# Heat liberated after connecting two plates

• Saitama
In summary: 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...
Saitama

(see attachment)

## 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!

#### Attachments

• capacitance.png
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• attachment 2.png
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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.

TSny said:
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?

Pranav-Arora said:
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.

TSny said:
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.

Can you use symmetry to decide on the total final charge for each plate?

TSny said:
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?

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

TSny said:
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?

Pranav-Arora said:
Anyone?

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.

Pranav-Arora said:
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.

haruspex said:
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?

Pranav-Arora said:
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?

Dick said:
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.

Pranav-Arora said:
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.

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?

Dick said:
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.

Pranav-Arora said:
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.

Dick said:
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?

Pranav-Arora said:
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?

Pranav-Arora said:
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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haruspex said:
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 canceled 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

• plates.png
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Pranav-Arora said:
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 canceled 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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Dick said:
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.

Pranav-Arora said:
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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Dick said:
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.

Thanks Dick, can you help me with finding the final charges? :)

Pranav, when the plates get connected, they become a single conductor, one side having more charge than the other. What happens if you touch a metal plate with a charged rod? The charges go over to the plate and disperse on it till the charge density is the same everywhere. The same happens here, the two plates are equivalent, the 4q charge will disperse evenly on the plates.

Now look at the picture. The two plates with their 4q charge look a single tiny charged object from far away. (At great distance, the field is like that of a point charge), and the field outside the plates does not depend on the field inside. The energy of the whole set-up is the sum of the field outside and the field inside. When the plates are connected there is no field inside. So what is the energy difference between the field before and after connecting the plates?

ehild

#### Attachments

• chplates.JPG
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ehild said:
Pranav, when the plates get connected, they become a single conductor, one side having more charge than the other. What happens if you touch a metal plate with a charged rod? The charges go over to the plate and disperse on it till the charge density is the same everywhere. The same happens here, the two plates are equivalent, the 4q charge will disperse evenly on the plates.

Now look at the picture. The two plates with their 4q charge look a single tiny charged object from far away. (At great distance, the field is like that of a point charge), and the field outside the plates does not depend on the field inside. The energy of the whole set-up is the sum of the field outside and the field inside. When the plates are connected there is no field inside. So what is the energy difference between the field before and after connecting the plates?

ehild

Nice explanation ehild. Thanks!

The energy difference is equal to the energy stored between the plates before the switch is closed.

Nice.

Before searching for equations which can be connected with a problem, try to imagine that you are involved in the situation. For example, in this problem, imagine two dance rooms of the same size, and there are 300 people in one of them and only 100 people in the other one. The door between the rooms is locked at the beginning. You are in the more crowded room. Somebody opens the door. What do you do? And how many people will be present in both rooms at the end?
Here the electrons are the people, the rooms are the plates, and you are an electron on the more crowded plate...

ehild

ehild said:
The same happens here, the two plates are equivalent, the 4q charge will disperse evenly on the plates.

Maybe more accurate to say that the 4q charge is evenly spread over the outside surfaces of the plates? There won't be any to speak of on the inside surfaces.

Dick said:
Maybe more accurate to say that the 4q charge is evenly spread over the outside surfaces of the plates? There won't be any to speak of on the inside surfaces.

Of course, you are right, evenly spread on the outside of the plates.

ehild

## 1. What is the definition of "heat liberated"?

"Heat liberated" refers to the amount of thermal energy released or given off when two objects with different temperatures are connected or brought into contact with each other.

## 2. How is the amount of heat liberated calculated?

The amount of heat liberated can be calculated using the formula Q = mcΔT, where Q is the heat liberated, m is the mass of the objects, c is the specific heat capacity, and ΔT is the change in temperature.

## 3. What factors affect the amount of heat liberated?

The amount of heat liberated is affected by the mass and specific heat capacity of the objects, as well as the temperature difference between the objects.

## 4. Why is it important to consider the heat liberated when connecting two plates?

Understanding the amount of heat liberated is important in many scientific and engineering applications, such as in thermodynamics, heat transfer, and material science. It can also help in predicting and controlling the temperature changes in systems.

## 5. How does the heat liberated after connecting two plates affect the surrounding environment?

The heat liberated after connecting two plates can affect the surrounding environment by increasing the temperature and potentially causing changes in the properties of the surrounding materials. This can also have an impact on the overall energy balance of the system.

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