There is an electric field $E$ in the $X$-direction. If the work done on moving a charge $0.2\,C$ through a distance of $2\,m$ along a line making an angle $60^\circ$ with the $X$-axis is $4.0\,J$,what is the value of $E$ in $N/C$?

Two charges $9e$ and $3e$ are placed at a distance $r$. At what distance will the electric field be zero?

Find the electric field at point $P$ (as shown in figure) on the perpendicular bisector of a uniformly charged thin wire of length $L$ carrying a charge $Q.$ The distance of the point $P$ from the centre of the rod is $a = \frac{\sqrt{3}}{2} L$.

Two point charges $A$ and $B$ of magnitude $+8 \times 10^{-6}\,C$ and $-8 \times 10^{-6}\,C$ respectively are placed at a distance $d$ apart. The electric field at the middle point $O$ between the charges is $6.4 \times 10^{4}\,NC^{-1}$. The distance $d$ between the point charges $A$ and $B$ is ............ $m$. (in $.0$)

For the given figure,the direction of the electric field at point $A$ will be:

Five charges,each $q$,are placed at the corners of a regular pentagon of side $a$ as shown in the figure.
$(a)$ $(i)$ What will be the electric field at $O$,the centre of the pentagon?
$(ii)$ What will be the electric field at $O$ if the charge from one of the corners (say $A$) is removed?
$(iii)$ What will be the electric field at $O$ if the charge $q$ at $A$ is replaced by $-q$?
$(b)$ How would your answer to $(a)$ be affected if the pentagon is replaced by an $n$-sided regular polygon with charge $q$ at each of its corners?