Explanation Given info: The electric potential through the parallel combination is 250.0 V and the value of the two capacitors are 6.00 μF and 2.00 μF respectively. Since, the capacitors are in parallel. So, the equivalent capacitance of the parallel connection is, C eq = C 1 + C 2 Here, C eq is the equivalent capacitance of the parallel connection. C 1 is the capacitance of the first capacitor. C 2 is the capacitance of the second capacitor. Substitute 6.00 μF for C 1 and 2.00 μF for C 2 in above equation to find C eq . C eq = 6.00 μF + 2.00 μF =8 .00 μF Thus, the equivalent capacitance of the parallel connection is 8.00 μF . Formula to calculate the charge store in the combination is, Q = C eq V Here, Q is the charge store in the combination. V is the potential through the parallel combination. Substitute 8.00 μF for C eq and 250.0 V for V in above equation to find Q . Q = ( 8.00 μF ( 10 − 6 F 1 μF ) ) ( 250.0 V ) = 2 × 10 − 3 C The charge store in the combination is 2 × 10 − 3 C . The charge store on the capacitor C 1 is, Q 1 = C 1 V Here, Q 1 is the charge store on the capacitor C 1 . Substitute 6.00 μF for C 1 and 250.0 V for V in above equation to find Q 1 . Q 1 = ( 6.00 μF ( 10 − 6 F 1 μF ) ) ( 250.0 V ) = 1.5 × 10 − 3 C Thus, the charge store on the capacitor C 1 is 1.5 × 10 − 3 C . The charge on the capacitor C 2 ( 2.00 μF ) is, Q 2 = Q − Q 1 Here, Q 2 is the charge store on the capacitor C 2 . Substitute 1.5 × 10 − 3 C for Q 1 and 2 × 10 − 3 C for Q in above equation to find Q 2 . Q 2 = 2 × 10 − 3 C − 1.5 × 10 − 3 C =0 .5 × 10 − 3 C The net charge on the capacitor after disconnecting from battery and connected to positive plate and negative plate and negative plate to positive plate is, q = Q 1 − Q 2 Here, q is the net charge on the capacitor after disconnecting from battery. Substitute 0 .5 × 10 − 3 C for Q 2 and 1

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN: 9781305116399

Author: Raymond A. Serway, John W. Jewett

Publisher: Cengage Learning

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Physics for Scientists and Engineers, Technology Update (No access codes included)

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The resulting charges on each capacitor.

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