Two charges $+4e$ and $+e$ are at a distance $x$ apart. At what distance must a charge $q$ be placed from charge $+e$ so that it is in equilibrium?

$(a)$ Consider an arbitrary electrostatic field configuration. $A$ small test charge is placed at a null point (i.e.,where $E = 0$) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.
$(b)$ Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.

$A$ thin semi-circular ring of radius $r$ has a positive charge $q$ distributed uniformly over it. The net electric field $\vec{E}$ at the centre $O$ is

Two point charges $a$ and $b$,whose magnitudes are the same,are positioned at a certain distance from each other with $a$ at the origin. $A$ graph is drawn between the electric field strength $E$ at points between $a$ and $b$ and the distance $x$ from $a$. $E$ is taken as positive if it is along the line joining $a$ to $b$. From the graph,it can be decided that:

The linear charge densities of two infinitely long thin and parallel wires are $4 \ Cm^{-1}$ and $8 \ Cm^{-1}$,and the separation between them is $4 \ cm$. The electric field intensity at the midpoint on the line joining them is:

The direction $(\theta)$ of the electric field $\vec{E}$ at point $P$ due to a uniformly charged finite rod is shown in the figure. Based on the geometry provided,determine the angle $\theta$.