Application of Gauss' law - point charge in hollow sphere

In summary, Gauss' law is true for any closed surface you care to consider. The field inside an ideal conductor must always be zero. The charges are free to move, so they will always move until the field is balanced and they feel no more force.
  • #1
pesslovany
4
0
Firstly I appologize, that I am not native english speaker and I don't study physics(but cybernetics we are getting just some general knowledge about physics), but hopeffuly I will write this right.

Homework Statement



We know that inside of a conductive object is protected from influence of outside electrical charges.2. Questions

a. Analogically, is outside of the conductive object protected from influence of inside electrical charges? For example from point charge Q situated inside the hollow sphere at a random spot.
b. The point charge is directly in the middle of the cavity of hollow sphere. Consider what is specific about this setup.
fyzika.jpg

The Attempt at a Solution



a. If I place a point charge Q+ inside the sphere, then because the sphere is conductive( therefore has free electrons ), the electrons will move to the inside of the shell to "cancel" electric field from inside charge. This will leave the outside surface of the shell positively charged, therefore creating electromagnetic field on the outside. Meaning outside of the shell is not protected from the inside charge. That is what I came with, but I am supposed to proof this by some equation, I don't really know where to start here...

b.I guess - by the picture - I am supposed to express E. I can divide this problem into three parts: inside the cavity of the shell, outside the shell and in the shell object itself.

1. Inside cavity: For inside cavity I used standard Gauss law therefore i got :
[tex]E = \frac{1}{4\pi\epsilon}\cdot\frac{Q}{r^2}[/tex]
2. In the shell: Here the E should be 0 after electrons cancel the point charge and stop moving. Again I should proof this but I don't really know where to start.

3. Outside: I am not really sure here, but I guess it should be:
[tex]E = \frac{1}{4\pi\epsilon}\cdot\frac{Q}{(r-R_2)^2}[/tex]

So I would be very happy if you could help me with the two proofs and if you could check, if the rest of my thoughts are allright. Thank you in advance.​
 
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  • #2
Gauss' law is true for any closed surface you care to consider. What does it tell you if you make a Gaussian sphere larger than your conductor?
 
  • #3
Ok, so outside the conductor it should be :
[tex]E = \frac{1}{4\pi\epsilon}\cdot\frac{Q}{r^2}[/tex]
Q will be same as inside the conductor.

And in the shell I think it will look like this:
[tex]E = \frac{1}{4\pi\epsilon}\cdot\frac{-Q_-+Q_+}{r^2} = 0[/tex]
 
  • #4
pesslovany said:
Ok, so outside the conductor it should be :
[tex]E = \frac{1}{4\pi\epsilon}\cdot\frac{Q}{r^2}[/tex]
Q will be same as inside the conductor.

And in the shell I think it will look like this:
[tex]E = \frac{1}{4\pi\epsilon}\cdot\frac{-Q_-+Q_+}{r^2} = 0[/tex]

Yep.
 
  • #5
Thank you...
I have got one supplementary:
"Would the E in the shell be also 0 if the point charge wouldn't be exactly in the middle?"
It should be, but there is a problem calculating
[tex]\vec{E}\cdot\vec{n} =
|\vec{E}|\cdot|\vec{n}|\cdot cos(\alpha)[/tex]
because the angles are all different, so I have to come up with some generalization. Here, I don't really know where to start at all..
 
  • #6
pesslovany said:
Thank you...
I have got one supplementary:
"Would the E in the shell be also 0 if the point charge wouldn't be exactly in the middle?"
It should be, but there is a problem calculating
[tex]\vec{E}\cdot\vec{n} =
|\vec{E}|\cdot|\vec{n}|\cdot cos(\alpha)[/tex]
because the angles are all different, so I have to come up with some generalization. Here, I don't really know where to start at all..

The field inside an ideal conductor must always be zero. The charges are free to move, so they will always move until the field is balanced and they feel no more force.

And yes the problem is harder to calculate when the symmetry is broken.
 
  • #7
pesslovany said:
Thank you...
I have got one supplementary:
"Would the E in the shell be also 0 if the point charge wouldn't be exactly in the middle?"
It should be, but there is a problem calculating
[tex]\vec{E}\cdot\vec{n} =
|\vec{E}|\cdot|\vec{n}|\cdot cos(\alpha)[/tex]
because the angles are all different, so I have to come up with some generalization. Here, I don't really know where to start at all..

Sorry, I didn't answer the rest of your question. With the charge off center I don't think there is a short cut. I think you have to solve the Poisson equation with Dirichlet boundary conditions. This procedure is one of the least favorite tasks for first year graduate students and I probably shouldn't try to describe it in this forum. To tell the truth I haven't thought about it since graduate school because I solve all such problems numerically. Differential equations all become simple when you have a computer!
 
  • #8
pesslovany said:
Thank you...
I have got one supplementary:
"Would the E in the shell be also 0 if the point charge wouldn't be exactly in the middle?"
It should be, but there is a problem calculating
[tex]\vec{E}\cdot\vec{n} =
|\vec{E}|\cdot|\vec{n}|\cdot cos(\alpha)[/tex]
because the angles are all different, so I have to come up with some generalization. Here, I don't really know where to start at all..

On the other hand you can pretty much guess what the equipotential lines will look like, and a lot of reasoning can be done without an exact solution.

OOOHHH! It's hard to calculate what happens inside the conductor, but given that it must be an equipotential you should be able to use Gauss law to say something very interesting about the field outside.
 
  • #9
Oh yeah, I forgot to write, that I am supposed to do it for the conductor. Dont really know what am I actually supposed to do, because I don't think Its possible to calculate it for the inside...
 
  • #10
pesslovany said:
Oh yeah, I forgot to write, that I am supposed to do it for the conductor. Dont really know what am I actually supposed to do, because I don't think Its possible to calculate it for the inside...

Good. Like I said, by definition it has to be an equipotential and Gauss law applied at the outside will tell you something very interesting.
 

FAQ: Application of Gauss' law - point charge in hollow sphere

1. What is Gauss' law and how is it applied?

Gauss' law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the enclosed charge. It is applied by using a closed Gaussian surface around a point charge to calculate the electric flux and determine the electric field at a point.

2. What is a point charge and how does it relate to Gauss' law?

A point charge is a theoretical concept used to represent a single charged particle with no physical size. It relates to Gauss' law because the law is based on the assumption that the charge is located at a single point within the Gaussian surface.

3. Why is a hollow sphere commonly used in the application of Gauss' law?

A hollow sphere is often used because it allows for a symmetrical distribution of charge, making it easier to calculate the electric field using Gauss' law. The electric field is constant at every point on the surface of a hollow sphere, making the calculations simpler.

4. What is the significance of using a closed surface in Gauss' law?

A closed surface is necessary in Gauss' law because it allows for the electric flux to be calculated by considering the electric field passing through the entire surface. This is important because it takes into account all the charges enclosed within the surface, which affects the electric field at a given point.

5. Can Gauss' law be applied to other shapes besides a point charge in a hollow sphere?

Yes, Gauss' law can be applied to other shapes and configurations of charges. However, the calculations become more complex and may require the use of calculus. The hollow sphere is a common and simplified example used to demonstrate the application of Gauss' law.

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