Six charges are placed around a regular hexagon of side length a as...

The given arrangements of the charges also form a regular hexagon with side length equal to& $\sqrt{3}a$

When $x = q$, the situation is symmetric

→ Electric field at $O$ would be zero.

→ \left(A\right)$is correct.</p><p>When$x = - q$, we can think of$x$as$q + \left(-\right)\left(2q\right)$. Therefore, the magnitude of the electric field will be equivalent to that if only charge of$- 2q$is kept at the the bottom point. Hence,</p><p>${E}_{O} = \dfrac{1}{4\pi{ϵ}_{0}}\left( 2q\right)\left(\sqrt{3}a\right)2= \dfrac{1}{4\pi{ϵ}_{0}}\dfrac{2q}{3a^{2}} = \dfrac{q}{6\pi{ϵ}_{0}a^{2}}$</p><p> → \left(B\right)$ is correct

For $x = 2q$, potential at $O$ is

${V}_{O} = 5 × \dfrac{1}{4\pi{ϵ}_{0}}\dfrac{q}{\sqrt{3}a} + \dfrac{1}{4\pi{ϵ}_{0}}\dfrac{2q}{\sqrt{3}a} = \dfrac{7q}{4\sqrt{3}\pi{ϵ}_{0}a}$

→ \left(C\right) is correct

For& $x = - 3q,$

${V}_{O} = 5 ×$\dfrac{1}{4\pi{ϵ}_{0}}\dfrac{q}{\sqrt{3}a} + \dfrac{1}{4\pi{ϵ}_{0}}\dfrac{- 3q}{\sqrt{3}a} = \dfrac{q}{2\sqrt{3}\pi{ϵ}_{0}a}

→ \left( D\right)$is not correct.</p>$