A capacitor has stored energy U. If the voltage and plate separation are both doubled, the new stored energy is ____.

Answer 2U
a) 2U b) U/2 c) U d)4U

When the voltage and plate separation of a capacitor change, the energy stored in the capacitor can be calculated using the formula:

U = (1/2) * C * V^2

Where:
U is the stored energy
C is the capacitance of the capacitor
V is the voltage

In this case, we are given that the voltage and plate separation are both doubled. Let's assume the original values of voltage and plate separation are V1 and d1, respectively. The new values are then 2V1 and 2d1.

To find the new stored energy, let's calculate the capacitance of the original capacitor, C1:

U = (1/2) * C1 * V1^2

Now, let's find the capacitance of the new capacitor, C2:

C2 = C1 * (A2 / A1)

Where:
A1 is the original area of the plate
A2 is the new area of the plate

Since the plate separation is doubled, the new area of the plate will be 4 times the original area:

A2 = 4 * A1

Substituting this back into the equation for C2, we get:

C2 = C1 * (4 * A1 / A1)
C2 = 4 * C1

Now, let's find the new stored energy, U2:

U2 = (1/2) * C2 * (2V1)^2
U2 = (1/2) * 4 * C1 * (2V1)^2
U2 = 4 * (1/2) * C1 * (2V1)^2
U2 = 2 * U

Therefore, the new stored energy is 2U. Answer choice a) 2U is correct.