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Charge on metal sheet

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  • #51
BvU
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Excellent, thank you.
Oops, I completely forget Pranav: Do you already see the Gauss volume they are dangling in front of you ? And then a symmetry plane on its 45 degree axis (oops...) ?
That's two oopses in one post. Now it's even three.
 
  • #52
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Hi ehild! :smile:

Great solution but I have a few doubts regarding it.

I would solve it as follows, see attachment. The charge Q induces -Q surface charge on the upper surface of the plates. The plates are divided into strips, the surface charge densities are q2 on the left plane, q1 in the strips symmetric to the charge Q, between x=0 and x=±D, and q2 on the rightmost strip. q1+q2=-Q/2
I am sorry if I am missing something but how do you get ##q_1+q_2=-Q/2##. You said that the positive charge induced on the surface beneath gets evenly distributed. I think it should be ##q_1+q_2=-Q##. :confused:
You get q1 by integrating ε0 in the range 0<x<D, -∞<y<∞. Integrate with respect to y first. The following integral is helpful (from Wolframalpha)

[tex]\int{\frac{1}{(\sqrt{y^2+a^2})^3}dy}=\frac{y}{a^2\sqrt{y^2+a^2}}dy[/tex]
Why integrate only for 0 to D only? In your later post, you said that you used the method of images. It is possible to find E at every point on the sheet, right? So why not integrate x from -D to ∞? :confused:

One more thing, if this problem is easily solved by method of images, why TSny and Chestermiller are using such complicated methods and I still have zero idea about the Gaussian surface, I feel I am only wasting the precious time of TSny and Chestermiller by asking more and more hints and making zero progress. :redface:
 
  • #53
ehild
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Hi ehild! :smile:

Great solution but I have a few doubts regarding it.


I am sorry if I am missing something but how do you get ##q_1+q_2=-Q/2##. You said that the positive charge induced on the surface beneath gets evenly distributed. I think it should be ##q_1+q_2=-Q##. :confused:
see my picture: there are two pairs of strips with induced charge q1 and q2 on the top surface of the two connected plates.The induced charge is -Q. -Q=2q1+2q2.

Why integrate only for 0 to D only? In your later post, you said that you used the method of images. It is possible to find E at every point on the sheet, right? So why not integrate x from -D to ∞? :confused:
I felt it easier. I tried to use symmetry. But you can avoid my strips and integrate to the whole plate.

One more thing, if this problem is easily solved by method of images, why TSny and Chestermiller are using such complicated methods and I still have zero idea about the Gaussian surface, I feel I am only wasting the precious time of TSny and Chestermiller by asking more and more hints and making zero progress. :redface:
To tell the truth, I do not understand their method, either. :redface: They suggest to extend the Gaussian surface to infinity.


ehild
 
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  • #54
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I felt it easier. I tried to use symmetry. But you can avoid my strips and integrate to the whole plate.
I think I see why you did that. Simply brilliant. :cool:

Thanks a lot ehild! :smile:
 
  • #55
TSny
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I've made an attempt to summarize Chestermiller's "no integration" method with a couple of pictures. Any errors are my own.

The method is to replace the original problem with a superposition of other problems that are easier to analyze but still produce the same net charge on the right half of the horizontal plate.

The first picture attached shows the original problem as Figure A. This figure is thought of as a superposition of Figure B and Figure C. Since Figure B does not contribute any net charge to the right plate, we can just consider Figure C. Note that the distribution of the net charge on the right plate in Figure C is not the same as the distribution in Figure A. But, that's OK since we are only interested in the net charge on the right plate, and the net charge on the right plate is the same in Figures A and C. (The distribution of charge in Figure A is the superposition of the distributions in Figures B and C.)

In the second picture attached, Figure C is replaced by a superposition of Figure 1 and Figure 2. Figure 1 provides no net charge on the right plate. So, the net charge on the right plate in Figure 2 is the same as the net charge on the right plate in Figure C which is the same as the net charge on the right plate of the original problem (Figure A).

A vertical conducting plate is added to Figure 2 to get Figure 3. This additional plate does not disturb the charge distribution on the horizontal plate. Then, using Gauss' law and symmetry you can deduce the surface charge on the plates in Figure 3 and arrive at the final answer.
 

