# What Happens When Two Charged Capacitors Are Connected?

• mer584
In summary, the conversation discusses a problem involving two capacitors that are charged to different voltages and then connected to each other. The question is to find the potential difference and charge on each capacitor. The conversation leads to the conclusion that the voltage across each capacitor will be the same and can be solved using the equation V = Q/C. The total charge is conserved and can be used to solve for the individual charges.
mer584
The problem.
A 2.50uF capacitor is charged to 857V and a 6.80uF capacitor is charged to 652V. These capacitors are then disconnected from their batteries. Next the positive plates are connected to each other and the negative plates are connected to each other. What will be the potential difference across each and the charge on each? (note that charge is conserved)

2. Homework Equations
PE= V/Q ; V=Ed; Q=CV; C= Eo (A/D); Vb-Va

Attempt:
What I started problem I used the forumla Q=CV to find the charge that is occurring in each case finding Q1 = .0021C and Q2= .0044 C. I really wasnt sure if this was even helpful or where to go from there. It appears to be a uniform field so I know I can use V=Ed but we don't have a distance or an area in this problem.

In order to get the potential difference I know you need to work with Vb-Va and possibly the potential energy. Should I use PE= V/Q then subtract the potential energies to find the work and then the poential difference?

Since the two capacitors are now connected in parallel, what must be true about the voltage across them?

What happens when the two caps are connected to each other? You know that some charge will flow (else the answer would be "no change") - so what's the condition for the charge to stop flowing?

Answering that should give you an equation relating the quantities you're asked solve for.

Doc Al said:
Since the two capacitors are now connected in parallel, what must be true about the voltage across them?

the voltage will be in half, split between the two, or i would have thought it was the average between the two which is 754V but the answer points towards 712V

Last edited:
No. How will the voltage across one capacitor compare to the voltage across the other?

Doc Al said:
No. How will the voltage across one capacitor compare to the voltage across the other?

i guess I don't really know
I would have said its based on the charge and the amount of capacitance but that will just bring me back to the original voltages given in the problem...I'm clearly missing a fundamental concept here.

I do get that once the battery is turned off they will maintain the same level of charge. true?

Last edited:
Nope - you're missing a key point. You know charge will flow ... which means there's a potential difference pushing it. The charge will stop flowing when there's no more potential difference. So, what's the potential between the two connected positive plates? And for the two connected negative plates?

mer584 said:
i guess I don't really know
I would have said its based on the charge and the amount of capacitance but that will just bring me back to the original voltages given in the problem...I'm clearly missing a fundamental concept here.

I do get that once the battery is turned off they will maintain the same level of charge. true?
Here's a hint: Think of it as a circuit with two parallel branches. Can one branch have a greater voltage than the other if they are in parallel?

Once this clicks you'll use the relationship between the branch voltages to set up an equation for finding the voltage. (Making use of the total charge that you've found, also.)

You might want to think about what is the total charge and total capacitance...

belliott4488 said:
The charge will stop flowing when there's no more potential difference. So, what's the potential between the two connected positive plates? And for the two connected negative plates?

OK so let me back up, because I'm fundamentally confused.
I know that the electrical potential is equal to V=Ed. which is also equal to V=PE/q. We know that E= kQ/r^2 but that we don't have a distance here so we can't use that. we also don't know d.

I'm assuming that there will be no more potential difference when the two plates are equal to each other because then Vb-Va =0 This doesn't seem right.

Can I set it up as though there is a test charge in between the two doing work..i can't really do this without a distance either i don't think

Doc Al said:
Here's a hint: Think of it as a circuit with two parallel branches. Can one branch have a greater voltage than the other if they are in parallel?)

won't it always be the same?? as in no one branch can not be greater? or is that just that the total voltage is conserved but that it can be different

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mer584 said:
won't it always be the same?? as in no one branch can not be greater?
Exactly! The voltage across each capacitor will be equal. Since C = Q/V, you can set up an equation for V = V1 = V2. (You know what the total charge is, Q1 + Q2, so you can solve for V and the two charges.)

Doc Al said:
Exactly! The voltage across each capacitor will be equal. Since C = Q/V, you can set up an equation for V = V1 = V2. (You know what the total charge is, Q1 + Q2, so you can solve for V and the two charges.)

Ok, so I completely understand how to get the different charges once I have the common voltage, and I understand why there is a common voltage. (doing better than before). But nothing I seem to set up is giving me the answer. If I want to solve for V and I know Q is .0065 the only thing i can set up is CV=CV which doesn't work.

Am I supposed to add the two capacities?

Start with C = Q/V, or V = Q/C.

Since the voltage must be the same:
V1 = V2
Q1/C1 = Q2/C2

You know C1 and C2. Combine this with what you know about the total charge (since you know the total charge is conserved): Q1 + Q2 = Q(total).

