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NEET Electrostatics MCQs with answers

452 questions on Electrostatics for NEET 2027, 1 of them from previous NEET papers. Below are 15 to try; tap a question to see its answer and explanation.

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15 sample questions

easyNEET 2022

1. Electric field inside a hollow conducting sphere:

  1. Maximum at center
  2. Zero
  3. Uniform
  4. Depends on charge
Show answer and explanation

Answer: (B) Zero

By Gauss's law, no enclosed charge → E = 0 inside conductor.

Quick tip: Inside conductor = E is ZERO always. Charges on outer surface.

NCERT Class 12, Chapter 1, page 18

easy

2. The magnitude of the electric field 1 m from a point charge of 9 μC in vacuum (1/(4πε₀) = 9 × 10⁹ N m² C⁻²) is:

  1. 40500 N/C
  2. 8.1 × 10¹⁰ N/C
  3. 81000 N/C
  4. 1.62 × 10⁵ N/C
Show answer and explanation

Answer: (C) 81000 N/C

E = kq/r² = 9 × 10⁹ × 9 × 10⁻⁶/1² = 81000 N/C.

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3. An electron orbits around a long straight line of uniform positive linear charge density λ in a circular path of radius r. The kinetic energy of the electron is:

  1. eλ / (2πε₀)
  2. eλ / (4πε₀)
  3. eλ / (8πε₀)
  4. eλ / (πε₀)
Show answer and explanation

Answer: (B) eλ / (4πε₀)

The electric field produced by the line of charge is E = λ / (2πε₀r). The centripetal force is m v² / r = e E = eλ / (2πε₀r), which yields m v² = eλ / (2πε₀). Therefore, the kinetic energy is K = (1/2) m v² = eλ / (4πε₀), which is notably independent of r.

hard

4. Three identical charges +Q are placed at the vertices of an equilateral triangle. A charge q is placed at the centroid of the triangle such that the entire system is in equilibrium. What must be the value of q?

  1. -Q / 3
  2. -Q / √3
  3. -√3 Q
  4. -Q / 2
Show answer and explanation

Answer: (B) -Q / √3

The centroid is at distance r = a/√3 from each vertex. For equilibrium of a corner charge Q, the net force from the other two corner charges (resultant = √3 k Q² / a² directed away from centroid) must be balanced by the attraction from charge q: k|q|Q / r² = 3 k|q|Q / a² directed towards the centroid. Equating these: √3 k Q² / a² = 3 k|q|Q / a², giving |q| = Q / √3. Since it must attract Q, q = -Q / √3.

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5. Two point charges carrying 9 μC and 8 μC are separated by 2 m in free space. With k = 9×10⁹ N m² C⁻², the electrostatic force between them is:

  1. 0.324 N
  2. 0.162 N
  3. 0.081 N
  4. 0.243 N
Show answer and explanation

Answer: (B) 0.162 N

F = kq₁q₂/r² = 9×10⁹×(9×10⁻⁶)×(8×10⁻⁶)/2² = 0.162 N.

Quick tip: Convert μC to C and use the inverse-square relation.

easy

6. A charge Q is to be divided into two parts, q and (Q - q). To maximize the electrostatic repulsive force between them at a given distance of separation, the ratio q / Q should be:

  1. 2 / 3
  2. 1 / 4
  3. 1 / 3
  4. 1 / 2
Show answer and explanation

Answer: (D) 1 / 2

The electrostatic force is proportional to f(q) = q(Q - q) = qQ - q². Differentiating with respect to q and setting to zero gives df/dq = Q - 2q = 0, which yields q = Q/2. Hence, q / Q = 1/2.

hard

7. At what angle θ between the position vector r and the dipole moment vector p is the electric field of a short dipole purely perpendicular to the dipole moment vector p?

  1. θ = 90°
  2. θ = cos⁻¹(1/√3)
  3. θ = cos⁻¹(√2/√3)
  4. θ = 45°
Show answer and explanation

Answer: (B) θ = cos⁻¹(1/√3)

The angle α between the electric field and the position vector is given by tan α = (1/2) tan θ. For the total electric field to be perpendicular to p, we must have α + θ = 90°, so tan α = cot θ. Thus, (1/2) tan θ = cot θ, which gives tan² θ = 2, so cos² θ = 1 / (1 + tan² θ) = 1/3, meaning θ = cos⁻¹(1/√3).

medium

8. Two large, parallel, non-conducting thin plates 1 and 2 carry uniform surface charge densities +2σ and -σ respectively. The magnitude of the net electric field in the region between the plates is:

  1. σ / (2ε₀)
  2. σ / ε₀
  3. 3σ / (2ε₀)
  4. 3σ / ε₀
Show answer and explanation

Answer: (C) 3σ / (2ε₀)

Each thin non-conducting plate produces a field of magnitude E = |σ_i| / (2ε₀) normal to its surface. Between the plates, the field due to the positively charged plate (+2σ) is directed away from it, and the field due to the negatively charged plate (-σ) is directed toward it, so both fields point in the same direction: E_net = (2σ / 2ε₀) + (σ / 2ε₀) = 3σ / (2ε₀).

medium

9. A uniformly charged solid insulating sphere of radius R has total charge Q distributed uniformly throughout its volume. At what distance r from the center of the sphere is the electric field intensity equal to half of its maximum value?

