Questions: Energy Storage and Forces in Capacitors

Question 1

An isolated capacitor (disconnected from any battery) holds fixed charge Q. The plates are slowly pulled farther apart. What happens to the stored energy?

• A. Energy decreases, because the plates move in the direction of the attractive force between them
• B. Energy stays the same, because charge Q is conserved
• C. Energy increases, because C decreases and U = Q²/2C grows as C shrinks
• D. Energy decreases, because the electric field weakens as the plate separation increases

Answer: (C)

With fixed charge Q, U = Q²/2C. As plates separate, capacitance C = ε₀A/d decreases (C ∝ 1/d), so U = Q²/2C increases. This tells you that an external force must do positive work to separate the plates — the system's energy increases. Meanwhile F = −∂U/∂d is negative (force opposes increasing d), confirming the plates attract each other. Option A confuses the direction of the force with the direction of energy change when an external agent does work against that force.

Question 2

A capacitor is connected to a battery (fixed voltage V). A dielectric slab is slowly inserted between the plates, increasing the capacitance C and therefore the stored energy U = ½CV². Why is the dielectric nonetheless pulled inward?

• A. Because the electric field exerts a direct attractive force on the bulk dielectric material
• B. Because the battery supplies extra charge to maintain V; the work done by the battery exceeds the energy stored, and the surplus goes to mechanical work pulling the dielectric inward
• C. Because the dielectric reduces the electric field between the plates, lowering energy
• D. Because the dielectric is attracted by the magnetic field generated during capacitor charging

Answer: (B)

At fixed V, inserting the dielectric increases C, which increases U = ½CV². This seems to argue against insertion — but the battery must supply extra charge (ΔQ = V·ΔC) to maintain V. The energy supplied by the battery is V·ΔQ = V²·ΔC, which is twice the energy increase ΔU = ½ΔC·V². The excess V²·ΔC/2 provides the mechanical work pulling the dielectric inward. The dielectric is attracted not by a direct field force on bulk material but by the system's tendency to lower its free energy — the energy available as mechanical work.

Question 3

The energy stored in a charged capacitor resides in the electric field between the plates, not on the surface charges themselves.

• T. True
• F. False

Answer: True

The energy density in the electric field is u = (ε₀/2)κᵣE². Integrating this over the volume between the plates recovers exactly U = ½CV². This is more than bookkeeping: in electrodynamics, fields can carry energy through vacuum (as in electromagnetic waves) independently of any charges. The field-energy picture is the correct fundamental view, with charge distributions as the sources that create the fields.

Question 4

The three energy expressions U = ½QV, U = ½CV², and U = ½Q²/C usually give the same numerical result, so it does not matter which one you use when analyzing how energy changes as capacitor plates are separated.

• T. True
• F. False

Answer: False

All three expressions are equivalent at any single moment (they all equal U), but they depend on different variables — Q, C, and V — and different quantities are held constant in different situations. When plates separate at fixed Q (isolated capacitor), use U = Q²/2C: Q is constant, C changes. Using U = ½CV² with varying V requires tracking how V changes simultaneously, which adds complexity without benefit. Choosing the form whose fixed variable matches your constraint makes the physics transparent and the algebra clean.

Question 5

Explain why the energy stored in a capacitor is U = ½QV rather than U = QV, even though the charge Q was moved across potential difference V.

Model answer: The factor of ½ arises because the potential difference builds up progressively as charge is transferred. The first small element of charge dq is moved across a near-zero potential (when the capacitor is nearly uncharged). As charge accumulates, the voltage rises, and later increments of charge are moved against a higher potential. The average potential during charging is V/2, not V. Integrating dU = v·dq from 0 to Q — where v = q/C at each stage — gives U = ∫₀Q (q/C)dq = Q²/2C = ½QV. If you moved all charge Q across the full final potential V, you'd compute QV — but that's not what happens during charging.

The ½ factor is the hallmark of energy stored in a quadratic process (like a spring: U = ½kx²). In both cases, the restoring force grows as the system charges up, so the average force (and thus work per unit displacement) is half the final value.