Question 245 of 311

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 $2 a$. For the charge particle at P not to experience any net force, which of the following correctly describes the situation?

A$\frac{a}{r} \sim 10$B$\frac{a}{r} \sim 20$C$\frac{a}{r} \sim 0.5$D$\frac{a}{r} \sim 3$

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Answer: (D) $\frac{a}{r} \sim 3$

$\begin{aligned}
& \Rightarrow \frac{1}{(r-a)^2}-\frac{1}{(r+a)^2}=\frac{2 a}{\left(a^2+r^2\right)^{3 / 2}} \\
& \frac{4 a r}{\left(r^2-a^2\right)^2}=\frac{2 a}{\left(a^2+r^2\right)^{3 / 2}} \\
& \left(r^2-a^2\right)^2=2 r\left(a^2+r^2\right)^{3 / 2} \\
& \left(1-\frac{a^2}{r^2}\right)^2=2\left(1+\frac{a^2}{r^2}\right)^{3 / 2} \\
& \left(1-x^2\right)^2=2\left(1+x^2\right)^{3 / 2}\left(x=\frac{a}{r}\right) \\
& \frac{\left(1-x^2\right)^2}{\left(1+x^2\right)^{3 / 2}}=2
\end{aligned}$

Now for $\mathrm{X}=3$
We get $\frac{64}{10 \sqrt{10}} \approx 2 \Rightarrow \frac{a}{r} \approx 3$
[But for $a \gt r$ point charge will between the dipole where $\vec{E} \neq 0$]