Question

If we double the distance , by what factor is the force on the point charge due to the dipole reduced

Answer 1

Answer:

$E'_{dipole}=\frac{1}{8} E_{dipole}$

Explanation:

The magnitude of the electric force exerted by a dipole on the point charge is given by:

$F=qE_{dipole}$

E is the electric field of the dipole, which are defined as:

$E_{dipole}\propto\frac{1}{d^3}$

We have $d'=2d$:

$E'_{dipole}\propto \frac{1}{d'^3}\\E'_{dipole}\propto \frac{1}{(2d)^3}\\E'_{dipole}\propto \frac{1}{8}\frac{1}{d^3}\\E'_{dipole}= \frac{1}{8} E_{dipole}$