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 x 10^-6 F and C2 = 6.00 x 10^-6 F. When fully charged, the charge on each capacitor is 8.00 x 10^-6 C. Then, without disconnecting either capacitor, a dielectric with dielectric constant K = 4.00 is inserted between the plates of capacitor C1, completely filling the space between the plates.

After the dielectric has been inserted, what is the magnitude of the charge on each plate of capacitor?

Explanation / Answer

Q = CV V = Q/C V = 8*10^-6 / (3*10^-6) + 8*10^-6 / (6*10^-6) = 4V new Capacitance = kC1 = 4*3*10^-6 = 12 * 10^-6 C1V1 = C2V2 12 * 10^-6 *V = 6.00 x 10^-6 *(4-V) V = 1.333 V Q = 1.333 * 12*10^-6 = 16 *10^-6 C

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