In the given figure, find the distance from point $A$ where the electric field is zero (in $cm$).

Two equal positive point charges are separated by a distance $2a$. The distance of a point from the centre of the line joining two charges on the equatorial line (perpendicular bisector) at which the force experienced by a test charge $q_0$ becomes maximum is $\frac{a}{\sqrt{x}}$. The value of $x$ is $................$

The magnitude of a point charge due to which the electric field $30 \ cm$ away has a magnitude of $2 \ N \ C^{-1}$ is:

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?

Electric field strength due to a point charge of $5\,\mu C$ at a distance of $80\, cm$ from the charge is:

An infinite number of electric charges,each with a magnitude of $5 \, nC$,are placed along the $X$-axis at $x = 1 \, cm$,$x = 2 \, cm$,$x = 4 \, cm$,$x = 8 \, cm$,and so on. If the consecutive charges have opposite signs,find the electric field in $N/C$ at $x = 0$. (Given: $\frac{1}{4\pi \varepsilon_0} = 9 \times 10^9 \, N \cdot m^2/C^2$)