An electron revolves around an infinite cylindrical wire having a uniform linear charge density of $2 \times 10^{-8} \, C \cdot m^{-1}$ in a circular path under the influence of an attractive electrostatic field,as shown in the figure. The velocity of the electron with which it is revolving is $......... \times 10^6 \, m \cdot s^{-1}$. (Given: mass of electron $= 9 \times 10^{-31} \, kg$)

$(a)$ Show that the normal component of the electrostatic field has a discontinuity from one side of a charged surface to another given by $(E_2 - E_1) \cdot \hat{n} = \frac{\sigma}{\varepsilon_0}$, where $\hat{n}$ is a unit vector normal to the surface at a point and $\sigma$ is the surface charge density at that point. (The direction of $\hat{n}$ is from side $1$ to side $2$.) Hence, show that just outside a conductor, the electric field is $\frac{\sigma \hat{n}}{\varepsilon_0}$. $(b)$ Show that the tangential component of the electrostatic field is continuous from one side of a charged surface to another.

Two parallel large thin metal sheets have equal surface charge densities $(\sigma = 26.4 \times 10^{-12} \, C/m^2)$ of opposite signs. The electric field between these sheets is

If the uniform surface charge density on an infinite plane sheet is $\sigma$,the electric field near the surface will be . . . . . . .

$A$ long cylindrical volume contains a uniformly distributed charge of density $\rho$. The radius of the cylindrical volume is $R$. $A$ charged particle $(q)$ revolves around the cylinder in a circular path at a distance $r$ from the axis. The kinetic energy of the particle is:

$A$ spherically symmetric charge distribution is characterized by a charge density $\rho(r) = \rho_0 \left( \frac{5}{4} - \frac{r}{R} \right)$ for $r \le R$ and $\rho(r) = 0$ for $r > R$,where $r$ is the distance from the origin. The electric field at a distance $r$ from the origin $(r < R)$ is given by: