Physics - Electrostatics Question with Solution | TestHub

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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

$\Rightarrow$ Electric field at $O$ would be zero.

$\Rightarrow \left(A\right)$ is correct.

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,

$E_O=\frac{1}{4\pi {ϵ}_0}\frac{\left(2q\right)}{{\left(\sqrt{3}a\right)}^2}$$=\frac{1}{4\pi {ϵ}_0}\frac{2q}{3a^2}=\frac{q}{6\pi {ϵ}_0{a}^2}$

$\Rightarrow \left(B\right)$ is correct

For $x=2q$, potential at $O$ is

$V_O=5\times \frac{1}{4\pi {ϵ}_0}\frac{q}{\sqrt{3}a}+\frac{1}{4\pi {ϵ}_0}\frac{2q}{\sqrt{3}a}=\frac{7q}{4\sqrt{3}\pi {ϵ}_{0}a}$

$\Rightarrow \left(C\right)$ is correct

For $x=-3q,$

$V_O=5\times \frac{1}{4\pi {ϵ}_0}\frac{q}{\sqrt{3}a}+\frac{1}{4\pi {ϵ}_0}\frac{-3q}{\sqrt{3}a}=\frac{q}{2\sqrt{3}\pi {ϵ}_{0}a}$

$\Rightarrow$$\left(D\right)$is not correct.