JEE (Advanced) · Physics, Class 12
Electrostatics: JEE Advanced PYQ, 2007–2026
Electrostatics (Physics, Class 12) has been asked 81 times in the 40 JEE (Advanced) papers from 2007–2026: about 2 questions a paper, 11% of all Physics questions. It is the most-asked of 19 Physics chapters. The two 2026 papers had 3 from it. Last checked 28 October 2026.
Questions per year, both papers
Year by year.
| Year | Questions |
|---|---|
| 2007 | 5in 2 papers |
| 2008 | 6in 2 papers |
| 2009 | 4in 2 papers |
| 2010 | 3in 2 papers |
| 2011 | 4in 2 papers |
| 2012 | 6in 2 papers |
| 2013 | 3in 2 papers |
| 2014 | 4in 2 papers |
| 2015 | 4in 2 papers |
| 2016 | 3in 2 papers |
| 2017 | 3in 2 papers |
| 2018 | 4in 2 papers |
| 2019 | 4in 2 papers |
| 2020 | 6in 2 papers |
| 2021 | 2in 2 papers |
| 2022 | 6in 2 papers |
| 2023 | 2in 2 papers |
| 2024 | 5in 2 papers |
| 2025 | 4in 2 papers |
| 2026 | 3in 2 papers |
Each year has two papers, and the number of questions has fallen over time, so compare per paper and expect some drift.
Inside the chapter
Topics asked most.
- Gauss's law25
- Electric field and potential22
- Capacitance17
- Electric dipole8
- Conductors and dielectrics6
- Coulomb's law3
Try it
10 Electrostatics questions to practise.
Our own questions, written in the JEE Advanced pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Physics · Electrostatics
An uncharged solid metal sphere is placed in a uniform external electric field E0 directed to the right, and electrostatic equilibrium is reached. What is the electric field at the centre of the sphere produced by the induced surface charges alone?
Answer: (A)
In equilibrium the net field inside a conductor is zero. The net field is the external field E0 plus the field of the induced charges, so E_induced + E0 = 0. Hence E_induced = -E0, which has magnitude E0 and points to the left, opposite to the applied field.
Q2 · Physics · Electrostatics
A solid conducting sphere of radius 2 cm carries a charge of +2 nC. It is surrounded concentrically by a thick, initially uncharged conducting spherical shell of inner radius 5 cm and outer radius 8 cm. The outer surface of the shell is then connected to the earth by a thin wire. Taking the potential at infinity as zero and k = 9 × 10⁹ N m² C⁻², what is the potential of the inner sphere?
Answer: (A)
The inner sphere's +2 nC induces −2 nC on the shell's inner surface (radius 5 cm) and +2 nC on its outer surface. Earthing the shell drives its potential to zero and leaves no charge on the outer surface. The field is nonzero only for 2 cm < r < 5 cm, where E = kQ/r². The shell material is at 0 V, so the sphere's potential equals the potential difference across that gap: V = kQ(1/0.02 − 1/0.05) = 18 × (50 − 20) = 540 V.
Q3 · Physics · Electrostatics
Four small charges are placed at the corners of a square of side a. Three of the corners carry an identical charge +q. What is the magnitude of the net electrostatic force on the +q charge at the fourth corner, in terms of k = 1/4πε₀?
Answer: (B)
Each of the two adjacent charges is at distance a and exerts a repulsion kq²/a². These two forces are at right angles, so their resultant is √2 kq²/a². It points along the diagonal, away from the opposite corner. The diagonally opposite charge is at distance √2 a and exerts kq²/(2a²) along the same line and in the same direction. The total is (√2 + 1/2) kq²/a².
Q4 · Physics · Electrostatics
An infinite ladder of identical capacitors, each of capacitance C, is built as follows. Input terminal A is joined through a capacitor C to a node P on the upper wire. Between P and the lower wire (terminal B) there is a capacitor C, and in parallel with it the next identical unit, starting with another series capacitor C on the upper wire followed by a shunt capacitor C, and so on without end. What is the equivalent capacitance between A and B?
Answer: (A)
Adding one more unit to an infinite ladder leaves the capacitance unchanged. Let it be X. The section to the right of node P is the shunt C in parallel with X, giving C + X. This is in series with the input capacitor C, so X = C(C + X)/(2C + X). Cross-multiplying gives X(2C + X) = C² + CX, so X² + CX − C² = 0. The positive root is X = C(√5 − 1)/2 ≈ 0.618C.
Q5 · Physics · Electrostatics
Compare the electrostatic force with the gravitational force between a proton and an electron at any given separation. Take k = 1/4πε₀ = 9×10⁹ N m² C⁻², e = 1.6×10⁻¹⁹ C, G = 6.67×10⁻¹¹ N m² kg⁻², mₑ = 9.1×10⁻³¹ kg and mₚ = 1.67×10⁻²⁷ kg. The ratio Fₑ/F_g is approximately:
Answer: (A)
Both forces vary as 1/r², so r cancels in the ratio. Fₑ = ke²/r² and F_g = G mₑ mₚ/r². So Fₑ/F_g = ke²/(G mₑ mₚ). The numerator is 9×10⁹ × (1.6×10⁻¹⁹)² = 2.30×10⁻²⁸. The denominator is 6.67×10⁻¹¹ × 9.1×10⁻³¹ × 1.67×10⁻²⁷ ≈ 1.01×10⁻⁶⁷. The ratio is about 2×10³⁹.
