3∫E.ds=Qen/ε
E=Qen/ε∫ds
which is constant only when Qen=0
prove that if the electric field in a region is uniform at all points then the net charge in the region is zero
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5 Answers
1∫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.
106Or 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
