# Direction of Voltage and E field

Ok, I'm a little confused about how Voltage works. So from my understanding is that Voltage is the amount of potential energy stored in one unit charge. Simple enough. So if you have an electric field from a positively charged ceiling to a negatively charged floor (with the same magnitude), there must be a point where the voltage is zero. So does this mean if you place a positive charge right at that point, it will not move? I'm very confused, please help.

berkeman
Mentor
Ok, I'm a little confused about how Voltage works. So from my understanding is that Voltage is the amount of potential energy stored in one unit charge. Simple enough. So if you have an electric field from a positively charged ceiling to a negatively charged floor (with the same magnitude), there must be a point where the voltage is zero. So does this mean if you place a positive charge right at that point, it will not move? I'm very confused, please help.

As you say, voltage is potential. You can define the zero voltage equipotential surface anywhere you want. You could define the voltage of the ceiling to be 0V, in which case the floor would have a negative voltage. Or you could define the floor as being at 0V, in which case the ceiling would have a positive voltage. There is an E field everywhere in the room, and it is basically uniform, so the equipotential surfaces are horizontal, and a test charge will feel the same force everywhere in the room.

jtbell
Mentor
Also note that the electric field and electric force on a charge at a certain location depend on how the the potential (voltage) varies with respect to position, at that location, not on the actual value of the potential. In a one-dimensional situation like you describe, the electric field is simply the (negative) gradient (slope) of the potential versus y:

$$E = - \frac{dV}{dy}$$