Question

A parallel-plate capacitor is connected to a battery. What happens to the stored energy if the plate separation is doubled while the capacitor remains connected to the battery?
(a) Remains the same
(b) It is doubled
(c) It decreases by a factor of 2
(d) It decreases by a factor of 4

Answer 1

The capacitance of the parallel plate capacitor is

$C=\varepsilon_r \varepsilon_0 \frac {S}{d},$

where $\varepsilon_r$ - the relative permittivity of the material between plates;

$\varepsilon_0$ - absolute dielectric permittivity of vacuum;

S - the area of the plate;

d - the plate separation.

If the plate separation is doubled, then $d_2=2d_1$, so $C_2=\frac{C_1}{2}.$

The stored energy is

$W=\frac{CU^2}{2},$

where U - the voltage on the capacitor.

While the capacitor remains connected to the battery, so U = const.

Then $W_2=\frac{W_1}{2}.$

Answer: (c) It decreases by a factor of 2.