Four equal point charges Q each are placed in the xy plane at (0, 2)… | JEE Main 2019 (10 Jan, Shift 2)

Q53JEE Main 2019 · 10 JAN · SHIFT 2Medium

Electrostatic Potential Energy

Four equal point charges Q each are placed in the xy plane at (0, 2), (4, 2), (4, –2) and (0, –2). The work required to put a fifth charge Q at the origin of the coordinate system will be -

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Answer: (d)

$W = Q\,V_{origin}$ .
Distances from origin: $(0,\pm2)$ are $2$ ; $(4,\pm2)$ are $\sqrt{20} = 2\sqrt{5}$ .
$V = \frac{Q}{4\pi\varepsilon_0}\left(\frac{2}{2} + \frac{2}{2\sqrt{5}}\right) = \frac{Q}{4\pi\varepsilon_0}\left(1 + \frac{1}{\sqrt{5}}\right)$
$W = \frac{Q^2}{4\pi\varepsilon_0}\left(1 + \frac{1}{\sqrt{5}}\right)$ , answer (d).

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Electrostatic Potential Energy question from JEE Main 2019 (10 Jan, Shift 2)

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