# Homework Help: Modeling conservation of charge

1. Mar 20, 2004

### SparkimusPrime

A group of physics students designed the following experiment to test the model of conservation of charge.

a) They first charged a capacitor C1 = 5600 micro F by applying a voltage V_0 = 10 volts (as shown in attachment Exam 2 - Problem 2 - Part a.JPG)

b) Then the connected a second capacitor of C2 = 1400 micro F in the circuit (as shown in Exam 2 - Problem 2 - Part b.JPG). They measured the voltage across C1 using a voltmeter. The reading of V_1 = 2 volts.

c) Then they calculated charge q_0 in the capacitor C1 in the circuit in part a and charges q_1 and q_2 in part b. Finally, they tried to compare the value of q_0 with the value of q_1 + q_2. They concluded that charge is conserved if q_0 = q_1 + q_2. Otherwise charge is not conserved.

The correct answer is supposed to be that all the charges are equal (q_0 = q_1 = q_2). But I get answers agreeing with the assessment of the experiment:

q_0 = C1 * V_0 = (10) * (5600) = 5.6e4 micro C

As shown in the figure, the total capacitance of the system decreases, because the capacitors are placed in series:

1 / C_total = 1 / C_1 + 1 / C_2
C_total = 1120 micro F

So the charge on the system should be decreased similarly (as the electric force between the plates decreases so too should the charge on the individual capacitors and thus the system). So it seems to me, by inspection that there should by such a relationship as expressed in the experiment.

If the voltage across C1 (in part b) is 2 volts, the charge on that capacitor should be:

C1 * V_1 = (5600)*(2) = 11200 micro C

Clearly less than the original charge. Of course this may be where my error occurs, I'm taking the charge across a single capacitor in a series, can I treat it as if it were alone or do I have to consider the entire system?

Also as an aside, I don't think the LaTeX system is working correctly. I tried to denote some Greek symbols (mu and delta) via latex and it showed some very strange behavior:

Edit:

Apparently I can't attach more than one image, nor can I attach an image after I've removed it. I'll reply to this thread with the appropriate images.

Last edited: Mar 20, 2004
2. Mar 20, 2004

### SparkimusPrime

part a:

#### Attached Files:

• ###### exam 2 - problem 2 - part a.jpg
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3. Mar 20, 2004

### SparkimusPrime

part b:

Edit:

If someone knows how to do this more elegantly please, enlighten me.

#### Attached Files:

• ###### exam 2 - problem 2 - part b.jpg
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Views:
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4. Mar 20, 2004

### Chen

It's been a while since I studied capacitors, but I'll give this a shot.

Are you sure this is the correct answer? It's obvious that $$q_1 = q_2$$ but I don't think it also equals the charge of the first capacitor when it was alone.

Ignoring for a minute that the number of capacitors changes, let's just look at the total capacity of the system as a whole in the two case. In (a) the capacity of the system is:
$$C_a = C_1$$
In (b) the capacity is of two capacitors connected in series, i.e:
$$C_b = \frac{1}{\frac{1}{C_1} + \frac{1}{C_2}}$$

It's easy to see that $$C_a > C_b$$. Since the voltage in both cases is equal $$V_a = V_b$$, and $$q = VC$$, the charge must change and $$q_a > q_b$$. So the charge quantity on the capacitors does change. In words, the voltage stays the same but the capacity of the system changes, and therefore the charge quantity has to change.

5. Mar 20, 2004

### SparkimusPrime

Yes, it seems to me that this should be the answer. Maybe it is a misprint, all the evidence seems to point that way.

Calculating the charges on the two capacitors:

Q_1 = C1 * V_1 = 5600*2 = 11200
Q_2 = C2 * V_1 = 1400*8 = 11200

It seems that the charges on the capacitors are equal, but that they don't equal the total charge on the system with just one capacitor (part a). Almost half as much charge on the system in part b as in part a, which jives with my intuition about the capacitence and therefore the charge decreasing. Have I missed anything?

6. Mar 20, 2004

### Chen

Argh, I am so stupid.

When talking about capacitors, we are not talking about the total charge of them. We are talking about the charge they hold in one of their plates. On one plate they hold +q charge, and on the other plate they have -q charge. So in total... they have no charge at all. So there is no conflict with the preservation of charge, it is just divided differently.

Last edited: Mar 20, 2004
7. Mar 20, 2004

### SparkimusPrime

I know there is charge present on each plate of a capacitor, as they store charge (by definition). The charges are equal and opposite and it is the electric forces generated by that charge that allows said charge to be stored. Taken as a system and preserving the signs, yes there is no net charge on the capacitor, given that it has reached electrostatic equilibrium.

So is my conclusion correct?

Peter

8. Mar 21, 2004

### Chen

What is your conclusion? If it is what you posted in the last post then yes, it is correct.