Definition
Equipotential surface
Surface on which all points have the same electric potential. Field lines are perpendicular to equipotential surfaces. No work done in moving a charge along an equipotential.
-- NCERT Class 12 Physics, Ch. 2, p. 54Equipotential Surfaces
8 MCQs9-step worked example
Source: NCERT ElectrostaticsPYQ coverage: NEET 2022Official key: NTA-verifiedLast updated: 21 Sep 2026
Equipotential Surfaces, explained for NEET
The trap: reading "the potential is the same everywhere on this surface" as "there is no field here." Constant V along a surface only means the field has no component along it. The field is still there, pointing straight through the surface, and it can be strong.
Definition. An equipotential surface is a surface with a constant value of potential at all points on it (NCERT Class 12 Physics Part I, Chapter 2, Section 2.6, page 54).
Why the field is normal to it (page 54). If the field had a component along the surface, moving a test charge against that component would need work. But between any two points on an equipotential surface there is no potential difference, so no work is needed to move a test charge on it. So, for any charge configuration, the field is normal to the equipotential surface at every point.
Shapes to know (page 54).
- Single point charge: concentric spherical surfaces centred at the charge; the field lines are radial.
- Uniform field along the x-axis: planes normal to the x-axis, i.e. parallel to the y–z plane.
- NCERT also draws the equipotentials of a dipole and of two identical positive charges (Fig. 2.11).
Field from potential (Section 2.6.1, page 55). For two close equipotential surfaces a perpendicular distance δl apart:
|E| = |δV| / δl (NCERT Chapter 2, Eq. 2.21, page 55)
NCERT's two conclusions: (i) the field points in the direction in which the potential decreases steepest; (ii) its magnitude is the change in potential per unit displacement normal to the equipotential surface.
Bridge to NEET. Questions ask for the angle between E and an equipotential (90°), the work done moving a charge on one (zero), or E from two labelled equipotentials. The number slips are multiplying ΔV by the spacing instead of dividing, using one potential instead of the difference, and measuring the distance along a slant instead of normal to the surfaces.
Watch out: for equal steps of potential, closer surfaces mean a stronger field (smaller δl).
Can you answer these Equipotential Surfaces MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
An equipotential surface is best described as a surface on which:
Show answer and why every option is right or wrong
Answer: D. NCERT Class 12 Physics Part I, Chapter 2, Section 2.6 (page 54) defines an equipotential surface as a surface with a constant value of potential at all points on it.
Why A is wrong: A is wrong because the definition fixes the potential, not the field strength; the field magnitude can differ from point to point on the same equipotential surface.
Why B is wrong: B is wrong because it reverses the geometry: the field is normal to an equipotential surface, not along it.
Why C is wrong: C is wrong because equipotential surfaces exist around charges; NCERT's first example is the set of spheres around a point charge.
MCQ 2Easy RecallPractice
The equipotential surfaces of a single isolated point charge are:
Show answer and why every option is right or wrong
Answer: B. For a point charge V depends only on r, so V is constant wherever r is constant; the equipotential surfaces are concentric spheres centred at the charge (NCERT Chapter 2, Section 2.6, page 54).
Why A is wrong: A is wrong because planes are the equipotentials of a uniform field, not of a point charge, whose potential depends only on the distance r.
Why C is wrong: C is wrong because radial lines are the electric field lines of a point charge; the equipotentials cross them at right angles.
Why D is wrong: D is wrong because a cylinder does not keep the distance from a single point constant, so the potential would vary over it.
MCQ 3Easy RecallPractice
At any point in an electrostatic field, the angle between the electric field and the equipotential surface passing through that point is:
Show answer and why every option is right or wrong
Answer: D. NCERT Chapter 2, Section 2.6 (page 54): for any charge configuration, the equipotential surface through a point is normal to the electric field at that point.
Why A is wrong: A is wrong because a field along the surface would have a component along it, and moving a charge against that component would need work, which cannot happen on an equipotential surface.
Why B is wrong: B is wrong because any angle other than 90° leaves a component of the field along the surface, which contradicts zero work on the surface.
