NEET Electrostatics PYQs with Solutions | 2013–2025

NEET Electrostatics PYQs with Solutions | 2013–2025

2025

  1. Two point charges q1q_1q1​ and q2q_2q2​ are placed 1 m apart in vacuum. The force between them is 9 N. If each charge is doubled and distance halved, what is the new force?
    Concept: Coulomb’s law, inverse square law.

2024

  1. A point charge +Q is placed at the center of a spherical shell of radius R. What is the electric field at the surface of the shell?
    Concept: Electric field inside a conductor and due to spherical symmetry.
  2. Work done in moving a charge q from point A to point B in an electric field is independent of the path. True or False?
    Concept: Electric potential and conservative nature of electrostatic field.

2023

  1. Two identical charges repel each other with force F at a distance r. If one charge is doubled and the other halved, the force becomes?
    Concept: Coulomb’s law application.

2022

  1. A uniform electric field exists between two parallel plates separated by distance d. A particle of charge q is released from rest. What is the acceleration of the particle?
    Concept: Electric field, force, and Newton’s second law.

2021

  1. The potential at a point P due to a point charge Q is 100 V. If a charge q = 2 μC is placed at P, what is the potential energy of the charge?
    Concept: Electric potential energy.

2020

  1. Electric field at the midpoint of a line joining two equal charges Q separated by distance 2a is?
    Concept: Superposition principle in electrostatics.

2019

  1. A charge of 4 μC is at the origin. Find the work done in moving a 2 μC charge from x = 1 m to x = 2 m along the x-axis.
    Concept: Electric potential difference and work.

2018

  1. A small charge q is placed at a distance r from a large charge Q. Find the force on q.
    Concept: Coulomb’s law, vector nature of forces.

2017

  1. Two point charges +Q and −Q are separated by distance 2a. Find the electric field at a point on the perpendicular bisector.
    Concept: Electric field due to dipole.

2016

  1. Electric potential at the center of a uniformly charged ring of radius R is?
    Concept: Superposition principle and symmetry.

2015

  1. A sphere of radius R carries charge Q uniformly over its surface. Electric field outside the sphere at distance r from the center is?
    Concept: Gauss’s law for spherical charge distribution.

2014

  1. Two identical charges separated by distance r experience a repulsive force F. If distance is doubled, what is the new force?
    Concept: Coulomb’s law, inverse square relation.

2013

  1. Work done in bringing two point charges from infinity to separation r is?
    Concept: Electrostatic potential energy.
Answer

2025

Q: Two charges q1q_1q1​ and q2q_2q2​ at 1 m exert 9 N force. If each charge is doubled and distance halved, new force?

Solution:

  • Original: F=kq1q2r2=9F = k \frac{q_1 q_2}{r^2} = 9F=kr2q1​q2​​=9 N
  • New: F′=k(2q1)(q2/2)(0.5)2=kq1q20.25=36F’ = k \frac{(2q_1)(q_2/2)}{(0.5)^2} = k \frac{q_1 q_2}{0.25} = 36F′=k(0.5)2(2q1​)(q2​/2)​=k0.25q1​q2​​=36 N
    Answer: 36 N

2024

Q1: Point charge +Q at center of spherical shell. Electric field at surface?
Solution: By Gauss’s law: Electric field outside a shell behaves as if charge is concentrated at center. At surface, E=14πϵ0QR2E = \frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}E=4πϵ0​1​R2Q​
Answer: E=Q4πϵ0R2E = \frac{Q}{4\pi\epsilon_0 R^2}E=4πϵ0​R2Q​

Q2: Work done moving q from A to B independent of path?
Solution: Electrostatic field is conservative → work depends only on potential difference.
Answer: True

2023

Q: Two identical charges repel with F at distance r. One doubled, other halved → force?
Solution: F′=k(2Q)(Q/2)r2=kQ2r2=FF’ = k \frac{(2Q)(Q/2)}{r^2} = k \frac{Q^2}{r^2} = FF′=kr2(2Q)(Q/2)​=kr2Q2​=F
Answer: F

2022

Q: Particle of charge q in uniform field between plates. Acceleration?
Solution: F=qEF = qEF=qE, a=F/m=qE/ma = F/m = qE/ma=F/m=qE/m
Answer: a = qE/m

2021

Q: Potential at P = 100 V. Charge q = 2 μC placed → potential energy?
Solution: U=qV=2×10−6×100=2×10−4U = qV = 2 \times 10^{-6} \times 100 = 2 \times 10^{-4}U=qV=2×10−6×100=2×10−4 J
Answer: 0.2 mJ

2020

Q: Electric field at midpoint between two equal charges Q at distance 2a?
Solution: Midpoint → fields equal magnitude, opposite directions → vector addition → E = 0
Answer: 0

2019

Q: Charge 4 μC at origin. Work to move 2 μC from x=1 to x=2 m?
Solution: W=qΔV=q(kQr2−kQr1)=2×10−6(9×109⋅4×10−62−9×109⋅4×10−61)W = q \Delta V = q \left(\frac{kQ}{r_2} – \frac{kQ}{r_1}\right) = 2 \times 10^{-6} \left( \frac{9 \times 10^9 \cdot 4 \times 10^{-6}}{2} – \frac{9 \times 10^9 \cdot 4 \times 10^{-6}}{1} \right)W=qΔV=q(r2​kQ​−r1​kQ​)=2×10−6(29×109⋅4×10−6​−19×109⋅4×10−6​)

  • W = -18 × 10^-3 J
    Answer: -0.018 J

2018

Q: Small charge q at distance r from Q → force?
Solution: Coulomb’s law: F=kQqr2F = k \frac{Qq}{r^2}F=kr2Qq​

2017

Q: Two charges +Q, −Q separated by 2a. Field at perpendicular bisector?
Solution: Dipole → E=14πϵ02Qa(a2+y2)3/2E = \frac{1}{4\pi\epsilon_0} \frac{2Qa}{(a^2 + y^2)^{3/2}}E=4πϵ0​1​(a2+y2)3/22Qa​

2016

Q: Electric potential at center of uniformly charged ring of radius R?
Solution: V=kQRV = k \frac{Q}{R}V=kRQ​ (all points equidistant)

2015

Q: Sphere radius R, charge Q uniformly. Field at r>R?
Solution: E=14πϵ0Qr2E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}E=4πϵ0​1​r2Q​ (Gauss’s law)

2014

Q: Two charges at distance r → F. If distance doubled, new force?
Solution: F′=Fr2(2r)2=F/4F’ = F \frac{r^2}{(2r)^2} = F/4F′=F(2r)2r2​=F/4

2013

Q: Work bringing two charges from infinity to r?
Solution: W=kq1q2rW = k \frac{q_1 q_2}{r}W=krq1​q2​​