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: Class 12 = af r AGAIN & AGAIN 2 2 dag a> = =EA It states that the total electric flux through a closed surface is equal to 1/é, times the net charge enclosed by the surface enclosed by the surface Formulae for Electric Field Strength Calculated From Gauss's Theorem =» Electric field due to infinitely long straight wire of charge per unit length A at a distance r from the wire is + + +/+ + + “s 9 v ty Formulae for Electric Field Strength Calculated From Gauss's Theorem => Electric field strength due to a uniformly charged thin spherical shell or conducting sphere of radius R having total charge q, at a distance r from centre is 1 q (i) at external point (Forr>R) Eext = ame ATE r2 (ii) at surf int (F R) E ae ii) at surface point (For r = — P °~ “4me9 R? (iii) at internal point (For r < R) Bint ='( THStee a condudss [E=°] ead = Gin) ics AV Ca, =9 C Cu R* As shown in the figure, q charge is placed at the open end of the cylinder with one end open. The total flux emerging from the surface of cylinder is When a 10 Uc charge is enclosed by a closed surface, the flux passing through the surface is ¢ . Now another -10 uC charge is placed inside the closed surface, then the flux passing through the surface is Two charges +q each are kept at 2a distance apart. A third charge -2q is placed midway between them. The potential energy of the system is: Y= ~ Akar aky* + hq? a a Ta ~ kg = kg + sm tT kg? = —3kq> = - 44". Va a UME ¥RA A charge of 6 iC is given to a hollow metallic sphere of radius 0.2 m. Find the potential at (i) the surface and (ii) the centre of the sphere. Equipotential Surface An equipotential surface is the surface having the same potential at each point. The surface of a charged conductor in equilibrium is a equipotential surface. e An equipotential surface and the electric field line meet at right angles. The region where E=0, Potential of the whole region must remain constant as no work is dong! in displacement of charge in it. V = Conatenk (ye Potentiah ‘Datfevmus =° AV= W 
