Homework Help: Two concentric conducting spherical shells, find outer Q

1. Aug 3, 2015

mattz205

1. The problem statement, all variables and given/known data
Two concentric conducting spherical shells produce a radially outward electric field of magnitude 49,000 N/C at a point 4.10 m from the center of the shells. The outer surface of the larger shell has a radius of 3.75 m. If the inner shell contains an excess charge of -5.30 μC, find the amount of charge on the outer surface of the larger shell. (k

2. Relevant equations
closed surface int(E dot dA) = Q encl/epsilon0 (gauss's law)

3. The attempt at a solution
I plugged in the given electric field into gauss's law in the form E=Q encl/ 4pir^2 epsilon0, with Q encl= q(inner)+ q(outer) and solved for q(outer) and arrived at ~ 96.9 μC but that is wrong according to MP (my homework site)

any help would be greatly appreciated

Last edited: Aug 3, 2015
2. Aug 3, 2015

SammyS

Staff Emeritus
First of all it's not a magnetic field (I suppose that was a typo).

How much charge is there on the inner surface of the outer larger shell?

3. Aug 3, 2015

mattz205

Yes, sorry about that i meant electric field, and wouldn't the charge on the inner surface of the outer shell just be the negative of the outer? If that's not it then I'm not entirely sure.

4. Aug 4, 2015

Qwertywerty

No , not necessarily . This would be true if the outer shell itself had no excess charge .

I couldn't understand your solution - have you equated electric field to kq/r2 (where q is charge on the outer surface of the outer sphere ) ?

5. Aug 4, 2015

SammyS

Staff Emeritus
The charge on the inner surface of the outer shell would just be the negative of the outer shell only if the outer shell has a net charge of zero. That's not stated anywhere. Besides if that was true, then the total charge inside the Gaussian surface (the closed surface you mention) would have to be -5.30 μC .

Consider a Gaussian surface totally embedded in the conducting material of the large sphere. What's the total charge enclosed by this surface ?

6. Aug 4, 2015

mattz205

the electric field inside a conductor is always zero according to my lecture... and the charge enclosed there would be the -5.3 + whatever the inner surface of the conductor is, which i dont know, nor do i know how to find it...

7. Aug 4, 2015

SammyS

Staff Emeritus
If the electric field is zero, then the flux through that Gaussian surface is zero. That tells you total charge enclosed is ___ .

8. Aug 4, 2015

mattz205

0? would that make the charge on the inside of the conductor +5.3?

9. Aug 4, 2015

Pyrus

I tried this by using E=V/r. I found potential at point at distance 4.1 m.
V=Er => V=20900 V
Then applying V= 1/(4πε) * {q/4.1 + Q/4.1}
V=(9*10^9 )[-5.3/4.1 + Q/4.1]*10^-6

Solving it I also got answer 96.6μC

10. Aug 4, 2015

mattz205

i tried that method too and got the same answer as i did using the other method, the answer is wrong regardless though

11. Aug 4, 2015

mattz205

ok, ok, i got it, thank you very much for your help!

12. Aug 4, 2015

Pyrus

So simple and I got it too complex....

13. Aug 4, 2015

SammyS

Staff Emeritus
Yes.
Notice that your final answer is independent of the amount of charge on the inner sphere. The conductor effectively shields the region outside of the sphere from the region inside the sphere.

14. Aug 4, 2015

mattz205

That makes sense, in lecture my instructor wasn't very clear about how conductors worked, but it makes sense now. Thnk you again for your help, i greatly appreciate it.