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Q & A: Gaussian surface

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Most recent answer: 07/09/2009
Q:
Sir.Is there any charge present in a gaussian surface or not.I mean is the total charge in a gaussian surface zero or not.Because while doing a sum I got a solution saying that the total electric field strength inside a gaussian is zero.So if the total charge is zero then the total field strength may be zero.
- Prudhvi Raj Borra (age 16)
Machilipatnam,Andhra,India.
A:
A Gaussian surface is a mathematical construct.  Its use comes in computing the total charge enclosed within a volume by performing an integral over the enclosing surface.  If you find the value of a Gaussian integral to be zero, that only tells you that the net enclosed charge is zero.  For example suppose you have an electric dipole at the origin.  The net charge is zero and the Gaussian integral over an enclosing sphere will be zero.  But the electric field inside the enclosed volume is definitely non-zero.  It just varies from point to point.
There is a nice, short, description of it in: 
LeeH

(published on 07/07/2009)

Follow-Up #1: Electric field of a dipole

Q:
Sir How can an electric field be present inside the gaussian surface when there is no charge..The Electric field is only due to the charge.How could it be possible to create a field with net charge zero
- Prudhvi Raj Borra (age 16)
Machilipatnam,Andhra,India.
A:
Hi Prudhvi,  good to see you are thinking about this problem.  You always ask
challenging questions. 

Consider an electric dipole, a positive charge and a equal negative charge separated by a small distance.  The electric field due to a dipole is well known.  It is not zero.
See:   
However the net charge within a spherical shell containing the dipole is zero.  Gauss's Law has to do with the surface integral off the dot product of the total electric field and the normal to the surface.   Actually doing the necessary surface integral with the enclosed dipole is mathematically challenging, however, if you are successful you will find zero.  That's Gauss's Law.  See:  

LeeH




(published on 07/09/2009)

Follow-up on this answer.