Two identical metal plates are given charges $q_1$ and $q_2$ $(q_2 < q_1)$ respectively. If they are now brought close together to form a parallel plate capacitor with capacitance $C$,the potential difference $V$ between the plates is

Two parallel air capacitors have their plate areas $100 \,cm^2$ and $500 \,cm^2$, respectively. They have the same charge and potential. If the distance between the plates of the first capacitor is $0.5 \,mm$, the distance between the plates of the second capacitor is (in $\,cm$)

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 $C$. What will be its capacity when the separation between the plates is halved?

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

Show that the force on each plate of a parallel plate capacitor has a magnitude equal to $\frac{1}{2} Q E$,where $Q$ is the charge on the capacitor,and $E$ is the magnitude of the electric field between the plates. Explain the origin of the factor $\frac{1}{2}$.