A two point charges 4q and -q are fixed on the x-axis at x = -d/2 and… | JEE Main 2020 (4 Sep, Shift 1)

Q47JEE Main 2020 · 4 SEP · SHIFT 1Medium

Electrostatic Potential Energy

A two point charges 4q and -q are fixed on the x-axis at x = $-\frac{d}{2}$ and x = $\frac{d}{2}$ respectively. If a third point charge 'q' is taken from the origin to x = d along the semicircle as shown in the figure, the energy of the charge will:

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Answer: (d)

At origin: $V_O = \frac{k\,4q}{d/2} - \frac{kq}{d/2} = \frac{6kq}{d}$
At $x = d$ : $V = \frac{k\,4q}{3d/2} - \frac{kq}{d/2} = \frac{2kq}{3d}$
$\Delta U = q(V - V_O) = -\frac{16kq^2}{3d} = -\frac{4q^2}{3\pi\varepsilon_0 d}$ (path does not matter)
Energy decreases by $\frac{4q^2}{3\pi\varepsilon_0 d}$ , answer (d).

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