Potentials in a Capacitor

1. Oct 21, 2007

robbondo

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

The capacitors in the figure are initially uncharged and are connected, as in the diagram, with switch S open. The applied potential difference is $$V_{ab}= + 210 {\rm V}$$.
What is the potential difference $$V_{cd}$$?
What is the potential difference $$V_{ad}$$ after the switch is closed?
What is the potential difference $$V_{db}$$ after the switch is closed?

2. Relevant equations
$$C = \frac{q}{V_{ab}}$$
$$C_{eq} = \frac{1}{C_{1}} + ...$$
$$C_{eq} = {C_{1}} + ...$$

3. The attempt at a solution

I have no clue how to approach this problem. All of the capacitors are attached in series before S is closed. And in half in series, half in paralell after. I also know the potentials all have to add up to 210. Other than that I'm completely lost on how to approach these types of problems. The only thing my book explains is how to determing the equivalent capacitance, and I don't see how those equations help me solve this problem.

2. Oct 21, 2007

learningphysics

Are you getting stuck trying to find equivalent capacitances? Two capacitors in series, are like two resistors in parallel... Ceq = C1*C2/(C1+C2)

Parallel capacitances are like series resisotrs... Ceq = C1 + C2.

Capacitors in series have the same charge...

To answer the first part... get the voltage across the 3.00microF capacitor between a and c (Vac)... and the 6.00microF cap between a and d (Vad).

What is the equivalent capacitance of the 3.00microF and the 6.00microF in the top line... knowing the equivalent capacitance and the voltage across this equiv. capacitance... what is the charge on the equivalent capacitance?

That's the charge on each of the capacitors in series... so you can then get the voltage across the capacitor of 3.00microF in the top line Vac...

using the same idea what's the voltage across the 6.00microF in the bottom line... you don't need to repeat the whole process because the capacitors in the bottom line are the same as the top just switched...

so you can get Vad and Vac. Vcd = Vad - Vac.

There is a shortcut to solving this using voltage dividor ideas for capacitors... but it is important to first get a feel for doing it this longer way...

Last edited: Oct 21, 2007
3. Oct 21, 2007

learningphysics

Voltage dividor for 2 capacitors in series... C1 and C2...

If total voltage across both capacitors is V, then voltage across C1 is $$\frac{C2}{C1+C2}*V$$

in other words the voltage across one capacitor is the ratio of the other capacitance to the sum of the capacitances... times voltage.

4. Oct 22, 2007

robbondo

OK, so the equivalent capacitance on both sides is the same, and it is 0.5 microF. So I now that $$C = \frac{q}{V_{ab}}$$

So $$0.5 = \frac{q}{210}$$ So then q=105 and then when I solve for Vad I use the same equation with the capacitance of 3 and q$$3 = \frac{105}{V_{ad}}$$ with the answer of Vad = 35. Thorough the same process I get Vac= 17.5 I know this is wrong because the correct answer is 70 for the voltage Vcd. Where'd I screw up. I think I'm having a hard time because I don't understand why the numbers I'm plugging in are related to each other. LIke Is the voltage is split up between the top and bottom? And even if I did that and used 105 for the voltage across the top, I still get a wrong answer.

5. Oct 22, 2007

robbondo

OK I'm dumb. I totally messed up the math determining the equivalent capacitance, so i got the correct answer now. Thanks! Now i just have to figure out what to do when the switch is closed...

6. Oct 22, 2007

robbondo

So now the question is... How much charge flowed through the switch when it is closed. Is it half of the total charge from all the potentials?

7. Oct 22, 2007

learningphysics

When the switch is closed, you have the 3.00microF on the top between a and d, in parallel with the 6.00microF on the bottom between a and c. same way the other 2 capacitors are in parallel...

And the 2 pairs of parallel capacitors are in series...

replace each pair of 2 parallel capacitors with the equivalent capacitance...

so instead of 4 capacitors, you now have 2 in series...

Now you can solve the problem, exactly the same way as the first part...

8. Oct 22, 2007

robbondo

where I'm getting confused is how to "work backwards" from one capacitor to two. So if I combine them, and find their capacitance and charge, then when I go backwards, what is their individual charges/potentials. Does it depend on whether they are connected in series or in parallel.

9. Oct 22, 2007

learningphysics

But you don't need to go backwards here... what did you get as the equivalent capacitances and charges?

10. Oct 22, 2007

pooface

Are you sure? It doesn't look like it.

11. Oct 22, 2007

robbondo

Pooface, yeah i know I was wrong about that. What I did was take the two parallel capacitors and make them into 2 9microF capacitors connected in series. Then I combined those two into one 9/2microF capacitor. Then I thought if you multiply that by the Potential 210 you would get the net charge. But my answer is WAY to big. The correct answer is 310.

12. Oct 22, 2007

pooface

Oh cool. Okay.

Once you get the charge on the first 9 microfarad capacitor go back to the original circuit. You dont want the net charge, you want the charge that went through the switch.

My knowledge is shady on this, but the 6mF is 2x bigger than the 3mF capacitor so 3/(6+3) charge will go through the switch. multiply 3/9 X 945 and i get 315.

5 more than the value you stated. Let's see what an expert says.

13. Oct 22, 2007

robbondo

I got 315 as well by using a different method. I subtracted the charges on the 6microF and the 3microF.

14. Oct 22, 2007

pooface

That works I believe only for this situation. If the other cap (6mF) was a different value then the subtracting wouldnt work.

15. Oct 22, 2007

robbondo

I hate capacitor problems. I was ok with Gauss' law and then so-so with potentials/electric fields and now this is for some reason insanely confusing to me. The answer you got of 310 may be correct though. This is the answer a friend of mine used and got correct on the computer program, but maybe it is accepting within a range of answers.

Here's the explanation he gave me which was a little confusing.

First, draw the circuit with the switch open. Find the charge on each capacitor (this requires you to reduce the entire circuit down to one capacitor, find the charge on the equivalent capacitor, and then work backwards to assign the charge to each capacitor).

Second, draw the circuit with the switch closed. Notice that this makes the two left capacitors in a parallel configuration and the two right capacitors in a parallel configuration. Find the charge on each one of these capacitors by the same method described above.

Then, if you take sum the absolute values of the differences between the charges on the capacitors, you will obtain the net charge flow when the switch is closed.

16. Oct 22, 2007

learningphysics

I'm confused. which part is 310?

When the switch closes... you have a 4.5microF from a to d/c... and a 4.5microF from d/c... to b.

by symmetry Vad = 105. Vdb = 105.

but I might be missing something...

17. Oct 22, 2007

learningphysics

Oh, I'm sorry guys... I didn't notice the next part of the question... "what is the total charge that passes..."

18. Oct 22, 2007

learningphysics

Well, I'm getting 315microC...

19. Oct 22, 2007

pooface

That is what i got initially but now i am thinking it should be Zero. because 315microC should be coming from the other direction as well no? This is a balanced bridge problem so it should be 0?

20. Oct 22, 2007

learningphysics

Initially the top line has a net charge of 0 between the - side of the left capacitor, and the + side of the right capacitor...

after the switch is closed... the - side of the left capacitor has a charge of -0.000315C. The + side of the right capacitor has a charge of +0.000630C. that's a net charge of +0.000315C.

same way on the bottom. the - side of the left capacitor has a charge of -.000630C and the + side of the right caacitor has a charge of +0.000315C. that's a net charge of -0.000315C.

so 0.000315C flowed from the bottom to the top.