Doc Al said:
Since the voltage must be the same:
V1 = V2
Q1/C1 = Q2/C2

You know C1 and C2. Combine this with what you know about the total charge (since you know the total charge is conserved): Q1 + Q2 = Q(total).

I understand Q total, I've solved for that using C1 and C2 and the two voltages given in the problem, but when solving for V total I can't just use that above equation...I know all 4 of the variables and they don't equal each other or leave me a V total to solve for. Am I misunderstanding you?

You have two equations that you can solve together to get Q1 and Q2 (those are the new charges on the capacitors, once they've been attached):

(1) Q1/C1 = Q2/C2
(2) Q1 + Q2 = Q(total)

Once you find Q1 and Q2, you can find V from V = Q/C.

You can find the charges for any of the two capacitors respectively. You can use that result to find the V around the capacitor or not?

Doc Al said:
You have two equations that you can solve together to get Q1 and Q2 (those are the new charges on the capacitors, once they've been attached):

(1) Q1/C1 = Q2/C2
(2) Q1 + Q2 = Q(total)

Once you find Q1 and Q2, you can find V from V = Q/C.

If you are saying that Q1 and Q2 are after they are rerouted and not before then we can't use the Q total from before...that leaves us with 3 unknowns and now I'm totally confused. Can I use the original Q values at all in this case? or are they just completely irrelevant. Sorry, something just isn't clicking here.

mer584 said:
If you are saying that Q1 and Q2 are after they are rerouted and not before then we can't use the Q total from before...
Why not?? Remember that total charge is conserved. Where else would any charge go?

belliott4488 said:
Why not?? Remember that total charge is conserved. Where else would any charge go?

What I know: the initial Q1 value: .0021
the initial Q2 value: .0044
the total Q value = .0066
then using the 2 equations above i can solve for Q1b = .00177
Q2b= .00481
which gives me V values of 707 and not 712 I carried it out as many digits in my calculator as I could so I guess I'm still concerned I'm doing something wrong, or just being too picky

Looks like you're doing OK to me. Using a bit more accuracy with the arithmetic, I get a total charge of 0.006576. Which leads to:
Q1b = .001768 C
Q2b = .004808 C
V = 707 V

It's usually a good idea to only round off your final answers--keep all intermediate calculations to as great a precision as your calculator can provide.

this is weird

hey so I seem to completely understand what I'm supposed to do... I set C1 = Q1/V and C2 = Q2/V
At that point I see the obvious V=Q1/C1=Q2/C2 so I turn it into
2V=Q1/C1 +Q2/C2 this is where I diverge from what I think you guys are doing. so I have
2V(C1 + C2)= Q1+Q2 I do this so I can relate them back to Q(total). This is where the arithmetic gets funky. I end up with a V that is half of what it is supposed to be?! If it wasn't 2V it would at the 707V as you got but I'm getting 353 V this is obviously an arithmetic error but I don't see where it could possibly be which is why I am concerned.

but assuming that is fixed i get the right charges for Q1 and Q2 so if anyone could help in the arithmetic for the potential difference that would be great

ALRIGHTY THEN! Anyone who would care to shoot me for thinking you could cross multiply like that may do so right now. K. new approach. Let's laugh together as I do this...
VC1 = Q1
VC2 = Q2

You with me so far because I am about to blow everyone's minds

VC1 + VC2 = Q1 + Q2

OMG!

V(C1 + C2) = .0065761 C Not 2V(C1 + C2)

.0065761/(9.3*10-6 F) = V = 707 V

and no i don't have insurance for your time.

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izforgoat said:
hey so I seem to completely understand what I'm supposed to do... I set C1 = Q1/V and C2 = Q2/V
At that point I see the obvious V=Q1/C1=Q2/C2 so I turn it into
2V=Q1/C1 +Q2/C2 this is where I diverge from what I think you guys are doing. so I have
2V(C1 + C2)= Q1+Q2
Nope. Q1/C1 + Q2/C2 does not equal (Q1 + Q2)/(C1 + C2). If it did, then 1/2 + 1/2 = 2/4 = 1/2, which I trust you would agree is not a good thing.

## What is capacitance?

Capacitance is the ability of an object to store electrical charge. It is measured in units of Farads (F).

## How is capacitance calculated?

Capacitance is calculated by dividing the amount of charge stored (Q) by the potential difference (V) across the object. This can be represented by the equation C = Q/V.

## What factors affect capacitance?

The capacitance of an object is affected by the distance between the two conductive surfaces, the surface area of the conductors, and the type of material used.

## What are some common uses of capacitance?

Capacitance is used in electronic circuits for storing charge, filtering signals, and smoothing power supplies. It is also used in touchscreens, sensors, and in medical devices such as pacemakers.

## How does capacitance differ from resistance?

Capacitance refers to the ability to store charge, while resistance refers to the opposition of current flow. Capacitance is measured in Farads (F), while resistance is measured in Ohms (Ω).

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