Six point charges are kept at the vertices of a regular hexagon of side $\mathrm{L}$ and centre $O$, as…
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?
The electric field at $O$ is $6 \mathrm{~K}$ along $O D$
The potential at $O$ is zero
The potential at all points on the line $P R$ is same
The potential at all points on the line $S T$ is same
Solution
Here
$\begin{aligned}
& \frac{\left|\overrightarrow{E_{A}}\right|}{2}=\left|\overrightarrow{E_{B}}\right|=\left|\overrightarrow{E_{C}}\right|=\frac{\left|\overrightarrow{E_{D}}\right|}{2}=\left|\overrightarrow{E_{E}}\right|=\left|\overrightarrow{E_{F}}\right|=K \\
\therefore & E_{O}=E_{A}+E_{D}+\left(E_{F}+E_{C}\right) \cos 60^{\circ}+\left(E_{B}+E_{C}\right) \cos 60^{\circ} \\
=& 2 \mathrm{~K}+2 \mathrm{~K}+(\mathrm{K}+\mathrm{K}) \times \frac{1}{2}+(\mathrm{K}+\mathrm{K}) \times \frac{1}{2}=6 \mathrm{~K}
\end{aligned}$
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 $P R$.
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