**GAUSS’S LAW**

The surface integral of
electrostatic field

**E**produce by any source over any closed surface S enclosing a volume V in vacuum i.e. total electric flux over the closed surface S in vacuum is**1/****epsilon**times the total charge Q contained inside S.
The surface chosen to
calculate the surface integral is called

**Gaussian surface**.**PROOF OF GAUSSIAN THEOREM**

The flux through area element ΔS is

The unit vector rˆ is along the radius vector from the centre to
the area element. Now, since the normal to a sphere at every point is along the
radius vector at that point, the area element ΔS and rˆ have the same
direction. Therefore,

**SIGNIFICANCE OF GAUSS’S LAW**

(i)
Gauss’s law is true for any
closed surface, no matter what its shape or size.

(ii)
The term q on the right side
of Gauss’s law includes the sum of all charges enclosed by the surface. The
charges may be located anywhere inside the surface.

(iii)
In the situation when the
surface is so chosen that there are some charges inside and some outside, the
electric field is due to all the charges, both inside and outside S. The term q
on the right side of Gauss’s law.

(iv)
Gauss’s law is often useful
towards a much easier calculation of the electrostatic field when the system
has some symmetry. This is facilitated by the choice of a suitable Gaussian
surface.

(v)
Finally, Gauss’s law is based
on the inverse square dependence on distance contained in the Coulomb’s law.
Any violation of Gauss’s law will indicate departure from the inverse square
law.

Good

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ReplyDeleteProve is using surface as sphere(Special Case). General Proof-http://web.mit.edu/viz/EM/visualizations/coursenotes/modules/guide04.pdf

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ReplyDeleteGood but it is of NCERT book

ReplyDeletenice notes

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ReplyDeletewhy flux through a closed surface containing charge q is equal to q by E0 for any surface??

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