- #1

- 45

- 0

## Homework Statement

Consider a hollow metallic sphere of finite thickness, with the inner radius a and the outer radius b. A point charge q is placed inside the sphere at a distance a/2 from the center of the sphere (the sphere is insulated).

a) What is the potential at a point r outside the sphere (r > b)?

b) What are the potentials at the inner (r = a) as well as the outer (r = b) surfaces of the sphere?

c) What is the potential at the center of the sphere? (Use the method of images to calculate this result.)

## Homework Equations

[tex]\Phi=\frac{q}{r}[/tex]

Law of cosines.

## The Attempt at a Solution

For part a I was thinking that [tex]\Phi=\frac{q}{r}[/tex], behaving as if the charge were a point charge concentrated at the origin. I'm not sure if this is right as I can't justify to myself why it would be the case. We've only been discussing properties of conductors in class and said very little about insulators.

In b, I assume the center of the sphere is at the origin, and d is the distance from q to a point on the inner surface at radius a. By the law of cosines,

[tex]d^2=a^2+\frac{a^2}{4}-a^2cos\theta, d=a\sqrt{\frac{5}{4}-cos\theta}[/tex]

so the potential on the surface at radius a is given by

[tex]\Phi=\frac{q}{a\sqrt{\frac{5}{4}-cos\theta}}[/tex]

Logically, I used the same process for the outer surface b and got

[tex]d^2=b^2+\frac{a^2}{4}-abcos\theta[/tex]

so the potential is just

[tex]\Phi=\frac{q}{\sqrt{b^2+\frac{a^2}{4}-abcos\theta}}[/tex]

I then generalized the expression for any r inside the sphere to

[tex]\Phi=\frac{q}{\sqrt{r^2+\frac{a^2}{4}-arcos\theta}}[/tex]

Using this, the potential at the origin should be simply [tex]\Phi=\frac{2q}{a}[/tex] and thus there is no need for the method of images.

I just have this feeling that at least some of the above answers are wrong, most likely because the sphere is insulated and I have not properly utilized that fact in the calculations. That's what has been throwing me off about this problem.