The energy density $u$ is plotted against the distance $r$ from the centre of a spherical charge distribution on a $log$-$log$ scale. The slope of the obtained straight line is:

The repulsive force between two point charges is $F$ when they are separated by a distance of $1 \, m$. Now,these point charges are replaced by spheres of radius $25 \, cm$ having the same charges. The distance between their centers is $1 \, m$. The repulsive force in the two cases will decrease according to:

$A$ tiny spherical oil drop carrying a net charge $q$ is balanced in still air with a vertical uniform electric field of strength $\frac{81 \pi}{7} \times 10^5 \text{ Vm}^{-1}$. When the field is switched off,the drop is observed to fall with a terminal velocity $2 \times 10^{-3} \text{ ms}^{-1}$. Given $g = 9.8 \text{ ms}^{-2}$,viscosity of the air $\eta = 1.8 \times 10^{-5} \text{ Ns m}^{-2}$,and the density of oil $\rho = 900 \text{ kg m}^{-3}$,the magnitude of $q$ is:

Identify the correct statements:
$A$. Electrostatic field lines form closed loops.
$B$. The electric field lines point radially outward when charge is greater than zero.
$C$. The Gauss-Law is valid only for inverse-square force.
$D$. The work done in moving a charged particle in a static electric field around a closed path is zero.
$E$. The motion of a particle under Coulomb's force must take place in a plane.
Choose the correct answer from the options given below:

Consider a system of three charges $\frac{q}{3}, \frac{q}{3}$ and $-\frac{2q}{3}$ placed at points $A, B$ and $C$,respectively,as shown in the figure. Take $O$ to be the centre of the circle of radius $R$ and angle $\angle CAB = 60^{\circ}$.

Two charged particles,each of mass $3 \ g$ and charge $0.2 \ \mu C$,stay in (vacuum) equilibrium on a horizontal surface with a separation of $20 \ cm$. The coefficient of friction is $\left[\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \ Nm^2 C^{-2}\right]$ and $\left(g=10 \ ms^{-2}\right)$.