What is the Charge on the Capacitor in a Circuit After Switching Positions?

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The discussion focuses on calculating the charge on a capacitor after switching a circuit from position "a" to "b" at t=0s. The initial calculations yield a charge of 36uC, but the user struggles to apply the equations correctly to find the charge at t=50s. Key equations include Q=CV, V=IR, and the exponential charge function q=Qinit(1-e^(-t/τ)), where τ=RC. Guidance suggests substituting τ into the charge equation and checking the behavior of the function at t=0 and as t approaches infinity to ensure intuitive understanding. The expected charge at t=50s is 22uC, indicating a need for careful application of the formulas.
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Homework Statement


The switch in the figure has been in position a for a long time. It is changed to position "b" at t=0s. What is the charge Q on the capacitor at t=50s?

(zeros denote empty space in the circuit)

|----(a)0000(b)--------- |
|00000\00000000000000|
9V00000\000000000000|
|00000000|00000000025 Ohm
|00000004uF000000000|
|________|___________|


Homework Equations


(1) C=Q/V
(2) V=IR
(3) q=Qinit.(1-e^(-t/τ))
(4) τ=RC


The Attempt at a Solution


Ok, so I start by using (1) to find that Q=36uC, then I use (2) to find that I=.36A, but after that is where i encounter problems. I am trying to use (3) to find the charge but it's not giving me the right answer. I'm not sure how to use (4), what is that formula telling me, and am i doing everything right?

The answer should be 22uC
 
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Notice that in equation (3) you have the parameter 1/T, this is exactly what equation (4) is telling you. Try substituting (4) into (3) and seeing if you get the right solution.
 
You may want to take a closer look at your equation (3). In particular, check what the function yields for t = 0 and in the limit as t --> infinity. Does it correspond to what you intuitively feel should happen to the charge on the capacitor?
 
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