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 read through the problem. See what they're asking us. I'm gonna struck diagram to kind of visualize what's going on here. So we have a solid copper ball of radius R. I. Carrying a charge of negative Q. It's placed inside a hollowed silver sphere of radius R. O. It has a charge of positive Q. Insulators hold this inner sphere in place. We are given this um equation and derivation for electric field magnitude. And we're told to use the expression for electric potential to derive an expression for the electric field magnitude between the two spheres. And we need to express that electric field magnitude. Using in terms of that electric potential V. I. O. Which they define as the potential of the copper ball to this inner sphere relative to the silver shell. That that outer sphere. So they're actually giving us a lot of information this problem and it may kind of seem tricky at first but we're just gonna break it down piece by piece and and we'll get through it. So the first thing that we need to do is recall that the electric electric potential at any given point is K. Q over R. Where K. Is a constant Q. Is the charge. And then R. Is the radius. Okay, so V. I. O. Which is this potential of the copper ball I interval relative to the silver shell O. Is going to be given or defined as V. I minus V. O. And so then from there we need to define the I. N. V. O. So V. I. Is going to be the sum of two spheres because this potentially electric potential inside this inner sphere sees two different services right sees the surface of the inner fear and the surface of the outer shell. So that's gonna be written as K. The charge that enters um sphere is minus Q. And that inner radius is R. I les que que for the outers positive Q. And R. O. We can factor out cake you here and we are left with K. Q. Times one over R. O minus one over R. I. Is our V. I. Term for V. O. Now we're looking at a point um in the outer sphere. So the radius at the surface of the shell, R. O. Is equal to R. Just kind of a not given radius a variable radius. And so this term vo looks like k minus Q. And instead of ri it's just our right this variable radius. And then because in the outer shell we are still aro is still valid radius. In the outer shell roo is going to be this that second term is gonna be the same cake you over our you can see that these two terms cancel because of this signs on the charges Q. So V O is zero. So now we have the io VI zero. So it's just me. I thank you. One minus R. O. For one minus our high. Okay, so that's the first part. Now, we're asked to find the magnitude of the electric field between the two spheres. So they do give us this E R derivative. So we're going to copy that down here. Oops, it's actually D V R. They are. So what is V of R? So V of R recall is given by V I plus V O. Where V I. So we're between the sphere so V I is outside of the small sphere. It's gonna look very similar to um what we had just solved for above. So K Q over our variable radius plus que que over R O. So we'll plug that back in here, T T R times E R. So we have K Q over R O minus K Q over R. So the derivative of K Q over R O is just zero. So then we are left with minus K Q time. Zero plus one over R squared. Which gives us minus K Q Over R squared. Okay, so we have this V I O solved for and it has this cake you term. And now we have E of our electric field magnitude that also has a K Q term. So we are going to solve B io in terms of cake you and then plug that in. That's our last step. So that becomes K Q equals B I 0/1 minus R 0/1 minus R. I. And we plug this whole term into there and we get E of R equals V i O over one minus r 01 over R o minus one over R I times are squared. So lots of um formula manipulation and some, you know, algebra here. But that is our final answer. And when we go back up to our solutions, that is answer C. All right. That's all we have for this problem. We'll see you in the next video.