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www.GOALias.blogspot.comPhysicsSolution(a) In the given network, C 1, C 2and C 3are connected in series. Theeffective capacitance C′ of these three capacitors is given by1 1 1 1= + +C′C1 C2 C3For C 1= C 2= C 3= 10 µF, C′ = (10/3) µF. The network has C′ and C 4connected in parallel. Thus, the equivalent capacitance C of thenetwork isEXAMPLE 2.9⎛10 ⎞C = C′ + C 4= ⎜ + 10 ⎟ µF =13.3µF⎝ 3 ⎠(b) Clearly, from the figure, the charge on each of the capacitors, C 1,C 2and C 3is the same, say Q. Let the charge on C 4be Q′. Now, sincethe potential difference across AB is Q/C 1, across BC is Q/C 2, acrossCD is Q/C 3, we haveQ Q Q+ + = 500 V.C C C1 2 3Also, Q′/C 4= 500 V.This gives for the given value of the capacitances,10−3Q = 500V× µ F = 1.7 × 10 C and3−3Q′= 500V× 10 µ F = 5.0 × 10 CFIGURE 2.30 (a) Work done in a smallstep of building charge on conductor 1from Q′ to Q′ + δ Q′. (b) Total work donein charging the capacitor may beviewed as stored in the energy ofelectric field between the plates.802.15 ENERGY STORED IN A CAPACITORA capacitor, as we have seen above, is a system of two conductors withcharge Q and –Q. To determine the energy stored in this configuration,consider initially two uncharged conductors 1 and 2. Imagine next aprocess of transferring charge from conductor 2 to conductor 1 bit bybit, so that at the end, conductor 1 gets charge Q. Bycharge conservation, conductor 2 has charge –Q atthe end (Fig 2.30 ).In transferring positive charge from conductor 2to conductor 1, work will be done externally, since atany stage conductor 1 is at a higher potential thanconductor 2. To calculate the total work done, we firstcalculate the work done in a small step involvingtransfer of an infinitesimal (i.e., vanishingly small)amount of charge. Consider the intermediate situationwhen the conductors 1 and 2 have charges Q′ and–Q′ respectively. At this stage, the potential differenceV′ between conductors 1 to 2 is Q′/C, where C is thecapacitance of the system. Next imagine that a smallcharge δ Q′ is transferred from conductor 2 to 1. Workdone in this step (δ W′ ), resulting in charge Q′ onconductor 1 increasing to Q′+ δ Q′, is given byQδW = V′ δQ′ =′ δQ′ (2.68)C