76. (a) The equivalent capacitance is Ceq= C1C2/(C1+ C2). Thus the charge qon each capacitor is 412eq(2.00 F)(8.00 F)(300V)4.80 10 C.2.00 F 8.00 FCCVqq q CVCCµµ−====== ×++(b) The potential difference is V1= q/C1= 4.80 ×10–4C/2.0 µF = 240 V. (c) As noted in part (a), 4214.80 10 C.qq−== ×(d) V2= V – V1= 300 V – 240 V = 60.0 V. Now we have q'1/C1q'2/C2V'(V'being the new potential difference across each capacitor) and q'1+ q'2= 2q. We solve for q'1, q'2and
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This note was uploaded on 01/25/2011 for the course PHYSICS 17029 taught by Professor Rebello,nobels during the Spring '10 term at Kansas State University.