An electron is moving under the influence of the electric field of a uniformly charged infinite plane sheet $S$ having surface charge density $+\sigma$. The electron at $t=0$ is at a distance of $1 \mathrm{~m}$ from $S$ and has a speed of $1 \mathrm{~m} / \mathrm{s}$. The maximum value of $\sigma$ if the electron strikes $S$ at $t=1 \mathrm{~s}$ is $\alpha\left[\frac{\mathrm{m} \in_0}{\mathrm{e}}\right] \frac{\mathrm{C}}{\mathrm{m}^2}$ the value of $\alpha$ is

Let $E_1(r), E_2(r)$ and $E_3(r)$ be the respective electric fields at a distance $r$ from a point charge $Q$, an infinitely long wire with constant linear charge density $\lambda$, and an infinite plane with uniform surface charge density $\sigma$. if $E_1\left(r_0\right)=E_2\left(r_0\right)=E_3\left(r_0\right)$ at a given distance $r_0$, then

A solid ball of radius $R$ has a charge density $\rho $ given by $\rho  = {\rho _0}\left( {1 - \frac{r}{R}} \right)$ for $0 \leq r \leq R$. The electric field outside the ball is

Two concentric conducting thin spherical shells $A$ and $B$ having radii ${r_A}$ and ${r_B}$ (${r_B} > {r_A})$ are charged to ${Q_A}$ and $ - {Q_B}$$(|{Q_B}|\, > \,|{Q_A}|)$. The electrical field along a line, (passing through the centre) 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 ?

An infinite line charge produce a field of $7.182 \times {10^8}\,N/C$ at a distance of $2\, cm$. The linear charge density is