Questions: Energy Storage in Capacitor Fields

Question 1

A capacitor with capacitance C is charged to voltage V. The voltage is then doubled to 2V. What happens to the stored energy?

• A. It doubles, since energy is proportional to voltage
• B. It quadruples, since U = ½CV² and voltage is squared
• C. It stays the same — only capacitance affects stored energy
• D. It halves — higher voltage means the same charge is stored more efficiently

Answer: (B)

U = ½CV², so doubling V gives U_new = ½C(2V)² = 4 × ½CV² — the energy quadruples. The quadratic dependence on voltage is why high-voltage capacitors store dramatically more energy than low-voltage ones of the same capacitance. This is not intuitive if you think of energy as simply proportional to voltage, which is the common error.

Question 2

Where does the energy in a charged parallel-plate capacitor actually reside?

• A. In the charges accumulated on the metal plates
• B. In the connecting wires, as electrical current flowing between plates
• C. In the electric field distributed throughout the volume between the plates
• D. Equally split between the surface charges and the space between the plates

Answer: (C)

The energy density u = ½ε₀E² is distributed throughout the volume where the electric field exists — the space between the plates. The plates carry charges that maintain the field, but the energy resides in the field itself. This field-centric view becomes essential for understanding electromagnetic waves, where energy propagates through empty space with no charges involved at all.

Question 3

The energy stored in a capacitor is ½QV rather than QV because the voltage builds gradually from zero as charge is added.

• T. True
• F. False

Answer: True

The first charge element is moved against essentially no field — little work required. Each subsequent element must fight the growing field established by earlier charge. The average voltage during the charging process is V/2 (not V), so total work = Q × (V/2) = ½QV. The factor of ½ is the signature of any energy-storage process where the 'back-pressure' increases linearly as the store fills — springs follow the same logic: F = kx increases linearly, giving stored energy ½kx².

Question 4

The energy density formula u = ½ε₀E² applies mainly to parallel-plate capacitors where the electric field is uniform.

• T. True
• F. False

Answer: False

The formula u = ½ε₀E² is a universal result for the energy density of any electric field in free space — not limited to uniform fields or parallel-plate geometry. For a nonuniform field (like that of a point charge), you integrate u over all space to find total stored energy. The parallel-plate case is simply the special situation where the field is uniform, making the integral trivial (u times volume). The generality of the formula is what makes it fundamental to electromagnetism.

Question 5

Explain why the energy stored in a capacitor is ½QV rather than QV, and what the factor of one-half tells you about energy-storage processes more generally.

Model answer: When charging begins, there is no field and the first charge moves easily. As charge accumulates, the electric field grows and each subsequent charge element must be pushed against a stronger opposing field, requiring more work. The voltage ramps linearly from 0 to V during charging, so the average voltage is V/2. Total work = charge × average voltage = Q × (V/2) = ½QV. The factor of ½ appears in any storage process where the 'resistance' grows linearly with the stored quantity — a spring stores ½kx² for the same reason.

This principle generalizes broadly. The quadratic energy-storage relationship (U ∝ V² for a capacitor, U ∝ x² for a spring, U ∝ I² for an inductor) always arises when the quantity being stored builds up the very force opposing further storage. In all these cases, the average 'back-pressure' during filling is half the final value, giving the factor of ½. Recognizing this pattern unifies capacitor, spring, and inductor energy storage under a single physical principle.