Question

Five-point charges, each of value $+q$ are placed on five vertices of a regular hexagon of side $L$. What is the magnitude of the force on a point charge of value $-q$ placed at the center of the hexagon?

Answer 1

Step 1: Construct the diagram

We know that a hexagon has $6$ sides.

It is given that $5$ point charges of value $+q$ are placed on five hexagon vertices.

We know that $+q$ has a radially outward electric field while $-q$ has a radially inward electric field.

Hence, from the figure, we can see that $4$ out of the $5$ force due to the charge $+q$ will cancel each other out since they are evaluated to be similar and in opposite directions.

Only the force due to charge $-q$ on one of the $+q$ will remain.

Step 2: Determine the net force

The magnitude of both the charges is $q$ , and the length of the polygon is $L$

We know that the force between two charges is given as,

$F=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{q_1{q}_2}{r^2}$

where, $q_1$ and $q_2$ are the two charges, and $r$ is the distance between them.

Thus, the net force,

$F=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{q.q}{L^2}$

$=9\times {10}^9\times \frac{q^2}{L^2}$ $\left[\because \frac{1}{4{\mathrm{\pi \epsilon }}_0}=9\times {10}^9\right]$

Hence the magnitude of the net force is $9\times {10}^9\times \frac{q^2}{L^2}$.