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Thanks TSny! :smile:

In the second picture attached, Figure C is replaced by a superposition of Figure 1 and Figure 2. Figure 1 provides no net charge on the right plate. So, the net charge on the right plate in Figure 2 is the same as the net charge on the right plate in Figure C which is the same as the net charge on the right plate of the original problem (Figure A).
Ok, this is going to be silly, why no charges are induced in the case of figure 1? :confused:

A vertical conducting plate is added to Figure 2 to get Figure 3. This additional plate does not disturb the charge distribution on the horizontal plate. Then, using Gauss' law and symmetry you can deduce the surface charge on the plates in Figure 3 and arrive at the final answer.
I still don't get the point of adding the conducting plate and why it doesn't affect the given setup. Is it not possible to find the induced charges directly without the conducting plate? Why does adding the conducting plate allows one to use these symmetrical arguments to reach the answer? :confused:

Is this method illustrated in any book? I have both Purcell and Griffiths, is this method present in any of the two books?
 
  • #57
ehild
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The method is to replace the original problem with a superposition of other problems that are easier to analyze but still produce the same net charge on the right half of the horizontal plate.


It is ingenious! Thank you TSny.

ehild
 
  • #58
TSny
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It is ingenious!
Yes. Hats off to Chestermiller.
 
  • #59
ehild
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My only concern is applying Gauss' Law. For that, it had to be known that the electric field tends to zero faster then 1/r2 (r is the distance from the charge in the quarter space).
Anyway, the symmetry considerations and superposition principle work well. And that was Chest's idea.


ehild
 
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  • #60
ehild
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Ok, this is going to be silly, why no charges are induced in the case of figure 1? :confused:

Charges are induced on the top surfaces of both plates, but they are equal because of symmetry, just like the opposite charge on the back surfaces. So the net charge of both plates is zero.

I still don't get the point of adding the conducting plate and why it doesn't affect the given setup. Is it not possible to find the induced charges directly without the conducting plate? Why does adding the conducting plate allows one to use these symmetrical arguments to reach the answer? :confused:

Is this method illustrated in any book? I have both Purcell and Griffiths, is this method present in any of the two books?


An equipotential surface and a metal surface are alike as the potential is the same and the electric field lines are normal to both. You get the same electric field of the four charges in each quadrant without the vertical plate and with it. In the latter case, when determining the field in one quadrant, the mirror charges are the same at the same places as the real ones.

The superposition principle is applied at the very beginning of studies of Electricity. When you determine the field of two charges, you just add the contributions.
And symmetry is very inherent property - if something is symmetric, its properties are also symmetric.

You can not learn every method of solution of problems. You have to practice and then you have the eye to see the solution. And it is no problem if you do not find the nicest method. In this problem, the integral method was also satisfactory.

In real life, you need to solve problems which do not have nice solutions.


ehild
 
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Yes. Hats off to Chestermiller.
Thanks TSny. You did a beautiful job of illustrating and articulating what I was trying to say. I might also add that, initially, I was unable to figure out how to do this, and it wasn't until I received the hints from yourself and Tiny Tim that I was able to put this analysis together. So, in reality, this was really a team effort.

Chet
 
  • #62
TSny
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My only concern is applying Gauss' Law. For that, it had to be known that the electric field tends to zero faster then 1/r2 (r is the distance from the charge in the quarter space).
I think that for the limit of infinite plates, you can argue that the field in the first quadrant for the left figure is the same as in the middle figure where the other three charges have been removed. That's because for infinite plates, the region of the first quadrant alone is a well-defined boundary valued problem. The solution of the field for the first quadrant can then be obtained by the method of images and would be the same field as produced in the first quadrant by the four point charges alone of the figure on the right. This is a quadrupole field that will fall off rapidly with distance.

But, maybe I'm overlooking something.

I've spent about ten times as much time trying to think through this problem without integration as it took to just do the integration. But it's fun.
 

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I am still going through all the replies in this thread, thanks a lot everyone! :)

That was an information overload for me this time. :rolleyes:
 
  • #64
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Thanks TSny. You did a beautiful job of illustrating and articulating what I was trying to say. I might also add that, initially, I was unable to figure out how to do this, and it wasn't until I received the hints from yourself and Tiny Tim …
yes, thanks TSny! :smile:

for the record, i had no idea until i saw TSny's hint at post #26!
 
  • #65
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Talk about a late entry! But, I'd go with:

1. Consider the two plates as one.
2. Image the charge below the plate at z = -d.
3. Compute the E field all along the plate based on the removal of the plate and its substitution by the (negative) image charge.
4. Use σ = ε0E to get σ(x,y).
5. Integrate σ to get the charge distribution along the plate, then divide it up per the location of Q.

Bottom line, I would base my approach on the image technique. I don't know to what extent this was done by all the other posters.
 

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