InterviewSolution
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InterviewSolution
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State Gauss's law. Derive an expression for electric intensity at a point outside the uniformly charged shell. |
Answer» Solution :Gauss.s theorem: "The total outward electric flux passing through a closed surface in air is `(1/epsilon_0)` times the total charge enclosed by it..  Let +Q. be the charge enclosed by a hollow conductor of RADIUS .R.. Let .p. be a point at a distance .r. from the centre of the conductor Let .ds be a small element of area SURROUNDING the point. A normal drawn from the point coincides with the direction of E. HENCE cos `0^@=1`. From Gauss. theorem, `TOEF=phi=(1/epsilon_0)Q` ...(1) by definition `phi=E cos theta sum ds`. where `sumds=4pir^2 , cos theta=cos 0^@ =1` Hence , `phi=E. 4pir^2` ...(2) Comparing (1) and (2) we write `E=(1/(4piepsilon_0))Q/r^2` For a point on the surface , r =R . `E=(1/(4piepsilon_0))(Q/R^2)` . This electric field INTENSITY is maximum. Since electric flux depends on the charge enclosed, and electric field intensity depends on the electric flux, electric field remains ZERO at all points inside the spherical hollow conductor. |
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