Dielectrics: A 6.0-μF air-filled capacitor is connected across a 100-V voltage source. After the source fully charges the capacitor, the capacitor is immersed in transformer oil (of dielectric constant 4.5). How much ADDITIONAL charge flows from the voltage source, which remained connected during the process?

Question

Quizplus · Accepted Answer

The initial charge stored in the capacitor is:

Q = CV = (6.0 x 10^-6 F) x (100 V) = 0.6 mC

When the capacitor is immersed in transformer oil, the capacitance increases by a factor of 4.5:

C' = kC = 4.5 x 6.0 x 10^-6 F = 27 x 10^-6 F

The voltage across the capacitor remains the same, since the voltage source is still connected:

V = 100 V

The additional charge that flows from the voltage source is:

ΔQ = C'V - Q = (27 x 10^-6 F) x (100 V) - 0.6 mC = 2.1 mC

Therefore, the answer is D.

Dielectrics: a 6