Skip to main content
Pearson+ LogoPearson+ Logo
Start typing, then use the up and down arrows to select an option from the list.

Anderson Video - Equipotential Surface due to a Point Charge

Professor Anderson
Was this helpful?
Okay, potential that we just talked about is a lot like altitude. Right, we made the analogy between gravity and electric potential and in gravity it was how high you were above the Earth. And so when you think about a map a lot of times if you're looking at a topological map it shows you altitude lines. It shows you where the mountains are and then it shows you altitude lines around those mountains. Well we can do the same thing for electric potential lines and those things are called equipotential surfaces. So, before we do that, we need to know a little something about the potential of a point charge. Okay, a point charge has a potential that looks like the following. V is equal to K Q over R. Now this we're not going to prove to you you can do it when you get to calculus and you can use some calculus to to show how this all works but this is just a definition you could just accept it as a definition. What it means is if I have a point charge Q sitting right there and I'm looking for the potential here a distance R away the potential is equal to KQ over R. So if it's a positive charge then we have a positive potential. And that potential is going to drop off as a function of R how far away from the charge you are. So it'll look like that. Positive charge and therefore V equals K Q over R. And we'll put the Q in magnitudes there because if it's a negative charge then the potential becomes negative. And so now it looks like this. That's a minus Q. V is equal to minus KQ over R. So as you go further away from a point charge the potential drops off and in fact if you go all the way out to infinity the potential goes to zero. Okay, R goes to infinity it's in the denominator and so the potential would go to zero. So R equals infinity is our reference point. That's where potential is equal to zero. But let's say we do the following questions. We just showed you what the potential is here V. It's KQ over R. But what about the potential here? Is that gonna be the same thing? Is that going to be KQ over R? Or is that going to be negative KQ over R? What do you guys think? Well, it kind of goes back to the question of what is this potential? Is potential a scalar or a vector? What do you guys think? Is this thing a scalar or a vector? Well, do you see any arrows anywhere? I don't see any arrows on top of anything right? There's nothing on top of V, there's no I hats, J hats, or K hats, or R hats or anything over here on the right side, so this is in fact a scalar quantity. It is not a vector. There's no direction associated with V. It's just a number. It's a hundred volts. It is a thousand volts, it's just a number. So this is not right, it's not negative KQ over R. It is in fact also positive KQ over R. And now by symmetry, I could put my point anywhere around this thing, and if I'm always at the same distance R away from it, then it's always gonna have the exact same potential, and this is the idea with an equipotential surface. It's where the potential is always equal, it's always the exact same thing. So for a point charge Q, I have an equipotential surface like that. And I can draw another one if I like. It's gonna have a different value but it is certainly equal all the way around and this is like your topological map, right if Q was the mountaintop this would be the next elevation step down this would be the next elevation step down from there and so forth. But if you walk around along that dashed line you're always at the same elevation. If you walk around these dashed lines you're always at the same potential. The E-field of course is always perpendicular to those. We know that that E goes radially outward and so the E field is always perpendicular to the equipotential surface. Surfaces of constant V that's what an equipotential surface is are always perpendicular to E. So let's take that under advisement and see if we can draw the equipotential surface for a dipole.