'A' Level question on capacitors

[jezza10181]

Homework Statement

Just for a laugh, I decided to try some A level physics questions (I am 50yrs old btw and did my physics A level over 30yrs ago :) ) & ran into this difficulty

Relevant Equations

Q = VC, 1/C = 1/C1 + 1/C2

I have been looking at this question:-

Now, I have found the charge in the whole system to be 36.0nC. I did this by 'condensing' the two 2.0nF into a single 4.0nF one, that then leaving me with an equivalent system of a 4.0nF capacitor & a 6.0nF one? I then found the equivalent capacitance of that ensemble using the 1/C = ... formula. I found this to be 2.4nF. From that I found the total charge in the system to be 36.0nC.

So, that would mean that each of the capacitors would have to have 18.0nC on them. So, as the 4.0nF 'equivalent' capacitor ,which is composed of the two 2.0nF capacitors, has 18.0nC on it, then that would mean each of the two 2.0nF capacitors has 9.0nC on it & not 18.0nC, as the question states.

What do you think?

 

[BvU]

Book answer is correct. Write the charges next to the capacitor plates to see why

 

[jezza10181]

18.0nC on the one 2nF, 18.0nC on the other, & 18.0nc on the 6.0nF? That is 54nC isn't it? I still don't see it, sorry

 

[BvU]

Over-all C is 2.4 nF, so -36 nC on rightmost plate: continue from there

 

[jezza10181]

Surely, there would only be -36nC on the right most plate, if you were dealing with the equivalent situation of having a single capacitor of 2.4nF there?

 

[ehild]

I have been looking at this question:-View attachment 257036Now, I have found the charge in the whole system to be 36.0nC. I did this by 'condensing' the two 2.0nF into a single 4.0nF one, that then leaving me with an equivalent system of a 4.0nF capacitor & a 6.0nF one? I then found the equivalent capacitance of that ensemble using the 1/C = ... formula. I found this to be 2.4nF. From that I found the total charge in the system to be 36.0nC.

that means, that replacing the whole system by a single capacitor, there is +36 nC charge on one plate and - 36 nC charge on the other plate.
When you connect two capacitors in series to a battery, the plates connected to the terminals of the battery get charge from the battery, but the plates connected to each other can get charge only from each other.
If the first plate of the first capacitor has Q charge,, there is -Q charge on
the opposite plate, and Q charge again on the first plate on the second capacitor, and -Q charge on the second plate on the second capacitor. It is analogous to the current through resistors connected in series.

 

[jezza10181]

that means, that replacing the whole system by a single capacitor, there is +36 nC charge on one plate and - 36 nC charge on the other plate.
When you connect two capacitors in series to a battery, the plates connected to the terminals of the battery get charge from the battery, but the plates connected to each other can get charge only from each other.
If the first plate of the first capacitor has Q charge,, there is -Q charge on
the opposite plate, and Q charge again on the first plate on the second capacitor, and -Q charge on the second plate on the second capacitor. It is analogous to the current through resistors connected in series.

That's a great explanation, thanks. It's finally sunk in lolll