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charges the capactor, the capactor is immersed in transformer oil (dielectric constant 45) How much addtional charge flows from the battery, O 12 mC O 25 mC which remained connected during the process? O 15 mC O 1.7mC
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Answer #1

Given that
Initial capacitance Ci = 6.0μF
Initial charge on capacitor Qi = Ci* V = 6.00 * 10^-6 *100 = 6.00 *10^-4 C
Final capacitance Cf = K* Ci = 4.5 * 6.00 *10^-6 = 2.7 *10^-5 F
Final charge Qf = Cf* V = 2.7 * 10^-5 *100 = 27 *10^-4 C
Charge flown in the capacitor ΔQ = Qf - Qi = 27* 10^-4 - 6.0 *10^-4
ΔQ = 2.1* 10^-3 C
Ans. 2.1 mC

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