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Prof of coulambs law from gauss …

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Prof of coulambs law from gauss theroum

    • Posted by Himanshu Rai 7 years, 1 month ago

    CBSE > Class 12 > Physics
    • 1 answers

    Prerna Kumari 7 years, 1 month ago

    Let us consider a spherical gaussian surface having charge 'q' at the centre and charge 'Q' at the circumference Since, electric flux is scalar product of electric intensity and area vector Therefore, Electric flux = E (due to q at Q)•A(at Q) =EAcos0 = EA = E(4πr^2) Now, from gauss law Electric flux =charge enclosed/8.85 ×10^(-12) = q/8.85×10^(-12) Therefore, E(4πr^2) = q/8.85×10^(-12) E = q/4πr^2 × 8.85×10^(-12) = Kq/r^2 Force expericed by Q , F = QE = Q(kq/r^2) =kqQ/r^2

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