Why C is wrong: C is wrong because 180° is still a direction along the surface (just reversed), so it has the same contradiction as 0°.
MCQ 4Direct ApplicationPractice
A charge of 3.0 µC is moved between two points 0.40 m apart that both lie on the same equipotential surface of potential 12 V. The work done by the electric field is:
Show answer and why every option is right or wrong
Answer: A. Both points are at 12 V, so the potential difference between them is zero and no work is required to move a test charge on an equipotential surface (NCERT Chapter 2, Section 2.6, page 54). W = q × 0 = 0.
Why B is wrong: B is wrong because 1.4 × 10⁻⁵ J = (3.0 × 10⁻⁶)(12)(0.40) multiplies the surface potential by the distance moved; the distance along an equipotential does not create a potential difference.
Why C is wrong: C is wrong because 9.0 × 10⁻⁵ J = (3.0 × 10⁻⁶)(12)/0.40 divides by the distance moved; the result still rests on the surface potential, and the potential difference is zero.
Why D is wrong: D is wrong because 3.6 × 10⁻⁵ J = (3.0 × 10⁻⁶ C)(12 V) uses the potential of the surface instead of the potential difference between the two points, which is zero.
MCQ 5Direct ApplicationPractice
In a region of uniform electric field, two equipotential planes at 18 V and 6.0 V are 0.040 m apart, measured perpendicular to the planes. The magnitude of the electric field is:
Show answer and why every option is right or wrong
Answer: C. |E| = |δV|/δl = (18 − 6.0) V / 0.040 m = 12/0.040 = 3.0 × 10² V/m (NCERT Chapter 2, Eq. 2.21, page 55).
Why A is wrong: A is wrong because 0.48 comes from 12 × 0.040, multiplying the potential difference by the spacing instead of dividing by it.
Why B is wrong: B is wrong because 4.5 × 10² comes from 18/0.040, using the potential of one plane instead of the difference between the two planes.
Why D is wrong: D is wrong because 3.3 × 10⁻³ comes from 0.040/12, inverting the ratio; the field is potential change per unit distance, not distance per unit potential.
MCQ 6Direct ApplicationPractice
In a uniform electric field, the equipotential surfaces are planes parallel to the y–z plane. The plane at x = 1.0 m is at 9.0 V and the plane at x = 3.0 m is at 3.0 V. The electric field is:
Show answer and why every option is right or wrong
Answer: C. |E| = (9.0 − 3.0) V / (3.0 − 1.0) m = 6.0/2.0 = 3.0 V/m. The field points the way the potential decreases steepest, which is from 9.0 V towards 3.0 V, i.e. along +x (NCERT Chapter 2, Section 2.6.1, page 55).
Why A is wrong: A is wrong because −x points from the 3.0 V plane towards the 9.0 V plane, the direction in which potential increases; the field points towards lower potential.
Why B is wrong: B is wrong because +y lies inside the equipotential planes; the field must be normal to them, i.e. along the x-axis.
Why D is wrong: D is wrong because 6.0 is the potential difference itself; it has not been divided by the 2.0 m spacing between the planes.
MCQ 7Concept TrapPractice
A small positive test charge sits at a point P in an electrostatic field. Compared with the equipotential surface through P, the electric force on the charge at P is directed:
Show answer and why every option is right or wrong
Answer: A. For a positive charge the force is along E. E is normal to the equipotential surface (page 54) and points in the direction in which the potential decreases steepest (page 55), so the force is normal to the surface, towards lower potential.
Why B is wrong: B is wrong because the field points towards decreasing potential, and a positive charge is pushed along the field, not against it.
Why C is wrong: C is wrong because it is the trap this lesson names: constant V along the surface only removes the field component along the surface; the normal component can be non-zero.
Why D is wrong: D is wrong because a force along the surface would mean a field component along it, which NCERT rules out: the field is normal to the equipotential surface.
