The capacity of a parallel plate capacitor is $15\,\mu F$,when the distance between its plates is $6\,cm$. If the distance between the plates is reduced to $2\,cm$,then the capacity of this parallel plate capacitor will be......$\mu F$.

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 $d << \sqrt{A}$):

Two identical thin metal plates have charges $q_{1}$ and $q_{2}$ respectively,such that $q_{1} > q_{2}$. The plates are brought close to each other to form a parallel plate capacitor of capacitance $C$. The potential difference between them is:

The distance between the plates of a parallel plate capacitor is $4 \ mm$ and the potential difference is $60 \ V$. If the distance between the plates is increased to $12 \ mm$,then:

$A$ parallel plate capacitor is charged. Then the battery is removed. Now,if the plates are pulled apart,

Two circular plates each of radius 'r' are kept parallel to each other distance 'd' apart. The capacitance of the capacitor formed is '$C_1$'. If the radius of each of the plates is increased to $\sqrt{3}$ times the earlier radius and their distance of separation decreased to half the initial value, the capacitance now becomes '$C_2$'. The ratio $C_1 : C_2$ is