Electrostatic Potential and Equipotential Surface – Electric potential and its calculation for a point charge: NEET MCQs with Answers

15 free questions on Electrostatic Potential and Equipotential Surface – Electric potential and its calculation for a point charge for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Electric potential and its calculation for a point charge, electric dipole and system of charges; potential difference, Equipotential surfaces

Practise all 134 Electrostatic Potential and Equipotential Surface – Electric potential and its calculation for a point charge questions free Take a free timed test

Question 1

Previous-year question
A short electric dipole has a dipole moment of 16 × 10⁻⁹ C m. The electric potential due to the dipole at a point at a distance of 0.6 m from the centre of the dipole, situated on a line making an angle of 60° with the dipole axis, is (1/4πε₀ = 9 × 10⁹ N m² C⁻²) [NEET 2020]
  1. 200 V
  2. 400 V
  3. zero
  4. 50 V
Show answer
Answer: (A) 200 V
Explanation: V = kp cos θ/r² = 9 × 10⁹ × 16 × 10⁻⁹ × cos 60°/(0.6)² = 144 × 0.5/0.36 = 200 V.

Question 2

Previous-year question
In a certain region of space with volume 0.2 m³, the electric potential is found to be 5 V throughout. The magnitude of the electric field in this region is [NEET 2020]
  1. 0.5 N C⁻¹
  2. 1 N C⁻¹
  3. 5 N C⁻¹
  4. zero
Show answer
Answer: (D) zero
Explanation: A constant potential means zero potential gradient in every direction, and since E = −dV/dr, the field is zero throughout that region.

Question 3

Previous-year question
The diagrams below show four regions of equipotentials, and a positive charge is moved from A to B in each. Then [NEET 2017]
  1. In all four cases the work done is the same
  2. Minimum work is required to move q in figure (i)
  3. Maximum work is required to move q in figure (ii)
  4. Maximum work is required to move q in figure (iii)
Show answer
Answer: (A) In all four cases the work done is the same
Explanation: The end points lie on the same two equipotential values in all four diagrams, so the potential difference — and hence the work — is identical. The shape and spacing of the intermediate lines do not matter.

Question 4

Previous-year question
If the potential (in volts) in a region is expressed as V(x, y, z) = 6xy − y + 2yz, the electric field (in N C⁻¹) at the point (1, 1, 0) is [AIPMT 2015]
  1. −(6î + 5ĵ + 2k̂)
  2. −(2î + 3ĵ + k̂)
  3. −(6î + 9ĵ + k̂)
  4. −(3î + 5ĵ + 3k̂)
Show answer
Answer: (A) −(6î + 5ĵ + 2k̂)
Explanation: ∂V/∂x = 6y = 6, ∂V/∂y = 6x − 1 + 2z = 5 and ∂V/∂z = 2y = 2. Since E⃗ = −∇V, the field is −(6î + 5ĵ + 2k̂).

Question 5

Previous-year question
In a region, the potential is represented by V(x, y, z) = 6x − 8xy − 8y + 6yz, where V is in volts and x, y, z are in metres. The electric force experienced by a charge of 2 coulomb situated at the point (1, 1, 1) is [AIPMT 2014]
  1. 6√5 N
  2. 30 N
  3. 24 N
  4. 4√35 N
Show answer
Answer: (D) 4√35 N
Explanation: The components are −∂V/∂x = 8y − 6 = 2, −∂V/∂y = 8x + 8 − 6z = 10 and −∂V/∂z = −6y = −6, so |E⃗| = √140. Hence F = 2√140 = 4√35 N.

Question 6

A charge Q is spread uniformly along a thin straight rod of length L. Point O lies on the line of the rod, beyond one end, at a distance 2L from the nearer end. The electric potential at O is
  1. (Q/8πε₀L) ln(3/2)
  2. Q/10πε₀L
  3. (Q/4πε₀L) ln(3/2)
  4. (Q/4πε₀L) ln 3
Show answer
Answer: (C) (Q/4πε₀L) ln(3/2)
Explanation: An element dx at a distance x from O carries charge (Q/L)dx and contributes dV = (Q/4πε₀L)(dx/x). The element positions run from x = 2L to x = 3L. Integrating gives V = (Q/4πε₀L) ln(3L/2L). Potential is a scalar, so no directions need to be considered.

