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

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Question

(A) Let $r$ be the distance from each vertex to the centre $O$. The electric field at $O$ due to a charge $q$ at a distance $r$ is $E = \frac{kq}{r^2}

Vedclass · Accepted Answer

$\frac{+q}{3}$

Vedclass · Answer

(A) Let $r$ be the distance from each vertex to the centre $O$. The electric field at $O$ due to a charge $q$ at a distance $r$ is $E = \frac{kq}{r^2}$.$1$. The charges at $A (+q)$ and $D (+q)$ produce electric fields at $O$ that are equal in magnitude but opposite in direction. Thus, they cancel each other.$2$. The charges at $F (-q)$ and $C (-q)$ produce electric fields at $O$ that are equal in magnitude but opposite in direction. Thus, they cancel each other.$3$. The only remaining charge from the set ${A, B, C, D, F}$ is the charge $-q$ at vertex $B$. The electric field at $O$ due to this charge is $E_{net} = \frac{kq}{r^2}$ directed towards $B$.$4$. The electric field at $O$ due to charge $+Q$ at $E$ is $E_Q = \frac{kQ}{r^2}$ directed away from $E$.$5$. According to the problem, $E_{net} = 3 E_Q$.$6$. Substituting the values: $\frac{kq}{r^2} = 3 \left( \frac{kQ}{r^2} \right)$.$7$. Solving for $Q$, we get $Q = \frac{q}{3}$.

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