Shown in the figure are two point charges $+Q$ and $-Q$ inside the cavity of a spherical shell. The charges are kept near the surface of the cavity on opposite sides of the centre of the shell. If $\sigma _1$ is the surface charge on the inner surface and $Q_1$ net charge on it and $\sigma _2$ the surface charge on the outer surface and $Q_2$ net charge on it then
$\begin{array}{l}
{\sigma _1}\, \ne \,0,\,\,{Q_1}\, = \,0\\
{\sigma _2}\, = \,0,\,\,{Q_2}\, = \,0
\end{array}$
$\begin{array}{l}
{\sigma _1}\, \ne \,0,\,\,{Q_1}\, = \,0\\
{\sigma _2}\, \ne \,0,\,\,{Q_2}\, = \,0
\end{array}$
$\begin{array}{l}
{\sigma _1}\, = \,0,\,\,{Q_1}\, = \,0\\
{\sigma _2}\, = \,0,\,\,{Q_2}\, = \,0
\end{array}$
$\begin{array}{l}
{\sigma _1}\, \ne \,0,\,\,{Q_1}\, \ne \,0\\
{\sigma _2}\, \ne \,0,\,\,{Q_2}\, \ne \,0
\end{array}$
An infinite line charge produces a field of $9 \times 10^4 \;N/C$ at a distance of $2\; cm$. Calculate the linear charge density in $\mu C / m$
According to Gauss’ Theorem, electric field of an infinitely long straight wire is proportional to
Obtain the formula for the electric field due to a long thin wire of uniform linear charge density $E$ without using Gauss’s law.
A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is $\left(\sigma / 2 \varepsilon_{0}\right) \hat{ n },$ where $\hat{ n }$ is the unit vector in the outward normal direction, and $\sigma$ is the surface charge density near the hole.
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