gauss

prove that if the electric field in a region is uniform at all points then the net charge in the region is zero

5 Answers

3
msp ·

∫E.ds=Qen/ε

E=Qen/ε∫ds

which is constant only when Qen=0

9
Celestine preetham ·

msp chk the integration type

ans is fundamenatlly wrong

1
vector ·

msp i ve a doubt plzzz tell how u brought E out of integral sign

1
metal ·

∫E.ds = 0 for uniform E because for any closed surface, if a field line enters it, it is bound to come out (for uniform E).
This is obvious isn't it------ a field line cannot suddenly get lost, neither can it emerge out of nowhere------ because if this were the case the field wouldn't be uniform any more.

So, ∫E.ds = 0 = Qenclosed/εo
Hence, Qenclosed = 0. This is true for any closed surface in the region, and so it can be concluded that net charge in the region is zero.

106
Asish Mahapatra ·

Or take a cubical Gaussian surface with two opposite faces exactly perpendicular to the electric field. Now the area vector of the remaining 4 faces is perpendicular to the field

Take the electric flux over the entire cube.
The flux through the latter 4 surfaces is zero (E.S = 0 as θ = 90°)
The flux through one surface whose area vector is parallel to the field is EScos0 and through the other is EScosÎ = -ES

So, total flux through cube = ES-ES = 0
Hence from Gauss Law, charge enclosed = 0

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