# Simple capacitors

1. Mar 24, 2009

### harvellt

1. The problem statement, all variables and given/known data

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.

2. Relevant 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?

#### Attached Files:

• ###### simplecircuit.pdf
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2. Mar 24, 2009

### Staff: Mentor

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

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.
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?