The capacitance of a capacitor made by a thin metal foil is $2\,\mu F$. If the foil is folded with paper of thickness $0.15\,mm$,the dielectric constant of the paper is $2.5$,and the width of the paper is $400\,mm$,then the length of the foil will be.....$m$.

The capacity of a parallel plate capacitor increases with the

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

The electrostatic force between the metal plates of an isolated parallel plate capacitor $C$ having a charge $Q$ and area $A$ is

The charges on the four surfaces of two isolated plates with capacitance $C$ are $Q_1$,$Q_2$,$Q_3$,and $Q_4$ as shown in the figure. The potential difference between the plates is:

What is the energy density of a parallel plate capacitor with area $A$ and charge $Q$?