A capacitor is charged to a voltage V. How does the stored energy change if the voltage is doubled, with the capacitance constant?

A capacitor is charged to a voltage V. How does the stored energy change if the voltage is doubled, with the capacitance constant?

• A. It doubles
• B. It remains constant
• C. It quadruples
• D. It halves

Answer: (C)

The key idea is that the energy stored in a capacitor with constant capacitance is proportional to the square of the voltage: E = 1/2 C V^2. If you double the voltage, V becomes 2V, and the energy becomes E' = 1/2 C (2V)^2 = 4 × (1/2 C V^2) = 4E. So the stored energy quadruples. This aligns with the fact that the electric field strength—and thus the energy density—grows with the square of the voltage. If the capacitance stayed the same, doubling voltage cannot leave the energy unchanged or reduce it; it must increase by a factor of four.

The key idea is that the energy stored in a capacitor with constant capacitance is proportional to the square of the voltage: E = 1/2 C V^2. If you double the voltage, V becomes 2V, and the energy becomes E' = 1/2 C (2V)^2 = 4 × (1/2 C V^2) = 4E. So the stored energy quadruples. This aligns with the fact that the electric field strength—and thus the energy density—grows with the square of the voltage. If the capacitance stayed the same, doubling voltage cannot leave the energy unchanged or reduce it; it must increase by a factor of four.