Answer
a. It doubles
b. It doubles
Work Step by Step
The capacitor is charged and not connected to anything, so the charge on the capacitor doesn’t change.
a. The stored energy is found using equation 17–10, $PE=\frac{1}{2}QV$.
Q does not change, and the potential difference V is doubled, so the stored energy doubles.
b. The stored energy is found using equation 17–10, $PE=\frac{Q^2}{2C} $.
As the plate separation is doubled, equations 17-8 and 17-9 shows that the capacitance will be halved. Q does not change, so the stored energy doubles.