How is the voltage along the equipotential line between the two equal-opposite charges ZERO?
Two common definitions of voltage between points A and B: (1) The net-work per unit charge against the net-electric field required to move a (+) test-charge from A to B; (2) The net-work per unit charge that the net-electric field will do on a (+) test-charge released, initially at rest, along a path from A to B.
The numbers in the image above were computed according to the following expression:
$$
V=\frac{U}{q_0}=\frac{1}{4\pi\epsilon_0}\sum_i\frac{q_i}{r_i} \qquad \text{(potential due to a collection of point charges)}
$$
This expression yields the potential of a test-charge at a point P a distance r_i away from a collection of point-charges 1 to i – with respect to infinity. However, this representation, IF TRUE, is hardly useful and massively misleading: a (+) test-charge placed anywhere along the $V=0$ line will move leftward until it touches the negative point-charge [assuming classical mechanics, or, treat the 'point' charges as macroscopic charged spheres]. Then, clearly, the particle indeed has a electric potential energy – and thereby electric potential energy per unit charge, or voltage – yet $V=0$?
Carefully investigating definitions (1) and (2), the only way this makes sense to me is if 'potential' here is computed purely along the path traced by the $V=0$ equipotential line; the vertical electric field components cancel to zero, and no work is done by the net-electric field as long as it stays on that path. Yet, this is also not entirely true – since work IS done on the particle BY the ELECTRIC FIELD, horizontally. A point charge will not move along the $V=0$ line unless an external force that is not the field impresses a force on the particle equal-opposite to that of the field.
I cannot think of a reference where the points along the $V=0$ line have zero potential. This holds even if we DEFINE the $V=0$ line as the zero voltage reference, because of a contradiction: the horizontal component of the net-electric field varies vertically, and no matter what point we set on the $V=0$ line as a reference, the points above and below it will have different potentials.
What's going on?
Best Answer
You are correct in your reasoning. However, that line does have a voltage equal to $0$. The trick comes in from the fact that $V=0$ doesn't mean $\vec{F}=\vec{0}$. The force depends on the change in the potential, and, as you noted, it is changing; the potential just happens to have a value of 0 along that line. Think of it like an inland hill, where the bottom of the hill happens to below sea level but the top of the hill is above sea level. Just because the altitude is $0$ in the middle of the hill doesn't mean a ball won't roll right past it all the way to the bottom.
Electric potential works the same way as height does in the context of gravity. The electric force will always accelerate positive charges toward lower and lower potentials. The reason force doesn't have to be $0$ when potential is $0$ is because potential can and does go negative. The potential is $0$ on that line, but it's negative left of the line, so a positive test charge will be accelerated to the left.
You're right in thinking that potential is all relative and that saying it has a potential of $0$ is arbitrary. The value of potential is irrelevant; only change in potential matters. A very convenient convention, though, is to define the potential at $r=\infty$ to be $0$. That's where the equation you have comes in, and when we use this "gauge", the potential on that line can be found to be $0$.
That line, since it has the same potential ($0$) along the whole thing, is an equipotential line. The point of the equipotential line is that if we take a test charge, it doesn't require work to move it along the line, i.e. vertically in the picture. You are correct that if the particle were to move horizontally, then there would be work done on it by the field. But it's not moving horizontally along the line. An equipotential line is always perpendicular to the electric field lines for this reason, which the line in the picture clearly is. Work is defined as $W=\vec{F}\dot{}\vec{d}$, and since $\vec{F}\perp{}\vec{d}$ along an equipotential, $W=0$.