In the diagram,the area of each plate is $A = 2 \,m^2$ and $d = 2 \times 10^{-3} \,m$. An electric charge of $q = 8.85 \times 10^{-8} \,C$ is given to the plate '$Q$'. Plates '$P$' and '$R$' are connected to the ground. Then the potential of plate '$Q$' 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:

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

Two parallel plates separated by $d \ mm$ are kept at a potential difference of $V \ V$. $A$ particle of mass $m$ and charge $q$ enters the region between them with some velocity. The acceleration of the particle will be

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

$A$ parallel plate capacitor is formed by connecting $n$ plates placed at equal distances alternately. If the capacitance between any two adjacent plates is $C$,find the resultant capacitance.