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?

Question

Aicharya · Accepted Answer

$\begin{aligned} & \Rightarrow \frac{1}{(r-a)^2}-\frac{1}{(r+a)^2}=\frac{2 a}{\left(a^2+r^2ight)^{3 / 2}} \ & \frac{4 a r}{\left(r^2-a^2ight)^2}=\frac{2 a}{\left(a^2+r^2ight)^{3 / 2}} \ & \left(r^2-a^2ight)^2=2 r\left(a^2+r^2ight)^{3 / 2} \ & \left(1-\frac{a^2}{r^2}ight)^2=2\left(1+\frac{a^2}{r^2}ight)^{3 / 2} \ & \left(1-x^2ight)^2=2\left(1+x^2ight)^{3 / 2}\left(x=\frac{a}{i}ight) \ & \frac{\left(1-x^2ight)^2}{\left(1+x^2ight)^{3 / 2}}=2 \end{aligned}$ 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 $\vec{E} 
eq 0$ ]

A point particle of charge $Q$ is located at $P$ along the axis of an electric dipole 1 at a distance $r$ as…

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