# Homework Help: Gauss' Law Help

1. Jan 31, 2015

### Sierra

1. The problem statement, all variables and given/known data
A solid insulating sphere of radius 5.0 cm shown in blue has a uniform charge density
throughout its volume. Concentric with this sphere is a conducting thick hollow spherical shell
of inner radius 20.0 cm and outer radius 25.0 cm shown in grey.
The electric field at a point 10 cm from the center is measured to be 3600 N/C directed radially
inward and the electric field at a point 50 cm from the center is measured to be 200 N/C
directed radially outward. Using this information, find:
a. The charge on the insulating sphere
b. The net charge on the hollow conducting spherical shell
c. The charge on the inner surface of the hollow conducting spherical shell
d. The charge on the outer surface of the hollow conducting spherical shell

2. Relevant equations
U=KQq/r
E= Kq/r^2

3. The attempt at a solution

So I know that you have to use Gauss's law and probably flux as well.

So this is what I tried to do:

a) q= 3600(0.02)^2 / (8.998x10^9)= 1.6x10^-10 C

b) E=0 because it is the conductor

and now i'm stuck since I am second guessing myself

Help would be great.

Thank you

2. Jan 31, 2015

### utkarsh009

Your answer to b part is wrong. There will be induced charge on the inner surface of the shell so that the field due to solid sphere and induced charge is 0 everywhere outside it. So the field at 50 cm is only due to the charge on outer surface. Use it to calculate net charge on shell.

3. Jan 31, 2015

### Sierra

So what equation would I use to get that? E or U?

4. Jan 31, 2015

### utkarsh009

Assume that the solid sphere and induced charges are not present at all (since the have no effect). Only a spherical conductor is present with charge on its outer surface.

5. Jan 31, 2015

### Sierra

So use flux? That would mean that Φ=EdS* Q/ε(o)?

6. Jan 31, 2015

### utkarsh009

Go ahead and use it. See if you can get the answer. (∫E.dS=q/ε(o)) or you could simply use the formula you mentioned it in relevant equations. I'm not supposed to spoonfeed you. Check the answer in your textbook.

7. Feb 2, 2015

### Sierra

For b I got Q= 5.56x10^-9 because I used E=kQ/r^2 and solved for Q to get Q= Er^2/k and used the charged found on 50 cm, the 200 N/C and plugged everything into its respectful places

8. Feb 2, 2015

### Sierra

Then for c it is 0 because it is the conductor and for d I just used the same as the other one and got 4x106-9

9. Feb 2, 2015

### utkarsh009

This Q is actually the charge on the outer surface (d) . For inner surface (c), it will be negative of the charge on solid sphere. Add both of these to get the net charge. That will be the answer to (b)

10. Feb 2, 2015

### Sierra

Ok, I changed my answer from b to answer d. I solved for C and got -7.96x10^-11 from q(inner)= -Q(i used my answer from a to get this Q) / 16pi(0.02)^2. And I got my net charge for b to be 5.48x10^-9 after adding c and d together

11. Feb 2, 2015

### BvU

Did you fix your (physical and calculation) errors in
(Physics: 10 cm $\ne$ 0.02 m, and: the field is pointing inwards. Math: power of 10. And k =8.988 109 Nm/C2).

Score so far:
a) 1.6 10-10 not correct
d) 5.56 10-9 good

things to do:
c) apply a gaussian spherical shell at r=22 cm. E=0 (you are inside a conductor)

Last edited: Feb 2, 2015
12. Feb 3, 2015

### Sierra

For a I got -4.00x10^-9 nC
B) 3800 N/C from 3600 N/C + 200 N/C
C) 0 nC from q= 0 N/C (0.2 m)^2 / (8.998x10^9)
D) 6 x 10^-9 nC from 200 N/C (0.5 m)^2 / (8.998 x 10^9)

13. Feb 3, 2015

### BvU

Note that 0 is NOT the answer to c) ! You have found that the total charge within that shell is 0.

And the adding up in b) went awry in a terrible manner: they ask for a charge and you add two field (absolute?) values from two far-apart locations ?!

Yet another detail: kCoulomb numerical value !

14. Feb 3, 2015

### Sierra

My mistake.

C) I got 1.6x10^-8.. changed it to 1.6x10^-9 to match D.
B) (1.6x10^-9) + (6.0x10^-9) = 7.6 x 10^-9 nC

15. Feb 3, 2015

### BvU

16. Feb 4, 2015

### Sierra

I found C by doing 3600(0.2m)^2/ (8.998x10^9) = 1.6x10^-8

17. Feb 4, 2015

### BvU

But the field is not 3600 N/C there at all !
Post #12 was much better! at 20 cm plus a little bit the field is zero, so the enclosed charge is zero. You know the charge on the insulating sphere, so ...

18. Feb 4, 2015

### Sierra

Omg so C is zero and the net charge for b 4.00??

19. Feb 4, 2015

### BvU

The answer to c is not zero! Are we going around in circles (post#13) or are you so surprised the words come out somewhat mixed up ?

Thank goodness for Gauss !

Have to run now. Later.

20. Feb 4, 2015

### Sierra

I might be confusing myself. There's only two electric fields to choose from and I've used both of them and they are both wrong apparently. I'm looking for the charge of c. But am I missing something? Like c is the opposite of a?

21. Feb 4, 2015

### BvU

Good idea! Charge on the inner sphere attracts opposite charge that distributes over the inner surface in such a way that that inner surface is an equipotential. So yes, c is the opposite of a. That way the gaussian surface at 21 cm can yield a zero field as becomes a conductor.