✔ 最佳答案
The answer is C.
Let's call the first capacitor be capacitor A (the one connected in series with the 4 kilo-ohm resistor), the second one be capacitor B.
Initially, K is closed, so capacitor B is shorted, and it does not have any charge on itself.
And at steady state, a capacitor acts as an open circuit.
So, the potential difference across capacitor A = potential difference across the 6 kilo-ohm resistor = 6 / (4 + 6) X 10 = 6 V
By Q = CV
Charge, Q = (2 X 10-6)(6) = 12 uC
Now, K is open, so the circuit is 'opened'. So, the two resistors can be ignored now.
It is in reality only two capacitors in parallel.
The equivalent capacitance = 2 + 2 = 4uF
And the potential difference = E.m.f. of the battery = 10 V
So, the number of charge on the capacitors, Q'
= C'V'
= (4 X 10-6)(10)
= 40 uC
Therefore, the additional number of charge
= Q' - Q
= 40 - 12
= 28 uC
