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www.GOALias.blogspot.comThe proof clearly goes through for any number ofcapacitors arranged in a similar way. Equation (2.55),for n capacitors arranged in series, generalises toQ Q QV = V1 + V2 + ... + Vn= + + ... +C1 C2 C(2.59)nFollowing the same steps as for the case of twocapacitors, we get the general formula for effectivecapacitance of a series combination of n capacitors:1 1 1 1 1= + + + ... +C C C C C(2.60)1 2 3 n2.14.2 Capacitors in parallelFigure 2.28 (a) shows two capacitors arranged inparallel. In this case, the same potential difference isapplied across both the capacitors. But the plate charges(±Q 1) on capacitor 1 and the plate charges (±Q 2) on thecapacitor 2 are not necessarily the same:Q 1= C 1V, Q 2= C 2V (2.61)The equivalent capacitor is one with chargeQ = Q 1+ Q 2(2.62)and potential difference V.Q = CV = C 1V + C 2V (2.63)The effective capacitance C is, from Eq. (2.63),C = C 1+ C 2(2.64)The general formula for effective capacitance C forparallel combination of n capacitors [Fig. 2.28 (b)]follows similarly,Q = Q 1+ Q 2+ ... + Q n(2.65)i.e., CV = C 1V + C 2V + ... C nV (2.66)which givesC = C 1+ C 2+ ... C n(2.67)Example 2.9 A network of four 10 µF capacitors is connected to a 500 Vsupply, as shown in Fig. 2.29. Determine (a) the equivalent capacitanceof the network and (b) the charge on each capacitor. (Note, the chargeon a capacitor is the charge on the plate with higher potential, equaland opposite to the charge on the plate with lower potential.)Electrostatic Potentialand CapacitanceFIGURE 2.28 Parallel combination of(a) two capacitors, (b) n capacitors.FIGURE 2.29EXAMPLE 2.979