Electric Potential of two conductors

[jesuslovesu]

Homework Statement

A hollow spherical conducting shell of inner radius b and outer radius c surrounds its concentric with a solid conducting sphere of radius a. The sphere carries a net charge -Q. The shell carries net +3Q. Find an expression for the potential difference from infinity to the surface of the sphere.
http://img185.imageshack.us/img185/3596/rabbitatesthekv2.th.jpg
Sorry it's a little cut off, but you get the idea.

Homework Equations

V = kq/r

The Attempt at a Solution

Well, I'm not quite sure if I need to use calculus or Gauss's law or what. The best so far that I've been able to come up with is V1 = k(-Q)/a, V2 = k(3Q)/(c-b)

I know that if the inner sphere were not there, then the electric field would be zero inside the shell.
If I recall correctly from the last chapter I had, the outside surface of the shell will be charged at +2q and the inner surface of the shell will be at q

 

[Andrew Mason]

Homework Statement

A hollow spherical conducting shell of inner radius b and outer radius c surrounds its concentric with a solid conducting sphere of radius a. The sphere carries a net charge -Q. The shell carries net +3Q. Find an expression for the potential difference from infinity to the surface of the sphere.
http://img185.imageshack.us/img185/3596/rabbitatesthekv2.th.jpg
Sorry it's a little cut off, but you get the idea.

Homework Equations

V = kq/r

The Attempt at a Solution

Well, I'm not quite sure if I need to use calculus or Gauss's law or what. The best so far that I've been able to come up with is V1 = k(-Q)/a, V2 = k(3Q)/(c-b)

I know that if the inner sphere were not there, then the electric field would be zero inside the shell.
If I recall correctly from the last chapter I had, the outside surface of the shell will be charged at +2q and the inner surface of the shell will be at q

Use Gauss' Law. That will give you the field outside the outer sphere, which is a function of the net enclosed charge.

AM

 

[Dick]

To extend Andrew's suggestion, there is no potential difference between the outside of the shell and the inside of the shell. So just pick up the same game inside the shell. Compute the potential difference between the surface of the sphere and the inside of the shell and add it to the outside.

 

[chaoseverlasting]

Will it be something like this:

Using Gauss's Law,

E=k2q/r^r for r>c. E=-dv/dr, integrating this from inf to r, and that's your potential?

 

[Andrew Mason]

Will it be something like this:

Using Gauss's Law,

E=k2q/r^r for r>c. E=-dv/dr, integrating this from inf to r, and that's your potential?

Yes. Gauss' law will give you the enclosed charge:

[tex]\phi = \oint E\cdot dA = \frac{q_{encl}}{\epsilon_0}[/tex]

By symmetry, E is uniform over a gaussian sphere of radius r > c

[tex]\oint E\cdot dA = E4\pi r^2 = \frac{q_{encl}}{\epsilon_0}[/tex]

[tex]E = \frac{q_{encl}}{4\pi\epsilon_0 r^2}[/tex]

The potential (work per unit charge) is:

[tex]\int_\infty^c E\cdot dr[/tex]

Work out that integral to get the potential on the surface c

AM