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. $\frac{Q}{4 \pi \varepsilon_0 a^2}$
• B. $\frac{\sqrt{2} Q}{4 \pi \varepsilon_0 a^2}$
• C. $\frac{\sqrt{3} Q}{4 \pi \varepsilon_0 a^2}$
• D. $\frac{2 Q}{4 \pi \varepsilon_0 a^2}$

Answer 1

Answer: (C)

(Regular hexagon)
In $\triangle A O M, \sin 30^{\circ}=\frac{A M}{A O}$
$\therefore A O=\frac{\frac{a}{2}}{1 / 2}=a$
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}+2 E_{A} E_{B} \cos 60^{\circ}} \left(\because E_{A}=B_{B}=E\right) $
$=\sqrt{E^{2}+E^{2}+2 E^{2} \cdot \frac{1}{2}} \left(\because \cos 60^{\circ}=\frac{1}{2}\right)$
$=E \sqrt{3}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{a^{2}} \sqrt{3} $