There is a uniform spherically symmetric surface charge density at a distance $R_0$ from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed $V(R(t))$ of the distribution as a function of its instantaneous radius $R(t)$ is

The work done in carrying a charge of $5\,\mu \,C$ from a point $A$ to a point $B$ in an electric field is $10\,mJ$. The potential difference $({V_B} - {V_A})$ is then

A solid sphere of radius $R$ carries a charge $(Q+q)$ distributed uniformly over its volume. A very small point like piece of it of mass $m$ gets detached from the bottom of the sphere and falls down vertically under gravity. This piece carries charge $q.$ If it acquires a speed $v$ when it has fallen through a vertical height $y$ (see figure), then :

(assume the remaining portion to be spherical).

The mean free path of electrons in a metal is $4 \times 10^{-8} \;m$. The electric field which can give on an average $2 \;eV$ energy to an electron in the metal will be in units of $V / m$

A point chargr $Q$ is fixed A small charge $q$ and mass $m$ is given a velocity $v_0$ from infinity & perpendicular distance $r_0$ as shown. If distance of closest approach is $r_0/2$. The value of $q$ is [Given $mv_0^2 = \frac{{{Q^2}}}{{4\pi { \in _0}\,{r_0}}}$]

An $\alpha $-particle is accelerated through a potential difference of $200\,V$. The increase in its kinetic energy is.......$eV$