Six charges $+q,-q,+q,-q,+q$ and $-q$ are fixed at the corners of a hexagon of side $d$ as shown in the…

Six charges $+q,-q,+q,-q,+q$ and $-q$ are fixed at the corners of a hexagon of side $d$ as shown in the figure. The work done in bringing a charge $q_0$ to the centre of the hexagon from infinity is $\left(\varepsilon_0-\right.$ permittivity of free space):
  1. Zero
  2. $\frac{-q^2}{4 \pi \varepsilon_0 d}$
  3. $\frac{-q^2}{4 \pi \varepsilon_0 d}\left(3-\frac{1}{\sqrt{2}}\right)$
  4. $\frac{-q^2}{4 \pi \varepsilon_0 d}\left(6-\frac{1}{\sqrt{2}}\right)$

Solution

Work done $=$ Change in potential energy $\begin{aligned} & \mathrm{W}=\mathrm{U}_f-\mathrm{U}_i \\ & \mathrm{~W}=0 \end{aligned}$ Hence, $W=0$

Asked in: NEET 2022 (Phase 2)

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