In the circuit shown in the figure, if switches `S_(1)` and `S_(2)` have been closed for a long time, then charge on the capacitor.

Options

A. is `100 muC`
B. increases to `120muC` if one-third of the gap of the capacitor's plates is filled with a dielectric (K=2) of same area.
C. both a. and b.
D. charge on the capacitor remains unchanged if one-third of the gap of the capacitor's plates is filled with a dielectric (K=2) of same area.

Answer

C

Explanation

c. After long time, current through the capacitor = 0
Therefore, current through the `6Omega` resistor, `I = (12-4).8 = 1A`
Voltage across capacitor, `V = 4+(6)(1) = 10V`
Charge on the capacitor, `Q = (10)(10) = 100muC`
After the insertion of the dielectric, we get
`C' = (A epsilon_0)/(d/3+d/3+d/(3(2))) = (Aepsilon_0)/d [1/(1/3+1/3+1/6)]`
`= (10) (1/((5//6))) = 5/6 (10) = 12 muF`
Hence, voltage across the capacitor remains the same.
Charge on the capacitor , `Q' = (12)(10) = 120 muC`.