CALC. A metal sphere with radius r_a is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius r_b. There is charge +q on the inner sphere and charge -q on the outer spherical shell. (a) Calculate the potential V(r) for (i) r < r_a; (ii) r_a < r < r_b; (iii) r > r_b. (Hint: The net potential is the sum of the potentials due to the individual spheres.) Take V to be zero when r is infinite. (b) Show that the potential of the inner sphere with respect to the outer is V_ab=q/(4πϵ_0 ) (1/r_a -1/r_b). (c) Use E_r=-∂V/∂r=(-∂/∂r) (1/(4πϵ_0 ) q/r)=[1/(4πϵ_0 )](q/r^2) and the result from part (a) to show that the electric field at any point between the spheres has magnitude E(r)=[V_ab/(1/r_a -1/r_b )](1/r^2) (d) Use E_r = [1/(4πϵ_0 )](q/r^2) and the result from part (a) to find the electric field at a point outside the larger sphere at a distance r from the center, where r > r_b. (e) Suppose the charge on the outer sphere is not -q but a negative charge of different magnitude, say -Q. Show that the answers for parts (b) and (c) are the same as before but the answer for part (d) is different.

Video transcript

Hey everyone. So this problem is working with electric potential. Let's see what they're asking from us. I'm just gonna draw a diagram as we work through the problem. So we have a solid conducting sphere of Radius R. one that carries a charge of minus Q. It's inside a hollowed sphere of Radius R. two that carries a charge of positive Q. Um We're told that, you know, insulating or non conductive supports hold the inner sphere in place and they're asking us to what would happen if we change the charge of the hollow sphere from positive lower case Q. Two, positive uppercase Q. So we're looking at two different scenarios. The first scenario, scenario one is where the charge in the hollow sphere is positive lowercase lowercase Q. And scenario two is where the charge of the Hollywood sphere is positive uppercase Q. All right. So um they're asking us to look at what will happen in three different um Scenario three different problems between these two scenarios if changing the charge will change the final value or if everything remains the same. So Part one is the potential of the solid sphere relative to the hollow sphere. Part two is the electric field magnitude between the two spheres and part three is the electric field outside of the hollow sphere. So they're asking us a lot, they're giving us a lot of information but we will break it down and take this step by step to get to our answers. So the first thing that we need to remember is that the electric potential inside a sphere of radius of a given radius is K. Q over R. Where r is that given radius. The electric potential outside. Actually write that as inside you'd like to potential outside of a sphere is given as que que over our. So that small R where R is the distance measured from the center center of the sphere. But we know that it's outside of the sphere. So as the distance from the sphere approaches infinity the potential goes to zero and that's what they're telling us here in this problem. Okay um in the problem we know that inside of the first sphere are uppercase R equals R. One. Inside the outer sphere it equals R. Two. So part one is asking us for the potential of the solid sphere or sphere, one relative to the hollow sphere or sphere to. So V 12 is given as V. I mine. Sorry V one minus V two. So we have to solve for both v. one and V two for when the outer charge is positive. Lower case Q. And then the outer charges positive uppercase Q. So let's just work on scenario one first. So B one is the sum of the potentials that the point. Um the surfaces that enclose the point. So there's two potential, there's two surfaces here. The inner surface of that of the solid sphere and then the surface of the hollow sphere. So we're looking at que the charge And the inter sphere is -6 Radius is R. one plus que que over R. Two. We can factor out que cues And we get one over R two minus one over R. One. So that is B one. In scenario one V two it's going to be K negative Q. And now here V two we are outside of the first sphere and we can um recalled that at the surface of the shell of the outer shell are two equals R. So for the inner sphere of the radius is going to be our and then for the outer sphere the radius is also going to be our So these two cancel out because of the signs of the charge. So V two is 0. So in scenario one V 12 is just V one minus V two. V two is zero. So it's just be one. So that's K. Q. One over R two minus one over R. One. We'll call that scenario one. So let's look at scenario to where uh the charge changes from positive Q. Positive lower case Q two, positive upper upper case Q. So how does that change things? So V one is going to be K negative Q over R. One. That doesn't change. And then the charge For the second sphere is gonna be positive Q. There's nothing here that you can really factor V two. So remember we said that at the surface are two equals R. So we're actually going to write these terms in these terms in terms of our two instead of just our so we have k minus Q over R plus k positive uppercase Q Over R two. Because both of those radius is R two equals R For the V two term. So now combine these V 12 equals V one minus V two. So we have K Q over R two minus K. Lower case Q over R one -KQ over R two plus minus a negative is a positive K Q over R two. Those terms canceled. We are left with B12 equals K Q over R two minus K Q over R one. We factor out the cake us One over R 2 -1 over R one. And so we can see here. That's so this is scenario two And it is the same as scenario one. So it does not change changing the the charge from one positive charge to a different magnitude. Still a positive charge does not change the potential. So looking at our answers, we can eliminate a couple of these right off the bat And then we will take it into part two. So part two of this question is asking about the electric field magnitude between the two squares between the between the two spheres. So this is where they're giving us this derivative here for E of R D M R equals negative D V of r D R. So we need to find V of R. For both scenarios. Then we'll take the derivatives and see to find the magnitude of the electric sphere, of the magnitude of the electric field. And so part two is asking us for between the two spheres. Okay, so VFR again, it's just the sum of the two electric potentials. So we have the one plus V two and so this is gonna be very similar to what we just did in part one. So because we are between the two spheres, this first radius is R and then we have cake. You are too. So KQ R K Q times one over R two minus one over R is going to be our V sub R in scenario one. and then in scenario two where we're changing from positive lower case Q two. Positive uppercase Q. The of our Yeah, B one Plus B two is K times negative Q over R. That's okay. Uppercase Q over R two. So there's nothing to factor out or or simplify that one. So we're just gonna go into the derivatives here. So negative D B R D. R. For scenario one looks like this plug in our B of our We have KQ Times 1 -2. One over R 2 -1 over R. We'll pull out the key Q here and then the derivative of one over R two is a constant. One over R two is a constant. So the derivative of a constant is zero. So we know that this looks like minus K Q times zero plus D. D are - over R. And then that further solves two minus K Q over R squared for scenario two very similar where we have KQ over R two minus K Q over R. So again, when we are deriving a constant, like this first term that's going to go to zero. Let's okay cute. We are are see of our peoples minus K Q over R squared And that is also equal to be of are up there and so there we can see that they are also the same. Going back up to our potential answers, we know that part one is the same. Part two is the same. So we get rid of one more potential choice and we'll go on to part three of the problem. So part three of this problem is asking us to look at the electric field outside of the hollow sphere. So we're going to um kind of approach this the same way as we did part two where we are going to solve for V of R for both scenarios and then take the derivatives. So V of our first scenario one again, V one plus V two where we have k minus Q over R And B two is going to be K Q over R and again because we are outside of the hollow sphere, both of those radius is are given as our So those two terms cancel. And be of our in scenario one is 0. Let me change our charge on the Hollow Sphere. Scenario two we get k minus Q over R L S k positive uppercase Q over R. There's two um charges do not cancel those two terms. So this is a non zero term. And from what we know about derivatives, we can actually figure out the answer from here. So E of our is de um D V R or D R. And so we know that the derivative of zero is 00 is a constant. So the derivative of constant is zero. So in scenario one, your magnitude of your electric field is going to be zero. And in scenario two this is a non zero term, it is not constant. And so it is not going to be zero. And so you know that those two values are going to change depending on the magnitude of that charge on a hollow sphere. And so from there we can determined that our answer is C. And that is all we have for this problem. We will see you in the next video