  1. R/2 and 2R
  2. R/2 and 4R
  3. R/4 and 2R
  4. R/2 and √2 R
Show answer and explanation

Answer: (D) R/2 and √2 R

The maximum electric field occurs at the surface (r = R) and is E_max = kQ/R². Inside the sphere (r < R), E = kQr/R³, so E = E_max / 2 gives r = R/2. Outside the sphere (r > R), E = kQ/r², so E = E_max / 2 gives kQ/r² = kQ/(2R²), which yields r = √2 R.

medium

10. Two point-sized charged bodies of 10 μC and 6 μC are placed 3 m apart in free space. Taking k = 9×10⁹ N m² C⁻², the magnitude of the electrostatic force between them is:

  1. 0.12 N
  2. 0.06 N
  3. 0.03 N
  4. 0.18 N
Show answer and explanation

Answer: (B) 0.06 N

F = kq₁q₂/r² = 9×10⁹×(10×10⁻⁶)×(6×10⁻⁶)/3² = 0.06 N.

Quick tip: Convert μC to C and use the inverse-square relation.

medium

11. Consider the following two statements regarding a parallel-plate capacitor: Statement I: The electric energy of a charged capacitor resides entirely in the electric field set up in the space between its plates. Statement II: The electrostatic force per unit area on each plate of a parallel-plate capacitor is proportional to the square of the electric field between the plates. In light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are correct.
  2. Statement I is correct but Statement II is incorrect.
  3. Statement I is incorrect but Statement II is correct.
  4. Both Statement I and Statement II are incorrect.
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct.

Statement I is correct because electrostatic potential energy is stored in the electric field localized between the plates with energy density u = (1/2)ε₀E². Statement II is also correct because the attractive force per unit area (electrostatic pressure) is given by F/A = (1/2)ε₀E², which is proportional to E².

medium

12. An electric field is given by E→ = (5x î + 2 ĵ) N/C. A closed cubical box of edge length 2 m is positioned with its edges parallel to the coordinate axes, spanning from x = 1 m to x = 3 m, y = 0 to y = 2 m, and z = 0 to z = 2 m. The net electric flux through this cube is:

  1. 20 N·m²/C
  2. 80 N·m²/C
  3. 40 N·m²/C
  4. 0 N·m²/C
Show answer and explanation

Answer: (C) 40 N·m²/C

The ĵ component of the field is uniform, so its net flux is zero. For the î component, the area of the faces perpendicular to x is A = 2 × 2 = 4 m². At x = 3 m, flux is E_x·A = (5 × 3)(4) = +60 N·m²/C. At x = 1 m, flux entering is E_x·A = (5 × 1)(4) = 20 N·m²/C, so net flux is 60 - 20 = 40 N·m²/C. Alternatively, by divergence theorem, ∫ (dE_x/dx) dV = 5 × (2)³ = 40 N·m²/C.

medium

13. A slab of isotropic dielectric material has an electric susceptibility χₑ = 4. The slab is placed in an external uniform electric field of magnitude 5.0 × 10⁵ V m⁻¹. What is the magnitude of the net electric field E inside the dielectric slab?

  1. 1.0 × 10⁵ V m⁻¹
  2. 1.25 × 10⁵ V m⁻¹
  3. 2.0 × 10⁵ V m⁻¹
  4. 2.5 × 10⁵ V m⁻¹
Show answer and explanation

Answer: (A) 1.0 × 10⁵ V m⁻¹

The dielectric constant is related to susceptibility by K = 1 + χₑ = 1 + 4 = 5. The net internal field inside the dielectric is E = E₀ / K = (5.0 × 10⁵ V m⁻¹) / 5 = 1.0 × 10⁵ V m⁻¹.

easy

14. Twenty-seven identical spherical drops of mercury, each charged to a potential V, coalesce to form a single larger drop. What is the electrostatic potential of the resulting big drop?

  1. 9 V
  2. 3 V
  3. 27 V
  4. V / 3
Show answer and explanation

Answer: (A) 9 V

When 27 identical drops of radius r coalesce to form a single drop of radius R, volume conservation gives (4/3)πR³ = 27 × (4/3)πr³, which leads to R = 3r. The total charge becomes Q = 27q. The potential of the big drop is V' = kQ / R = k(27q) / (3r) = 9(kq / r) = 9V.

medium

15. A dielectric slab of dielectric constant K = 4 is placed inside a uniform external electric field E₀ = 8.0 × 10⁵ V m⁻¹. What is the magnitude of the electric field produced solely by the bound polarization charges inside the slab?

  1. 2.0 × 10⁵ V m⁻¹
  2. 6.0 × 10⁵ V m⁻¹
  3. 4.0 × 10⁵ V m⁻¹
  4. 8.0 × 10⁵ V m⁻¹
Show answer and explanation

Answer: (B) 6.0 × 10⁵ V m⁻¹

The net electric field inside the dielectric is E = E₀ / K = (8.0 × 10⁵) / 4 = 2.0 × 10⁵ V m⁻¹. Since E = E₀ - E_induced, the field due to polarization charges is E_induced = E₀ - E = (8.0 - 2.0) × 10⁵ = 6.0 × 10⁵ V m⁻¹.

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