Question 7

A uniform electric field of 200 V/m points along the +x direction. Points A, B and C have co-ordinates (1 m, 0), (1 m, 4 m) and (5 m, 3 m) respectively. Which statement about their potentials is correct?
  1. V_A = V_B, and V_C exceeds V_A by 800 V
  2. V_A = V_B, and V_A exceeds V_C by 800 V
  3. V_C > V_B > V_A, and V_C − V_A = 1000 V
  4. V_A > V_B > V_C, and V_A − V_C = 1000 V
Show answer
Answer: (B) V_A = V_B, and V_A exceeds V_C by 800 V
Explanation: Potential falls along the field direction and does not change along a direction perpendicular to it. A and B have the same x, so they lie on one equipotential plane and V_A = V_B. C is 4 m further along the field, so V_A − V_C = 200 × 4 = 800 V. Only the component of displacement along the field counts, not the full 5 m distance from A to C.
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Question 8

An isolated metal sphere of radius 10 cm carries a charge of +1 nC. A fixed point charge of −20 nC lies at a distance of 1 m from the centre of the sphere. Taking the potential at infinity as zero and 1/4πε₀ = 9 × 10⁹ N m² C⁻², what is the potential of the sphere?
  1. −180 V
  2. +270 V
  3. +90 V
  4. −90 V
Show answer
Answer: (D) −90 V
Explanation: A conductor is an equipotential, so its potential equals that at its centre. All charge on the sphere, including the induced charges that sum to zero, is at distance R from the centre, so it gives 9 × 10⁹ × 10⁻⁹ / 0.1 = +90 V. The point charge gives 9 × 10⁹ × (−20 × 10⁻⁹) / 1 = −180 V. The total is −90 V, so a positively charged conductor can be at negative potential.

Question 9

Along the x-axis the electric potential is 12 V everywhere from x = 0 to x = 2 cm. It then falls uniformly to 4 V at x = 6 cm. What is the electric force on a charge of +3 μC placed in turn at x = 1 cm and at x = 4 cm?
  1. Zero at x = 1 cm, and 6 × 10⁻⁴ N along −x at x = 4 cm
  2. Zero at x = 1 cm, and 6 × 10⁻⁴ N along +x at x = 4 cm
  3. Zero at both points, because the charge is positive and the potentials are positive
  4. 3.6 × 10⁻³ N along +x at x = 1 cm, and 6 × 10⁻⁴ N along +x at x = 4 cm
Show answer
Answer: (B) Zero at x = 1 cm, and 6 × 10⁻⁴ N along +x at x = 4 cm
Explanation: Where the potential is constant, dV/dx = 0, so the field and the force are zero. Between 2 cm and 6 cm, the field is E = −dV/dx = 8 V / 0.04 m = 200 V/m, directed towards lower potential, along +x. The force is qE = 3 × 10⁻⁶ × 200 = 6 × 10⁻⁴ N along +x.

Question 10

Two large parallel plates are held at potentials of +100 V and 0 V. A proton and an electron are placed, one at a time, at the same point between the plates, where the potential is +80 V. Which statement correctly compares the two particles at that point?
  1. The force on the proton is much smaller because of its larger mass; both have potential energy +80 eV
  2. The forces on them are equal in magnitude and opposite in direction; both have potential energy +80 eV
  3. The forces on them are equal in magnitude and opposite in direction; the proton has potential energy +80 eV and the electron −80 eV
  4. The forces on them are equal in magnitude and in the same direction; the proton has potential energy −80 eV and the electron +80 eV
Show answer
Answer: (C) The forces on them are equal in magnitude and opposite in direction; the proton has potential energy +80 eV and the electron −80 eV
Explanation: The field between the plates is uniform, and the force qE depends only on charge, not mass, so the two forces are equal in magnitude and opposite in direction. Potential energy is U = qV. With q = +e and V = +80 V the proton has +80 eV, while with q = −e the electron has −80 eV.

