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 `200 mu C` 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)` above
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
Current through `6 Omega` resistor, `i= (12 - 4)/(8) 1A`
Voltage across capacitor, `V = 4 + (6) (1) = 10V`
Change on the capacitor, `Q = (10) (10) = 100 mu C`.
After the insertion of dielectic
`C' (A epsilon_(0))/((d)/(3) + (d)/(3) + (d)/(3 (2))) = (A epsilon_(0))/(d) [(1)/(((1)/(3) + (1)/(3) + (1)/(6)))]`
`= (10) ((1)/((5 // 6))) = (6)/(5) (10) = 12 mu F`
Here voltage across capacitor remains same
Charge on the capacitor, `Q' = (12) (10) = 120 muC`.