Electrostatics

22 NEET questions on Electrostatics. A sample is shown below with full explanations; the rest are available free after signing up.

Electrostatics2025hard

Q1.The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constant K₁ and K₂ with thickness 3d/8 and d/2, respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If K₁ = 1.25 K₂, the value of K₁ is:

  • A2.66Answer
  • B2.33
  • C1.60
  • D1.33

Explanation

Slabs stacked along the field direction behave as capacitors in series, so C = ε₀A / Σ(tᵢ/Kᵢ), and note the leftover air gap of thickness d/8. Setting this equal to 2ε₀A/d and substituting K₂ = 0.8K₁ gives 1/K₁ = 3/8, so K₁ = 8/3 ≈ 2.66.

Source: NEET 2025 (4 May), Code 45, Q20

Electrostatics2025easy

Q2.Two identical charged conducting spheres A and B have their centres separated by a certain distance. Charge on each sphere is q and the force of repulsion between them is F. A third identical uncharged conducting sphere is brought in contact with sphere A first and then with B and finally removed from both. New force of repulsion between spheres A and B (radii of A and B are negligible compared to the distance of separation, so for calculating the force between them they can be considered as point charges) is best given as:

  • A3F/5
  • B2F/3
  • CF/2
  • D3F/8Answer

Explanation

Identical conductors in contact share their total charge equally. Touching A leaves q/2 on A and q/2 on the third sphere; then touching B shares q/2 + q equally, leaving 3q/4 on B. Coulomb force goes as the product of the charges, so F' = (1/2)(3/4)F = 3F/8.

Source: NEET 2025 (4 May), Code 45, Q28

Electrostatics2025easy

Q3.An electric dipole with dipole moment 5 x 10⁻⁶ C m is aligned with the direction of a uniform electric field of magnitude 4 x 10⁵ N/C. The dipole is then rotated through an angle of 60° with respect to the electric field. The change in the potential energy of the dipole is:

  • A0.8 J
  • B1.0 JAnswer
  • C1.2 J
  • D1.5 J

Explanation

The potential energy of a dipole in a uniform field is U = −pE cos θ, so the change on rotating from 0° to 60° is ΔU = pE(cos 0° − cos 60°) = pE/2. Here pE = 5 x 10⁻⁶ x 4 x 10⁵ = 2 J, giving ΔU = 1.0 J.

Source: NEET 2025 (4 May), Code 45, Q35

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Questions are reformatted from NEET previous year papers with original explanations and topic tagging. NEET Prep is not affiliated with the NEET examining body.