RC circuit - charge on capacitor after a long time

[Linus Pauling]

RC circuit - charge on capacitor after a "long time"

Circuit: 10V goes to 100ohm resistor, then there's a branch point. One side leads to a a resistor R2 = 100ohms and the other to a capacitor with C = 1uF. The switch is located just before R1, and is closed for a "long time."

Solution: Ok, I first solved for the current through R1. After a long time, the capacitor becomes saturated with charge, so the current flows through R1 then R2 then back to the battery. Thus, R1 and R2 are in series and R_eq = 200ohms. By ohms law, the current must be the same through both resistors, I = V/R = 10V/200ohms = 0.05A.

Question: Now I am asked what the charge on the capacitor is in this situation (after a long time). The answer is 5uC. Can someone explain why?

 

[Jmf]

an ideal capacitor is usually defined by:

Q = CV (charge on plates = capacitance * voltage across plates)

After a long time, when the capacitor is saturated with charge (as you said), we have 5V across the capacitor due to potential division between the two resistors (or applying ohms law to a resistor, if you like). So the charge on the capacitor is 5V * 1uF.

Hope this helps. :)