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. So we know we have a conducting solid sphere with radius R. I. The charge of negative Q. And that's located at the center of another conducting but it's a hollow sphere with radius R. O. And a charge of positive cube. Um insulating or non conducting supports hold the inner sphere in place and we are told that potential is zero at R equals infinity. So we are asked to determine the electric potential at three different spots. Okay, so the key here is to first recall the electric potential at any point for a point charge is given as KQ over. R K is a constant Q. Is the charge and r is the radius of the area of the sphere in question. Okay, so next we have to recall that this um total potential is the sum of potentials and it's actually nice here because they actually give us that hint so, potential at a point is the sum of potentials from each sphere. So we can rewrite that as V R equals V I plus V. O. Alright, so we're gonna use this kind of over and over again this equation right here to um Look at each of the three different scenarios for the three parts of the problem. So let's just take part 1 1st. And that's asking us inside the solid sphere. Okay, so V I inside the solid sphere is going to be K. The charge is negative Q. The radius is R. I. And then V. O. Is going to be K charges positive Q over R. O. We can rewrite this factoring out cake you as one over R. O minus one over R. I. And that is Our Answy for Part one. Okay, so right off the bat we can eliminate let's see B. C. And E. Because those answers for part one are not correct. Part two is looking at the potential between the two spheres. Okay, so I'm gonna write that again. V. R equals V. I plus Vo except this time I is outside of the small sphere. So while the first part of the term is the same instead of R. I. It's going to be just our because it is some given radius that is outside of that sphere. Um for V. O the term is the same as in part one, it's KQ R O. Because the point is still within that outer sphere. And so We can again pull out the KQ term and we get one over R -1 over R. So that is the answer for the second part. So here see we can eliminate D. Is not correct. So let's take the third part is outside of the hollowed sphere. So now we're looking at some distance. Australia's like this Are just some unknown distance are outside of the sphere and were told to recall in the problem we are reminded that potential is zero as our approaches infinity. So that's a nice hint that they give us there for part three. Um, so it's the same kind of starting equation V I plus V. O. But as we are outside of the sphere, both radius is our, this our term. So we have K charges negative Q. R. And then KQ positive charge are in those terms canceled because of the Q signs. And that is zero. So that is the answer for part three, which means that choice a is eliminated. And your correct answer for this problem is f. That's all we have for this problem. We'll see you in the next video.