An infinitely long solid cylinder of radius $R$ has a uniform volume charge density $\rho $. It has a spherical cavity of radius $R/2$ with its centre on the axis of the cylinder, as shown in the figure. The magnitude of the electric field at the point $P$, which is at a distance $2R$ from the axis of the cylinder, is given by the expression $\frac{{23\rho R}}{{16K{\varepsilon _0}}}$ .The value of $K$ is

Let $\rho (r) =\frac{Q}{{\pi {R^4}}}r$ be the charge density distribution for a solid sphere of radius $R$ and total charge $Q$. For a point '$p$' inside the sphere at distance $r_1$ from the centre of the sphere, the magnitude of electric field is

Obtain the formula for the electric field due to a long thin wire of uniform linear charge density $E$ without using Gauss’s law.

An electrostatic field in a region is radially outward with magnitude $E$ = $\alpha r$ , where $\alpha $ is a constant and $r$ is radial distance. The charge contained in a sphere of radius $R$ in this region (centred at the origin) is

$\sigma$ is the uniform surface charge density of a thin spherical shell of radius $R$. The electric field at any point on the surface of the spherical shell is:

There is a solid sphere of radius $‘R’$ having uniformly distributed charge throughout it. What is the relation between electric field $‘E’$ and distance $‘r’$ from the centre ( $r$ is less than R ) ?