Five charges, $\mathrm{q}$ each are placed at the corners of a regular pentagon of side $\mathrm{'a'}$ as in 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 pentagon is replaced by $n\,-$ sided regular polygon with charge $q$ at each of its corners ?

$(a)$ $(i)$ The point $\mathrm{O}$, the centre of the pentagon is equidistant from all the charges at the end point of pentagon. Thus, due to symmetry the electric field due to all the charges are cancelled out. As a result electric field at $\mathrm{O}$ is zero.

$(ii)$ When charge $q$ is removed from A net electric field at the centre due to remaining charges $\mathrm{E}=\frac{k q}{r^{2}}$ along $\mathrm{OA}$.

$(iii)$ If charge $q$ at $\mathrm{A}$ is replaced by $-q$ then, electric field due to this negative charge $\overrightarrow{\mathrm{E}}_{-q}=\frac{k q}{r^{2}}$ along $\mathrm{OA}$.

$(b)$ If pentagon is replaced by $\mathrm{n}$-sided regular polygon with charge $q$ at each of its corners. Here, again charges are symmetrical about the centre. The net electric field at $\mathrm{O}$ would continue to be zero, it doesn't depend on the number of sides or the number of charges. Hence, the answer of (a) would not be affected.