For the Circuit Shown in the Figure $V = 60 \mathrm {~V} , C = 40 \mu \mathrm { F } , R = 0.90 \mathrm { M } \Omega$

For the circuit shown in the figure, $V = 60 \mathrm {~V} , C = 40 \mu \mathrm { F } , R = 0.90 \mathrm { M } \Omega$ , and the battery is ideal. Initially the switch $\mathrm { S }$ is open and the capacitor is uncharged. The switch is then closed at time $t =$ $0.00 \mathrm {~s}$ . 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) $1200 \mu \mathrm { C }$
B) $890 \mu \mathrm { C }$
C) $1400 \mu \mathrm { C }$
D) $1600 \mu \mathrm { C }$
E) $600 \mu \mathrm { C }$

Correct Answer:

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