‘At the surface of a charged conductor electrostatic field must be normal to the surface at every point’. Explain.

Vedclass pdf generator app on play store
Vedclass iOS app on app store

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).

$\left[\because 0=\frac{\sigma}{\epsilon_{0}}\right]$

898-s105

Similar Questions

A metallic spherical shell has an inner radius $R_1$ and outer radius $R_2$. A charge $Q$ is placed at the centre of the spherical cavity. What will be surface charge density on the inner surface

$IAn$ empty thick conducting shell of inner radius $a$ and outer radius $b$ is shown in figure.If it is observed that the inner face of the shell carries a uniform charge density $-\sigma$ and the surface carries a uniform charge density $ '\sigma '$

Choose the correct statement related to the potential of the shell in absence of $q_B$

The adjacent diagram shows a charge $+Q$ held on an insulating support $S$ and enclosed by a hollow spherical conductor. $O$ represents the centre of the spherical conductor. and $P$ is a point such that $OP = x $ and $SP = r$ . The electric field at point $P$  will be

A positive charge $q$ is placed at the centre of a neutral hollow cylindrical conducting shell with its cross-section as shown in the figure below. Which one of the following figures correctly indicates the induced charge distribution on the conductor? (Ignore edge effects)

  • [KVPY 2017]

For a spherical shell