Question

Five point charges each $+q$ are placed on five vertices of a regular hexagon of side $a$ unit. What is the magnitude of force experienced by a charge $-q$ placed at the centre of the hexagon ?

Answer 1

The regular hexagon shown in figure the distance between centre and vertex $=a$
Thus, each pair of interaction force will be same i.e., $F$
$|F|=\frac{k{q}^2}{a^2}$
If there had been a sixth charge at the remaining vertex $B$ of the hexagon, then at the charge placed at centre
${\overrightarrow{F}}_{\text{net}}=0$
i.e., the sixth charge will exert force $F$ along $OB$.
$\therefore$ unbalanced force is the net force on the central charge
$\Rightarrow F_{\text{net}}=F=\frac{k{q}^2}{a^2}$
Direction of $F_{\text{net}}$ is along $BO$
$\therefore |F_{\text{net}}|=\frac{1}{4\pi {ϵ}_0}\frac{q^2}{a^2}$

$\begin{array}{c}\text{Tip: When identical charges are placed symmetrically at the vertices of a regular}\\ \text{polygon, then net force,}{\overrightarrow{F}}_{\text{net}}=0\\ \text{at geometric centre of polygon.}\end{array}$