# Electric field of hollow ball

1. Feb 20, 2006

### cherrios

I've been having some trouble with this problem: A hollow ball that has radius=X, has uniform volume charge density of $$\rho$$.

1) Gauss's Law->Find electric field at .1X and 3X from the center

2)What is electric potential at .1X and 3X from center?

3) Find electric potential energy of charge q, with mass = m, when it is released 3X from center of the hollow ball. When it hits the ball, what is its speed?

For 3, would I have to equate electric potential energy to kinetic energy, and then solve for velocity?

Any tips would be much appreciated!

2. Feb 20, 2006

### dicerandom

How does the ball have a uniform volume charge density if it's hollow? I find that confusing, I think it's meant to be a solid sphere of charge.

Your method for #3 should work fine, did you do the other two okay?

3. Feb 20, 2006

### cherrios

oh sorry, it's filled with uniform volume charge

4. Feb 20, 2006

### topsquark

1) Generally, when you need to find an electric field use Gauss' Law. The trick here is in the answer to the question, how much charge is inside your Gaussian surface? In this case, the charge distribution in uniform, so just integrate over the volume of the surface: $$q_{in} = \int_0^r \rho r^2 sin \theta dr d \theta d\phi = \rho * (4/3) \pi r^3$$, where r is the radius of your (hopefully spherical!) Gaussian surface. Now apply Gauss' Law. For the outside calculation remember that the object's radius ends at r = X. So we have a total charge $$Q = \rho * (4/3) \pi X^3$$.

2) Recall the relationship between the Electric Field and the Electric Potential. Now, since E is spherically symmetric, the E and V are related by a simple derivative of r.

3) Your method sounds like a plan to me. Unless you really WANT to find the force on the ball, use F = ma, then use the uniform linear acceleration equations. I'd rather use your method.

-Dan

5. Feb 20, 2006

### cherrios

Thank guys! I think I should be fine for now; I believe I had an integration error. I might have some questions later... but thanks again!