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This depends on the sign of the charge. Positive charge flows from high to low electric potential. Negative charge, alas, flows from _low_ potential to high potential. This apparent inconsistency comes from the actual fact, which is that particles tend to move toward a situation of minimized _potential energy_. That is, a charge q will move from point A to point B if U(B) < U(A). (Note that this is necessary. For a charge to spontaneously move from A to B would require its kinetic energy to be negative ... not possible.) Here's the confusion: The potential energy of a charge q in a potential V is given by U = q V. If q>0, everything's peachy: this forces V(B) < V(A) as hoped for. But say q < 0. In fact, assume q = -q0 (where q0>0). This allows us to treat the negative sign explicitly. In this case, since U = q V, we have -q0 V(B) < -q0 V(A). Since q0 > 0, we can divide without changing sense, so -V(B) < -V(A). Now we divide by -1 _and flip the sense_: V(B) > V(A). So the negative charge moves from low potential V(A) to a higher potential V(B). Ta-da!

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