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

For two electrons separated by a distance of $10 \, cm$,let $F_g$ and $F_e$ represent the gravitational force and electrostatic force between them,respectively. The ratio $F_g / F_e$ is of the order of:

Four point charges,each of $+q$,are rigidly fixed at the four corners of a square planar soap film of side $a$. The surface tension of the soap film is $\gamma$. The system of charges and the planar film are in equilibrium,and $a = k \left[ \frac{q^2}{\gamma} \right]^{1/N}$,where $k$ is a constant. Then $N$ is:

Two identical charges $+Q$ are kept fixed at some distance apart. $A$ small particle $P$ with charge $q$ is placed midway between them. If $P$ is given a small displacement $\Delta$,it will undergo simple harmonic motion if:

Five balls marked $a$ to $e$ are suspended using separate threads. Pairs $(b, c)$ and $(d, e)$ show electrostatic repulsion,while pairs $(a, b)$,$(c, e)$,and $(a, e)$ show electrostatic attraction. The ball marked $a$ must be:

Two spherical conductors $B$ and $C$ having equal radii and carrying equal charges repel each other with a force $F$ when kept apart at some distance. $A$ third spherical conductor $A$ having the same radius as that of $B$ but uncharged,is brought in contact with $B$,then brought in contact with $C$,and finally removed away from both. The new force of repulsion between $B$ and $C$ is-