coulomd law from gauss theorm 
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Asked by Seema Jain | 11th Apr, 2015, 04:52: PM

Expert Answer:

Consider a positive charge q at O. Imagine a sphere of radius r with centre O as shown in the figure. The magnitude of electric field intensity at every point on the surface of the sphere is the same and is directed radially outwards. Let ds represent a small area element on the surface of the sphere. The direction of the area element vector will be along begin mathsize 12px style straight E with rightwards arrow on top end style. i.e. θ = 0°.

begin mathsize 12px style According space to space gauss apostrophe straight s space theorem comma contour integral subscript straight S space stack straight E space with rightwards arrow on top. ds with rightwards arrow on top equals straight q over straight epsilon subscript 0 contour integral subscript straight S space straight E space ds space cos space 0 degree equals straight q over straight epsilon subscript 0 or space straight E contour integral subscript straight S space ds equals straight q over straight epsilon subscript 0 straight E open parentheses 4 πr squared close parentheses space equals straight q over straight epsilon subscript 0 straight E equals fraction numerator straight q over denominator 4 πε subscript 0 straight r squared end fraction Thus space the space electric space field space intensity space at space any space point left parenthesis straight P right parenthesis straight r space distant space from space an space isolated space point space charge space straight q space at space the space centre space of space the space sphere space is colon space straight E equals fraction numerator straight q over denominator 4 πε subscript 0 straight r squared end fraction If space another space point space charge space straight q subscript 0 space were space placed space at space straight P. space then space the space force space on space straight q subscript 0 space would space be colon straight F space equals space straight E cross times straight q subscript 0 straight F equals fraction numerator straight q over denominator 4 πε subscript 0 straight r squared end fraction cross times straight q subscript 0 straight F equals fraction numerator qq subscript 0 over denominator 4 πε subscript 0 straight r squared end fraction This space is space the space Coloumb apostrophe straight s space law. end style

Answered by Jyothi Nair | 13th Apr, 2015, 01:00: PM

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