Question 11

With potential taken as zero at infinity, two points P and Q in an electrostatic field have potentials +120 V and +70 V. The zero of potential is now redefined so that P is at 0 V. What are the new potential of Q and the work done by the electric field when a +2 μC charge moves from P to Q?
  1. −50 V and 100 μJ
  2. +50 V and −100 μJ
  3. −50 V and zero
  4. +70 V and 140 μJ
Show answer
Answer: (A) −50 V and 100 μJ
Explanation: Moving the zero of potential shifts every potential by the same amount, here 120 V, so Q is now at 70 − 120 = −50 V. The potential difference V_P − V_Q remains 50 V, since only differences have physical meaning. The work done by the field is q(V_P − V_Q) = 2 μC × 50 V = 100 μJ, the same as before the change of reference.

Question 12

A short electric dipole of moment 2 × 10⁻⁹ C m is at the origin with its moment along the polar axis. A charge of +3 nC is carried slowly along a circular arc of radius 30 cm centred on the dipole, from the point at 120° from the dipole moment to the point at 60° from it. What work is done by the external agent? (Take 1/4πε₀ = 9 × 10⁹ N m² C⁻²)
  1. −6 × 10⁻⁷ J
  2. 6 × 10⁻⁷ J
  3. Zero, since the distance from the dipole does not change
  4. 3 × 10⁻⁷ J
Show answer
Answer: (B) 6 × 10⁻⁷ J
Explanation: The potential of a short dipole is V = kp cos θ/r², so an arc of constant r is not an equipotential. Here kp/r² = 9 × 10⁹ × 2 × 10⁻⁹/0.09 = 200 V. The potential rises from −100 V at 120° to +100 V at 60°, a change of 200 V. The work done is W = qΔV = 3 × 10⁻⁹ × 200 = 6 × 10⁻⁷ J.

Question 13

Four charges of equal magnitude q are placed at the corners A, B, C and D of a square, taken in order. Which sign arrangement gives both zero electric field and zero electric potential at the centre of the square?
  1. +q at A, +q at C, −q at B, −q at D
  2. +q at A, +q at B, +q at C, −q at D
  3. +q at all four corners
  4. +q at A, +q at B, −q at C, −q at D
Show answer
Answer: (A) +q at A, +q at C, −q at B, −q at D
Explanation: Potential is a scalar sum, so it vanishes only if two charges are positive and two are negative. The field vanishes only if the charges on each diagonal are of the same sign, so that their fields cancel. Equal positive charges on one diagonal and equal negative charges on the other satisfy both conditions. With all charges positive the field is zero but the potential is not, and with adjacent like charges the potential is zero but the field is not.

Question 14

A charge of +2 µC is fixed at point O. A test charge of +1 µC is carried from P to Q along a circular arc of radius 10 cm centred at O, and then radially outward from Q to R, where OR = 30 cm. What is the work done by the electric field on the test charge over the whole path? (Take 1/4πε0 = 9 × 10^9 N m^2 C^−2)
  1. Zero
  2. +0.12 J
  3. −0.12 J
  4. +0.18 J
Show answer
Answer: (B) +0.12 J
Explanation: Along the arc the test charge stays on an equipotential surface, so no work is done there. On the radial part the work done by the field is k q q0 (1/r1 − 1/r2). This equals 0.018 × (1/0.1 − 1/0.3) = 0.018 × 6.67 = 0.12 J. The sign is positive because the repulsive force acts along the outward displacement.

Question 15

A large plane sheet has a uniform surface charge density +8.85 nC m⁻². A point charge of +2 μC is moved slowly along the normal to the sheet, from a distance of 5 cm to a distance of 1 cm. The work done by the external agent is (ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²):
  1. +2 × 10⁻⁵ J
  2. +8 × 10⁻⁵ J
  3. −4 × 10⁻⁵ J
  4. +4 × 10⁻⁵ J
Show answer
Answer: (D) +4 × 10⁻⁵ J
Explanation: The field of the sheet is uniform with E = σ/2ε₀ = 500 V m⁻¹, directed away from the sheet. The potential difference between the two points is Ed = 500 × 0.04 = 20 V, the nearer point being at the higher potential. The agent pushes the positive charge against the field, so the work is positive and equals qV = 2 × 10⁻⁶ × 20 = 4 × 10⁻⁵ J.

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