Physics problem, please help.

[Shades McCool]

Need a little help please. Dang BCA.

Capacitors C1(5.5 microF) and C2(2.5 microF) are charged as a series combination across a 200-V battery. The two capacitors are disconnected from the battery and from each other. They are then connected positive plate to positive plate and negative plate to negative plate. Calculate the resulting charge on each capacitor.

Round your answer to two significant figures.

What would be the charges on C1 and C2?


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[Nutmeg]

Q=V*C

and

1/Ctot = 1/C1 + 1/C2 = > Ctot = 1.7 microF

So Q = 100 V * 1.7 microF = 170 micro Coloumbs per capacitor

Put the two together, and it would be 170 + 170 = 340 micro Coloumbs total.


I think that's right. Can anyone back that up?

 

[DD214_DOC]

Here's one for you guys.

THis is a really simple problem, but I think I'm screwing up somewhere in the calculations.


Q1---r---Q2---r---Q3


Q1=1.9 microC Q2=-Q1 Q3=-Q1
r=1.7m

Find the total force (in N) upon Q2

I know the equation is F=k[(|Qx||Qx|)/r^2)]. Maybe you guys won't make the same mistake I did.

 

[Gbemi24]

Originally posted by Nutmeg
Q=V*C

and

1/Ctot = 1/C1 + 1/C2 = > Ctot = 1.7 microF

So Q = 100 V * 1.7 microF = 170 micro Coloumbs per capacitor

Put the two together, and it would be 170 + 170 = 340 micro Coloumbs total.


I think that's right. Can anyone back that up?

Actually Q = 340 microColoumbs per capacitor, not 170.

Since the capacitors are in series, the charges on them must be equal. This follows intuitionally from the fact that all current entering and leaving one capacitor must enter and leave the other capacitor. You can prove this mathematically by showing that for Qtot/Ctot=Q1/C1+Q2/C2 AND 1/Ctot = 1/C1 + 1/C2 to be true at the same time, Qtot=Q1=Q2 must be true.

The sticky part of the problem asks for the new Qs after new capacitors are created using one plate from each of the original capacitors, with the plates of the new capacitors having the same sign. I suspect the answer is still 340 microColoumbs but I am not sure.

The strange thing about this answer is that IF it is true, the capacitances of the new arrangements would be infinite because the potential difference between the plates would be zero. This is why I am uncomfortable with the answer, but it could still be true.

 

[Gbemi24]

Originally posted by JKDMed
Here's one for you guys.

THis is a really simple problem, but I think I'm screwing up somewhere in the calculations.


Q1---r---Q2---r---Q3


Q1=1.9 microC Q2=-Q1 Q3=-Q1
r=1.7m

Find the total force (in N) upon Q2

I know the equation is F=k[(|Qx||Qx|)/r^2)]. Maybe you guys won't make the same mistake I did.

The total force on Q2 is 2kQ1^2/r^2 in the direction of Q1

 

[rugbyboy]

Hey guys,
Unfortunately the computer I am using does not have a calculator or I could have backed up the numbers.

The way I understand it is this. Since initially they both are in series ( negative plate to positive plate ) their capacitance would add reciprocally

1/Ctot= 1/C1 + 1/C2

connect the battery and they should both have a charge equal to

Q= Ctot * V

Now the battery is disconnected and they are connected in parallel ( negative to negative and positive to positive). I think the battery would still be connected. Now there capacitance would just add

Ctot= C1 + C2

The charge should be Q=Ctot* V

If the battery is not connected, there charges would repel and essentially be the same.

Disconnecting plates and creating new capacitors dosen't reallly make a whole lot of sense to me.The repulsion would be so great that they can't make a capacitor.

 

[Gbemi24]

Originally posted by rugbyboy
Hey guys,
Unfortunately the computer I am using does not have a calculator or I could have backed up the numbers.

The way I understand it is this. Since initially they both are in series ( negative plate to positive plate ) their capacitance would add reciprocally

1/Ctot= 1/C1 + 1/C2

connect the battery and they should both have a charge equal to

Q= Ctot * V

Now the battery is disconnected and they are connected in parallel ( negative to negative and positive to positive). I think the battery would still be connected. Now there capacitance would just add

Ctot= C1 + C2

The charge should be Q=Ctot* V

If the battery is not connected, there charges would repel and essentially be the same.

You just said the samething I said in different words.

Originally posted by rugbyboy

Disconnecting plates and creating new capacitors dosen't reallly make a whole lot of sense to me.The repulsion would be so great that they can't make a capacitor.

I agree, but lets suppose it could be done theoretically. Then what? This part of the question seems too easy and strange to be correct.

 

[nrosigh]

You probably won't need to get into that much detail on the test.

 

[crazee8]

I got the same answer as Nutmeg but I never split it up.

a) 1/Ctot= 1/C1 + 1/C2

1/Ctot = 1/5.5 microF + 1/2.5 microF

1/.582 = 1.71 microF= c

b) q=CV
q= (1.71microF) (200V)
q= 340 microColoumbs

(Anyway, I am mainly relying on my AP Physics C, electricity and magnetism background here.)

