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(C) Let the charges at $A, B,$ and $C$ be $+Q, -Q,$ and $+Q$ respectively. The distance between any two vertices is $a$.The force on charge at $A$ due

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Zero

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(C) Let the charges at $A, B,$ and $C$ be $+Q, -Q,$ and $+Q$ respectively. The distance between any two vertices is $a$.The force on charge at $A$ due to charge at $B$ is $F_B = \frac{1}{4\pi \varepsilon_0} \frac{Q^2}{a^2}$ (attractive,directed towards $B$).The force on charge at $A$ due to charge at $C$ is $F_C = \frac{1}{4\pi \varepsilon_0} \frac{Q^2}{a^2}$ (repulsive,directed away from $C$).Resolving these forces into components parallel and normal to $BC$:$1$. The components of $F_B$ and $F_C$ parallel to $BC$ are $F_B \cos 60^\circ$ and $F_C \cos 60^\circ$ respectively,both directed towards the left.$2$. The components of $F_B$ and $F_C$ normal to $BC$ are $F_B \sin 60^\circ$ (downwards) and $F_C \sin 60^\circ$ (upwards) respectively.Since $|F_B| = |F_C|$,the net force in the direction normal to $BC$ is $F_C \sin 60^\circ - F_B \sin 60^\circ = 0$.

Three charges are placed at the vertices of an equilateral triangle of side '$a$' as shown in the figure. The force experienced by the charge placed at the vertex $A$ in a direction normal to $BC$ is

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