Statement : The net electric flux through any closed surface is equal to 1/εο times the charge enclosed by the surface. Φ = q/εο where, q = charge enclosed by the surface PROOF consider, a positive point charge q which is enclosed by an imaginary spherical shell of radius r with 'q' at the centre. ds is a very very small area element of the shell. As we know, the electric field of a positive charge is always radially outward. Pyaare bacchon, agar online study se headache hota hai toh dekh lena, thoda aaram milega tumhe Now, small electric flux through 'ds' is - But, E = q/(4πεο.r²) Therefore, dΦ = [ q/(4πεο.r²) ].ds Now, Net electric flux through the whole spherical surface is - (Integrating both sides and taking limits) Question Find the value of x in the figure if the net flux through the Gaussian surface is '1/2εο' Nm²/C. For "electric field near an uniformly...
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