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Six charges are placed around a regular hexagon of side length a as shown in the figure. Five of them have charge $q$, and the remaining one has charge $x$. The perpendicular from each charge to the nearest hexagon side passes through the center $O$ of the hexagon and is bisected by the side.
Which of the following statement($s$) is(are) correct in SI units?
$(A)$ When $x=q$, the magnitude of the electric field at $O$ is zero.
$(B)$ When $x=-q$, the magnitude of the electric field at $O$ is $\frac{q}{6 \pi \epsilon_0 a^2}$.
$(C)$ When $x=2 q$, the potential at $O$ is $\frac{7 q}{4 \sqrt{3} \pi \epsilon_0 a}$.
$(D)$ When $x=-3 q$, the potential at $O$ is $\frac{3 q}{4 \sqrt{3} \pi \epsilon_0 a}$.
$A,B,C$
$A,B,D$
$A,B$
$A,C$
Solution
$(A)$ Due to symmetry $\overrightarrow{ E }_0=0$
$E _{\text {net }}=\frac{ kq }{(2 d )^2} \times 2=\frac{2 q \times 4}{4 \pi \varepsilon_0 \cdot 4 \cdot 3 a ^2}$
$=\frac{ q }{6 \pi \varepsilon_0 a ^2}$
$(C)$ $v =\frac{7 kq }{2 d }=\frac{7 q }{4 \pi \varepsilon_0 \cdot \sqrt{3} a }=\frac{7 q }{4 \sqrt{3} \pi \varepsilon_0 q }$
$(D)$ $v =\frac{2 kq }{2 d }=\frac{2 q }{4 \pi \varepsilon_0 \cdot \sqrt{3 a }}=\frac{ q }{2 \sqrt{3} \pi \varepsilon_0 q }$
