The point charges +q, -q, -q, +q, +Q and -q a | vedclass

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Question

(A) Let $a$ be the side length of the regular hexagon. The distance from the centre $O$ to each vertex is $a$.The electric field due to a charge $q$ a

Vedclass · Accepted Answer

$\frac{q}{3}$

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(A) Let $a$ be the side length of the regular hexagon. The distance from the centre $O$ to each vertex is $a$.The electric field due to a charge $q$ at a distance $a$ is $E = \frac{kq}{a^2}$ directed away from the charge if $q > 0$ and towards the charge if $q Let the charges be: $A(+q), B(-q), C(-q), D(+q), E(+Q), F(-q)$.Electric field at $O$ due to charges at $A$ and $D$:$E_A = \frac{kq}{a^2}$ (away from $A$, towards $D$)$E_D = \frac{kq}{a^2}$ (away from $D$, towards $A$)Since these are equal and opposite, they cancel each other out: $\vec{E}_A + \vec{E}_D = 0$.Electric field at $O$ due to charges at $B$ and $E$ (ignoring $E$ for now) and $C$ and $F$:$E_B = \frac{kq}{a^2}$ (towards $B$)$E_F = \frac{kq}{a^2}$ (towards $F$)$E_C = \frac{kq}{a^2}$ (towards $C$)Resultant field due to $B, C, F$ at $O$:By symmetry, the resultant of $E_B, E_C, E_F$ is $\frac{kq}{a^2}$ directed towards $E$.Given: $|\vec{E}_{A,B,C,D,F}| = 3 |\vec{E}_E|$$|\vec{E}_{A,B,C,D,F}| = \frac{kq}{a^2}$$|\vec{E}_E| = \frac{kQ}{a^2}$So, $\frac{kq}{a^2} = 3 \left( \frac{kQ}{a^2} \right) \implies Q = \frac{q}{3}$.

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