# A parellel plate capacitor with plate speration d is charged with a battery so that energy stored is "U". If a plate of dielectric "K" and thickness d is placed betweeen the plates of capacitor and battery remains connected. The work by battery is?

## Options: $\text{KU}$ $\text{2U(K - 1)}$ $\text{U(K - 1)}$ $\text{2KU}$

Mar 14, 2018

$2 U \left(K - 1\right)$

#### Explanation:

The energy stored in a capacitor is given by $U = \frac{1}{2} C {V}^{2}$ where $C$ is the capacitance and $V$ is the potential.

When the dielectric is introduced, the capacitance increases by a factor of $K$, but since the battery is still connected, the potential $V$ does not change.

So, the charge stored in the capacitor grows from $C V$ to $K \times C V$. The battery must pump in this extra charge $\left(K - 1\right) C V$ at the constant voltage $V$ - and the work it must do for this is
$\left(K - 1\right) C V \times V = \left(K - 1\right) C {V}^{2} = 2 \left(K - 1\right) U$

Note that this is twice the energy gained by the capacitor.