$\sigma$ 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:

According to Gauss’ Theorem, electric field of an infinitely long straight wire is proportional to

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

The volume charge density of a sphere of radius $6 \,m$ is $2 \,\mu cm ^{-3}$. The number of lines of force per unit surface area coming out from the surface of the sphere is $....\times 10^{10}\, NC ^{-1}$.  [Given : Permittivity of vacuum  $\left.\epsilon_{0}=8.85 \times 10^{-12} C ^{2} N ^{-1}- m ^{-2}\right]$

$(a)$ Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by

$\left( E _{2}- E _{1}\right) \cdot \hat{ n }=\frac{\sigma}{\varepsilon_{0}}$

where $\hat{ n }$ is a unit vector normal to the surface at a point and $\sigma$ is the surface charge density at that point. (The direction of $\hat { n }$ is from side $1$ to side $2 .$ ) Hence, show that just outside a conductor, the electric field is $\sigma \hat{ n } / \varepsilon_{0}$

$(b)$ Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.

A conducting sphere of radius $R = 20$ $cm$ is given a charge $Q = 16\,\mu C$. What is $\overrightarrow E $ at centre