A point particle of charge Q is located at P along the axis of an electric dipole 1 at a distance r as shown in the figure. The point P is also on the equatorial plane of a second electric dipole 2 at a distance r. The dipoles are made of opposite charge q separated by a distance 2a. For the charge particle at P not to experience any net force, which of the following correctly describes the situation?

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$\frac{a}{r} ~ 3$

$\frac{a}{r} ~ 10$

$\frac{a}{r} ~ 0.5$

$\frac{a}{r} ~ 20$

answer is D.

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Detailed Solution

$\begin{array}{l} \frac{kq}{(r - a)^{2}} = \frac{kq}{(r + a)^{2}} + \frac{2 kq}{(r^{2} + a^{2})} \cos θ \\ \frac{1}{(r - a)^{2}} = \frac{1}{(r + a)^{2}} + \frac{2 a}{{(r^{2} + a^{2})}^{\frac{3}{2}}} \\ \frac{1}{(r - a)^{2}} - \frac{1}{(r + a)^{2}} = \frac{2 a}{{(r^{2} + a^{2})}^{\frac{3}{2}}} \\ \frac{4 ra}{{(r^{2} - a^{2})}^{2}} = \frac{2 a}{{(r^{2} + a^{2})}^{\frac{3}{2}}} \end{array}$
$\begin{array}{l} \Rightarrow \frac{2 r}{{(r^{2} - a^{2})}^{2}} = \frac{1}{{(r^{2} + a^{2})}^{\frac{3}{2}}} \\ \frac{4 r^{2}}{{(r^{2} - a^{2})}^{4}} = \frac{1}{{(r^{2} + a^{2})}^{3}} \\ \Rightarrow 4 r^{2} {(r^{2} + a^{2})}^{3} = {(r^{2} - a^{2})}^{4} \\ 4 r^{8} {(1 + \frac{a^{2}}{r^{2}})}^{3} = r^{8} {(1 - \frac{a^{2}}{r^{2}})}^{4} \\ 4 {(1 + \frac{a^{2}}{r^{2}})}^{3} = {(1 - \frac{a^{2}}{r^{2}})}^{4} \end{array}$
by solving , we are getting a/r $\approx$ 3.