Heat liberated after connecting two plates

Saitama

1. Homework Statement
(see attachment)

2. Homework Equations

3. The Attempt at a Solution
When the switch is open, the charges on the plates will rearrange and the final charges would be as shown in attachment 2. The energy stored due to the electric field between the plates (sorry if I said it incorrectly) is $U=\frac{1}{2}ε_o E^2 \cdot (Ad)$, where E is the electric field between the plates. Solving I get, $U=\frac{q^2d}{2Aε_o}$. I don't know how to proceed from here.

Any help is appreciated. Thanks!

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TSny

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You need to determine how the charges will be arranged after the switch is closed. What do you think would happen in the figure of attachment 2 if you connected a wire from the inside surface of the left plate to the inside surface of the right plate?

Note that the expression for U that you gave is only for the energy of the field between the plates before the switch is closed. There would also be energy in the field outside the plates before the switch is closed. (Far away form the plates the field will be similar to a point charge of charge 4q.) But don't worry about that for now.

Saitama

You need to determine how the charges will be arranged after the switch is closed. What do you think would happen in the figure of attachment 2 if you connected a wire from the inside surface of the left plate to the inside surface of the right plate?
That's something I am having trouble with. Its been long I have touched this.

The charges rearrange again?

TSny

Homework Helper
Gold Member
That's something I am having trouble with. Its been long I have touched this.

The charges rearrange again?
Yes, when the switch is closed there is some rearrangement of the charge.

Saitama

Yes, when the switch is closed there is some rearrangement of the charge.
I still can't think of any equation that would help me to find the final charges. TSny

Homework Helper
Gold Member
Can you use symmetry to decide on the total final charge for each plate?

Saitama

Can you use symmetry to decide on the total final charge for each plate?
Still clueless. :(

Would the charges be 2Q and 2Q on each plate?

TSny

Homework Helper
Gold Member
Yes. By symmetry the plates end up with the same net charge and the total charge in the system must be conserved.

Saitama

Yes. By symmetry the plates end up with the same net charge and the total charge in the system must be conserved.
lol, that was just a wild guess. Can you teach me how can I find the value of final charges in a proper manner?

Anyone?

sankalpmittal

Well, according to me you have to use symmetry only. When switch is closed, there will be no net transfer of charges and charges on both side parallel plates will be same. This is because of established charge equilibrium between two plates, and coulomb law of conservation of charge is obeyed.

haruspex

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2018 Award
Can you teach me how can I find the value of final charges in a proper manner?
If the charges are not equal, there will be a potential difference, so a current will flow in a direction tending to equalise them. To answer it rigorously, you would have to assume a certain set of circuit characteristics (resistance, induction, capacitance), find the equations of state, and show that as time tends to infinity it converges to equal charge. In the real world, just assume a steady state is reached.

Saitama

If the charges are not equal, there will be a potential difference, so a current will flow in a direction tending to equalise them. To answer it rigorously, you would have to assume a certain set of circuit characteristics (resistance, induction, capacitance), find the equations of state, and show that as time tends to infinity it converges to equal charge. In the real world, just assume a steady state is reached.
Thanks haruspex! What should be my next step? Find the charges on each side of plate?

Dick

Homework Helper
Thanks haruspex! What should be my next step? Find the charges on each side of plate?
Well, yes. Did you derive the charges on each side of the plate? Or was that given to you? Do you agree the charges will equalize on each plate?

Saitama

Well, yes. Did you derive the charges on each side of the plate?
Before closing the switch?
Dick said:
Do you agree the charges will equalize on each plate?
I do know that the charges flow until the plates are at same potential but can't really form an equation to find the final charges. :uhh:

Dick

Homework Helper
Before closing the switch?

I do know that the charges flow until the plates are at same potential but can't really form an equation to find the final charges. :uhh:
Let's bypass the other questions for now. If you agree the plates are at the same potential after the switch is closed, what's the E field between them after the switch is closed?

Saitama

Let's bypass the other questions for now. If you agree the plates are at the same potential after the switch is closed, what's the E field between them after the switch is closed?
How? I don't have the final charges to find the electric field. Dick

Homework Helper
How? I don't have the final charges to find the electric field. You don't need to. The E field is the gradient of the potential between the two plates, right? If they are at the same potential, what does that mean? If you want to backtrack you can certainly find the charge on the inner surface between the two plates if you need to.

Saitama

You don't need to. The E field is the gradient of the potential between the two plates, right? If they are at the same potential, what does that mean? If you want to backtrack you can certainly find the charge on the inner surface between the two plates if you need to.
Oh yes, there would no electric field between the plates.

I think I have got it but what about the energy TSny pointed out in his post #2. Don't I need to worry about that?

haruspex

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2018 Award
I think I have got it but what about the energy TSny pointed out in his post #2. Don't I need to worry about that?
Yes, you need to determine the drop in potential energy. What's the energy of the E field when the charges are equal?

Dick

Homework Helper
Oh yes, there would no electric field between the plates.

I think I have got it but what about the energy TSny pointed out in his post #2. Don't I need to worry about that?
TSny said to ignore that for now. And I agree. It's not important in a realistic capacitor where the area is large and the separation is small. What is important that the energy contained in the E field between the plates vanished after the switch was closed. That's what they expect you to realize must have been turned into heat.

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Saitama

Yes, you need to determine the drop in potential energy. What's the energy of the E field when the charges are equal?
There will be no electric field in the middle of plates, therefore there will be no energy.

I tried to calculate the final charges, not sure if I am doing it right. I assumed that the charges on each side of left plate changes by an amount q_1 and on the right changes by q_2. (see attachment). Since electric field is zero in the middle, electric field due to right side of the left plate is cancelled by the electric field due to the left side of right plate. From here I get $q_1-q_2=2q$. I don't know how to proceed from here.

Attachments

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Dick

Homework Helper
There will be no electric field in the middle of plates, therefore there will be no energy.

I tried to calculate the final charges, not sure if I am doing it right. I assumed that the charges on each side of left plate changes by an amount q_1 and on the right changes by q_2. (see attachment). Since electric field is zero in the middle, electric field due to right side of the left plate is cancelled by the electric field due to the left side of right plate. From here I get $q_1-q_2=2q$. I don't know how to proceed from here.
You are doing the thing I told you that you don't have to do. You already know the E field is zero. That's the important thing. Do you actually want to solve this or solve the heat problem part?

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Saitama

Do you actually want to solve this or solve the heat problem part?
Both. I am still not able to understand why do I need to consider only the energy stored between the plates? For instance take the space right to the right plate. The electric field before opening and closing the switch would be different. This would mean that the energy due to electric field is also different.

Dick

Homework Helper
Both. I am still not able to understand why do I need to consider only the energy stored between the plates? For instance take the space right to the right plate. The electric field before opening and closing the switch would be different. This would mean that the energy due to electric field is also different.
The electric field is basically nonzero everywhere except inside of the conductors. The LARGE electric field is between the two plates. I think this is the approximation they are expecting you to make. You are getting distracted by details. Just consider the approximate E field between the plates.

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