Two capacitors connected in series

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Two capacitors with capacitances of 30 μF and 60 μF are initially charged with 50 μC and 70 μC, respectively. When connected in series, the charge on capacitors should be the same, but confusion arises over the initial charge values. It is clarified that only one plate of each capacitor is connected, resulting in no current flow and the capacitors retaining their original charge and voltage. The discussion concludes with the realization that the calculated voltages do not match any of the provided answer choices, indicating a potential issue with the question's setup. Understanding the series connection concept is crucial for solving the problem correctly.
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Homework Statement


Two capacitors, capacitance of 30 μF and 60 μF respectively, each has charge of 50 μF and 70 μF. The capacitors then connected in series. Find the potential difference on each capacitor now


Homework Equations


Q = CV


The Attempt at a Solution


I know that the charge should be same for series circuit but I don't know how to analyse this question. Each capacitors has its own charge and voltage and after connected in series, their charge should be the same. How to find it?
 
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songoku said:
... each has charge of 50 μF and 70 μF?

uF is a measure of capacitance, not charge, so I don't understand what you are saying here.
 
Think about what "The capacitors then connected in series" means. Presumably only one plate of each is connected together otherwise they would be in parallel. If only one side of each capacitor is connected together how much current can flow?
 
phinds said:
uF is a measure of capacitance, not charge, so I don't understand what you are saying here.

sorry, I meant 50 μC and 70 μC

CWatters said:
Think about what "The capacitors then connected in series" means. Presumably only one plate of each is connected together otherwise they would be in parallel. If only one side of each capacitor is connected together how much current can flow?

Sorry I really don't know the concept here of how to determine the current when only one side of each capacitor is connected...I assume that one positive plate is connected to other negative plate. Will the case be the same if the series circuit connects positive plate to positive plate?

Thanks
 
If the only one end of each capacitor is connected to anything there is no closed circuit so no current flow. Therefore the capacitors retain whatever charge and voltage they had before they were connected.
 
CWatters said:
If the only one end of each capacitor is connected to anything there is no closed circuit so no current flow. Therefore the capacitors retain whatever charge and voltage they had before they were connected.

Sorry, actually there are choices for this question:
a. 4 and 8 Volt
b. 8 and 4 Volt
c. 5 and 5 Volt
d. 3 and 6 Volt
e. 6 and 3 Volt

If the capacitors retain their voltage, their voltage should be 5/3 and 7/6 Volt.

So there is no correct choice?

Thanks
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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