Help with Gauss's Law Quick Quiz

In summary, the conversation discusses the concept of electrostatic equilibrium and Gauss's law in relation to charges on the surface of a hollow conductor. It is explained that in electrostatic equilibrium, the electric field within the conductor is zero, and thus the charge is concentrated on the outer surface. The conversation also explores the difference between options a and b in a quiz, with option b resulting in a shock due to the induced charge on the outer surface. Additionally, the concept of Gaussian surfaces and their relation to net charge within the conductor is discussed.
  • #1
jdefrancesco
9
0
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

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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  • #2
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.
 
  • #3
Thanks for reply :biggrin:
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 :confused:
 
  • #4
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'.
 
  • #5
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.
 
  • #6
jdefrancesco said:
Thanks for reply :biggrin:
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 :confused:
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.
 
  • #7
Thanks guys. I have made this solution, it's ok??

http://img179.imageshack.us/img179/3599/solutiondo0.png

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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  • #8
jdefrancesco said:
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.
 
  • #9
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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  • #10
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?
 
  • #11
jdefrancesco said:
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.
 
  • #12
Thanks Doc Al now I understand :biggrin: . 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."
 
  • #13
jdefrancesco said:
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.
 
  • #14
: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?
 
  • #15
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.
 
  • #16
Thanks a lot! :biggrin:
 

1. What is Gauss's Law?

Gauss's Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the charge enclosed within that surface. It states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space.

2. How is Gauss's Law used?

Gauss's Law is used to calculate the electric field due to a charge distribution. It is particularly useful in solving problems involving symmetrical charge distributions, as it allows for a simpler and more intuitive calculation of the electric field.

3. What is the equation for Gauss's Law?

The equation for Gauss's Law is ΦE = q/ε0, where ΦE represents the electric flux through a closed surface, q is the charge enclosed within that surface, and ε0 is the permittivity of free space.

4. What are the units of Gauss's Law?

The units of Gauss's Law are N*m2/C2 or V*m, which represent the units of electric flux. Other common units for electric flux include C (Coulombs) and m3/s2 (meters cubed per second squared).

5. What are some real-world applications of Gauss's Law?

Gauss's Law has many practical applications, including calculating the electric field around charged particles, designing electrical circuits, and understanding the behavior of lightning. It is also used in various engineering fields, such as in the design of capacitors and in the analysis of electromagnetic fields in radio and television broadcasting.

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