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

Calculate the capacitance of two plates of equal area of $1 \ m^2$ separated by a distance of $1 \ mm$.

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

The plates of a parallel plate capacitor of capacity $C_1$ are moved closer together until they are half their original separation. The new capacitance $C_2$ is

The area of the plates of a parallel plate capacitor is $A$ and the gap between them is $d$. The gap is filled with a non-homogeneous dielectric whose dielectric constant varies with the distance $y$ from one plate as: $K = \lambda \sec(\pi y/2d)$,where $\lambda$ is a dimensionless constant. The capacitance of this capacitor is