Question

A test charge q0 is brought from infinity along the perpendicular bisector of an electric dipole.the work done on q0 by the electric field of the dipole is​

Answer 1

we know, potential due to an electric dipole is given by, $V=\frac{\kappa Pcos\theta}{r^2}$

where, $\theta$ is angle between point of observation and dipole, P is magnitude of dipole and r is separation between point of observation to dipole.

if, test charge $q_0$ brought from infinity along the perpendicular bisector of an electric dipole. then, $\theta$ = 90°

so, potential = $\frac{\kappa Pcos90^{\circ}}{r^2}=0$

we know, workdone = charge × potential

= $q_0\times0=0$

hence, workdone on $q_0$by the electric field of the dipole is 0.