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Physics25. Electric PotentialEquipotential Surfaces

Hello class, Professor Anderson here. Let's say we do an example problem with equipotential surfaces. Okay, so let's say that we have these surfaces, we're going to draw them, and then we want to calculate something about moving charge from one surface to another. So we're going to draw some region of space here and let's draw these equipotential surfaces. So let's say we have one like this, one like this, one like that, and one there, okay and each of these surfaces represents a voltage. Okay, and let's say that this is zero volts, 1 volt, 2 volts, 3 volts, and let's pick a few points. Let's say that we're going to look at, how about this position? We'll call that A, this position B, this position C, and this position D. Okay. Let's figure out how much work it would take to move charge between these different points. All right, and we're going to make it really simple, we're gonna say that we're gonna move one Coulomb of charge between these different points. Alright? And let's ask how much work is done? All right, let's take a look at A to B. If I move a charge from A to B it's going directly along one of those exponential surfaces, so how much work would be done in moving it from A to B? Well we know what work is. Work is equal to charge times the change in the potential. The charge here is of course one Coulomb but the change in the potential going from A to B is zero. And so there's no work moving from A to B and that's exactly what an equipotential surface means, it's free for the charge to move along an equipotential surface. All right, that one's not too bad, what about going from A to C? All right we can do that, again we know that the work is equal to Q Delta V, we have one Coulomb and we're gonna move a Delta V of 1 volt. Okay, A was that 2 volts, C is that 1 volt, and so you've got 1 Coulomb times 1 volt, and let's not worry about positive or negative here let's just worry about how much work the magnitude of it takes, and so you get one Joule. A coulomb volt is a Joule. All right, and finally let's look at the work going from D to C. Well the work going D to C is going to be Q delta v, one Coulomb times Delta V, which is now 2 volts, because they've got some 3 volts to 1 volt and so we get 2 joules. All right, now let's worry about the signs, is it positive or is it negative? One way to think about this is the following: if I have a high potential here and I have a low potential there, that's just like a whole bunch of positive charge there and a whole bunch of charge there, and so the electric field is in fact pointing in that direction, okay? Electric field remember is negative the gradient of the potential, so if the potentials increasing this way the electric field must be pointing the other way. Well now if I take it positive Q and I put it in this field, what does it want to do? Remember the force is equal to Q times the electric field, so it's gonna move that way, it will feel a positive force in that direction and so in fact these should be positive quantities. What's the work done, it's positive one Joule or D to C, it's positive 2 joules, in other words you just put the charge there and you let it go, the work done on it by the field is going to increase the energy of that particle by 2 joules, in that case it will start moving very fast. All right, equipotentials and work. Cheers!