Six point charges are kept at the vertices of a regular hexagon of side \(\mathrm{L}\) and centre \(O\), as shown in the figure.
Given that \(K=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{L^{2}}\), which of the following statement(s) is (are) correct?

Here
\(\frac{\left|\overrightarrow{E_{A}}\right|}{2}=\left|\overrightarrow{E_{B}}\right|=\left|\overrightarrow{E_{C}}\right|=\frac{\left|\overrightarrow{E_{D}}\right|}{2}=\left|\overrightarrow{E_{E}}\right|\) \(=\left|\overrightarrow{E_{F}}\right|=K \)
\( \therefore E_{O}=E_{A}+E_{D}+\left(E_{F}+E_{C}\right) \cos 60^{\circ}+\) \(\left(E_{B}+E_{C}\right) \cos 60^{\circ} \)
\( =2 \mathrm{~K}+2 \mathrm{~K}+(\mathrm{K}+\mathrm{K}) \times \frac{1}{2}+(\mathrm{K}+\mathrm{K})\) \(\times \frac{1}{2}=6 \mathrm{~K}\)

Electric potential at \(O\)
\(V_{O}=\frac{1}{4 \pi \varepsilon_{0} L}[2 q+q+q-q-q-2 q]=0\)
Potential at all points on the line PR is same not on line ST. \(P R\) is perpendicular bisector (the equatorial line) for the electric dipoles \(A B, F E\) and \(B C\). Therefore the electric potential will be zero at any point on \(P R\).