Question

If you increase the charge on a parallel-plate capacitor from 3 mu or micro CC to 9 mu or micro CC and increase the plate separation from 1.8 mm to 5.4 mm, the energy stored in the capacitor changes by a factor of:__________

Answer 1

Explanation:

The energy stored in a capacitor is given by

$U = \dfrac{1}{2}QV = \dfrac{Q^2}{2C}$

In the case of a parallel plate capacitor, the capacitance C is given by

$C = \dfrac{\epsilon_0A}{d}$

so we can rewrite the expression for the energy as

$U = \dfrac{Q^2d}{2\epsilon_0 A}$

Increasing the charge from $3\:\mu\text{C}\:\text{to}\:9\:\mu\text{C}$ means that you're tripling the charge. The same thing is true when you increase the distance from 1.8 mm to 5.4 mm, i.e., you triple the separation distance. So the new energy U' is given by

$U' = \dfrac{Q'^2d'}{2\epsilon_0 A}$

$\:\:\:\:\:\:= \dfrac{(3Q)^2(3d)}{2\epsilon_0 A}$

$\:\:\:\:\:\:= 27\left(\dfrac{Q^2d}{2\epsilon_0 A}\right)$

$\:\:\:\:\:\:= 27U$

As we can see, tripling both the charge and the separation distance result in the 27-fold increase in its stored energy U.