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!
Best Answer
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.