Question

Four equal charges of value $+Q$ are placed at any four vertices of a regular hexagon of side ‘$a$’. By suitably choosing the vertices, what can be the maximum possible magnitude of electric field at the centre of the hexagon ?

• A. $\frac{Q}{4\pi {\epsilon }_0{a}^2}$
• B. $\frac{\sqrt{2}Q}{4\pi {\epsilon }_0{a}^2}$
• C. $\frac{\sqrt{3}Q}{4\pi {\epsilon }_0{a}^2}$
• D. $\frac{2Q}{4\pi {\epsilon }_0{a}^2}$

Answer 1

Answer: (C)

(Regular hexagon)
In $△AOM,\sin 30^{\circ }=\frac{AM}{AO}$
$\therefore AO=\frac{\frac{a}{2}}{1/2}=a$
For maximum electric field at centre $O$ charges should be placed at $F,A,B$ and $C$.
$\because$ Electric field due to charges at $F$ and $C$ is equal and opposite at $O$.
$\therefore$ Net electric field at centre $O$ due to charges at $A$ and $B$.
Angle between $E_A$ and $E_B$ is $60^{\circ }$.
$\therefore E_{\text{net }}$ at $O$,
$E_{\text{net }}=\sqrt{E_A^2+E_B^2+2E_A{E}_B\cos 60^{\circ }}\left(\because E_A=B_B=E\right)$
$=\sqrt{E^2+E^2+2E^2\cdot \frac{1}{2}}\left(\because \cos 60^{\circ }=\frac{1}{2}\right)$
$=E\sqrt{3}=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Q}{a^2}\sqrt{3}$