If a dielectric is inserted into a capacitor with a fixed voltage, the stored energy changes by what factor?

If a dielectric is inserted into a capacitor with a fixed voltage, the stored energy changes by what factor?

• A. It decreases by κ
• B. It remains the same
• C. It increases by κ^2
• D. It increases by κ

Answer: (D)

The key idea is that the energy stored in a capacitor at a fixed voltage depends on its capacitance via U = 1/2 C V^2. When a dielectric with dielectric constant κ is fully inserted, the capacitance increases by that factor: C' = κC. With the voltage kept constant, the new energy is U' = 1/2 C' V^2 = 1/2 (κC) V^2 = κ (1/2 C V^2) = κU. So the stored energy increases by κ. It’s helpful to contrast with a fixed-charge scenario: if the charge were fixed, energy would be U = Q^2/(2C), so increasing C would decrease energy by κ. Here, because voltage is fixed, the energy scales linearly with the capacitance, giving the factor κ.

The key idea is that the energy stored in a capacitor at a fixed voltage depends on its capacitance via U = 1/2 C V^2. When a dielectric with dielectric constant κ is fully inserted, the capacitance increases by that factor: C' = κC. With the voltage kept constant, the new energy is U' = 1/2 C' V^2 = 1/2 (κC) V^2 = κ (1/2 C V^2) = κU. So the stored energy increases by κ.

It’s helpful to contrast with a fixed-charge scenario: if the charge were fixed, energy would be U = Q^2/(2C), so increasing C would decrease energy by κ. Here, because voltage is fixed, the energy scales linearly with the capacitance, giving the factor κ.