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Deduction of Coulombs Inverse Square Law From Gauss Law


Gauss law to an isolated point charge + q lying in vacuum. The Gaussian surface is a spherical surface of radius r having its centre at the point charge. E and dS at any point on the surface are directed radially outward. Angle between E and dS is zero.

. :  E . dS = E dS cos 0 = E dS

From Gauss’ law,
    ф E . dS = ф E dS = q/ εo

E us constant at all points on the surface. Thus

    E ф dS = q/ εo
 
or    E ( 4 π r2 ) q/ εo

. :    E = 1 / (4 π εo) q / r2

Let us put a second point charge q’ on this surface. The force F that acts on this charge q’ is

    F = q’ E = 1 / (4 π εo) q q’ / r2 r

The direction of this force is along the direction of E, i.e., along the radius vector drawn from q to q’. This is coulomb’s law.




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