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?

\(\begin{aligned}
& \Rightarrow \frac{1}{(r-a)^2}-\frac{1}{(r+a)^2}=\frac{2 a}{\left(a^2+r^2\right)^{3 / 2}} \\
& \frac{4 a r}{\left(r^2-a^2\right)^2}=\frac{2 a}{\left(a^2+r^2\right)^{3 / 2}} \\
& \left(r^2-a^2\right)^2=2 r\left(a^2+r^2\right)^{3 / 2} \\
& \left(1-\frac{a^2}{r^2}\right)^2=2\left(1+\frac{a^2}{r^2}\right)^{3 / 2} \\
& \left(1-x^2\right)^2=2\left(1+x^2\right)^{3 / 2}\left(x=\frac{a}{r}\right) \\
& \frac{\left(1-x^2\right)^2}{\left(1+x^2\right)^{3 / 2}}=2
\end{aligned}\)
Now for \(\mathrm{X}=3\)
We get \(\frac{64}{10 \sqrt{10}} \approx 2 \Rightarrow \frac{a}{r} \approx 3\)
[But for \(a > r\) point charge will between the dipole where \(\vec{E} \neq 0\)]