Charge stored in a capacitor

[ShaddollDa9u]

Homework Statement

Question : How much charge is stored on the capacitor C2 ?

2. Homework Equations

Q = C . U

The Attempt at a Solution

[/B]
Since Q2, the charge stored on the capacitor, is C2 . (Va - Vb) , I thought that I just had to multiply 1.0 x 60 = 60.
But in fact, the expected result is 40, how ??

 

[CWatters]

Since Q2, the charge stored on the capacitor, is C2 . (Va - Vb) ,

Your equation Q=CV is correct but the voltage V is the voltage across the capacitor. Does all of the 60V appear across C2? How much of the 60V appears across C3//C5 ?

You haven't mentioned C1 but it can be ignored. Why?

 

[ShaddollDa9u]

The potential difference across C1 is 0 since C2 = C4 and C3 = C5.
I have thought about the fact that the voltage across C2 may be incorrect, but in that case how can I find it ? Kirchhoff's law doesn't seem to help in this case

 

[CWatters]

The potential difference across C1 is 0 since C2 = C4 and C3 = C5.

Correct.

I have thought about the fact that the voltage across C2 may be incorrect, but in that case how can I find it ? Kirchhoff's law doesn't seem to help in this case

You can simplify the circuit to two capacitors in series C2//C4 + C3//C5. Then you assume the capacitors start off discharged when the voltage is applied. There will be a rush of current as they become charged. Since they are in series the current (eg charge) that flows through both capacitors is the same.

 

[ShaddollDa9u]

Correct.
You can simplify the circuit to two capacitors in series C2//C4 + C3//C5. Then you assume the capacitors start off discharged when the voltage is applied. There will be a rush of current as they become charged. Since they are in series the current (eg charge) that flows through both capacitors is the same.

So if we have Ca = C2 + C4, the charge stored by that capacitor will be q = 2 x 60 = 120, right ?? But then what do I do ?? Isn't it the same thing that I had tried in my first attempt ?

 

[epenguin]

Since you have realized there is no charge on C₁ you can ignore it, rub it out, everything is the same as if it were not there.
Then in parallel you have two branches which are exactly the same. Essentially they are independent. You are only asked something about one of them. So the problem is exactly the same as if you had only that branch. Two capacitors in series.

Then you can mechanically use the formula for that which I don't encourage, except maybe as a check at the end to see everything is the same however you work it out. Or you could think: the charges aon the two capacitors are the same, so from Q = CV, as the Q's are the same what's the ratio of the voltages across them? That will give you the voltage across the capacitor C₂, and the charge on it.

You can practically do this problem in your head. In more than half of the capacitor or resistor networks given as problems on this forum, there is simplifying feature of a symmetry. Which would cut down on calculation. But the students have been intimidated or brainwashed or something and can only think of doing it in a heavy plod way, formulaically, missing out even some of the simpe physical insight, even after solving.

I would also disagree with the question when it says "write out as a decimal"... The component parameters are usually given as whole numbers, so write answers as rational numbers. Because that is exact! Then at the last convert into a decimal if required.

 

[ShaddollDa9u]

I see now, so at the end, we will have the two capacitors C2 and C3 in series, which we can consider as a single capacitor Ca = 2/3.
By applying Q = Ca . V, we have :
Q = (2/3) . 60 = 40.
And since the two capacitors does have the same charge, C2 has a charge of 40. Is it correct ??
Thank you very much

 

[epenguin]

Yes I0W, in fact IW, since C₂ has half as much capacitance as C₃, to hold the same charge it needs twice the voltage across it as C₃, so two thirds of the total voltage, so 40 V, so gets (with 1 μF capacitance) 40 µC of charge.

Re another point I made, you see that even the "rounding off" request was intended to confuse.

 

[CWatters]

You can also think of this circuit as a "voltage/potential divider" ..

https://en.wikipedia.org/wiki/Voltage_divider