If a charge is brought from some distance to a point which lies on the equatorial line of an electric dipole, the work done is 0 and so is the electric potential. But how should I imagine this?

  • 1
    $\begingroup$ You should imagine it as a symmetry plane where (+) and (-) cancel, leaving you to roam freely. $\endgroup$
    – Jon Custer
    Commented Jul 20, 2021 at 12:46

1 Answer 1


There are several ways to think about this. Maybe you find the case of gravity more intuitive.

In that case, we can imagine the dipole as a hill (positive charge) and a valley (negative charge) lying next to each other. Now, if you climb the hill, you will gain potential energy. If you go down to the valley, you would lose potential energy.

However, if there are no cliffs (discontinuities), there must exist a path in which you don't have to move either upwards or downwards (the mean value theorem). This is the equatorial line. If you are on this path, you have not gained or lost any potential energy. Hence, you spent a net zero amount of work.

See this image for an example of such a path.

  • $\begingroup$ Thank you, it helped but can u please share a link or an image to visualize this. So that I can view that line or area where there are no ups or downs in curve $\endgroup$ Commented Jul 20, 2021 at 11:30
  • $\begingroup$ But when the charge is at equatorial line, it is in electric field which is in opposite direction of dipole then how can it stay or roam freely. $\endgroup$ Commented Jul 21, 2021 at 3:31
  • $\begingroup$ You are correct. There is an electric field perpendicular to the equipotential line. It will act with a force on the charge and it can not "stay" or "roam freely" in any direction. However, if the charge moves along the line, then this does not require any work. In such a case, the displacement is perpendicular to the force. $\endgroup$
    – B. Brekke
    Commented Jul 21, 2021 at 7:04
  • $\begingroup$ But in that case the charge will not reach that point where I want it to be, I have to apply a variable force opposite to the direction of field. $\endgroup$ Commented Jul 21, 2021 at 9:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.