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At the point horizontally across and equidistant from the centers of the two charges (also oriented horizontally), what is the electric potential? At that point, the electric field of the first charge cancels with that from the second charge, so there is no net electric field. An charge placed at that point will not move. Does this make the electric potential 0? But calculations indicate, with $V_{net}=\frac{kQ}{r}+\frac{kQ}{r}$ that the potential is certainly not zero, but double the potential from each charge. If the potential at the middle were larger than the potential outside, would a test charge not be thrust outside? Yet there is no net electric field there; in terms of forces, the test charge will remain in the center.

Also, in the diagram below, why do the potential lines criss cross at that middle point? What does this signify, and why does it occur?

enter image description here
(image taken from: http://www.physics.sjsu.edu/becker/physics51/images/24_17%20Equipotential_surfaces_and_E.jpg)

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The electric field shows the gradient (slope and direction of change) of the potential. If the electric field is high magnitude, the potential is changing quickly with a change in position. If its magnitude is small, the potential is changing slowly with a change in position.

If the electric field is zero, the potential is either at a maximum or a minimum. This means a third particle can be at either a stable or unstable equilibrium (let's fix the positions of the first 2 charges). The geometry you have given creates what we call a saddle point. This means the potential is at a maximum in one direction and a minimum in another.

For a third charge with the same sign as the first two at the zero field point, the third will be stable in the "horizontal" direction and unstable in the "perpendicular" direction. For an opposite sign third charge, it will be unstable in the "horizontal" direction but stable in the "perpendicular".

As @Paul says, it's okay for equipotential surfaces to cross each other, but E-field lines can't cross (they can converge and diverge).

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Potential is not same as electric field,electric field is zero doesn't mean potential is zero too. your calculation is right,total potential is double the potential of each charge. Edit:For the 2nd part of your question ,there is nothing wrong in potential surfaces criss crossing like that(but electric field lines shouldnt criss cross like that).

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  • $\begingroup$ But won’t a test charge move from an area of high potential to low potential? Yet in terms of forces, if there is no electric field, the charge won’t move. $\endgroup$ – lightweaver Jul 22 '15 at 2:46
  • $\begingroup$ Imagine a ball at the top of a flat hill. It won't move. But won't it want to move from high height to low height? Yes. These statements don't conflict. $\endgroup$ – knzhou Jul 22 '15 at 3:07
  • $\begingroup$ @stygian : electric potential at that point is how much energy needed to bring a unit charge to that point from infinity (clearly it is not zero,you need energy to bring the unit charge to the middle point),but electric field can be zero. also see the comment by Kevin. $\endgroup$ – Paul Jul 22 '15 at 3:17

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