Question

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

• A. $\frac{1}{\pi {\epsilon }_0}{\left(\frac{q}{L}\right)}^2$
• B. $\frac{2}{\pi {\epsilon }_0}{\left(\frac{q}{L}\right)}^2$
• C. $\frac{1}{2\pi {\epsilon }_0}{\left(\frac{q}{L}\right)}^2$
• D. $\frac{1}{4\pi {\epsilon }_0}{\left(\frac{q}{L}\right)}^2$

Answer 1

Answer: (D)

If there had been a sixth charge $+q$ at the remaining vertex of hexagon, force due to all the six charges on $-q$ at $O$ will be zero.
Now if $f$ is the force due to the sixth charge and $F$ due to the remaining five charges, then
$\vec{F}+\vec{f}=0,\text{ i.e., }\vec{F}=-\vec{f}$
$|\vec{F}|=|\vec{f}|=\frac{1}{4\pi {\epsilon }_0}\frac{q\times q}{L^2}=\frac{1}{4\pi {\epsilon }_0}{\left(\frac{q}{L}\right)}^2$