# Gauss's Law question

Hi there, I was reading the Serway's physics book and I don't understand this quiz neither the solution. I mean, if the solution is a, what's the difference with option b?

http://img140.imageshack.us/img140/6812/gaussslawud7.png [Broken]

The "Quick Quiz" is in the chapter "Gauss's Law", under the section "Conductors in Electrostatic Equilibrium" Can somebody help me??

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Doc Al
Mentor
In b you are touching the outer surface of the conductor, which is where the charge is. But in a you are touching the inner surface, where no charge resides, thus you are safe.

But why the charge is in the outer surface?
I thought that the charge of the conductor resides on its surface whether inner or outer

The conductor's charge will only exist on the outer surface when it's in electrostatic equilibrium.Otherwise charge will move to the out surface to keep the 'ELECTROSTATIC EQUILIBRIUM'.

When a conductor is in electrostatic equilibrium,there is no electric field inside the conductor(in fact ,the inner surface also belongs to the 'inside' part).According to 'Gauss law',as a result, there shouldn't be any charge inside the conductor whatsoever.

Doc Al
Mentor
But why the charge is in the outer surface?
I thought that the charge of the conductor resides on its surface whether inner or outer
Yes, it's true that the charge on the conductor resides only on its surface, inner or outer. But you'll only have charge on the inner surface if you have a charge inside the hollow space--that's not the case in a. When your charged brother touches the outside of the conductor, no charge is created on the inner surface.

The reason for the charge being on the outer surface is a combination of Gauss's law and the fact that in electrostatic equilibrium (as scienture explained) the field within the conducting material is zero.

Thanks guys. I have made this solution, it's ok??

http://img179.imageshack.us/img179/3599/solutiondo0.png [Broken]

And in the case b, when the charged brother touches the inner surface the charges will move and will give a shock. it's ok?

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Doc Al
Mentor
Thanks guys. I have made this solution, it's ok??
Not exactly. The first sentence is incorrect: The inner surface of a hollow conductor is not within the conducting material and thus you cannot immediately conclude that the field there is zero and that there is no charge on the inner surface. (Your argument implies that there can never be a charge on the inner surface, which is false.)
And in the case b, when the charged brother touches the inner surface the charges will move and will give a shock. it's ok?
Actually, even before he touches the inner surface, he will have induced a charge on the outer surface of the conductor.

I wanted to say:"The gaussian surface is inside the conductor, so the electric field on the gaussian surface is zero. Then the flux through the gaussian surface is zero. This implies that q_in must be zero (q_in is the net charge inside the gaussian surface). So the fact that charges resides on the inner surface is absurd."

Actually, even before he touches the inner surface, he will have induced a charge on the outer surface of the conductor.

And this part I don't get it :tongue:

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Actually, even before he touches the inner surface, he will have induced a charge on the outer surface of the conductor.
This is because the charged boy is an electric field, and the free charges on the conductor will move?

Doc Al
Mentor
This is because the charged boy is an electric field, and the free charges on the conductor will move?
It's because the boy has a charge. Since a Gaussian surface within the conductor must have flux = 0, the net charge within that surface must be zero. (This is Gauss's law.) That means that a charge will be induced on the inner surface of the conductor to "cancel out" the boy's charge, and an opposite charge will be induced on the outer surface.

Thanks Doc Al now I understand . But tell me what is wrong in this?
I wanted to say:"The gaussian surface is inside the conductor, so the electric field on the gaussian surface is zero. Then the flux through the gaussian surface is zero. This implies that q_in must be zero (q_in is the net charge inside the gaussian surface). So the fact that charges resides on the inner surface is absurd."

Doc Al
Mentor
I wanted to say:"The gaussian surface is inside the conductor, so the electric field on the gaussian surface is zero. Then the flux through the gaussian surface is zero. This implies that q_in must be zero (q_in is the net charge inside the gaussian surface).
Perfectly correct.
So the fact that charges resides on the inner surface is absurd."
This doesn't follow. Just because the net charge within that surface is zero, doesn't mean that there's no charge.

:tongue:
So in the inner surface will be charge but the net charge will be zero because:
...That means that a charge will be induced on the inner surface of the conductor to "cancel out" the boy's charge, and an opposite charge will be induced on the outer surface...
And this opposite charge induced is the reason of the shock in the option b. it's ok?

Doc Al
Mentor
Good.

Let's assume the boy has a charge of +Q. As soon as he steps inside the hollow conductor, the inner surface will establish a charge of -Q and the outer surface will have a charge of +Q.

Thanks a lot!!!