Five identical capacitor plates are arranged as shown in the figure. They form capacitors each of $2 \mu F$. The plates are connected to a source of $emf$ $10 \ V$. The charge on plate $C$ is......$\mu C$.

$A$ parallel plate capacitor is charged and then disconnected from the charging battery. If the plates are now moved further apart by pulling them by means of insulating handles,then

$A$ parallel plate capacitor is made by stacking $n$ equally spaced plates connected alternately. If the capacitance between any two plates is $C$, then the resultant capacitance is:

$A$ metal slab of thickness $b = d/2$ is inserted between the plates of a parallel plate capacitor of plate separation $d$. What is the ratio of the new capacitance to the original capacitance?

Four metallic plates are arranged as shown in the figure. If the distance between each plate is $d$ and the area of each plate is $A$, then the capacitance of the given system between points $A$ and $B$ is (Given $d << \sqrt{A}$):

$A$ 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 $\epsilon_0 = 9 \times 10^{-12} \ F/m, \pi = 22/7$) . . . . . . $\mu m$.