Questions: Equipotential Surfaces and Their Properties

Question 1

A positive charge of 2 μC is moved from point A to point B along a curved path that stays entirely on an equipotential surface. How much work does the electric field do on the charge?

• A. A positive amount of work, since the charge is positive and moving through an electric field
• B. A negative amount of work, since the charge is moving against the field lines
• C. Zero, because ΔV = 0 along an equipotential
• D. It depends on the length and shape of the path taken

Answer: (C)

Work done by the electric field is W = qΔV. Since the path stays on an equipotential, ΔV = 0 by definition, so W = 0 regardless of the charge's sign, the path length, or the direction of travel. This is the defining property of an equipotential surface — it is the electric analogue of moving horizontally on a hillside where no gravitational potential energy changes. The charge's sign and the path shape are irrelevant when ΔV = 0.

Question 2

Near a sharp conductor tip (like a lightning rod), the equipotential surfaces are packed densely together. What does this tell you about the electric field in that region?

• A. The field is weak there because the conductor redistributes charge to minimize energy
• B. The field is strong there because densely packed equipotentials correspond to a large potential gradient
• C. The field is zero just outside the tip because the conductor surface is an equipotential
• D. The closely packed surfaces indicate the potential changes slowly near the tip

Answer: (B)

The electric field magnitude is the rate of change of potential with distance: E = |∇V|. Closely packed equipotential surfaces mean the potential changes by the same amount over a shorter distance — a steep potential gradient, which means a large field. Near a sharp conductor tip, charge concentrates, equipotentials crowd together, and the field is very strong — strong enough to ionize air and produce sparking. This is why lightning rods work: they concentrate the field at their tip, providing a preferred discharge path.

Question 3

A conductor in electrostatic equilibrium is an equipotential throughout its entire volume, not just on its surface.

• T. True
• F. False

Answer: True

In electrostatic equilibrium, the electric field inside a conductor is zero (otherwise free charges would accelerate, contradicting equilibrium). Since E⃗ = −∇V and E⃗ = 0 everywhere inside, the potential gradient is zero throughout — V is constant everywhere in the conductor's volume, not just on its surface. The surface is an equipotential, and so is every interior point. This is why a Faraday cage screens its interior: the entire conducting shell sits at one potential, with zero field inside.

Question 4

If a field line makes a 45° angle with an equipotential surface at some point, it means the electric field has a component both perpendicular to and along the equipotential at that location.

• T. True
• F. False

Answer: False

Electric field lines are always perpendicular (90°) to equipotential surfaces — never at 45° or any other angle. This follows from E⃗ = −∇V: the gradient of a scalar function always points perpendicular to its level surfaces. If the field had any component parallel to an equipotential, that component would do work on a charge moving along the surface, implying ΔV ≠ 0 along the surface — contradicting it being equipotential. A 45° angle is physically impossible in electrostatics.

Question 5

Why must electric field lines be perpendicular to equipotential surfaces? Explain the reasoning from first principles.

Model answer: The electric field is defined as E⃗ = −∇V, the negative gradient of potential. The gradient of any scalar function points in the direction of steepest change and is always perpendicular to the level surfaces (surfaces where the function is constant). Since equipotential surfaces are exactly the level surfaces of V, E⃗ must be perpendicular to them. Equivalently: if E⃗ had a component parallel to an equipotential, moving a charge along that surface would require work, implying ΔV ≠ 0 — contradicting the surface being equipotential.

This is a mathematical consequence of the gradient being perpendicular to level sets, applied to the specific identification of E⃗ as −∇V. The physical intuition parallels gravity: water flows straight downhill (perpendicular to contour lines), never along a contour. The electric field similarly points perpendicular to surfaces of constant potential, in the direction of steepest potential drop. The perpendicularity is not an empirical regularity — it follows necessarily from the definition of field as the gradient of potential.