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Question
State Gauss’s law on electrostatics and drive expression for the electric field due to a long straight thin uniformly charged wire (linear charge density λ) at a point lying at a distance r from the wire.
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Solution
Gauss' Law states that the net electric flux through any closed surface is equal to `1/epsilon_0` times the net electric charge within that closed surface.
`oint vec" E".d vec" s" = (q_(enclosed))/epsilon_o`
In the diagram, we have taken a cylindrical gaussian surface of radius = r and length = l.
The net charge enclosed inside the gaussian surface `q_(enclosed) = lambdal`
By symmetry, we can say that the Electric field will be in radially outward direction.
According to gauss' law,
`oint vec"E".d vec"s" = q_(enclosed)/epsilon_o`
`int_1 vec"E" .d vec"s" + int_2 vec"E" .d vec"s" + int_3 vec"E". d vec"s" = (lambdal)/epsilon_o`
`int_1 vec"E". d vec"s" & int_3 vec"E". d vec"s" "are zero", "Since" vec"E" "is perpendicular to" d vec"s"`
`int_2 vec"E" . d vec"s" = (lambdal)/epsilon_o`
`"at" 2, vec"E" and d vec"s" "are in the same direction, we can write"`
`E.2pirl = (lambdal)/epsilon_o`
`E = lambda/(2piepsilon_o r)`
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 (i) |
 (ii) |
 (iii) |
 (iv) |
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