Questions: Electric Potential and Potential Energy

Question 1

A positive test charge is placed between two parallel plates: the left plate is at +200V and the right plate is at 0V. In which direction does the electric force on the charge point?

• A. Toward the left plate (toward higher potential), because positive charges are attracted to positive regions
• B. Toward the right plate (toward lower potential), because E = −∇V means the field points from high to low potential
• C. The charge feels no net force, because it is midway between the plates
• D. Perpendicular to both plates, along the equipotential surfaces

Answer: (B)

The relation E = −∇V means the electric field points in the direction of steepest *decrease* in potential — like water flowing downhill. With the left plate at +200V and the right at 0V, potential decreases from left to right, so E points rightward. The force on a positive test charge is F = qE, so it is pushed toward the right (lower potential) plate. The common misconception is that positive charges are attracted to regions of higher (positive) potential, but it is the gradient of V — the rate of change — that determines force direction, and the force is in the direction of decreasing V.

Question 2

A positive charge q is moved from point A (V = 50 V) to point B (V = 20 V). What happens to its electric potential energy?

• A. It increases, because the charge moved to a region of lower potential
• B. It decreases, because ΔU = q(V_B − V_A) = q(20 − 50) < 0 for positive q
• C. It stays the same, because the electric force is conservative
• D. It changes by 30 J regardless of the value of q

Answer: (B)

Potential energy is U = qV, so the change is ΔU = q(V_B − V_A) = q(20 − 50) = −30q. For a positive charge, this is a decrease in potential energy. The work done by the electric field equals −ΔU, so the field does positive work on the charge as it moves from high to low potential — analogous to gravity doing positive work as a mass falls from high to low altitude. Option C confuses conservation of total energy with constancy of potential energy; option D ignores that ΔU depends on both the potential difference and the charge.

Question 3

Electric potential V and electric potential energy U are different quantities — V is a property of the point in space, while U depends on both the potential at that point and the charge placed there.

• T. True
• F. False

Answer: True

This distinction is essential. V = W/q is a property of the location — it describes the energy landscape per unit charge, independent of any charge placed there. U = qV depends on both the potential and the specific charge q. A proton and an electron at the same location have potential energies of opposite sign (same V, but opposite q). The field, equipotential surfaces, and all geometric properties belong to V; U only appears when you introduce a specific charge.

Question 4

The electric field usually points from regions of low potential to regions of high potential, since positive charges are attracted toward higher potentials.

• T. True
• F. False

Answer: False

The relationship is E = −∇V, where the minus sign is critical. The field points in the direction of *decreasing* potential — from high to low V. Think of V as altitude: the electric field is like the gravitational field, which points downhill (toward decreasing altitude), not uphill. A positive charge released in an electric field accelerates toward lower potential (losing potential energy, gaining kinetic energy) — the field points in that same direction of lower potential.

Question 5

Why is working with electric potential V often simpler than working directly with the electric field E when solving problems involving multiple point charges?

Model answer: Electric potential V is a scalar — it has magnitude but no direction. To find the total potential at a point due to multiple charges, you add the scalar contributions: V_total = kQ₁/r₁ + kQ₂/r₂ + ... Each term is a number, and numbers add simply. The electric field E is a vector — each charge's contribution has both magnitude and direction, requiring separate x, y, and z components that must be combined geometrically. Once V is found by scalar addition, the field can be recovered by differentiation (E = −∇V), which is often less work than direct vector addition throughout.

The practical power of the potential formalism is computational: scalar addition is far simpler than vector addition. Students who understand this recognize why physicists reach for potentials first in electrostatics problems. It is not that potential is more fundamental than the field — it is that calculating one scalar function and then differentiating it once is usually easier than tracking three field components at every point.