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What is the voltage for each capacitor ifplates of opposite sign are connected?

ID: 1672084 • Letter: W

Question

What is the voltage for each capacitor ifplates of opposite sign are connected?

What is the charge on each capacitor if plates of opposite sign areconnected?
A 2.80 mu F capacitor is charged to 460 V and a3.75 mu F capacitor is charged to 525 V. These capacitors are then disconnected from their batteries, andthe positive plates are now connected to each other and thenegative plates are connected to each other. What will be thepotential difference across each capacitor? What will be the charge on each capacitor? What is the voltage for each capacitor ifplates of opposite sign are connected? What is the charge on each capacitor if plates of opposite sign areconnected?

Explanation / Answer

   Given   C1   =   2.80 * 10-6   F                C2   =   3.75 * 10-6   F                V1   =   460   V                V2   =   525   V    a.   When capacitors are connected such that + ve plate connects to the +ve plate, equilibrium potential difference          V++   =   (C1 * V1   +   C2 * V2) / (C1   +   C2)                   =   (2.80 * 10-6 * 460   +   3.75 * 10-6 * 525) / (2.80 * 10-6   +   3.75 * 10-6)                   =   497.21   V    b.   Charge on   C1,         Q1   =   C1 * V++                                                   =   2.80 * 10-6 * 497.21                                                    =   1.39 * 10-3   C                                                     =   1.39   mC          Charge on C2            Q2            =   C2 * V++                                                       =   3.75 * 10-6 * 497.21                                                       =   1.86 mC                                                       =   3.75 * 10-6 * 497.21                                                       =   1.86 mC    c.   When capacitors are connected with opposite polarity, equilibrium potential difference is          V+-   =   (C1 * V1   ~   C2 * V2) / (C1   +   C2)                   =   (2.80 * 10-6 * 460   ~   3.75 * 10-6 * 525) / (2.80 * 10-6   +   3.75 * 10-6)                   =   103.93   V                                                   =   2.80 * 10-6 * 103.93                                                    =   2.91 * 10-4   C                                                     =   0.291   mC          Charge on C2            Q2'            =   C2 * V+-                                                       =   3.75 * 10-6 * 103.93                                                       =   3.90 * 10-4   C                                                       =   0.390 mC                                                       =   3.75 * 10-6 * 103.93                                                       =   3.90 * 10-4   C                                                       =   0.390 mC
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