Question

A charged parallel plate capacitor has an energy $U$ in the system. A slab of dielectric constant $K$ is inserted, which completely fills the space between the plates. The new energy of the system will be

• A. $KU$
• B. $K^{2}U$
• C. $U/K$
• D. $U/K^2$

Answer 1

The correct option is C $U/K$

Let the capacitance of the capacitor be $C$ and initial charge on it be $Q$.

So, initial energy stored in the capacitor,

$U=\frac{Q^2}{2C}$

Now, upon insertion of the dielectric, the charge on the capacitor will not change (since it is isolated and not connected to a battery).
However, the capacitance will change and become equal to $KC(=C^{\prime })$.

Hence, final stored energy will be

$U=\frac{Q^2}{2\left(KC\right)}=\frac{U}{K}$

Hence, option $\left(c\right)$ is correct.

Key Concept:- $U=\frac{q^2}{2C}$