Let $P\left( r \right) = \frac{Q}{{\pi {R^4}}}r$ be the charge density distribution for a solid sphere of radius $R$ and total charge $Q$. For a point $P$ inside the sphere at distance $r_1$ from the centre of the sphere, the magnitude of electric field is

Consider a sphere of radius $\mathrm{R}$ which carries a uniform charge density $\rho .$ If a sphere of radius $\frac{\mathrm{R}}{2}$ is carved out of it, as shown, the ratio $\frac{\left|\overrightarrow{\mathrm{E}}_{\mathrm{A}}\right|}{\left|\overrightarrow{\mathrm{E}}_{\mathrm{B}}\right|}$ of magnitude of electric field $\overrightarrow{\mathrm{E}}_{\mathrm{A}}$ and $\overrightarrow{\mathrm{E}}_{\mathrm{B}}$ respectively, at points $\mathrm{A}$ and $\mathrm{B}$ due to the remaining portion is

Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude $17.0\times 10^{-22}\; C/m^2$. What is $E$:

$(a)$ in the outer region of the first plate,

$(b)$ in the outer region of the second plate, and

$(c)$ between the plates?

If an insulated non-conducting sphere of radius $R$ has charge density $\rho $. The electric field at a distance $r$ from the centre of sphere $(r < R)$ will be

$\sigma$ 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:

The dimensions of an atom are of the order of an Angstrom. Thus there must be large electric fields between the protons and electrons. Why, then is the electrostatic field inside a conductor zero ?