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I am having trouble finding the charge on the capacitor for $t\rightarrow \infty$ after the switch has been closed at $t=0$. I already know the current $I$. I also have to find out on which sides the positive/negative charges are.

I have a feeling, that the positive charges are located on the left side of the capacitor but finding the charge $Q$ troubles me.

The circuit

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You can use either

$$Q=CV$$

or

$$Q(t)= Q(t_0) + \int_{t_0}^t I(t')dt'$$

to find the charge on the capacitor, depending which of $I(t)$ and $V(t)$ you already know.

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  • $\begingroup$ BTW, $I(t)$ here is the current through the capacitor, not the $I$ indicated in OP's diagram. $\endgroup$ – The Photon Apr 25 at 16:37
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At t = infinity the capacitor looks like an open circuit because all voltages are non time changing when transients die out. The current in a capacitor is given by

$$i(t)=C\frac{dV(t)}{dt}$$

So if V=constant, current is zero .

Assuming no charge is initially on the capacitor (you didn’t say) replace capacitor and series resistor with an open circuit and determine the voltage between a and the other side of the series resistor as if the capacitor and resistor weren’t there since there is no current. That voltage will be the voltage across the capacitor. From there you can calculate the charge on the capacitor from

$$C=\frac{Q}{V}$$

Hope this helps

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enter image description here

In steady state, $t \to \infty $ the capacitor and resistor in the red ellipse can be ignored as no current flows in that part of the circuit.

You now have a network of resistors and need to find in terms of $\mathcal E$: $V_{\rm BF}\,\Rightarrow \, V_{\rm AF}$, $V_{\rm DF} \Rightarrow V_{\rm AD}$

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  • $\begingroup$ Ok, so I calculated the following $V_{BF}=\frac{4}{7}\mathcal{E}\land V_{AF}=V_{BA}=\frac{2}{7}\mathcal{E}\land V_{BD}=\frac{3}{7}\mathcal{E}\land V_{DF}=\frac{1}{7}\mathcal{E}$. Now, is $V_{AD}=V_{AF}-V_{DF}$? $\endgroup$ – Kasper Larsen Apr 25 at 18:23
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Well, we can first of all write:

$$\text{v}_\text{in}\left(\text{s}\right)=\text{i}_\text{in}\left(\text{s}\right)\cdot\left(\text{R}+\frac{1}{\frac{1}{\text{R}+3\text{R}}+\frac{1}{\frac{1}{\text{sC}}+\text{R}}+\frac{1}{\text{R}+\text{R}}}\right)=\text{i}_\text{in}\left(\text{s}\right)\cdot\left(\text{R}+\frac{1}{\frac{3}{4\text{R}}+\frac{1}{\frac{1}{\text{sC}}+\text{R}}}\right)\tag1$$

So, the current trough the capacitor is given by:

$$\text{i}_\text{C}\left(\text{s}\right)=\frac{\frac{1}{\text{sC}}+\text{R}}{\frac{1}{\text{sC}}+\text{R}+\frac{1}{\frac{1}{\text{R}+3\text{R}}+\frac{1}{\text{R}+\text{R}}}}\cdot\text{i}_\text{in}\left(\text{s}\right)=\frac{\frac{1}{\text{sC}}+\text{R}}{\frac{1}{\text{sC}}+\text{R}+\frac{4\text{R}}{3}}\cdot\text{i}_\text{in}\left(\text{s}\right)\tag2$$

So:

$$\text{i}_\text{C}\left(\text{s}\right)=\text{v}_\text{in}\left(\text{s}\right)\cdot\frac{\frac{1}{\text{sC}}+\text{R}}{\frac{1}{\text{sC}}+\text{R}+\frac{4\text{R}}{3}}\cdot\frac{1}{\text{R}+\frac{1}{\frac{3}{4\text{R}}+\frac{1}{\frac{1}{\text{sC}}+\text{R}}}}\tag3$$

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