College Physics

Figure 19.21(a) Capacitors connected in series. The magnitude of the charge on each plate isQ. (b) An equivalent capacitor has a larger plate separationd. Series

connections produce a total capacitance that is less than that of any of the individual capacitors.

We can find an expression for the total capacitance by considering the voltage across the individual capacitors shown inFigure 19.21. Solving

C=

Q

V

forVgivesV=

Q

C

. The voltages across the individual capacitors are thusV 1 =

Q

C 1

,V 2 =

Q

C 2

, andV 3 =

Q

C 3

. The total voltage is

the sum of the individual voltages:

V=V 1 +V 2 +V 3. (19.60)

Now, calling the total capacitanceCSfor series capacitance, consider that

(19.61)

V=

Q

CS

=V 1 +V 2 +V 3.

Entering the expressions forV 1 ,V 2 , andV 3 , we get

Q (19.62)

CS

=

Q

C 1

+

Q

C 2

+

Q

C 3

.

Canceling theQs, we obtain the equation for the total capacitance in seriesCSto be

1 (19.63)

CS

=^1

C 1

+^1

C 2

+^1

C 3

+ ...,

where “...” indicates that the expression is valid for any number of capacitors connected in series. An expression of this form always results in a total

capacitanceCSthat is less than any of the individual capacitancesC 1 ,C 2 , ..., as the next example illustrates.

Total Capacitance in Series,Cs

Total capacitance in series:^1

CS

=^1

C 1

+^1

C 2

+^1

C 3

+ ...

684 CHAPTER 19 | ELECTRIC POTENTIAL AND ELECTRIC FIELD

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