Electric Fields and Capacitors 89 Worked Example 3.9 Q Three capacitors of value 4.7 F, 3.9 F and 2.2 F are connected in parallel. Calculate the resulting capacitance of this combination. A In this case, since all of the capacitor values are in F then it is not necessary to show the 10 6 multiplier in each case, since the answer is best quoted in F. However it must be made clear that this is what has been done. CCCCCC112347392208 microfaradso F ....Ans 3.13 Capacitors in Series Three parallel plate capacitors are shown connected in series in Fig. 3.10 . Each capacitor will receive a charge. However, you may wonder how capacitor C2 can receive any charging current since it is sandwiched between the other two, and of course the charging current cannot ﬂ ow through the dielectrics of these. The answer lies in the explanation of the charging process described in section 3.6 earlier. To assist the explanation, the plates of capacitors C1 to C3 have been labelled with letters. C1ABC DEEFC2C3 Fig. 3.10 Plate A will lose electrons to the positive terminal of the supply, and so acquires a positive charge. This creates an electric ﬁ eld in the dielectric of C1 which will cause plate B to attract electrons from plate C of C2 . The resulting electric ﬁ eld in C2 in turn causes plate D to attract electrons from plate E . Finally, plate F attracts electrons from the negative terminal of the supply. Thus all three capacitors become charged to the same value. Having established that all three capacitors will receive the same amount of charge, let us now determine the total capacitance of the arrangement. Since the capacitors are of different values then each will acquire a different p.d. between its plates. This is illustrated in Fig. 3.11 .