MCQ 8CalculationPractice
A map of equipotential surfaces is drawn at equal steps of 2.0 V. In region 1 neighbouring surfaces are 0.010 m apart; in region 2 they are 0.040 m apart (both spacings measured normal to the surfaces, field roughly uniform within each region). The ratio of field magnitudes E₁ : E₂ is:
Show answer and why every option is right or wrong
Answer: B. E₁ = 2.0/0.010 = 2.0 × 10² V/m and E₂ = 2.0/0.040 = 5.0 × 10¹ V/m, so E₁ : E₂ = 4 : 1. Closer equipotentials mean a stronger field (NCERT Chapter 2, Eq. 2.21, page 55).
Why A is wrong: A is wrong because 1 : 4 treats the field as proportional to the spacing; |E| = |δV|/δl puts the spacing in the denominator, so the ratio inverts.
Why C is wrong: C is wrong because it assumes an equal potential step gives an equal field; the field depends on the step divided by the spacing, and the spacings differ.
Why D is wrong: D is wrong because 16 : 1 squares the spacing ratio (0.040/0.010 = 4, then 4² = 16); the field depends on 1/δl, not 1/δl².
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How do you solve a Equipotential Surfaces question? A worked example
- 1
Given
A region has a uniform electric field. Its equipotential surfaces are planes parallel to the y–z plane. The plane at x = 0.10 m is at 48 V and the plane at x = 0.40 m is at 12 V. A charge of 2.5 µC is then moved 0.15 m along the 48 V plane.
- 2
Required
(a) The magnitude and direction of the electric field. (b) The potential of the equipotential plane at x = 0.20 m. (c) The work done by the field on the 2.5 µC charge during its move along the 48 V plane.
- 3
Concept
The field is normal to the equipotential planes (so along the x-axis), points in the direction in which potential decreases, and has magnitude |δV|/δl with δl measured normal to the planes. Moving along one equipotential plane involves no potential difference, so no work.
- 4
Formula
|E| = |δV| / δl (NCERT Chapter 2, Eq. 2.21, page 55); for a uniform field, the potential drop over a normal distance d is E·d; W = q × (potential difference) = 0 on one equipotential surface.
- 5
Substitution
(a) |E| = (48 V − 12 V) / (0.40 m − 0.10 m)
(b) V(0.20 m) = 48 V − |E| × (0.20 m − 0.10 m)
(c) W = (2.5 × 10⁻⁶ C) × (48 V − 48 V) - 6
Calculation
(a) |E| = 36 / 0.30 = 1.2 × 10² V/m. The potential falls as x increases, so E points along +x.
(b) V = 48 − (1.2 × 10²)(0.10) = 48 − 12 = 36 V
(c) W = 2.5 × 10⁻⁶ × 0 = 0 J
No exact constants (counting numbers or mathematical constants) are used here; every input is a measured value with two significant figures, so the answers are given to two significant figures. The zero in (c) is exact because the potential difference along one equipotential is exactly zero. - 7
Final answer
(a) E = 1.2 × 10² V/m along +x; (b) 36 V; (c) zero work.
- 8
Common trap
Two slips give wrong answers here. First, computing 36 × 0.30 = 10.8 (multiplying the potential difference by the spacing instead of dividing). Second, using the 0.15 m move along the 48 V plane to get a non-zero work, for example 2.5 × 10⁻⁶ × 48 = 1.2 × 10⁻⁴ J. That uses the potential of the plane, but only a potential difference does work, and along one equipotential it is zero.
- 9
Similar NEET-style question
"Two equipotential planes in a uniform field differ in potential by 4.0 V and are 0.020 m apart, measured normal to the planes. Find the magnitude of the field, and state the angle between the field and either plane."
Strategy: |E| = 4.0/0.020 = 2.0 × 10² V/m, and the angle is 90°. Do not multiply (0.080) or invert (5.0 × 10⁻³).
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What to remember before solving Equipotential Surfaces questions
More in Electrostatics: 3 exam traps and mistakes · 11 formulas · 6 question patterns from its other lessons.
Equipotential Surfaces questions from past NEET papers
1 question from NEET 2022. Answers verified against NTA official keys. — click to collapse
NEET 2022
The angle between the electric lines of force and the equipotential surface is
1180°
20°
345°
490°
NTA Answer: Option 4(final)
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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