Homework Help: Electric Fields, Insulators, Conductors

1. Feb 10, 2004

Aerospace

I am going to attach the question along with this post so that you guys can see the questions and the diagram and be able to help me in answering my question better.
I found the ccharge on the inner sphere and it is -3.928e-9 and the charge on the outer sphere is 8.15698e-9.
Now I have to find the total charge on the inner and outer surface of the hollow conducting sphere.
I'm not sure why it is any different from the charge on the outer sphere written above. If someone could help me by explaining and helping me set up the problem, it would be DEFINITELY appreciated. Thanks.

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2. Feb 10, 2004

Staff: Mentor

How did you find the charges on the spheres?

Consider a guassian surface at three different radii:
- just outside the insulator (I assume this is how you found the charge on the inner surface)
- outside the conducting sphere (What charge would that give?)
- and one inside the conducting sphere (What's the field there?)

3. Feb 10, 2004

Aerospace

I still don't understand...
I found the charge on the insulating sphere using the formula (E)(Eo)(SA of sphere)=-q
E is the electric field, Eo is Epsilon naught, SA is Surface Area, and the negative in front of q (charge) is because the electric field is radially inward.

For the conducting sphere, I used (Ec)(Eo)(SAc of sphere)=Q-q
Where Ec is Electric field for the conductor, Eo is epsilon naught, SAc is Surface Area of conducting sphere, Q is charge on the conductor and q is the inside charge of the conducting sphere...

So my initial question was, what's the difference between the charge on the conductor and the charge on the inner and outer surface of the conducting sphere? because I have to find the total charge on both of those.

4. Feb 10, 2004

Staff: Mentor

Gauss's law gives you the total charge within the surface. Setting this equal to Q - q, means that Q is the total charge on the conducting sphere: including both inner and outer surfaces.
I don't understand your question. The total charge on anything is the sum of all the charges. For a conductor, this is the sum of the charge on its inner and outer surfaces.

What's the charge on the inner surface? Use a Gaussian surface in the interior of the conductor and see what the net charge must be. (It's trivial.)

5. Feb 10, 2004

Aerospace

Figured it out :) Thanks

Thank you so much. I used your idea of using the Gaussian surface and therefore, the total charge on the inner surface is therefore q, which is 3.928e-9 and the total charge on the outer surface is Q-q which is 8.15698e-9 - 3.928e-9 = 4.22898e-9

Yay...haha...