Questions: Energy Density in Electric Fields

5 questions to test your understanding

Score: 0 / 5
Question 1 Multiple Choice

A parallel-plate capacitor is charged to voltage V and then disconnected from the battery. The plates are pulled apart, doubling the separation d. What happens to the stored energy?

AIt stays the same because the charge Q on the plates is conserved
BIt doubles, because U = Q²/(2C) and C = ε₀A/d, so halving C doubles U
CIt halves, because the electric field E = Q/(ε₀A) weakens as the plates move further apart
DIt stays the same because U = ½CV² and V doesn't change after disconnection
Question 2 Multiple Choice

Where is the energy of a charged capacitor physically stored?

AIn the chemical potential of the battery that originally charged it
BIn the electric field occupying the space between the plates, as energy per unit volume u = ½ε₀E²
CIn the surface charge distribution on the capacitor plates
DIn the connecting wires as kinetic energy of electrons
Question 3 True / False

Any region of space that contains an electric field contains real, physically stored energy proportional to the square of the field strength.

TTrue
FFalse
Question 4 True / False

The three formulas U = ½CV², U = ½QV, and U = Q²/(2C) give different values for the same capacitor and the user should choose the most accurate one.

TTrue
FFalse
Question 5 Short Answer

Why is it significant that energy is stored in the electric field rather than in the charge configuration? What does this perspective enable?

Think about your answer, then reveal below.