Two charges $+5\,\mu C$ and $+10\,\mu C$ are placed $20\,cm$ apart. The net electric field at the midpoint between the two charges is

$(a)$ 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.

The number of electrons to be put on a spherical conductor of radius $0.1\,m$ to produce an electric field of $0.036\,N/C$ just above its surface is:

Four point charges are placed at the vertices of a regular hexagon as shown in the figure below. What will be the net electric field at the centre?

Two point charges $(+Q)$ and $(-2Q)$ are placed at distances $a$ and $2a$ respectively from the origin on the $X$-axis. At what distance from the origin is the electric field zero?

Suppose a uniformly charged wall provides a uniform electric field of $2 \times 10^4 \ N/C$ normally. $A$ charged particle of mass $2 \ g$ is suspended by a silk thread of length $20 \ cm$ and remains at a distance of $10 \ cm$ from the wall. Then the charge on the particle will be $\frac{1}{\sqrt{x}} \ \mu C$ where $x=$ . . . . . . . (Use $g=10 \ m/s^2$)