The capacity of a parallel plate capacitor is $C$. What will be its capacity when the separation between the plates is halved?

$A$ parallel plate capacitor with area $200\,cm^2$ and separation between the plates $1.5\,cm$,is connected across a battery of $emf$ $V$. If the force of attraction between the plates is $25\times10^{-6}\,N$,the value of $V$ is approximately........$V$ $\left( {{\varepsilon _0} = 8.85 \times {{10}^{ - 12}}\,\frac{{{C^2}}}{{N{m^2}}}} \right)$

Three plates $A, B$ and $C$ each of area $0.1 \ m^2$ are separated by $0.885 \ mm$ from each other as shown in the figure. A $10 \ V$ battery is used to charge the system. The energy stored in the system is

$A$ thin aluminum sheet of negligible thickness is placed between the plates of a parallel plate capacitor. The capacitance of the capacitor:

Seven identical plates each of area $A$ and successive separation $d$ are arranged as shown in the figure. The effective capacitance of the system between $P$ and $Q$ is

The capacity of a parallel plate capacitor depends on