Physics - Electrostatics Question with Solution | TestHub

The force experienced by $A$ due to $B$ is $F_1=\frac{1}{4\pi {\epsilon }_0}\frac{QQ}{a^2}$ along $\overrightarrow{AB}$ (attractive) The force experienced by $A$ due to $C$ is $F_2=\frac{1}{4\pi {\epsilon }_0}\frac{Q^2}{a^2}$ along $\overrightarrow{CA}$ produced (repulsive)
Now the force ${\vec{F}}_1$ is having a component $F_1\cos 30^{\circ }$ along and the force ${\vec{F}}_2$ is having a component $F_2\cos 30^{\circ }$ along $\overrightarrow{AE},\overrightarrow{AE}$ and $\overrightarrow{AF}$ are both normal to $\overrightarrow{BC}$ but they are mutually opposite to each other. And as $\left|{\vec{F}}_1\right|=\left|{\vec{F}}_2\right|$, so the forces along $\overrightarrow{AE}$ and $\overrightarrow{AF}$ will both cancel each other. And so the force experienced by the charge at $A$ in the direction normal to $\overrightarrow{BC}$ is zero. $$ \begin{aligned}