$A$ 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?

If $Q$ is the charge on the plates of a capacitor of capacitance $C$,$V$ is the potential difference between the plates,$A$ is the area of each plate,and $d$ is the distance between the plates,then the force of attraction between the plates is:

Two charged capacitors have their outer plates fixed and inner plates connected by a spring of force constant $k$. The charge on each capacitor is $q$. Find the extension in the spring at equilibrium.

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

The capacitance of a parallel plate capacitor does not depend on which of the following?

Between the plates of a parallel plate capacitor,there is a $1\,mm$ thick paper of dielectric constant $4$. It is charged to $100\,V$. The electric field in $V/m$ between the plates of the capacitor is: