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I was doing this excercise and, while I did it, I feel that I didn't truly understand how I got the solution.

The excercice had a simple serial circuit of two capacitors, $C_1, C_2=4C_1$, resistor $R$ and a switch.
When the switch is off, there is a charge $Q_0$ on the capacitor $C_1$ and the capacitor $C_2$ is uncharged. I had to find the voltage on the capacitor $C_1$ after the switch in turned on.

I know that the amount of charge after the switch is turned on has to stay the same. So
$$Q_0=Q_1+Q_2$$ I continued by writing $Q_2=4Q_1$ so it was easy to find that $Q_1=\frac{Q_0}{5}$, and the wanted voltage is $U_1=\frac{Q_1}{C_1}$.

I felt that $Q_2=4Q_1$ because capacitance is the ability to store electrical charge so it's logical that bigger the capacitance, more charge can be stored. Therefore it's stored in a 1:4 manner, $Q_1$ will get one quarter of the whole charge.

I don't really know how to show that, how to get to the point of $Q_2=4Q_1$, without just throwing it in? Could the answer be that's beacuse $U_1=U_2$ because there is no current through $R$?

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First you need to be sure you are getting the final equilibrium state, not the value of each charge as a function of time. That function of time, the discharge problem, is more fun. But not what you seem to be asking for here.

Where does the Q0=Q1+Q2 equation come from? It's conservation of charge, right? So it means that charge moves around after you close the switch, but does not go away. It is conservation of charge.

So, after the switch is closed, things have changed, and charge starts to move. So, when does the charge stop moving? What is the condition under which charge will not be moving? That is equilibrium. When does that happen? When the voltage on each capacitor is the same as the other.

But what is the voltage across a capacitor? It is just Q/C, right?

So you need a situation in which the Q/C for each capacitor is the same. But you have that C2 = 4C1. And you now have Q1/C1 = Q2/C2. So what will you get for Q1 and Q2? You will get Q2=4Q1.

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After the charge gets redistributed, there would be no current in the circuit, therefore, and no voltage drop on the resistor, so the resistor could be replaced by a wire and you can view the two capacitors as connected in parallel.

If so, the final voltage on the two capacitors will be the same and, from there, you can see that $Q_2$ should be equal to $4Q_1$.

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