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Anderson Video - Work to Move Charge Across Equipotential Surfaces

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