Please help How to find the potential difference of a spherical shell

[stunner5000pt]

I can frankly say I'm totally confused on how to solve this problem. Here it is:

A think spherical shell of charge Q and uniform volume charge density p is bounded by radii r1 and r1 where r2>r1. WIth V=0 at infinity find the electric potential V as a function of the distance from the centre of the distribution considering the regions:

a) r > r2 Ans. V = kQ/r because the spherical distribution will act like a point charge when any point is taken outside the shell, by Gauss Law

b) r2 > r > r1 Ans. completely confused here...Using the concept in the first part a) i would think from the iner radii point of view

Vsmall = kQ/r

but since it is enclosed in a bigger radii i have no idea how to proceed

c) r < r1 if the previous confused me then this one is so above my head it's orbiting the earth

d) do these results agree at r = r2 and r = r1 ... Well if i could answer b and c then i might be able to answer this one

 

[Claude Bile]

c) should be easy, if the Electric field is zero inside the shell, what does this say about the potential?

b) is a little trickier. How does the Electric field behave inside the shell? Does it go up as r^2? Use Gauss' Law if you must.

d) The answers should agree, if not, then a mistake has been made.

Claude.

 

[M98Ranger]

First of all, don't worry if you are confused about this problem. It can be a bit tricky to understand at first, but with some practice and understanding of the concepts, you will be able to solve it.

To find the potential difference of a spherical shell, we need to use the formula V = kQ/r, where k is the Coulomb's constant, Q is the total charge of the shell, and r is the distance from the center of the shell.

a) In this case, r > r2, which means that the point is outside the shell. As you correctly mentioned, the spherical distribution will act like a point charge and the potential difference can be calculated using the formula V = kQ/r.

b) In this region, r2 > r > r1, we need to take into account the fact that the point is now inside the shell. We can still use the same formula, but we need to consider the charge enclosed within the smaller radius r. This can be done by using Gauss's Law, which states that the electric flux through a closed surface is equal to the charge enclosed by that surface. So, in this case, the potential difference can be calculated as V = kQenc/r, where Qenc is the charge enclosed within the smaller radius r. This can be calculated by subtracting the charge within r1 from the total charge Q.

c) In this region, r < r1, the point is now inside the shell and we need to take into account the charge enclosed within the smaller radius r. The potential difference can be calculated using the same formula as in part b) but with the charge enclosed within r.

d) Yes, the results should agree at r = r2 and r = r1. At these points, the potential difference should be the same since the point is either on the surface or inside the shell.

I hope this helps to clarify the concept. Remember, practice makes perfect, so keep practicing and don't hesitate to ask for help if you are still confused.