The surface charge density of a thin charged disc of radius $R$ is $\sigma$. The value of the electric field at the centre of the disc is $\frac{\sigma}{2\epsilon_0}$. With respect to the field at the centre,the electric field along the axis at a distance $R$ from the centre of the disc:

$X$ and $Y$ are large,parallel conducting plates close to each other. Each face has an area $A$. $X$ is given a charge $Q$. $Y$ is without any charge. Points $A, B$,and $C$ are as shown in the figure.

$A$ large metal plate has a surface charge density of $8.85 \times 10^{-6} \ C \ m^{-2}$. An electron having initial kinetic energy of $8 \times 10^{-17} \ J$ is moving towards the center of the plate. If the electron stops just before reaching the plate,then the initial distance between the electron and the plate is [Take $\epsilon_{0} = 8.85 \times 10^{-12} \ C^{2} \ N^{-1} \ m^{-2}$]

$A$ solid sphere of radius $r_1=1 \text{ cm}$ carries charge distributed uniformly over it with density $\rho_1=-3 \text{ C/cm}^3$. It is surrounded by a concentric spherical shell of radius $r_2=2 \text{ cm}$ carrying uniform charge density $\rho_2=0.5 \text{ C/cm}^3$. If $E_d$ denotes the magnitude of the electric field at distance $d$ from the common centre of the spheres,then

For an infinite line of charge having charge density $\lambda$ lying along the $x$-axis,the work required in moving a charge $q$ from point $C$ to point $A$ along the arc $CA$ is:

If the atmospheric electric field is approximately $150 \,V/m$ and the radius of the Earth is $6400 \,km$,then the total charge on the Earth's surface is .......... coulomb.