# Non conducting hollow ball

1. Oct 21, 2014

### NYK

1. The problem statement, all variables and given/known data
https://fbcdn-sphotos-c-a.akamaihd.net/hphotos-ak-xaf1/v/t1.0-9/10155063_1485929725016579_7557468199677155885_n.jpg?oh=d575aa48176de7ecda27201b7ce35a5b&oe=54F7F368&__gda__=1425195231_44abd00cc231109df2fbe8d44cef9869

2. Relevant equations
∫E⋅da⋅n = 4πkQenc
ΔV(voltage) = -∫Edr
3. The attempt at a solution
To be honest I am having trouble starting the problem and defining the Qenc for the three areas of the sphere.

I do believe that for r < R ⇒ E(r) = 2QK/r2

and for r > R ⇒ E(r) = KQ/r2

I know that the answer is listed for R<r<2R but i cant seem to come up with that answer.

I think im just having problems with defining the Qenc

for R<r<2R does Qenc = 2Qpt charge - Qinduced = Q sound right?

2. Oct 24, 2014

### vela

Staff Emeritus
That's right.

No. The ball is non-conducting, so there won't be any induced charges. You have a charge -Q spread out uniformly within the ball. When R < r < 2R, you have to figure out what fraction of the ball is enclosed inside of the sphere of radius r. It's a geometry problem.

3. Oct 24, 2014

### NYK

I understand that, so if it was a conducting sphere the charge would be Qenc = Qpt charge +λ(π(r2-R2))?

4. Oct 24, 2014

### vela

Staff Emeritus
No. How did you come up with that expression?