‘At the surface of a charged conductor electrostatic field must be normal to the surface at every point’. Explain.
If $\overrightarrow{\mathrm{E}}$ were not normal to the surface, it would have some non-zero component along the surface. Free charges on the surface of the conductor would then experience force and move. Hence, conductor does not remains in stable situation.
Therefore, $\vec{E}$ should have no tangential component parallel to the surface in stable situation. Thus, electrostatic field at the surface of a charged conductor must be normal to the surface at every point. (For a conductor without any surface charge density, field is zero even at the surface).
$\left[\because 0=\frac{\sigma}{\epsilon_{0}}\right]$
Three concentric metallic spherical shells of radii $R, 2R, 3R$, are given charges $Q_1, Q_2, Q_3$, respectively. It is found that the surface charge densities on the outer surfaces of the shells are equal. Then, the ratio of the charges given to the shells, $Q_1 : Q_2 : Q_3$ is
‘The interior of a conductor can have no excess charge in the static situation’. Explain.
Two thin conducting shells of radii $R$ and $3R$ are shown in the figure. The outer shell carries a charge $+ Q$ and the inner shell is neutral. The inner shell is earthed with the help of a switch $S$.
A conducting sphere of radius $r$ has a charge. Then
A thin metallic spherical shell contains a charge $Q$ on it. A point charge $+q$ is placed at the centre of the shell and another charge $q'$ is placed outside it as shown in fig. All the three charges are positive. The force on the central charge due to the shell is :-