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Question

(C) In a regular hexagon, the distance from the centre $O$ to any vertex is equal to the side length $a$. To maximize the electric field at the centre

Vedclass · Accepted Answer

$\frac{\sqrt{3}Q}{4 \pi \epsilon _{0} a^{2}}$

Vedclass · Answer

(C) In a regular hexagon, the distance from the centre $O$ to any vertex is equal to the side length $a$. To maximize the electric field at the centre, we place the four charges at vertices $A, B, C,$ and $F$ as shown in the figure. The electric field due to charges at $F$ and $C$ are equal in magnitude and opposite in direction, so they cancel each other out $(E_F + E_C = 0)$. The net electric field at the centre $O$ is due to the charges at $A$ and $B$. The magnitude of the electric field due to each charge is $E = \frac{1}{4 \pi \epsilon_0} \frac{Q}{a^2}$. The angle between the electric field vectors $E_A$ and $E_B$ is $60^{\circ}$. The resultant electric field $E_{	ext{net}}$ is given by: $E_{	ext{net}} = \sqrt{E_A^2 + E_B^2 + 2 E_A E_B \cos 60^{\circ}}$ Since $E_A = E_B = E$, we have: $E_{	ext{net}} = \sqrt{E^2 + E^2 + 2 E^2 (1/2)} = \sqrt{3E^2} = E\sqrt{3}$ Substituting the value of $E$: $E_{	ext{net}} = \frac{\sqrt{3}Q}{4 \pi \epsilon_0 a^2}$

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

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