What is the explanation for the discrepancy in energy stored in a capacitor?

In summary: I had a go doing the calculation without keeping v constant, but my maths is not up to it. With v constant we have: (say the +ve plate is travelling, E parallel with J, ignore edge fields)P/Vol = E.JdP = E I dl = E dq/dt dl = E dq v/+ve dq=E dq/−ve dt dF = E dq Since q remains constant, so does F therefore:F = E q W = int F dl = q int E dl = qU
  • #1
Per Oni
261
1
In the thread about permanent magnets it is stated that power per unit volume is E.J As you perhaps saw it is quite a job to prove that fact in the case of magnets. I thought it should be a lot easier to prove that in the electrical equivalent case of 2 opposite charged plates.

My back of envelope calculation went as follows: Let one plate approach the other with a constant (low) velocity and collide. Now, according to Gauss’ law E between the plates is q/Aε. E remains constant until the gap is nearly closed, I will ignore the last micro meter of distance where E vanishes. Next: J=q/At. Put together: (E is parallel to J) P=E.J x vol=E x q/At x Vol so that energy W=E x q x d , where Vol=A x d. But here E x d = U, then W=qU.

So at first sight not a bad result except that the result should be 1/2qU. Where’s the rub? I think I know but what do you think?
 
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  • #2
Per Oni said:
In the thread about permanent magnets it is stated that power per unit volume is E.J As you perhaps saw it is quite a job to prove that fact in the case of magnets. I thought it should be a lot easier to prove that in the electrical equivalent case of 2 opposite charged plates.

My back of envelope calculation went as follows: Let one plate approach the other with a constant (low) velocity and collide. Now, according to Gauss’ law E between the plates is q/Aε. E remains constant until the gap is nearly closed, I will ignore the last micro meter of distance where E vanishes. Next: J=q/At. Put together: (E is parallel to J) P=E.J x vol=E x q/At x Vol so that energy W=E x q x d , where Vol=A x d. But here E x d = U, then W=qU.

So at first sight not a bad result except that the result should be 1/2qU. Where’s the rub? I think I know but what do you think?

The problem is, you are putting energy in the system unknowingly, when you are making the plate to move at constant velocity !

Otherwise you would have to integrate to get the result, which would have put 1/2, that your calculation is missing.
 
  • #3
universal_101 said:
The problem is, you are putting energy in the system unknowingly, when you are making the plate to move at constant velocity !

Otherwise you would have to integrate to get the result, which would have put 1/2, that your calculation is missing.

That was one of my thoughts as well. But consider the fall of raindrops. They are falling after a short while with a constant velocity without putting energy in the system!

However, I should of course have done this calculation properly and used integration. Perhaps I will do in the weekend. In the meantime I’m convinced that the answer lays elsewhere.
 
  • #4
I had a go doing the calculation without keeping v constant, but my maths is not up to it. With v constant we have: (say the +ve plate is travelling, E parallel with J, ignore edge fields)

P/Vol = E.J
dP = E I dl = E dq/dt dl = E dq v
dF = E dq
Since q remains constant, so does F therefore:
F = E q
W = int F dl = q int E dl = qU

I hope somebody will do this calculation with v as a function of time or distance, perhaps it will give the correct answer after all.
 
  • #5

For the explanation.

The lecturer states that the charge is emerged in an E field which goes from E max to zero and therefore the average value is ½ E, which is one way of looking at it. I prefer the view that the +ve charge can only be attracted by the –ve plate and the field of the –ve plate is only ½ E max.
 
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1. What is a capacitor?

A capacitor is an electronic component that stores electrical charge. It consists of two conductive plates separated by an insulating material, and is commonly used in electronic circuits to temporarily hold and release electrical energy.

2. How is energy stored in a capacitor?

When a capacitor is connected to a power source, one plate accumulates a positive charge and the other plate accumulates a negative charge. This separation of charges creates an electric field between the plates, which stores the energy in the capacitor.

3. What affects the amount of energy stored in a capacitor?

The amount of energy stored in a capacitor is directly proportional to the capacitance (C) of the capacitor and the square of the voltage (V) applied. In other words, a higher capacitance or voltage will result in a greater amount of energy stored in the capacitor.

4. How is energy released from a capacitor?

When a capacitor is connected to a circuit, the stored energy is released as the charges flow from one plate to the other. This flow of charges creates an electric current, which can be used to power devices in the circuit.

5. Are there any factors that can affect the storage and release of energy in a capacitor?

Yes, there are several factors that can affect the storage and release of energy in a capacitor. These include the material and distance between the plates, the dielectric constant of the insulating material, and the temperature of the capacitor. Additionally, the type of circuit and the frequency of the power source can also impact the storage and release of energy in a capacitor.

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