ch25-p019

# ch25-p019 - × 10 − 6 F(100 V = 3 × 10 − 4 C so the...

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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 . 0 × 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 . 0 × 10 6 F)(100 V) = 1 . 0 × 10 4 C and the charge on capacitor 2 is q 2 = C 2 V =(3
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Unformatted text preview: . × 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 1 = C 1 V ab = (1 . × 10 − 6 F)(50 V) = 5 . × 10 − 5 C. (c) The charge on capacitor 2 is now q 2 = C 2 V ab = (3 . × 10 − 6 F)(50 V) = 1 . 5 × 10 − 4 C....
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