Electric dipole

Consider an electric dipole as in the figure. There is a vertical equipotential line/surface. If I understand correctly, electric potential is the amount of work done per unit charge in bringing a charge from infinity to a distance r from a charge.

But consider moving along the equipotential line from infinity to a point along the equipotential line. Surely the work done to move the charge must be zero, since we are moving along an equipotential line. However, the electrostatic field is conservative, so work done in moving from infinity to a given point should be the same along any path. Therefore, work done in moving a charge from infinity to a point close to a dipole is zero.

This is clearly not the case, where have I gone wrong?? Thank you!

  • 1
    $\begingroup$ I think if the charges forming the dipole are fixed then the only equipotential line that you can follow all the way to infinity is the $\phi=0$ vertical line. Moving a charge between any two points in this line will require no work, including bringing it from infinity $y \rightarrow \pm \infty$ regardless of the trajectory you follow. But notice that the potential at infinity in any other direction also vanishes, which meas as long as you bring it to the vertical line following any path, the work will still be $0$ regardless and it will be consistent with the potential difference. $\endgroup$
    – secavara
    Feb 19, 2018 at 17:39
  • $\begingroup$ Adding to @secavara, let's say you're off to left at infinity, where the potential is zero, and you move to the vertical line (narrowly missing the pole): all the loops you cross are closed, so at first you cross them going against the arrows for most of the journey, and then you cross the same lines again with the arrows at your back as the "+" enters the rearview : net delta is zero for all paths. $\endgroup$
    – JEB
    Feb 19, 2018 at 17:44
  • $\begingroup$ You talk about "bringing a charge ... to a distance r from a charge". This only applies in the case with where a single point charge causes the potential (or other cases with the same symmetry). In other cases you should think about bringing a charge to a particular point in space. $\endgroup$
    – The Photon
    Feb 19, 2018 at 17:44

1 Answer 1


Equipotential lines are always at right angles with the electric field (most clearly shown in the centermost equipotential line). This implies that if a charge were to move along an equipotential line then throughout the entire journey $F_{electric} \perp dr $ and hence, $F_{electric} \cdot dr = 0$.

To move a charge along an equipotential line, you'd need to supply two forces: one to cancel out the net force from the two charges, and the other to move it along the equipotential. So: $F = -F_{electric} + F_{tangential}$. Calculating the total work done: $$W = \int{F \cdot dr} = \int{(-F_{electric} \cdot dr)} + \int{F_{tangential} \cdot dr}$$ As we previously argued, $F_{electric} \perp dr$ so $$W = \int{F_{tangential} \cdot dr} $$

which is nonzero.

EDIT: So just to clarify, the electric field's contribution to the total work is zero, but the electric field on its own will never be able to pull the particle down from infinity.

  • $\begingroup$ Ah, I see, thank you! So rather than saying the work done along an equipotential line is zero, I should rather say the work done by the electric field along an equipotential line is zero? $\endgroup$
    – martin
    Feb 19, 2018 at 18:00
  • $\begingroup$ You're welcome. Yup you got it! $\endgroup$
    – talrefae
    Feb 19, 2018 at 18:06
  • $\begingroup$ @talrefae This clarification is interesting. I guess rectilinear equipotential lines can be followed with constant speed, without the need of tangential forces, actually requiring no work at all. Curved equipotentials will require forces coming from the constraint of following a specific trajectory. But these are normal to the trajectory as well, right? Hence it would seem that they don't contribute to the work anyway. I can have tangential forces but they don't seem necessary. $\endgroup$
    – secavara
    Feb 19, 2018 at 18:20
  • $\begingroup$ @secavara You raise an interesting point, but I still think that (even in the case of rectilinear equipotentials) we would still require a tangential force to get the particle started along the path. Once we have established the initial velocity then you're 100% correct - the net work after that point can be zero. $\endgroup$
    – talrefae
    Feb 19, 2018 at 19:07

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