If $E$ is the electric field intensity of an electrostatic field, then the electrostatic energy density is proportional to

A $2 \ \mu F$ capacitor is charged as shown in figure. The percentage of its stored energy dissipated after the switch $S$ is turned to position $2$ is

A series combination of $n_1$ capacitors, each of value $C_1$ is charged by a source of potential difference $4\, V.$ When another parallel combination of $n_2$ capacitors, each of value $C_2,$ is charged by a source of potential difference $V$, it has the same (total) energy stored in it, as the first combination has. The value of $C_2,$ in terms of $C_1$ is then

A capacitor of capacitance $\mathrm{C}$ and potential $\mathrm{V}$ has energy $E$. It is connected to another capacitor of capacitance $2 \mathrm{C}$ and potential $2 \mathrm{~V}$. Then the loss of energy is $\frac{x}{3} E$, where $\mathrm{x}$ is____________.

A capacitor of capacitance $50 \; pF$ is charged by $100 \; V$ source. It is then connected to another uncharged identical capacitor. Electrostatic energy loss in the process is $\dots \; nJ$.

A battery does $200 \,J$ of work in charging a capacitor. The energy stored in the capacitor is ......... $J$