10⁹ N m² C⁻² and ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻².
Average level
Find the force between charges of 2 μC and 3 μC placed 0.3 m apart in air.
Show answer & solution
Concept: F = kq₁q₂/r².
- q₁q₂ = 6 × 10⁻¹² C²; r² = 0.09 m².
- F = 9 × 10⁹ × 6 × 10⁻¹²/0.09 = 0.6 N (repulsive).
Answer: 0.6 N
Find the electric field 10 cm from a 5 nC point charge.
Show answer & solution
Concept: E = kq/r².
- E = 9 × 10⁹ × 5 × 10⁻⁹/(0.1)².
- = 45/0.01 = 4500 N C⁻¹, directed away from the charge.
Answer: 4500 N C⁻¹
Find the potential 10 cm from a 5 nC charge, and the work needed to bring a 2 μC charge there from infinity.
Show answer & solution
Concept: V = kq/r; W = qV.
- V = 9 × 10⁹ × 5 × 10⁻⁹/0.1 = 450 V.
- W = 2 × 10⁻⁶ × 450 = 9 × 10⁻⁴ J.
Answer: 450 V; 9 × 10⁻⁴ J
A charge of 8.85 nC sits at the centre of a cube. Find the flux through the whole cube and through one face.
Show answer & solution
Concept: Gauss's law Φ = q/ε₀; by symmetry each face gets one-sixth.
- Φ = 8.85 × 10⁻⁹/8.85 × 10⁻¹² = 1000 N m² C⁻¹.
- Per face = 1000/6 ≈ 167 N m² C⁻¹.
Answer: 1000 and about 167 N m² C⁻¹
An electron starting from rest is accelerated through a potential difference of 100 V. Find its kinetic energy and speed (me = 9.1 × 10⁻³¹ kg).
Show answer & solution
Concept: KE = eV.
- KE = 100 eV = 100 × 1.6 × 10⁻¹⁹ = 1.6 × 10⁻¹⁷ J.
- v = √(2KE/m) = √(3.2 × 10⁻¹⁷/9.1 × 10⁻³¹).
- ≈ 5.9 × 10⁶ m s⁻¹.
Answer: 100 eV; about 5.9 × 10⁶ m s⁻¹
Advanced level
Charges +4q and +q are fixed 3 m apart. Where between them is the net electric field zero?
Show answer & solution
Concept: the fields from like charges point opposite each other between them; set magnitudes equal.
- Let the point be x from 4q: k(4q)/x² = kq/(3 − x)².
- Take square roots: 2/x = 1/(3 − x) ⇒ 6 − 2x = x.
- x = 2 m from the 4q charge (1 m from q).
Answer: 2 m from +4q
A dipole of moment 2 × 10⁻⁸ C m is short compared with 0.1 m. Find the field at 0.1 m on its axis and on its equatorial line. Find also the torque on it in a uniform field of 10⁴ N C⁻¹ when p makes 30° with E.
Show answer & solution
Concept: Eaxial = 2kp/r³, Eeq = kp/r³, τ = pE sinθ.
- kp/r³ = 9 × 10⁹ × 2 × 10⁻⁸/10⁻³ = 1.8 × 10⁵ N C⁻¹.
- Axial: 3.6 × 10⁵ N C⁻¹ (along p); equatorial: 1.8 × 10⁵ N C⁻¹ (opposite p).
- τ = 2 × 10⁻⁸ × 10⁴ × 0.5 = 1 × 10⁻⁴ N m.
Answer: 3.6 × 10⁵ and 1.8 × 10⁵ N C⁻¹; 10⁻⁴ N m
Three charges of 1 μC each are placed at the corners of an equilateral triangle of side 10 cm. How much work is needed to assemble them from infinity?
Show answer & solution
Concept: U = sum over all pairs of kqiqj/r.
- There are 3 pairs, each at 0.1 m.
- Each pair: 9 × 10⁹ × 10⁻¹²/0.1 = 0.09 J.
