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 are asking us. We have a hollow conducting sphere of radius R. R. O. And a charge of positive Q. And inside of that hollow sphere we have a solid sphere of radius R. I. Carrying a charge of negative Q. So we're told that there's insulating supports that are holding this solid sphere inside of the hollow sphere. And they are asking us to derive an expression for the I. O. Which is defined as the potential of the solid sphere relative to the hollow sphere. So I'm just gonna draw here a little diagram to kind of show what we're working with. And so our inner sphere we know is um radius ri with charge negative Q. The outer sphere is radius R. O. With a charge of positive Q. And so then the next thing we want to do is recall that the electric potential at any given point is given by V equals K. Q over R. Okay, is a constant Q. Is a charge and R. Is the radius. So we are asked to derive this expression for V. I. O. And so we know then that that is equal to V. I minus V. O. Because it's the potential of the solid sphere. This term relative to the hollow sphere. Okay, so now we need to solve for the eye and video, we need to figure out what those equations will look like. So the next thing we need to recall is that any electric potential where we're working with multiple surfaces is the sum of all of those potentials. So V I is because the inside sphere is inside or any point on the inside sphere is inside two of these services. It's actually going to be the sum of the potentials. So we're going to have que que I over our eye plus que que o over R O. And we were given Q. I. And Q. O. In the problem where we know that the inner charge is negative Q. And the outer charges positive Q. And so we can factor out K and Q flip around the term for the negative sign and get que que times one over R zero minus one over R I. So I run a one over R O. For the outside our Os for the outside. All right. And so then the second thing we need to figure out. So that's what this is our term for V. I. Now we need to find the term for V O. Okay. And at the surface of the shell we need to recognize that R O is equal to R. So this looks like okay, times negative Q over R plus K. Just two positive Q over our We can see here that these terms cancel. We have negative positive Q. And so vo is actually zero. So, if we go back up to this V I minus V I O term V I minus V O. V I O. When the V one V O goes to zero is just V I. Which is just this K Q times one over. R O minus one over R. Hi. Alright. And that is our answer. So we go into our um possible answers and that aligns with choice B. So the correct answer for this problem its feet. That's all we have for this one. We'll see you in the next video.