Question

1. A 6.0-μF air capacitor is connected across a 100-V battery. After the battery fully charges the capacitor, the capacitor is immersed in oil (dielectric constant = 4.5). How much additional charge flows from the battery, which remained connected to the battery?. (5 points)

Answer 1

ΔQ = 2.1 x 10^(-3) C

Step-by-step explanation:

The initial charge stored on the capacitor is given by;

Q = C_o•V

Where;

C_o is initial capacitance

V is potential difference across capacitor.

From the question,

C_o = 6.0-μF = 6 x 10^(-6) F

V = 100 V

Thus,

Q = 6 x 10^(-6) x 100 = 6 x 10^(-4) C

Now, the charge stored in the capacitor when inserting the dielectric is given by the formula;

Q = k•Q_o

Where k is the dielectric constant.

Thus,

Q = 4.5 x 6 x 10^(-4) = 2.7 x 10^(-3) C

Thus, the additional charge would be;

ΔQ = Q - Q_o = 2.7 x 10^(-3) - 6 x 10^(-4)

ΔQ = 2.1 x 10^(-3) C