# Charged sphere and charged conducting shell

pedro97

## Homework Statement

A + q = 5 pC charge is uniformly distributed on a non-conducting sphere of radius a= 5 cm , which is placed in the center of a spherical conducting shell of inner radius b = 10 cm and outer radius c = 12 cm. The outer conducting shell is charged with a -q charge. Determine:
1) the charges on the inner and outer surfaces of the shell;
2) the electric field (module, direction) everywhere;
3) the electrostatic potential on the external surface of the conducting shell (r = c), on the internal surface of the shell (r = b) and on the outer surface of the internal sphere of radius a (r = a).
Suppose now to replace the inner sphere with a spherical conductor of radius a charged with the same + q charge:
4) Which of the previous answers will change and how?
5) Determine the potential everywhere;
6) Determine the capacitance of the spherical capacitor formed from the internal conducting sphere of radius a and the outer conducting shell;
7) If a proton (m = 1.67·10-27 Kg) starts from rest from the spherical conductor of radius a, which will be its speed when it hit the inner surface of the outer spherical shell?

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## The Attempt at a Solution

So, I would like to know if my attempt to solve the problem is correct, and how can I continue it in the parts I didn't manage to solve...
Thank you!

haruspex
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I would like to know if my attempt to solve the problem is correct,
No.

pedro97
For the first point my reasoning is: since on the inner sphere we have a + q charge, because of the induction phenomenon, the inner surface of the outer shell has to "balance" that charge, and thus has a charge -q. Since the problem statement tells us that -q is the total shell charge, we know that there is no charge on its outer surface...

rude man
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Seems to me his answer to (1) is correct.
Placing gaussian sphere at b < r < c forces qb = -q.
Placing gaussian sphere at r > c forces qc = 0.

SammyS
rude man
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3) the electrostatic potential on the external surface of the conducting shell (r = c), on the internal surface of the shell (r = b) and on the outer surface of the internal sphere of radius a (r = a).
You can just integrate the E field using your data from (2).

haruspex
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For the first point my reasoning is: since on the inner sphere we have a + q charge, because of the induction phenomenon, the inner surface of the outer shell has to "balance" that charge, and thus has a charge -q. Since the problem statement tells us that -q is the total shell charge, we know that there is no charge on its outer surface...
My mistake - I read it as +5q instead of +q = 5pC.

I don't understand your attempt at 5). Looks more like an attempt at 6).

Last edited:
haruspex
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You can just integrate the E field using your data from (2).
Yes, but we're not told where to set the zero potential. The usual convention is to set that at infinity.

rude man
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Yes, but we're not told where to set the zero potential. The usual convention is to set that at infinity.
Yes, and that is where he should and can integrate the E field from. What's the problem?

The custom is to set potential to zero at infinity unless that is impossible - as with a charged wire of infinite length.

haruspex