Question

Six point charges are placed at the vertices of a hexagon of side $a$ as shown in the figure.

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

• A. $\frac{3kQq}{a^2}$
• B. $\frac{2kQq}{a^2}$
• C. $\frac{6kQq}{a^2}$
• D. zero

Answer 1

The correct option is D zero

As we know that like charges repels each other and unlike charges attracts. So, the force experienced by the charge $+Q$ at the centre of the hexagon is shown in the figure.

So, force experienced by $+Q$ due to vertex charges is


For hexagon,
side of hexagon = distance between centre and vertex $=a$

From Coulomb's law,

$|F_{01}|=\frac{kQq}{a^2}$

Since, magnitude of charges and distance between them is same for each pair.

$\therefore |F_{01}|=|F_{02}|=|F_{03}|=|F_{04}|=|F_{05}|=|F_{06}|$

So, $+Q$ charge experienced three forces of magnitude $\frac{2kQq}{a^2}$ in three different direction at angle ${120}^{\circ }$.

$\therefore$ Net force experienced by $+Q$ charge is zero.

Hence, option $\left(d\right)$ is the correct answer.


$\begin{array}{c}\text{Why this question ?}\\ \text{Tip: In hexagon, force experienced by charge placed}\\ \text{at centre will be zero, if}\\ \text{(a) charges at vertices of hexagon have equal in magnitude}\\ \text{and same in nature.}\\ \text{(b) equal charge of different nature placed at}\\ \text{alternate vertices of hexagon.}\\ \end{array}$