# Electric potential energy and equipotential lines

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!

• 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. – secavara Feb 19 '18 at 17:39
• 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. – JEB Feb 19 '18 at 17:44
• 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. – The Photon Feb 19 '18 at 17:44

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 $$W = \int{F_{tangential} \cdot dr}$$