$A$ and $B$ are two concentric spheres. If $A$ is given a charge $Q$ while $B$ is earthed as shown.

The potential due to an electrostatic charge distribution is $V(r) = \frac{q e^{-\alpha r}}{4 \pi \varepsilon_{0} r}$,where $\alpha$ is positive. The net charge within a sphere centred at the origin and of radius $1/\alpha$ is

The charge density of a solid sphere of radius $R$ and total charge $Q$ is given by $\rho(r) = \frac{Q}{\pi R^4} r$. The value of the electric field at a point $P$ inside the sphere at a distance $r_1$ from the center is .......

$A$ positive point charge is released from rest at a distance $r_0$ from a positive line charge with uniform density. The speed $(v)$ of the point charge,as a function of instantaneous distance $r$ from the line charge,is proportional to

Obtain the expression for the electric field due to:
$(i)$ An infinite plane sheet with uniform charge distribution.
$(ii)$ $A$ thin spherical shell with uniform charge distribution at a point outside it.
$(iii)$ $A$ thin spherical shell with uniform charge distribution at a point inside it.

In the figure,a very large plane sheet of positive charge is shown. $P_{1}$ and $P_{2}$ are two points at distances $l$ and $2l$ from the charge distribution. If $\sigma$ is the surface charge density,then the magnitude of electric fields $E_{1}$ and $E_{2}$ at $P_{1}$ and $P_{2}$ respectively are: