For the circuit shown in the figure,V = 60 V,C = 40 µF,R = 0.90 MΩ,and the battery is ideal.Initially the switch S is open and the capacitor is uncharged.The switch is then closed at time t = 0.00 s.At a given instant after closing the switch,the potential difference across the capacitor is twice the potential difference across the resistor.At that instant,what is the charge on the capacitor? 
A) 1,600 µC
B) 1,400 µC
C) 1,200 µC
D) 890 µC
E) 600 µC
Correct Answer:
Verified
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