ssm_ch25 - Chapter 25 5 (a) The capacitance of a...

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Unformatted text preview: Chapter 25 5 (a) The capacitance of a parallel-plate capacitor is given by C = A=d , where A is the area of each plate and d is the plate separation. Since the plates are circular, the plate area is A = R 2 , where R is the radius of a plate. Thus C = R 2 d = (8 : 85 10 12 F = m) (8 : 20 10 2 m) 2 1 : 30 10 3 m = 1 : 44 10 10 F = 144 pF : (b) The charge on the positive plate is given by q = CV , where V is the potential difference across the plates. Thus q = (1 : 44 10 10 F)(120 V) = 1 : 73 10 8 C = 17 : 3 nC. 15 The charge initially on the charged capacitor is given by q = C 1 V , where C 1 (= 100 pF) is the capacitance and V (= 50 V) is the initial potential difference. After the battery is disconnected and the second capacitor wired in parallel to the first, the charge on the first capacitor is q 1 = C 1 V , where v (= 35 V) is the new potential difference. Since charge is conserved in the process, the charge on the second capacitor is q 2 = q q 1 , where C 2 is the capacitance of the second capacitor. Substitute C 1 V for q and C 1 V for q 1 to obtain q 2 = C 1 ( V V ). The potential difference across the second capacitor is also V , so the capacitance is C 2 = q 2 V = V V V C 1 = 50 V 35 V 35 V (100 pF) = 43 pF : 19 (a) After the switches are closed, the potential differences across the capacitors are the same and the two capacitors are in parallel. The potential difference from a to b is given by V ab = Q=C eq , where Q is the net charge on the combination and C eq is the equivalent capacitance. The equivalent capacitance is C eq = C 1 + C 2 = 4 : 10 6 F. The total charge on the combination is the net charge on either pair of connected plates. The charge on capacitor 1 is q 1 = C 1 V = (1 : 10 6 F)(100 V) = 1 : 10 4 C and the charge on capacitor 2 is q 2 = C 2 V = (3 : 10 6 F)(100 V) = 3 : 10 4 C ; so the net charge on the combination is 3 : 10 4 C 1 : 10 4 C = 2 : 10 4 C. The potential difference is V ab = 2 : 10 4 C 4 : 10 6 F = 50 V : (b) The charge on capacitor 1 is now q...
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This note was uploaded on 01/19/2009 for the course PHYS 171.102 taught by Professor Heckman during the Fall '98 term at Johns Hopkins.

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ssm_ch25 - Chapter 25 5 (a) The capacitance of a...

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