Page 330 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 26] CAPACITANCE 315

SOLVED PROBLEM 26.14
A 20-µF capacitor is connected to a 45-V battery through a circuit whose resistance is 2000 . (a) What
is the final charge on the capacitor? (b) How long does it take for the charge to reach 63 percent of its
final value?

(a) Q = CV = (20 × 10 −6 F)(45 V) = 9 × 10 −4 C
(b) t = RC = (2000 )(20 × 10 −6 F) = 0.04 s

SOLVED PROBLEM 26.15
Find the charge on the capacitor of Prob. 26.14 at 0.01 and 0.1 s after the connection to the battery is
made.
When t = 0.01 s, t/T = (0.01 s)/(0.04 s) = 0.25. Using a calculator or table of exponentials gives

e −t/T = e −0.25 = 0.78

Hence

Q = Q 0 (1 − e −t/T ) = (9 × 10 −4 C)(1 − 0.78) = (9 × 10 −4 C)(0.22) = 2.0 × 10 −4 C

Similarly, when t = 0.1s, t/T = (0.1s)/(0.04 s) = 2.5 and

e −t/T = e −2.5 = 0.082

Hence

Q = Q 0 (1 − e −t/T ) = (9 × 10 −4 C)(1 − 0.082) = (9 × 10 −4 C)(0.918) = 8.3 × 10 −4 C

SOLVED PROBLEM 26.16
A5-µF capacitor is charged by being connected to a 3-V battery. The battery is then disconnected. (a)If
9
the resistance of the dielectric material between the capacitor plates is 10 , find the time required for
the charge on the capacitor to drop to 37 percent of its original value. (b) What is the charge remaining
on the capacitor 1 h after it has been disconnected? What is the charge 10 h afterward?
9
(a) T = RC = (10 )(5 × 10 −6 F) = 5 × 10 −3 s
which is 5 × 10 −3 s 5000 s
= = 1.4h
(60 s/min)(60 min/h) 3600 s/h
(b) The initial charge on the capacitor is

Q 0 = CV = (5 × 10 −6 F)(3V) = 1.5 × 10 −5 C

After t = 1h, t/T = (1h)/(1.4h) = 0.71, and

Q = Q 0 e −t/T = (1.5 × 10 −5 C)(e −0.71 ) = (1.5 × 10 −5 C)(0.49) = 7.4 × 10 −6 C

After t = 10 h, t/T = (10 h)/(1.4h) = 7.1, and

−4
Q = Q 0 e −t/T = (1.5 × 10 −5 C)(e −7.1 ) = (1.5 × 10 −5 C)(8.3 × 10 ) = 1.2 × 10 −8 C