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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 char

Best of Luck STUDENTS Que1. Two small conducting spheres of equal radius have charges 10μC & -20μC respectively placed at a distance R from each other. They experience force F1 . If they are brought in contact and separated to the same distance, they experience force F2 .   Ratio of F1 to F2 is ___ a) 1:2 b) 1:1 c) 2:1 d) 8:1 Pyaare bacchon, agar online study se headache hota hai toh dekh lena, aaram milega Que2. Electric field intensity at a point in between two parallel sheets with like charges of same surface charge densities σ is ___ a) σ/εo b) zero c) σ/2εo d) none of these Que3. There exists an electric field of 2 N/C along y direction. The flux passing through the square of side 1 m placed in xy plane inside the electric field is - a) Zero b) 2 Nm²/C c) 1 Nm²/C d) √3 Nm²/C Que4. A mass of 20g has a charge of 3mC . It moves with a velocity of 20m/s and enters a region of electric field of 80N/C in the same direction as the velocity of the mass. The velocity of t