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Three charges $- q, + q and + q$ are situated in $X - Y$ plane at points $(0, - a), (0, 0) and (0, a)$ respectively. The potential in xy plane at a distant point r (r > > a) in a direction making an angle $θ$ from the Y-axis :

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$\frac{q}{4 {πε}_{0} r} + \frac{qasin θ}{4 {πε}_{0} r^{2}}$

$\frac{q}{4 {πε}_{0} r} + \frac{qacos θ}{4 {πε}_{0} r^{2}}$

$\frac{q}{4 {πε}_{0} r} + \frac{qacos θ}{2 {πε}_{0} r^{2}}$

$\frac{q}{4 {πε}_{0} r} + \frac{qasin θ}{2 {πε}_{0} r^{2}}$

answer is B.

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Detailed Solution

V = potential due to the point charge q at the origin+ potential due to dipole

$= \frac{Kq}{r} + \frac{KPcosθ}{r^{2}} = \frac{q}{4 π \in_{0} r} + \frac{1}{4 π \in_{0}} \frac{2 a q \cos θ}{r^{2}}$

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