In an isolated parallel plate capacitor of capacitance $C$,the four surfaces have charges $Q_1$,$Q_2$,$Q_3$,and $Q_4$ as shown. The potential difference between the plates is

$A$ parallel plate capacitor consisting of two circular plates of radius $10 \ cm$ is being charged by a constant current of $0.15 \ A$. If the rate of change of potential difference between the plates is $7 \times 10^8 \ V/s$,then the integer value of the distance between the parallel plates is $—$ (Take $\epsilon_0 = 9 \times 10^{-12} \ F/m, \pi = 22/7$) . . . . . . $\mu m$.

Four identical plates each of area $a$ are separated by a distance $d$. The connection is shown below. What is the capacitance between $P$ and $Q$?

$A$ parallel plate capacitor system is formed by connecting $n$ plates placed at equal distances alternately. If the capacitance between any two adjacent plates is $C$, find the resultant capacitance.

The capacity of a parallel plate capacitor is $10\,\mu F$,when the distance between its plates is $8\,cm$. If the distance between the plates is reduced to $4\,cm$,then the capacity of this parallel plate capacitor will be.........$\mu F$.

As shown in the figure,two parallel plate capacitors having equal plate area of $200 \, cm^2$ are joined in such a way that $a \neq b$. The equivalent capacitance of the combination is $x \varepsilon_0 \, F$. The value of $x$ is $..........$.