# Charged sphere and charged conducting shell

• pedro97
In summary, 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. We can determine the charges on the inner and outer surfaces of the shell by considering the induction phenomenon, which results in a charge of -q on the inner surface and no charge on the outer surface of the shell. We can also determine the electric field at all points using Gauss's Law, and we can use this to integrate and find the electrostatic potential at the external surface of the 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!

pedro97 said:
I would like to know if my attempt to solve the problem is correct,
No.

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...

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
pedro97 said:
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).

pedro97 said:
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:
rude man said:
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.

haruspex said:
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.

rude man said:
Yes, and that is where he should and can integrate the E field from. What's the problem?

haruspex said:
10-4 ( means "OK", from an old US television show called "Highway Patrol" from which hopefully you were spared! )

## 1. What is a charged sphere and charged conducting shell?

A charged sphere and charged conducting shell are two different objects used in the study of electrostatics. A charged sphere is a spherical object that carries a net electric charge. A charged conducting shell is a spherical shell made of a conducting material, such as metal, that also carries a net electric charge.

## 2. How does charge distribution affect the electric field inside a charged sphere and charged conducting shell?

In a charged sphere, the electric field is zero at all points inside the sphere. In a charged conducting shell, the electric field is also zero at all points inside the shell. This is because the charges on the surface of the sphere or shell distribute themselves in a way that cancels out the electric field inside the object.

## 3. Can a charged sphere and charged conducting shell have different amounts of charge?

Yes, a charged sphere and charged conducting shell can have different amounts of charge. The amount of charge on a charged sphere is determined by the net charge of the object, while the amount of charge on a charged conducting shell is determined by the distribution of charge on its surface.

## 4. What happens to the electric field outside of a charged sphere and charged conducting shell?

Outside of a charged sphere, the electric field behaves as if all the charge is concentrated at the center of the sphere. Outside of a charged conducting shell, the electric field behaves as if all the charge is concentrated at the center of the shell. In both cases, the electric field follows the inverse square law, decreasing with distance from the center.

## 5. How do charged spheres and charged conducting shells interact with each other?

If the two objects are brought into contact, electrons will transfer between them until they both have the same charge distribution. If they are not in contact, the electric field of one object will influence the charge distribution on the other, causing an attraction or repulsion between them.

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