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......$\mu F$.

In the given system,the area of each plate is $A$ and the distance between two consecutive plates is $d$. What is the charge on plates $1$ and $4$?

The diameter of each plate of an air capacitor is $4\,cm$. To make the capacity of this parallel plate capacitor equal to that of a sphere of diameter $20\,cm$,the distance between the plates will be:

$A$ parallel plate capacitor is made up of a stair-like structure with a plate area $A$ for each stair,and the plates are separated by distances $b$,$2b$,and $3b$ as shown in the figure. The capacitance of the arrangement is $\frac{ x }{15} \frac{\varepsilon_{0} A }{ b }$. The value of $x$ is ............

Why is a $1$ $F$ unit so big in practice?

$A$ 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?