Electric lines of force about a negative point charge are:

Two point charges $(+Q)$ and $(-2Q)$ are fixed on the $X$-axis at positions $a$ and $2a$ from the origin,respectively. At what position on the axis is the resultant electric field zero?

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?

Three charges $2q, -q$ and $-q$ are located at the vertices of an equilateral triangle. At the center of the triangle,

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: $\frac{1}{4\pi \varepsilon_0} = 9 \times 10^9 \, N \cdot m^2/C^2$)

The electric field lines on the left have twice the separation of those on the right as shown in the figure. If the magnitude of the electric field at $A$ is $40 \ Vm^{-1}$,what is the force on a $20 \ \mu C$ charge kept at $B$?