 

[Gbemi24]

Originally posted by crazee8
I got the same answer as Nutmeg but I never split it up.

a) 1/Ctot= 1/C1 + 1/C2

1/Ctot = 1/5.5 microF + 1/2.5 microF

1/.582 = 1.71 microF= c

b) q=CV
q= (1.71microF) (200V)
q= 340 microColoumbs

(Anyway, I am mainly relying on my AP Physics C, electricity and magnetism background here.)

You are correct, but you did not address the second part of the question. I feel like the answer is 340 regardless but the second part of the question is weird. What do yo think?

 

[DD214_DOC]

Originally posted by Gbemi24
The total force on Q2 is 2kQ1^2/r^2 in the direction of Q1

Ah, I see what I did wrong. I forgot to put the (-) in front of my answer. I knew it was something stupid.

 

[Nutmeg]

Originally posted by crazee8
I got the same answer as Nutmeg but I never split it up.

a) 1/Ctot= 1/C1 + 1/C2

1/Ctot = 1/5.5 microF + 1/2.5 microF

1/.582 = 1.71 microF= c

b) q=CV
q= (1.71microF) (200V)
q= 340 microColoumbs

(Anyway, I am mainly relying on my AP Physics C, electricity and magnetism background here.)

That's actually what I did at first, but then I became uncertain and I went back in and split 'em. But splitting doesn't work I now realize, I'm mismatching my charges and capacitors.

But for the second part, use have essentially put them in series, so the voltage drop is the same over each of the capacitors. So now you have a total charge of 340 divided between two capacitors in series. The voltage over the two capacitors is the same, but since the capacitences are different, the charge will be unequally divided. So you get Q1/C1 = Q2/C2, and Q1 + Q2 = 340. Two eqns, two UKs.

Q2 = 106.25, Q1 = 233.75

 

[Gbemi24]

Originally posted by Nutmeg
But for the second part, use have essentially put them in series, so the voltage drop is the same over each of the capacitors. So now you have a total charge of 340 divided between two capacitors in series. The voltage over the two capacitors is the same, but since the capacitences are different, the charge will be unequally divided. So you get Q1/C1 = Q2/C2, and Q1 + Q2 = 340. Two eqns, two UKs.

Q2 = 106.25, Q1 = 233.75

Q1+Q2 is NOT 340. Q1=Q2=340 in the original circuit. Read my response to your earlier post to see why.

Also, you ignored an important part of the second part of the question. The capacitors in the second part are hybrid with one plate coming from the C1 capacitor and the other coming from the C2 capacitor. Another thing is that both plates of the new capacitors have the same type of charge(+ or -). In light of these facts I doubt your solution is correct. Part of the problem is that the question is too vague with regard to the second part. The OP does not say whether the hybrid capacitors are connected in series or not.

 

[Gbemi24]

Originally posted by Shades McCool
Need a little help please. Dang BCA.

Capacitors C1(5.5 microF) and C2(2.5 microF) are charged as a series combination across a 200-V battery. The two capacitors are disconnected from the battery and from each other. They are then connected positive plate to positive plate and negative plate to negative plate. Calculate the resulting charge on each capacitor.

Round your answer to two significant figures.

What would be the charges on C1 and C2?

Are you stating the question directly from a book or are you paraphrasing?

 

[Shades McCool]

That is the question straight up.

 

[Nutmeg]

Originally posted by Gbemi24
Q1+Q2 is NOT 340. Q1=Q2=340 in the original circuit. Read my response to your earlier post to see why.

Also, you ignored an important part of the second part of the question. The capacitors in the second part are hybrid with one plate coming from the C1 capacitor and the other coming from the C2 capacitor. Another thing is that both plates of the new capacitors have the same type of charge(+ or -). In light of these facts I doubt your solution is correct. Part of the problem is that the question is too vague with regard to the second part. The OP does not say whether the hybrid capacitors are connected in series or not.

Series was a typo--I meant to write parallel, and it was on the parallel assumption that I calculated the results.

 

[Nutmeg]

Originally posted by Gbemi24
Q1+Q2 is NOT 340. Q1=Q2=340 in the original circuit. Read my response to your earlier post to see why.

Also, you ignored an important part of the second part of the question. The capacitors in the second part are hybrid with one plate coming from the C1 capacitor and the other coming from the C2 capacitor. Another thing is that both plates of the new capacitors have the same type of charge(+ or -). In light of these facts I doubt your solution is correct. Part of the problem is that the question is too vague with regard to the second part. The OP does not say whether the hybrid capacitors are connected in series or not.

I had skipped past your post earlier, and went to the post where they said they'd decided it was Qtot = 340, but I see that you're right, so multiply that last pair o' answers I gave by 2 and you've got the answer. Of course, it was probably due yesterday, but hey.