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Electric potential, potential difference, equipotential surfaces and potential energy: NEET MCQs with answers

48 questions on this topic of Electrostatics (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. Which of the following statements is INCORRECT regarding equipotential surfaces?

  1. Two equipotential surfaces corresponding to different potentials can intersect along a line.
  2. Equipotential surfaces are closer together in regions of stronger electric fields.
  3. Electric field lines are always perpendicular to equipotential surfaces at every point.
  4. For a uniform electric field, the equipotential surfaces are a family of parallel planes.
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Answer: (A) Two equipotential surfaces corresponding to different potentials can intersect along a line.

If two equipotential surfaces of different potentials were to intersect, the point of intersection would have two different values of electric potential and two distinct directions of electric field, which is physically impossible. Hence, equipotential surfaces can never intersect.

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2. The electric potential V at any point (x, y, z) in space is given by V = 6xy - y² - 2yz volts, where x, y, and z are in metres. The magnitude of the electric field at the point (1, 1, 0) m is:

  1. 2√10 V m⁻¹
  2. 2√11 V m⁻¹
  3. √56 V m⁻¹
  4. √44 V m⁻¹
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Answer: (C) √56 V m⁻¹

The components of the electric field are: E_x = -∂V/∂x = -6y; E_y = -∂V/∂y = -(6x - 2y - 2z); E_z = -∂V/∂z = -(-2y) = 2y. At the point (1, 1, 0): E_x = -6(1) = -6 V m⁻¹, E_y = -(6(1) - 2(1) - 0) = -4 V m⁻¹, E_z = 2(1) = 2 V m⁻¹. The magnitude is E = √((-6)² + (-4)² + 2²) = √(36 + 16 + 4) = √56 V m⁻¹.

hard

3. A point charge +q is placed at the centre O of an uncharged, isolated conducting spherical shell of inner radius R₁ and outer radius R₂. A point P lies at a distance r from the centre, such that R₁ < r < R₂. The electric potential at point P is:

  1. 0
  2. q / (4πε₀ R₁)
  3. q / (4πε₀ r)
  4. q / (4πε₀ R₂)
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Answer: (D) q / (4πε₀ R₂)

The point P lies inside the conducting material of the shell. Since the electric field inside a conductor is zero, the entire conducting shell (from r = R₁ to r = R₂) is an equipotential volume whose potential equals the potential at its outer surface r = R₂. The potential at the outer surface is entirely due to the induced outer charge +q, which is V = q / (4πε₀ R₂).

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4. Three point charges +q, +q and -q are located at the vertices of an equilateral triangle of side length a. What is the total electrostatic potential energy of this system?

  1. -1/(4πε₀) × (q² / a)
  2. +1/(4πε₀) × (q² / a)
  3. -1/(4πε₀) × (3q² / a)
  4. 0
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Answer: (A) -1/(4πε₀) × (q² / a)

The system contains three pairs of interacting charges: (+q, +q), (+q, -q), and (+q, -q). The total potential energy is U = [1/(4πε₀a)] × [(q)(q) + (q)(-q) + (q)(-q)] = [1/(4πε₀a)] × [q² - q² - q²] = -q² / (4πε₀a).

easy

5. Four identical point charges +q are placed at the four corners of a square of side L. The electrostatic potential energy of this four-charge system is:

  1. (1 / 4πε₀) · (q² / L) · (4 + √2)
  2. (1 / 4πε₀) · (4q² / L)
  3. (1 / 4πε₀) · (q² / L) · (4 + 1/√2)
  4. (1 / 4πε₀) · (2q² / L) · (2 + √2)
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Answer: (A) (1 / 4πε₀) · (q² / L) · (4 + √2)

For four charges at the corners of a square, there are 4 pairs along the edges (separation L) and 2 pairs along the diagonals (separation √2 L). The total potential energy is U = 4 · (1 / 4πε₀)(q² / L) + 2 · (1 / 4πε₀)(q² / (√2 L)) = (1 / 4πε₀)(q² / L)(4 + 2/√2) = (1 / 4πε₀)(q² / L)(4 + √2).

hard

6. A solid conducting sphere of radius R₁ is surrounded by a concentric, thin conducting spherical shell of radius R₂ (R₂ > R₁). The inner sphere is given a charge +Q, while the outer shell is grounded. The potential of the inner sphere is:

  1. Q/(4πε₀R₁)
  2. Q/(4πε₀) · [1/R₁ - 1/R₂]
  3. Q/(4πε₀R₂)
  4. Zero
Show answer and explanation

Answer: (B) Q/(4πε₀) · [1/R₁ - 1/R₂]

Since the outer shell is grounded, its potential is zero: V₂ = (1/4πε₀)[Q/R₂ + Q'/R₂] = 0, which implies induced charge Q' = -Q. The potential of the inner sphere is then V₁ = (1/4πε₀)[Q/R₁ + Q'/R₂] = Q/(4πε₀) · [1/R₁ - 1/R₂].

