How Do Voltage Changes Affect Capacitors in a Circuit?

[robbondo]

Homework Statement

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?

Homework Equations

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

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 parallel 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.

 

[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...

 

[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.

 

[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.

 

[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...

 

[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?

 

[learningphysics]

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?

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...

 

[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.

 

[learningphysics]

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.

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

 

[pooface]

All of the capacitors are attached in series before S is closed.

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

 

[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.

 

[pooface]

Oh cool. Okay.

Once you get the charge on the first 9 microfarad capacitor go back to the original circuit. You don't 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.

 

[robbondo]

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

 

[pooface]

That works I believe only for this situation. If the other cap (6mF) was a different value then the subtracting wouldn't work.

 

[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.

 

[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...

 

[learningphysics]

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

 

[learningphysics]

Well, I'm getting 315microC...

 

[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?

 

[learningphysics]

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?

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.

 

[pooface]

I see. But what about after the switch is closed. What is the charge through the switch?

 

[learningphysics]

I see. But what about after the switch is closed. What is the charge through the switch?

I don't understand... before the switch is closed the charge above the switch (in the segment between the two capacitors) is 0C. same thing below the switch. 0C

after the switch is closed... the charge above the switch is 0.000315C. The charge below the switch is -0.000315C.

That means that when the switch closed, 0.000315C flowed from the bottom to the top.

 

[pooface]

That is according to the equivalent circuit and you are correct. but if you step back to the original circuit the switch is a wire connecting in between the 4 capacitors. after it is closed how much charge is going through it?

I think that is the question that was being asked.

 

[robbondo]

reply... well the correct answer is 315, 310 was an error on the part of the person I spoke to. So how did you get a negative value for Q? That may be beyond my understanding of capacitors. Which is the - side and which is the + side? The method that I used of subtracting the charges got the correct answer, was that just dumb luck?

 

[learningphysics]

reply... well the correct answer is 315, 310 was an error on the part of the person I spoke to. So how did you get a negative value for Q? That may be beyond my understanding of capacitors. Which is the - side and which is the + side? The method that I used of subtracting the charges got the correct answer, was that just dumb luck?

The + side is at a higher voltage than the - side of a capacitor... in the first case when the capacitors are connected in series at the top... the 3.00microF has a voltage of 140V across it... a voltage drop of 140V from left to right. (left side is 140V above right side). so the left side is +... right side is -.

as far as the charge that flows when the switch is closed, I think they just wanted the magnitude... they only asked how much flowed through... not how much flowed from the bottom to the top... or top to the bottom...