How will the voltage $(V)$ between the two plates of a parallel plate capacitor depend on the distance $(d)$ between the plates,if the charge on the capacitor remains the same?

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

$A$ parallel plate capacitor with plates of area $1\,m^2$ each,are at a separation of $0.1\,m.$ If the electric field between the plates is $100\,N/C,$ what is the magnitude of the charge on each plate?

Seven identical plates each of area $A$ and successive separation $d$ are arranged as shown in the figure. The effective capacitance of the system between $P$ and $Q$ is

Two identical thin metal plates have charges $q_{1}$ and $q_{2}$ respectively,such that $q_{1} > q_{2}$. The plates are brought close to each other to form a parallel plate capacitor of capacitance $C$. The potential difference between them is:

The area of each plate of a parallel plate capacitor is $20 \, cm^2$ and the separation between the plates is $2 \, mm$. If the dielectric strength of air is $3 \times 10^6 \, V/m$,the maximum possible value of the emf of the battery that can be connected across the plates of this capacitor and the corresponding charge on the plates are: