miamirulz29 said:
Sorry, this a non-calculus based physics course and I do not know calculus yet. However, I have figured out the answers. For 10: it is kQ/a and For 11: it is the one with three terms. I don't understand how the answer those two answers work. Could you explain them to me please? Thanks in advance.
Ooh. No calculus it is then.
Okay, for problem 10, point a is on the outside of the whole thing. Whenever you end up on the
outside of something with spherically symmetric charge distribution, simply treat the whole thing as a point charge. Just make sure the charge distribution is spherically symmetric, and the point of interest (the test charge) is on the outside. You can do this as a result of Guass' law. So that's where you get the kQ/a.
Problem 11 is a little trickier. The electric field inside a conductor is always zero. What's more, the electric potential inside a conductor is always
constant. There are two conductors here to worry about. There is the solid sphere (where the test charge is at) and the spherical shell.
Let me start by commenting on the solid sphere by itself. You know that the potential anywhere inside the sphere is a constant. So simply measure the potential at the perimeter of the sphere, where you can treat the sphere as a point charge, and you know that the potential will be the same value anywhere inside.
If there was no conducting shell, this would be just like the previous problem. So as a first step, find the potential at R
1, ignoring the spherical shell for the moment.
For the next step we need to deal with the spherical shell. The potential is the same anywhere between R
2 and R
3 (the potential within a conductor is a constant). So we need to subtract its contribution. We know that the potential at point R
2 is the same as the potential at point R
3. If the shell was not there in the first place, what would this potential difference be between these two points? It would be
V(R
2) - V(R
3)
So calculate what that is. This is what we need to subtract before we finish the problem.
So now put everything together. Start with the potential of the solid sphere, as if there was no shell around it. Then from that, subtract the potential caused by inserting the conducting shell (i.e. subtract the result obtained in the above paragraph).
[Edit: Depending on how you approach this problem, you can either subtract [V(R
2) - V(R
3)] as discussed above, or add [V(R
3) - V(R
2)], which gives you the same answer. Both are valid methods. But
subtracting the [V(R
2) - V(R
3)] might be more appropriate depending on how your text and instructor present this material. I went back and forth on editing this post, and decided on the subtracting method in the end, since I feel it is more in line with the calculus approach.]