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:

For a thin spherical shell of radius $R$ with a uniform surface charge density, which of the following graphs best represents the variation of the electric field magnitude $|\vec{E}(r)|$ and the electric potential $V(r)$ with the distance $r$ from the center?

In the given square $ABCD$,charges $q$,$2q$,$3q$,and $4q$ are placed at corners $A$,$B$,$C$,and $D$ respectively. What is the direction of the net electric field at the center $O$?

Two charges each of magnitude $Q$ are fixed at a distance of $2a$ apart. $A$ third charge ($-q$ of mass $m$) is placed at the midpoint of the two charges. If the $-q$ charge is slightly displaced perpendicular to the line joining the charges,find its time period.

In $1959$,Lyttleton and Bondi suggested that the expansion of the Universe could be explained if matter carried a net charge. Suppose that the Universe is made up of hydrogen atoms with a number density $N$,which is maintained constant. Let the charge on the proton be: $e_p = -(1 + y)e$,where $e$ is the electronic charge.
$(a)$ Find the critical value of $y$ such that expansion may start.
$(b)$ Show that the velocity of expansion is proportional to the distance from the centre.

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 $= 6.37 \times 10^{6} \; m$.)