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Applying Gauss's Law Several Times 
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#1
Feb1213, 10:04 AM

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1. The problem statement, all variables and given/known data
I am working on a problem that is similar to this one: http://www.physicsforums.com/showthread.php?t=598914 2. Relevant equations 3. The attempt at a solution When r < a, why doesn't the electric fields generated by the concentric conducting, hollow sphere affect the Gaussian surface at that point?? Why is the electric field at a point in the concentric conducting, hollow sphere zero? After all, it doesn't specify that it is in electrostatic equilibrium. Also, what is the difference between the entire system being in electrostatic equilibrium and just the concentric conducting, hollow sphere being in electrostatic equilibrium? Can an insulator be in electrostatic equilibrium? 


#2
Feb1213, 04:54 PM

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#3
Feb1313, 07:14 AM

P: 946

So, you are saying that if I was to generate a Gaussian surface inside of the conductor (say, a sphere), then the flux through this surface is zero, because the electric field through this surface is zero? How can this be? I know that the charge isn't distributed inside of the conductor, the charge resides on the surface, but the Gaussian surface would still enclose the insulator, which has a charge.
EDIT: Something just came to my mind: does the insulator induce the electrostatic equilibrium? And if the insulator had a positive charge, it would cause the electrons in the conductor to move to the surface closest to the insulator? Leaving the other surface "less" negative? 


#4
Feb1313, 07:41 AM

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Applying Gauss's Law Several Times



#5
Feb1313, 08:00 AM

P: 946

So, using the diagram given in the link, if I was to place a Gaussian sphere directly between the insulator and conductor, the electric field lines emanating from the insulator will "hit" the sphere from the inside, and the electric field lines emanating from the conductor "strike" the very surface of the Gaussian sphere; and so, the electric fields combine to cancel out?
And is this the reason why we don't have to consider the electric field of the conductor when generating a Gaussian surface that meets the condition of r < a, because the electric fields lines never even make it that far? EDIT: Also, is there a way of quantifying the amount of induction that the insulator affects? Would it just be equal and opposite to the charge of the insulator? And the remaining charge would reside on the outside? 


#6
Feb1313, 09:31 AM

P: 946

For example, say the insulator has a charge of Q1 = 3.60 C, and the conductor had a total charge of 2.00 C. Would the inner surface of the conductor would have a charge of Q2 = Q1 > Q2 = 3.60 C; and would this then imply that Q3, the outer surface of the conductor, would have a charge of Q2 + Q3 = 2.00 C > Q3 = 1.6 C?



#8
Feb1313, 09:59 AM

P: 946

Why is it correct, if you don't mind me asking?



#9
Feb1313, 01:18 PM

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Is there some law or principle of induction?



#10
Feb1313, 01:55 PM

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#11
Feb1313, 02:02 PM

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So, would electric field at a radius of a < r < b be zero?
EDIT: And would it be zero because the number of electric field lines radiating out of the Gaussian surface equals the number of electric field lines coming in? 


#12
Feb1313, 02:16 PM

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#13
Feb1313, 02:19 PM

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#14
Feb1313, 02:19 PM

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So, the flux through a Gaussian only depends on the charge enclosed, further implying that it only depends on the electric field emanating from that enclosed charge?



#15
Feb1313, 02:19 PM

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E field lines originate from Positive and terminate on Negative charge. The negative charge on the inner surface of the conductor does not interfere with the electric field at a < r < b . So the field at a radius of a < r < b is not zero. Gauss's Law says the flux though a sphere of such radius is not zero. 


#16
Feb1313, 02:20 PM

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#17
Feb1313, 02:34 PM

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