# Conductor/Image Charge, Volume Charge Distribution

1. Oct 9, 2009

### tyj8ti

1. The problem statement, all variables and given/known data

http://uberlycool.com/2.jpg

2. Relevant equations
Working on the conceptual first...

3. The attempt at a solution
2a. My understand is that the electric potential inside the shell is only affected by the enclosed q? Is that correct? Because if I was to place an image charge to solve the problem, the image charge wouldn't contribute anything and since the sphere is grounded, that doesn't contribute anything either.
2b. Total induced charge would be -q to cancel out the q's e-field. Correct?
2c. Total force on the conducting shell would be zero since the shell is not moving, other wise the shell would start moving around. I'm pretty sure that's wrong...but I can't think of anything else...Unless I have to use an image charge (read 2d).
2d. I'm pretty sure this one I'll have to find the image charge for the system and compute the potential energy according to the image charge(s) that I find. But I don't know how to start off to find the image charge, any pointer?

3. I have no idea how to start this problem at all, any input is welcome.

P.S. 1st time posting on the physics forums...please don't bash me if I didn't do something right...=/

2. Oct 9, 2009

### gabbagabbahey

Hi tyj8i, welcome to PF!

No. In the actual problem, you have a point charge and a conductor. The point charge will induce some unknown charge density onto the conductor, and the resulting potential will be the potential due to the point charge and the induced charge on the conducting shell.

If the charge distribution was spherically symmetric, then you could use Gauss' Law to find the field and then integrate it to get the potential, and you would find that the potential is that of just the point charge. However, since the point charge is off-center, there is no reason to assume spherical symmetry and a different method must be used.

This is where the method of images comes into play. You forget about the conductor and instead add an image charge (not in the region you are trying to find the potential) outside the now forgotten shell (r>R) in such a place that the total potential due to the point charge an the image charge will be zero on the shell (r=R). When you do that, the uniqueness theorem can be used to show that the total potential due to these two charges (point and image) is the same as the potential of your original problem, inside the shell. (How?)

Correct

NO, just because an object isn't moving (has zero velocity) at a given instant, doesn't mean it can't have a non-zero acceleration (force acting on it) at that same instant.

You'll want to find the electric field (inside and out) using MOI, and the induced surface charge density $\sigma(\theta,\phi)$ on the shell. From there, you can take a tiny piece of the shell $dq=\sigma da$ and calculate the force on it, by treating it as a point charge and using the Lorentz force law...to find the total force on the shell, you simply integrate this over the entire shell.