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www.GOALias.blogspot.comPhysicsEXAMPLE 2.81 E03V = E0( d) + ( d)4 K 41 3 K + 3= Ed0( + ) = V04 4K4KThe potential difference decreases by the factor (K + 3)/4K while thefree charge Q 0on the plates remains unchanged. The capacitancethus increasesQ0 4KQ04KC = = = C0V K + 3 V K + 3078FIGURE 2.26 Combination of twocapacitors in series.FIGURE 2.27 Combination of ncapacitors in series.2.14 COMBINATION OF CAPACITORSWe can combine several capacitors of capacitance C 1, C 2,…, C nto obtaina system with some effective capacitance C. The effective capacitancedepends on the way the individual capacitors are combined. Two simplepossibilities are discussed below.2.14.1 Capacitors in seriesFigure 2.26 shows capacitors C 1and C 2combined in series.The left plate of C 1and the right plate of C 2are connected to twoterminals of a battery and have charges Q and –Q ,respectively. It then follows that the right plate of C 1has charge –Q and the left plate of C 2has charge Q.If this was not so, the net charge on each capacitorwould not be zero. This would result in an electricfield in the conductor connecting C 1and C 2. Chargewould flow until the net charge on both C 1and C 2is zero and there is no electric field in the conductorconnecting C 1and C 2. Thus, in the seriescombination, charges on the two plates (±Q) are thesame on each capacitor. The total potential drop Vacross the combination is the sum of the potentialdrops V 1and V 2across C 1and C 2, respectively.V = V 1+ V 2=QCQ+C(2.55)1 2V 1 1i.e., = +Q C1 C, (2.56)2Now we can regard the combination as aneffective capacitor with charge Q and potentialdifference V. The effective capacitance of thecombination isQC = (2.57)VWe compare Eq. (2.57) with Eq. (2.56), andobtain1 1 1= +C C C(2.58)1 2