Q6 · Physics · Electrostatics
Four point charges are fixed at the corners of a square of side a, taken in order around the square: +q, +q, −q, −q. Let k = 1/(4πε₀) and take the potential at infinity as zero. Which option correctly gives the electric potential V and the magnitude of the electric field E at the centre of the square?
Answer: (A)
Each corner is at distance r = a/√2 from the centre, so each charge contributes a field of magnitude kq/r² = 2kq/a². Potential is a scalar: 2(kq/r) − 2(kq/r) = 0. For the field, place +q at (a/2, a/2) and (−a/2, a/2), and −q at (−a/2, −a/2) and (a/2, −a/2). Each +q pushes the field away from itself, and each −q pulls it toward itself. The x-components cancel in pairs, and every charge contributes a component in the −y direction equal to 2kq/a² × cos45°. The total is 4 × (2kq/a²)/√2 = 4√2 kq/a². So V = 0 but E = 4√2 kq/a².
Q7 · Physics · Electrostatics
A dipole in a uniform electric field E has a potential energy of −2.0 mJ when its axis makes 60° with the field direction (taking U = 0 at 90°). The magnitude of the torque on the dipole at this orientation is:
Answer: (D)
U = −pE cosθ, so −2.0 mJ = −pE cos60° and pE = 4.0 × 10⁻³ J. The torque magnitude is τ = pE sinθ = 4.0 × 10⁻³ × (√3/2) = 2√3 × 10⁻³ ≈ 3.46 × 10⁻³ N m.
Q8 · Physics · Electrostatics
Two short electric dipoles of moments p₁ = 2.0 × 10⁻⁹ C m and p₂ = 3.0 × 10⁻⁹ C m lie along the same straight line, with their centres 0.30 m apart. Both moments point in the same direction along this line (head to tail), and the dipole lengths are much smaller than 0.30 m. Taking 1/(4πε₀) = 9 × 10⁹ N m²/C², the force exerted by the first dipole on the second is:
Answer: (A)
The field of p₁ on its axis is E = 2kp₁/r³ along p₁. The field is non-uniform, so a dipole p₂ aligned with it feels a net force F = p₂ (dE/dr) in the direction of increasing field. Here dE/dr = −6kp₁/r⁴, so |F| = 6kp₁p₂/r⁴. Substituting, F = 6 × 9×10⁹ × (2×10⁻⁹)(3×10⁻⁹)/(0.30)⁴ = 3.24×10⁻⁷/8.1×10⁻³ = 4.0×10⁻⁵ N. The field is larger nearer p₁, so the force points toward p₁ and is attractive. The repulsive option reverses the direction. The 1.3×10⁻⁵ N option uses the wrong coefficient 2. The 4.0×10⁻⁶ N option is the interaction energy 2kp₁p₂/r³ (a quantity in joules), not the force.
Q9 · Physics · Electrostatics
A conducting sphere of radius 3 cm carries a charge of 12 nC. It is connected by a long thin conducting wire to a distant, initially uncharged conducting sphere of radius 1 cm. After equilibrium is reached, what is the charge on the smaller sphere? (Ignore the charge on the wire and any interaction between the spheres.)
Answer: (A)
Connected conductors reach the same potential. For far-apart spheres V = kQ/R, so Q1/R1 = Q2/R2, giving Q1:Q2 = 3:1. The total charge is conserved at 12 nC. The smaller sphere therefore gets 12 × 1/4 = 3 nC and the larger gets 9 nC.
Q10 · Physics · Electrostatics
A point charge Q = +4 nC is fixed in space. A short dipole of dipole moment p = 2 × 10⁻¹⁰ C m is placed at a distance r = 0.2 m from Q, with its dipole moment along the line joining the dipole centre to Q and directed away from Q. The dipole length is much smaller than r. Taking 1/(4πε₀) = 9 × 10⁹ N m²/C², the net force on the dipole is:
Answer: (A)
A dipole in a non-uniform field feels a net force even though its net charge is zero. For a dipole aligned with the field, F = p dE/dr. Here E = kQ/r², so dE/dr = -2kQ/r³, and the force is directed toward Q. The magnitude is F = 2kQp/r³ = 2(9×10⁹)(4×10⁻⁹)(2×10⁻¹⁰)/(0.2)³ = 1.44×10⁻⁸/8×10⁻³ = 1.8×10⁻⁶ N. This is attractive, because the end of the dipole nearer Q carries the negative charge and sits in the stronger field.
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All 81 real JEE Advanced questions from Electrostatics.
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About Electrostatics in JEE Advanced.
How many questions come from Electrostatics in JEE Advanced?
Electrostatics (Physics, Class 12) has been asked 81 times in the 40 JEE (Advanced) papers from 2007–2026: about 2 questions a paper, 11% of all Physics questions. It is the most-asked of 19 Physics chapters. The two 2026 papers had 3 from it.
Which topics of Electrostatics are asked most in JEE Advanced?
Gauss's law (25), Electric field and potential (22), Capacitance (17), Electric dipole (8), counted over the papers from 2007–2026.
Is Electrostatics important for JEE Advanced 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 1 of 19 in Physics; use that to decide how much time it gets, not whether to study it.
Where can I solve Electrostatics JEE Advanced previous year questions?
On Nikaalle, free: sign up and practise every past JEE Advanced question from Electrostatics with an explanation for each, or sit the full papers timed. The ten questions on this page are our own, written in the same pattern.
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