A condenser is charged to a potential difference of $120V$.Its energy is $1 \times {10^{ - 5}}J$.If battery is there and the space between the plates is filled up with a dielectric medium $\left( {{ \in _r} = 5} \right)$.Its new energy is
A. ${10^{ - 5}}J$
B. $2 \times {10^{ - 5}}J$
C. $3 \times {10^{ - 5}}J$
D. $5 \times {10^{ - 5}}J$

Hint:In this question, first we will discuss the energy of a condenser when there is no battery connected and no medium is filled up in the space between the plates. Then discuss the new energy of a condenser when a battery is connected and the space between the plates is filled up with any medium.

Complete step by step answer:
In this question, we have given the potential difference of the condenser which is charged is $120V$.The energy of the condenser is $1 \times {10^{ - 5}}J$.
Now a battery is connected and the space between the plates is filled up with a dielectric medium in which $\left( {{ \in _r} = 5} \right)$.
So when the battery is connected the voltage across the capacitor will be constant and when the dielectric is filled in the space between the plates then the energy of the capacitor will be increased.Let say $E$ is the energy of the condenser, now when the dielectric is filled up in the space between the plates whose dielectric constant is $\left( {{ \in _r} = 5} \right)$then the energy of the condenser will be ${ \in _r}E$.We know the energy of the condenser when there is no dielectric medium which is $1 \times {10^{ - 5}}J$.Now the dielectric medium is added then the energy will be
${ \in _r}E = 5 \times \left( {1 \times {{10}^{ - 5}}} \right) = 5 \times {10^{ - 5}}J$.

So, its new energy is $5 \times {10^{ - 5}}J$.Hence, option (D) is the correct option.

Note:The capacitance of a set of charged parallel plates is increased when a dielectric material is inserted in the space between plates. The capacitance is inversely proportional to the electric field between the plates.