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?

AtPP, force on charge Q is zero if net electric field is zero.

Field due to dipole 1 (axial point):

$$E_1=\frac{1}{4\pi\epsilon_0}\frac{2pr}{(r^2-a^2)^2}$$

where

p=2aq

Direction is leftward (toward −q).

Field due to dipole 2 (equatorial point):

$$E_2=\frac{1}{4\pi\epsilon_0}\frac{p}{(r^2+a^2)^{3/2}}$$

Direction is rightward (opposite to dipole moment).

For zero net force,

$$E_1=E_2$$

So,

$$\frac{2r}{(r^2-a^2)^2}=\frac{1}{(r^2+a^2)^{3/2}}$$

Let

$$x=\frac{a}{r}$$

Then

$$r^2-a^2=r^2(1-x^2)$$

$$r^2+a^2=r^2(1+x^2)$$

Substitute:

$$\frac{2}{(1-x^2)^2}=\frac{1}{(1+x^2)^{3/2}}$$

Solving,

$$x=\sqrt{5}-2$$