Final answer:
If the charge on a parallel-plate capacitor is doubled, the new energy stored in the capacitor is four times the initial energy, as the energy stored in a capacitor is proportional to the square of the charge.
Step-by-step explanation:
If the energy stored in a parallel-plate capacitor is E, and the charge on the capacitor is doubled, the new energy stored in the capacitor can be determined by considering the relationship between the charge and the stored energy. The energy (U) stored in a capacitor is given by the formula U = (1/2)QV, where Q is the charge and V is the voltage across the plates. Since a parallel plate capacitor also has the relationship V = Q/C, where C is the capacitance, we can express energy as U = (1/2)(Q^2/C).
By doubling the charge (Q to 2Q), the energy becomes U' = (1/2)(4Q^2/C) = 4 * (1/2)(Q^2/C) = 4U. This means that the new stored energy will be four times greater than the initial energy (E). Therefore, if the charge is doubled, the new energy stored in the capacitor is 4E.