Page 325 - Schaum's Outline of Theory and Problems of Applied Physics

310 CAPACITANCE [CHAP. 26

(a) The capacitor’s charge is
Q = CV = (10 × 10 −6 F)(50 V) = 5 × 10 −4 C

The potential difference across it remains 50 V after it is disconnected.
(b) The presence of another dielectric does not change the charge on the capacitor. Since its capacitance is now

K 2
C 2 = C 1
K 1
and V = Q/C, the new potential difference is
Q K 1 Q K 1 1
V 2 = = = V 1 = (50 V) = 23.8V
C 2 K 2 C 1 K 2 2.1

CAPACITORS IN COMBINATION
The equivalent capacitance of a set of connected capacitors is the capacitance of the single capacitor that can
replace the set without changing the properties of any circuit it is part of. The equivalent capacitance of a set of
capacitors joined in series (Fig. 26-2) is
1 1 1 1
= + + +· · · capacitors in series
C C 1 C 2 C 3
When using a calculator with a reciprocal (1/X) key, it is easiest to proceed in the way described in
Chapter 25 in the case of the similar formula for the equivalent resistance of a set of resistors in parallel. The
key sequence here would be
[C 1 ][1/X][+][C 2 ][1/X][+][C 3 ][1/X][+] ··· [=][1/X]

If there are only two capacitors in series,
1 1 1 C 1 + C 2 C 1 C 2
= + = and so C =
C C 1 C 2 C 1 C 2 C 1 + C 2
In a parallel set of capacitors (Fig. 26-3), the equivalent capacitance is the sum of the individual capacitances:

C = C 1 + C 2 + C 3 +· · · capacitors in parallel

Fig. 26-2 Fig. 26-3

SOLVED PROBLEM 26.6
Show that the equivalent capacitance C of three capacitors connected in series is given by 1/C =
1/C 1 + 1/C 2 + 1/C 3 .
Each capacitor in a series connection has charges of the same magnitude Q on its plates, so the voltages across
them are, respectively,
Q Q Q
V 1 = V 2 = V 3 =
C 1 C 2 C 3