Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I have read that Gaussian surface cannot pass through discrete charges. Why is it so? I have even seen in application of Gauss' Law when we imagine a Gaussian Surface passing through a charge distribution, e.g. in case of infinite plane charge carrying sheet .

If it cannot pass through discrete charges how do we use it in continuous charge distributions as same 'objection' must be there for it also.

Please explain the reason.

Here $E \rightarrow \infty$ as, $r\rightarrow 0$

If this is ambiguity then this must be same in continuous charge distribution , otherwise please state it more clearly because we can define charge to be a spherical ball and half charge can be considered inside surface (as in pic and even agreed by @JoshuaBarr).

share|improve this question
    
There is a good discussion of this in the first chapter of Purcell and Morin, Electricity and Magnetism. –  Ben Crowell Mar 22 '13 at 15:21
1  
@BenCrowell If a pdf version of this exist please provide me that , i don't have book –  Mr.ØØ7 Mar 22 '13 at 15:51
2  
There's this thing called a library. –  Ben Crowell Mar 22 '13 at 22:42
add comment

1 Answer 1

If there is a point charge located on the Gaussian surface, then it is ambiguous how much of that charge should be counted as inside the surface. The same ambiguity does not arise for a three dimensional charge distribution, or when a Gaussian surface intersects a two dimensional charge distribution only over a one dimensional region.

The same ambiguity would arise if a two dimensional charge distribution coincided with the Gaussian surface over a two dimensional region.

share|improve this answer
    
But i have seen a question where they specified that gaussian surface cuts the charge in half then what is the flux. Answer was still can't be told. WHy is it so if the ambiquity is just that we can't tell amount of charge inside the gaussian surface –  Mr.ØØ7 Mar 22 '13 at 15:53
    
@exploringnet If the charge is truly cut in "half" then you can just treat this by allowing the charge to have some extent (e.g. a sphere) and then taking the limit where its size goes to zero. The flux would simply be (1/2) Q / epsilon_0. –  Joshua Barr Mar 22 '13 at 16:01
    
Please see my extended question. –  Mr.ØØ7 Mar 23 '13 at 6:04
    
The limit of the flux will converge in the picture you drew. This is so because for any finite value of the radius, Gauss' law will hold just fine. If you take the limit in such a manner that the sphere is always split in half, then the flux will always be (1/2) Q / epsilon_0. Remember that the limit depends on the behavior of the flux with the radius near R = 0 but not on the value at R = 0. –  Joshua Barr Mar 24 '13 at 3:55
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.