Physics - Electrostatics Question with Solution | TestHub

Here $\begin{array}{rl}& \frac{\left|\overrightarrow{E_A}\right|}{2}=\left|\overrightarrow{E_B}\right|=\left|\overrightarrow{E_C}\right|=\frac{\left|\overrightarrow{E_D}\right|}{2}=\left|\overrightarrow{E_E}\right|=\left|\overrightarrow{E_F}\right|=K\\ \therefore & E_O=E_A+E_D+\left(E_F+E_C\right)\cos 60^{\circ }+\left(E_B+E_C\right)\cos 60^{\circ }\\ =& 2\mathrm{K}+2\mathrm{K}+(\mathrm{K}+\mathrm{K})\times \frac{1}{2}+(\mathrm{K}+\mathrm{K})\times \frac{1}{2}=6\mathrm{K}\end{array}$
Electric potential at $O$ $V_O=\frac{1}{4\pi {\epsilon }_{0}L}[2q+q+q-q-q-2q]=0$ Potential at all points on the line PR is same not on line ST. $PR$ is perpendicular bisector (the equatorial line) for the electric dipoles $AB,FE$ and $BC$. Therefore the electric potential will be zero at any point on $PR$.