# Capacitance problem

1. Oct 22, 2005

In the figure attached, the battery has a potential difference of 10 V and the five capacitors each have a capacitance of 20 µF. What is the excess charge on capacitor 2?

I go inside out from the capacitors in series $$C_2$$ and $$C_3$$. My notation simply denotes the equivalent resistance depending on the type of connection (series/parallel).

$$C_{S1}= \left( \frac{1}{C_2} + \frac{1}{C_3} \right) ^{-1}$$

$$C_{P1}= C_{S1} + C_4$$

$$C_{S2}= \left( \frac{1}{C_{P1}} + \frac{1}{C_5} \right) ^{-1}$$

$$C_{P2}= C_{S2} + C_1$$

It also follows that

$$Q_2 = Q_3 = Q_{S1}$$

$$V_4 = V_{S1}$$

$$Q_5 = Q_{P1} = Q_{S2}$$

$$V_1 = V_{S2}$$

So, I get

$$Q_2 = Q_{S1} = V_{S1} C_{S1} = 1.0\bar{6} \times 10^{-4} \mbox{ C}$$

Any help is highly appreciated.

#### Attached Files:

• ###### circuit_capacitors.bmp
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2. Oct 23, 2005

### daniel_i_l

How did you get Vs1? It isn't 10V, because some of the voltage goes to C5,
if you correctly found the Vs1 and just didn't show the work then it looks right to me.
I would find the Qs2 = Qp1 and then find how much Q goes to the different parts of p1 - how much of Qp1 goes to C4 and how much goes to s1.

3. Oct 23, 2005

I think I finally have it!!! I worked backwards from the desired result and used some relationships to guide me through.
$$Q_2 = V_2 C_2 = Q_{S1} = V_{P1} C_{S1} = \frac{Q_{S2}}{C_{P1}} C_{S1} = \frac{VC_{S2}}{C_{P1}} C_{S1} = 4 \times 10^{-5} \mathrm{ C}$$
Is this correct?
Thank you

4. Oct 23, 2005