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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

Do   your best Q1. If a charged body A having charge +2q touched another uncharged body B . Then A touched another charged body C having charge +2q . Find the force between A and C if they are separated by 1 m. Q2. A system has two charges Q1 = 3μC and Q2 = -3μC . The coordinates of Q1 and Q2 are (10,4,2)m and (5,-9,-10)m respectively. (a) Name the system. (b) Find the net charge of the system. (c) Find the dipole moment of the system. Q3. find the force between the total negative and the total positive charge extracted from 100 ml of water and then kept at 10 cm apart. Q4. Find the nature and magnitude of  x  so that "P" is a null point. Q6. find the amount of two equal charges separated by 10 cm so that the force between them can support the weight of a 40 kg person. ( Take, g = 10 m/s² ) ---------------------------------------------------------------------------------------------- ANSWERS Ans1. , Ans2. , Ans3. (( all questions I have discu