‘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).
$\left[\because 0=\frac{\sigma}{\epsilon_{0}}\right]$
Inside a hollow charged spherical conductor, the potential
A point charge $q$ is placed in a cavity in a metal block. If a charge $Q$ is brought outside the metal, then the electric force experienced by $q$ is
A thin conducting spherical shell (center at $O$ ) having charge $Q_0$ , radius $R$ and three point charges $Q_0$ , $-2Q_0$ , $3Q_0$ are also kept at point $A$ , $B$ and $C$ respectively as shown. Find the potential at any point on the conducting shell. (Potential at infinity is assumed to be zero)
Two metal spheres, one of radius $R$ and the other of radius $2R$, both have same surface charge density $\sigma $. They are brought in contact and separated. What will be new surface charge densities on them ?
A conducting sphere of radius $10\, cm$ is charged $10\,\mu \,C$. Another uncharged sphere of radius $20\, cm$ is allowed to touch it for some time. After that if the sphere are separated, then surface density of charges, on the spheres will be in the ratio of