Question

Six point charges are placed at the vertices of a hexagon of side $1\text{m}$ as shown in figure.


What will be the force experienced by a charge $+q$ placed at centre of hexagon ?

• A. Zero
• B. $4k{q}^2$
• C. $3k{q}^2$
• D. $6k{q}^2$

Answer 1

The correct option is B $4k{q}^2$

The force experienced by the charge $+q$ placed at center of the hexagon is shown in the figure.


For the regular hexagon of side $a=1\text{m}$, the distance between centre $\left(O\right)$ and each vertex will be also $a$.
Thus, the magnitude of force acting between each pair of charges will be equal i.e. $F$.
($\because$ magnitude of charge and distance between them is same)
$|F|=\frac{k{q}^2}{a^2}$


The resultant of two forces $2F$ will lie along the angle bisector and along the direction of another force $2F$.
$F^{\prime }=\sqrt{(2F{)}^2+(2F{)}^2+2\times (2F{)}^2\cos {120}^{\circ }}$
$\Rightarrow F^{\prime }=\sqrt{8F^2-4F^2}=2F$
Thus net force on charge $+q$ at $O$.
$F_{\text{net}}=F^{\prime }+2F$
$\Rightarrow F_{\text{net}}=4F$
$\Rightarrow F_{\text{net}}=\frac{4k{q}^2}{a^2}$
$(\because a=1\text{m})$
$\therefore F_{\text{net}}=4k{q}^2$

$\begin{array}{c}\text{Why this question ?}\\ \text{Tip: Charge +q at centre will experience repulsive}\\ \text{and attractive forces of same magnitude along the}\\ \text{same direction, from charges at diagonally opposite end.}\end{array}$