Problem #1
The two metal objects in Fig. 01 have net charges of +70 pC and ─70 pC, which result in a 20 V potential difference between them. (a) What is the capacitance of the system? (b) If the charges are changed to 200 pC and ─200 pC, what does the capacitance become? (c) What does the potential difference become?
Known:
net charges q = +70 pC = 70 x 10─12 C
potential difference, V = 20 V
(a) the capacitance of the system is
C = q/V = (70 x 10─12 C)/(20 V) = 3.5 x 10─12 F = 3.5 pF
(b) Capacitance C is contant and does not depend on the charge: C = 3.5 pF
(c) the potential difference become
V = q/C = 200 pC/3.5 pF = 57 V
Problem #2
The capacitor in Fig. 02 has a capacitance of 25 mF and is initially uncharged. The battery provides a potential difference of 120 V. After switch S is closed, how much charge will pass through it?
Known:
capacitance capacitor C = 25 mF = 25 x 10─3 F
potential difference, V = 120 V
then,
q = CV = (25 x 10─3 F)(120 V) = 3 C
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