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?

(N/A) $(i)$ The point $O$,the centre of the pentagon,is equidistant from all the charges at the vertices. Due to the rotational symmetry of the regular pentagon,the electric field vectors produced by each charge at $O$ cancel each other out. Thus,the net electric field at $O$ is $0$.
$(ii)$ Let $\vec{E}_A$ be the electric field at $O$ due to the charge at $A$. The sum of the electric fields due to all five charges is $\vec{E}_{total} = \vec{E}_A + \vec{E}_B + \vec{E}_C + \vec{E}_D + \vec{E}_E = 0$. If the charge at $A$ is removed,the remaining field is $\vec{E}' = \vec{E}_B + \vec{E}_C + \vec{E}_D + \vec{E}_E = -\vec{E}_A$. Since $\vec{E}_A$ points from $A$ to $O$,$-\vec{E}_A$ points from $O$ to $A$. The magnitude is $E = \frac{kq}{r^2}$,where $r$ is the distance from the centre to a vertex.
$(iii)$ If the charge at $A$ is replaced by $-q$,the new field is $\vec{E}'' = \vec{E}' + \vec{E}_{-q}$. Since $\vec{E}' = -\vec{E}_A$ and $\vec{E}_{-q} = -\vec{E}_A$,the total field is $\vec{E}'' = -2\vec{E}_A$. The magnitude is $\frac{2kq}{r^2}$ directed from $O$ to $A$.
$(b)$ For an $n$-sided regular polygon with equal charges $q$ at each vertex,the symmetry ensures that the vector sum of the electric fields at the centre is always $0$. Therefore,the logic remains the same,and the answers to $(a)$ are not affected.