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Electrostatic Potential Question - Basic physics

  • Thread starter Zubilaemo
  • Start date
  • #1
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Hi Everyone! I'm new to this forum, and English isn't my first language - I would really appreciate your help with this question :)

Homework Statement



[PLAIN]http://img225.imageshack.us/img225/3807/1milliontotheground.jpg [Broken]

there are given 2 thin shells with radiuses b and c, and a full sphere with radius a as shown in the image.

the inner sphere is isolated, its charge is Q and it is divided equally.
the middle shell is made from a conductive material and it is grqounded.
the outer shell is made from isolated material and its charge is Q*10^6 and it is divided equally.


Homework Equations



the potential at inf is 0. what is the potential in the center of the sphere?

pick one option
[PLAIN]http://img706.imageshack.us/img706/2631/1option.jpg [Broken]


The Attempt at a Solution



As you can see I was wrong - but I don't understand why. Can someone please help me?

Sorry for the bad english. Would appreciate a lot!

Zubi
 
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Answers and Replies

  • #2
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As far as I know, this looks exactly like a Gaussian Law problem with a spherical, symmetric charge distribution. You should consult your text for this formula. It should be fairly straight forward from there...
 
  • #3
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thanks for your help! It's just that I don't understand what the result of the grounding is in this case - what does this mean about the charge of the middle shell?

What I thought is that the charge on the inside of the middle shell is -1Q, and thus the potential in the centre is Q/a-Q/b, but as you can see I was wrong...
 
  • #4
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thanks for your help! It's just that I don't understand what the result of the grounding is in this case - what does this mean about the charge of the middle shell?

What I thought is that the charge on the inside of the middle shell is -1Q, and thus the potential in the centre is Q/a-Q/b, but as you can see I was wrong...
 
  • #5
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Well, explain to me your thought process, why did you think that the charge on the second surface is equal to -Q? This may bring up your problem, perhaps where you went wrong. Just out of curiosity, did you take into account the 1[MC], or did you use 1[C]?
 
  • #6
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Hey!
I thought that the 1MC charge didn't have any effect, as it surrounds a grounded sphere. Perhaps I was wrong in this assumption?
 
  • #7
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Whoa there, read your assumptions. There is only ONE sphere (the inner) and then there are two shells. Grounding one shell shouldn't alter the other. It may change the electric flux density depending on its location, but the charge shouldn't change...I think that's right...
 
  • #8
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Hmm.. Yeah, I meant Shell.

But - Why should the outer shell have any influence on the potential if the middle shell is grounded?
 
  • #9
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Where is charge located in this problem?
 
  • #10
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This was a matter approached in my previous semester, and my "guess" (because that's the first problem of this kind where I see a sphere with a ground connection and I can't be sure) is that the middle sphere has charge 0 Q on its surface (due to that ground connection). It is an interesting problem.

You should pay attention to wether the charges are located on the surface or distributed equally inside the spheres, since depending on the situation there are certain assumptions that you can make about the electric field and the potencial. These assumptions can be derived from the Gaussian Law and from an extra equation that relates the electric field with the flux given by the Gaussian Law.

Just to give you an example, if you have a sphere were the charges are distributed on its surface, the electric field inside the sphere is zero (you can show this by the formulas that I and satchmo05 mentioned), and from this you can conclude that the potential inside the sphere is constant.

Also, English is not my first language. If there's something you didn't understand, say something :biggrin:

PS: I was posting at the same time as you guys, so I might be a little late
 
  • #11
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I can calculate the potential from the charge... how else would i calculate the potential?
 
  • #12
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from the E-field, but I think that may be a bit past the scope of this problem.
 
  • #13
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This was a matter approached in my previous semester, and my "guess" (because that's the first problem of this kind where I see a sphere with a ground connection and I can't be sure) is that the middle sphere has charge 0 Q on its surface (due to that ground connection). It is an interesting problem.
Like i corrected myself earlier, the middle is a shell not the sphere. the only sphere is the middle one.

You should pay attention to wether the charges are located on the surface or distributed equally inside the spheres, since depending on the situation there are certain assumptions that you can make about the electric field and the potencial. These assumptions can be derived from the Gaussian Law and from an extra equation that relates the electric field with the flux given by the Gaussian Law.
The inner sphere is equally distibuted. the outer shells are contuctive.

Just to give you an example, if you have a sphere were the charges are distributed on its surface, the electric field inside the sphere is zero (you can show this by the formulas that I and satchmo05 mentioned), and from this you can conclude that the potential inside the sphere is constant.
Hmm. You're seriously making me wonder if the answer isn't simply Zero.. :)
 
  • #14
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from the E-field, but I think that may be a bit past the scope of this problem.
you mean Pot=-Integral(E) ?

I suppose I could do that, but then in order to know E I need to know the charge... :)
 
  • #15
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exactly, you are given no vectors in this problem so the del operator would be difficult to use without it, haha.
 
  • #16
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Hmm. You're seriously making me wonder if the answer isn't simply Zero.. :)
If the outside shell only has charges on its surface, the potential inside will be constant. On a first guess it can't be zero. But then again there's that middle sphere.

I remember seeing in my book an equation that calculates the electric field of a sphere whose charge is distributed equally inside the object. Again, you'd have to demonstrate it using the Gaussian Law.

Anyway, I've looked at this over and over and don't have the slightest idea of how to solve it. If you manage to solve it, please post your work :)
 
Last edited:
  • #17
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OK so this is how I solved it (with some help from a friend):

because the middle shell is grounded the total potential of it is Zero. This potential is also calculated from the "donations" of Sphere 1 and Shell 3 and this is what determines the charge that exists on the shell. So:

(Q*10^6)/C + Q/B + q2 /B = 0

and thus q2 (the charge that is on shell 2) = (-b*Q*10^6)/C - Q
Now we can calculate the total potential in the middle:

Phi = 3Q/2a (from the isolated sphere) + (q2/b) (from grounded shell 2) + Q*10^6/c (from shell 3) = ... = 3Q/2a - Q/b = Q(3/2a - 1/b) and the answer is g.
 

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