Two parallel-plate capacitors C1 and C2 are connected in series to a battery. In

Question

Two parallel-plate capacitors C1 and C2 are connected in series to a battery. Initially both capacitors have air between the plates and C1 = 3.00 * 10^-6 F and C2 = 6.00 * 10^-6 F. When fully charged, the charge on each capacitor is 8.00 * 10^-6 C. Then, without disconnecting either capacitor, a dielectric with K=4.00 is inserted between the plates of Capacitor C1, completely filling the space between the plates. After the dielectic has been inserted, what is the magnitude of the charge on each plate of C1?

Explanation / Answer

given C1 = 3.00 * 10^-6 F
        C2 = 6.00 * 10^-6 F When fully charged, the charge on each capacitor is q = 8.00 * 10^-6 C dielectric constant K=4.00 C1 and C2 are in series.So, resultant capacitance C = ( C1*C2) / ( C1+ C2)                                                                                = 2 * 10 ^-6 F So,Potential difference between the plates of C1 is V1 = q / C1 = 2.6666 volt Potential difference between the plates of C2 is V2 = q / C2 = 1.3333 volt potential difference between the terminals of the battery V = V1+ V2 = 4 volt Capacitance of C1 after inserting dielectric is C1 ' = K * C1 = 12* 10 ^-6 F resulatant capacitance of C1' and C2 is C ' = ( C1'*C2 ) / ( C1 ' + C2 )                                                                   = 4 * 10^-6 F So, charge on the circuit after insertion of the dielectric Q = C ' * V = 16 * 10 ^-6 C Therefore the magnitude of the charge on each plate of C1 after inserting dielectric is          Q = 16 * 10^-6 C

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