Simple Capacitors: 6.0uC Charge, 1.5V Source

[harvellt]

Homework Statement

See attached PDF for circuit.The arangement of capacitors is attached to a voltage source. Each of the 4.0uF compactors stores a charge of 6.0uC. What charge is stored on the 6.0uF capacitor? What is the voltage of the source.

Homework Equations

Q=C(V)



3. The Attempt at a Solution

$$\frac{Q}{C}$$=V=$$\frac{6.0uC}{4.0uF}$$=1.5V
This is the voltage drop across the two capacitors in series.

Q=CV=1.5V(6.0uF)=9uC
But this is not correct the charge on the larger capacitor should be 12uC correct?

 

[Doc Al]

$$\frac{Q}{C}$$=V=$$\frac{6.0uC}{4.0uF}$$=1.5V
This is the voltage drop across the two capacitors in series.

That's the voltage across the two 4μF capacitors, which are in parallel. It's not the voltage across the 6μF capacitor.

Q=CV=1.5V(6.0uF)=9uC

Don't assume a voltage and use it to try to calculate the charge. Instead, you should be able to deduce the charge from the given information. Then you'll be able to calculate the voltage, in order to answer the second part of the question.

But this is not correct the charge on the larger capacitor should be 12uC correct?

That's true. The way to see that is to ask yourself what's the total charge on the two capacitors in parallel. They, in turn, are in series with the other capacitor. And the total charge on capacitor plates in series is always what?

 

[nlightner]

Yes, you are correct. The charge stored on the 6.0uF capacitor should be 12uC. This can be found by using the same equation, Q=C(V), but with the voltage drop of 1.5V across the two capacitors in series. This results in a charge of 12uC on the larger capacitor. The voltage of the source in this circuit is still 1.5V.