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7. In a certain region of space, the electrostatic potential is given by V(x, y, z) = 3x²y - y³ + 2z volts, with coordinates in metres. The magnitude of the electric field at the point (1, 1, 0) is:

  1. √36 V m⁻¹
  2. √37 V m⁻¹
  3. √40 V m⁻¹
  4. √45 V m⁻¹
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Answer: (C) √40 V m⁻¹

The components of the electric field are E_x = -∂V/∂x = -6xy, E_y = -∂V/∂y = -(3x² - 3y²), and E_z = -∂V/∂z = -2. Evaluating at (1, 1, 0) gives E_x = -6(1)(1) = -6 V m⁻¹, E_y = -(3 - 3) = 0, and E_z = -2 V m⁻¹. The magnitude is E = √((-6)² + 0² + (-2)²) = √(36 + 4) = √40 V m⁻¹.

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8. Two identical thin conducting spherical shells of radii r₁ and r₂ (with r₁ < r₂) are separated by a large distance. Initially, each shell has a charge Q. They are now connected by a thin conducting wire. The total electrostatic energy of the system:

  1. Increases
  2. Becomes zero
  3. Remains the same
  4. Decreases
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Answer: (D) Decreases

Before connection, the potentials are V₁ = Q/(4πε₀r₁) and V₂ = Q/(4πε₀r₂). Since r₁ < r₂, V₁ > V₂. When connected, charge flows from the smaller shell to the larger shell until their potentials equalize, dissipating energy as heat through the wire. Thus, total electrostatic energy decreases.

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9. Two concentric thin spherical conducting shells have radii R and 2R. The inner shell carries a charge +2Q and the outer shell carries a charge -Q. What is the electrostatic potential at the surface of the inner shell?

  1. Q / (8πε₀R)
  2. 5Q / (8πε₀R)
  3. Q / (4πε₀R)
  4. 3Q / (8πε₀R)
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Answer: (D) 3Q / (8πε₀R)

The potential at the surface of the inner shell is the sum of potentials due to both shells: V = [1/(4πε₀)] × [(2Q / R) + (-Q / 2R)]. Simplifying gives V = [1/(4πε₀)] × [(4Q - Q) / 2R] = 3Q / (8πε₀R).

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10. An electric field in a region is given by E = (20x) î V m⁻¹, where x is in metres. The potential difference V_O - V_A, where V_O is the potential at origin (0, 0) and V_A is the potential at (2 m, 0), is:

  1. -40 V
  2. -20 V
  3. +40 V
  4. +20 V
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Answer: (C) +40 V

The potential difference is given by V_A - V_O = -∫ E dx from 0 to 2. Evaluating this gives -[10x²] from 0 to 2 = -40 V. Therefore, V_O - V_A = -(-40 V) = +40 V.

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11. Consider the following two statements: Statement I: The electric potential is necessarily zero at any point where the electric field is zero. Statement II: The electric field inside a charged hollow spherical conductor is zero, but the electric potential is non-zero and constant throughout its interior. In the light of the above statements, choose the most appropriate 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.
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Answer: (C) Statement I is incorrect but Statement II is correct.

Statement I is incorrect because electric field is the negative gradient of potential (E = -dV/dr); E = 0 only implies that V is constant, not necessarily zero. Statement II is correct because inside a charged hollow conductor, E = 0 and V has a constant, non-zero value equal to its surface potential.

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12. A particle of mass m and charge +q is projected from a large distance (infinity) with an initial kinetic energy K directly toward a fixed point charge +Q. If the initial kinetic energy of the particle is doubled, its distance of closest approach to the charge +Q will:

  1. become four times
  2. become two times
  3. remain unchanged
  4. be halved
Show answer and explanation

Answer: (D) be halved

By conservation of mechanical energy, the initial kinetic energy K transforms entirely into electrostatic potential energy at the distance of closest approach r_min: K = (1 / 4πε₀)(qQ / r_min). Thus, r_min is inversely proportional to K. Doubling the kinetic energy halves the distance of closest approach.

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