# Calculate the charge in a hollow spherical conductor

• Renaldo
In summary, the homework statement is about calculating the charge on the sphere's inner and outer surfaces if a charge of -3e is placed at the center of the sphere.
Renaldo

## Homework Statement

Consider a hollow spherical conductor with total charge +5e. The outer and inner radii are a and b, respectively.

(a) Calculate the charge on the sphere's inner and outer surfaces if a charge of -3e is placed at the center of the sphere. (Use the following as necessary: e.)

Gauss's Law?

## The Attempt at a Solution

I honestly don't have much of a clue. I feel like the problem should be pretty simple, but I'm not grasping all the aspects of Gauss's Law.

I hypothesized that the charge was 0 on the inner surface, and surmised that the charge on the outer surface should be equal to the total charge. This was wrong. I then hypothesized that the inner charge would be -3e and that the outer charge would need to be added to this to result in 5e, so 8e. Also not correct.

Renaldo said:
I then hypothesized that the inner charge would be -3e
Close, but you made a small mistake. Think that through again.

The inner surface would be +3e because the positive charge would be attracted by the negative field, accumulating on the outside of the sphere. The charge on the outer surface would be +2e because the field is weaker there, and we know the total charge is +5e. Is this correct?

Renaldo said:
The inner surface would be +3e because the positive charge would be attracted by the negative field, accumulating on the outside of the sphere. The charge on the outer surface would be +2e because the field is weaker there, and we know the total charge is +5e. Is this correct?
Sounds right to me. But it also sounds as though you are basing this on a hunch rather than on known principles. Consider the field in the conductor, i.e. between the inner and outer surfaces. What do you know about that? How would you calculate it from the isolated charge in the cavity, the charge on the inner surface, and the charge on the outer surface?

All right, I will give it a shot.

q/εo = ∫Eda
q/εo = E4∏r2

E = kq/r2

E1 = kq1/b2

E2 = kq2/a2

E1+E2 = kq1/b2 + kq2/a2

E3 = E1+E2
E3 = kq3/r2

kq3/r2 = kq1/b2 + kq2/a2

This leaves me with three variables.

Renaldo said:
All right, I will give it a shot.

q/εo = ∫Eda
q/εo = E4∏r2

E = kq/r2

E1 = kq1/b2

E2 = kq2/a2

E1+E2 = kq1/b2 + kq2/a2

E3 = E1+E2
E3 = kq3/r2

kq3/r2 = kq1/b2 + kq2/a2

This leaves me with three variables.

The electric field inside of a conductor is zero, right? Pick a radius r where a<r<b. So it's inside of the conductor. Then use Gauss' law at radius r. What does that tell you?

Last edited:
Dick said:
The electric field inside of a conductor is zero, right? Pick a radius r where a<r<b. So it's inside of the conductor. Then use Gauss' law at radius r. What does that tell you?

Perhaps I am confused as to the nature of hollow spherical conductors. If I understand the problem correctly, a point where a<r<b would be inside the metal of the shell. What is the purpose of using Gauss's Law here? Would not the field at this point be = to 0?

Renaldo said:
Perhaps I am confused as to the nature of hollow spherical conductors. If I understand the problem correctly, a point where a<r<b would be inside the metal of the shell. What is the purpose of using Gauss's Law here? Would not the field at this point be = to 0?

Yes, it would. So the enclosed charge must be zero, yes?

Dick said:
Yes, it would. So the enclosed charge must be zero, yes?

(a) Calculate the charge on the sphere's inner and outer surfaces if a charge of -3e is placed at the center of the sphere.

So the enclosed charge is not equal to -3e? What does -3e represent?

Renaldo said:
(a) Calculate the charge on the sphere's inner and outer surfaces if a charge of -3e is placed at the center of the sphere.

So the enclosed charge is not equal to -3e? What does -3e represent?

The enclosed charge is the sum of -3e plus the charge on the inner surface of the conductor. The Gauss integral is zero. Hence, what's the charge on the inner surface on the conductor? Help me out here.

Dick said:
The enclosed charge is the sum of -3e plus the charge on the inner surface of the conductor. The Gauss integral is zero. Hence, what's the charge on the inner surface on the conductor? Help me out here.

Oh, I see. The enclosed charge is equal to the charge on the inner radius, b, plus the charge in the center. It's obviously 3e, since the field is equal to 0.

As for the outer radius, I know that the spherical conductor has total charge, +5e. In order for it to have this total charge, I add the charges on the two surfaces together:

3e + 2e = 5e

It's really quite simple, but I was failing to understand the logic behind it. Thanks for the help.

## 1. How do you calculate the charge in a hollow spherical conductor?

To calculate the charge in a hollow spherical conductor, you can use the formula Q = 4πε0RiVi, where Q is the charge, ε0 is the permittivity of free space, Ri is the radius of the inner surface of the conductor, and Vi is the potential at the inner surface.

## 2. What is a hollow spherical conductor?

A hollow spherical conductor is a spherical object made of a conductive material, such as metal, that has a hollow interior. It is usually used to store or conduct electrical charge.

## 3. What does it mean for a conductor to be hollow?

A conductor is considered to be hollow if it has a hollow interior, meaning that it is empty or has a void space inside. In the case of a hollow spherical conductor, the interior of the sphere is empty.

## 4. What factors affect the charge in a hollow spherical conductor?

The charge in a hollow spherical conductor is affected by the radius of the inner surface, the potential at the inner surface, and the permittivity of free space. As these factors change, the amount of charge in the conductor will also change.

## 5. How is the charge distributed in a hollow spherical conductor?

In a hollow spherical conductor, the charge is evenly distributed on the outer surface of the sphere. This means that the electric field inside the conductor is zero, and the charge is concentrated on the outer surface due to the repulsion of like charges.

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