$A$ charge $Q$ is uniformly distributed over the volume of an insulating sphere of radius $R$. $A$ thin metal shell of radius $b$ $(b > R)$ with a charge $-Q$ is placed around the sphere. The space between the sphere and the shell is filled with air. Which of the following graphs correctly represents the electric field $E$ as a function of distance $r$ from the center?

Four identical point masses, each of mass $m$ and carrying charge $+q$, are placed at the corners of a square of side $a$ on a frictionless horizontal surface. If the particles are released simultaneously, the kinetic energy of the system when they are infinitely far apart is

Three point charges are placed at the corners of an equilateral triangle of side $L$ as shown in the figure.

In free space,a particle $A$ of charge $1\,\mu C$ is held fixed at a point $P.$ Another particle $B$ of the same charge and mass $4\,\mu g$ is kept at a distance of $1\,mm$ from $P$. If $B$ is released,then its velocity at a distance of $9\,mm$ from $P$ is [ Take $\frac{1}{4\pi \varepsilon_0} = 9 \times 10^9\,N m^2 C^{-2}$ ]

$A$ positively charged thin metal ring of radius $R$ is fixed in the $xy$-plane with its centre at the origin $O$. $A$ negatively charged particle $P$ is released from rest at the point $(0, 0, z_0)$,where $z_0 > 0$. Then the motion of $P$ is:

The earth's surface has a negative surface charge density of $10^{-9} \; C\;m^{-2}$. The potential difference of $400 \; kV$ between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only $1800 \; A$ over the entire globe. If there were no mechanism of sustaining the atmospheric electric field,how much time (roughly) would be required to neutralize the earth's surface (in $; s$)? (This never happens in practice because there is a mechanism to replenish electric charges,namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth $= 6.37 \times 10^{6} \; m$.)