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# Capacitors Watch

1. Two uncharged capacitors of capacitance 300 μF and 600 μF, are
joined in series.
(i) Calculate their combined total capacitance.
This part is easy total capacitance is 200μF

(ii) This arrangement is then connected to a 15 V supply. Calculate
the voltage across the 300 μF capacitor.

So the charge across the capacitors is 3*10^-3C I would have then split up the charge in proportion to the size of each capacitor. Then recalculated the voltage across the 300 μF capacitor. However the answer uses 3*10^-3C=V*300 μF to calculate V. Why is my thinking incorrect?
2. Strange as it may seem, the charge on series capacitors is equal, no matter what their capacitance.
(Think that the current through series resistors is the same no matter what their resistance.)
The charge on any one of the series capacitors is equal to the charge on the equivalent capacitor you get by combining them in series.
That's why they use the charge calculated on the equivalent capacitor as being also the charge on the individual capacitor.

This might help a bit.
http://hyperphysics.phy-astr.gsu.edu.../capac.html#c2
3. (Original post by Stonebridge)
Strange as it may seem, the charge on series capacitors is equal, no matter what their capacitance.
(Think that the current through series resistors is the same no matter what their resistance.)
The charge on any one of the series capacitors is equal to the charge on the equivalent capacitor you get by combining them in series.
That's why they use the charge calculated on the equivalent capacitor as being also the charge on the individual capacitor.

This might help a bit.
http://hyperphysics.phy-astr.gsu.edu.../capac.html#c2
Thanks so much that's a really good explanation. What would happen with charge then in capacitors that were in parallel? Would it then be appropriate to split the charge, or would it still follow the same principle?
4. (Original post by Tobeadoc)
Thanks so much that's a really good explanation. What would happen with charge then in capacitors that were in parallel? Would it then be appropriate to split the charge, or would it still follow the same principle?
When in parallel, the pd is the same across both, and the charge is given by Q=CV
So it's different charges on different capacitors. The sum of the charges on the individual capacitors equates to the charge across the combined capacitor.
(Think of resistors in parallel, same pd but different currents. Total current is the sum of that through the individual resistors.)

This is also done on that link I gave you.
5. (Original post by Stonebridge)
Strange as it may seem, the charge on series capacitors is equal, no matter what their capacitance.
(Think that the current through series resistors is the same no matter what their resistance.)
The charge on any one of the series capacitors is equal to the charge on the equivalent capacitor you get by combining them in series.
That's why they use the charge calculated on the equivalent capacitor as being also the charge on the individual capacitor.

This might help a bit.
http://hyperphysics.phy-astr.gsu.edu.../capac.html#c2
You are right . charges on capacitors in a series combination are indeed equal this is due to the fact that as the left plate of the first capacitor gets charged it induces an equal and opposite charge on the right plate which in turn induces equal and opposite charge on left plate of next capacitor in the combo and thus all plates have equal magnitude of charge . while in parallel combination charge get distributed according to their capacitance

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