- U = 3 × 0.09 = 0.27 J.
Answer: 0.27 J
(a) Find the field 10 cm from a long straight wire carrying 2 μC per metre. (b) A conducting sphere of radius 10 cm carries 10 nC. Find the field just outside its surface, the field at its centre and the potential at its centre.
Show answer & solution
Concept: line E = 2kλ/r; conducting sphere: outside kQ/r², inside E = 0, V = kQ/R throughout.
- (a) E = 2 × 9 × 10⁹ × 2 × 10⁻⁶/0.1 = 3.6 × 10⁵ N C⁻¹.
- (b) Surface field = 9 × 10⁹ × 10⁻⁸/0.01 = 9000 N C⁻¹.
- At the centre E = 0.
- Vcentre = kQ/R = 90/0.1 = 900 V.
Answer: 3.6 × 10⁵ N C⁻¹; 9000 N C⁻¹, 0, 900 V
Two identical pith balls, each of mass 1 g, hang from a common point on silk threads 50 cm long. When given equal charges, they come to rest 10 cm apart. Find the charge on each (g = 10 m s⁻²).
Show answer & solution
Concept: each ball is in equilibrium under weight, tension and electric force: F = mg tanθ.
- sinθ = 5/50 = 0.1 ⇒ tanθ ≈ 0.1005.
- F = 10⁻³ × 10 × 0.1005 ≈ 1.0 × 10⁻³ N.
- kq²/r² = F ⇒ q = r√(F/k) = 0.1 × √(1.0 × 10⁻³/9 × 10⁹).
- q ≈ 3.3 × 10⁻⁸ C.
Answer: about 3.3 × 10⁻⁸ C each
12. PYQ-pattern Questions with Solutions
These questions are original and modelled on concepts frequently tested in JEE Main.
The electric field inside a charged hollow conducting sphere is:
- Maximum at the centre
- Zero
- kQ/R²
- Infinite
Show solution
A Gaussian surface inside encloses no charge, so E = 0.
Answer: (B)
If the distance between two point charges is halved, the force between them becomes:
- Half
- Double
- Four times
- One-fourth
Show solution
F ∝ 1/r², so halving r multiplies F by 4.
Answer: (C)
An electric dipole is placed in a uniform electric field. It experiences:
- A net force and a torque
- A net force but no torque
- A torque but no net force (in general)
- Neither
Show solution
Forces qE and −qE cancel, but they act at different points, giving a torque pE sinθ (zero only when p ∥ E).
Answer: (C)
Four equal charges +q sit at the corners of a square of side a. At the centre:
- E = 0 and V = 0
- E = 0, V = 4√2 kq/a
- E ≠ 0, V = 0
- E ≠ 0, V ≠ 0
Show solution
Fields from opposite corners cancel, so E = 0. Each charge is a/√2 away: V = 4kq/(a/√2) = 4√2 kq/a.
Answer: (B)
How much work (in J) is needed to move a 5 μC charge between two points on the same equipotential surface of 100 V?
Show solution
W = qΔV = 0, because ΔV = 0 on an equipotential.
Answer: 0
The electric field due to an infinite plane sheet of charge density σ:
- Falls as 1/r
- Falls as 1/r²
- Is σ/2ε₀, independent of distance
- Is σ/ε₀ on both sides
Show solution
By Gauss's law with a pillbox, E = σ/2ε₀ on each side, uniform.
Answer: (C)
13. Quick Revision Summary
- F = kq₁q₂/r²; E = kq/r²; V = kq/r; E = −dV/dr.
- Dipole: axial 2kp/r³, equatorial kp/r³; τ = pE sinθ; U = −pE cosθ.
- Gauss: Φ = q/ε₀; line 2kλ/r; sheet σ/2ε₀; conductor surface σ/ε₀.
- Conductor: E = 0 inside, charge on surface, V constant.
- System energy: sum of kqiqj/r over all pairs.