Energy stored in capacitor w/ dielectric

In summary, when a dielectric material is inserted between the plates of a charged parallel plate capacitor, the energy stored decreases. This is because the force applied to push the dielectric does work on the system, extracting or dissipating some of the energy. The final energy of the capacitor is lower due to this work. If the capacitor is disconnected from the charging source, the dielectric will oscillate back and forth, but eventually the energy will be dissipated due to EM radiation. However, if the capacitor is connected to the source, a current will flow and the dielectric will not oscillate. The force exerted on the dielectric is given by the formula {εb(v^2)(k-1)}/d, but
  • #1
kbar1
15
0
Suppose a (parallel plate) capacitor of capacitance C is charged to a potential difference V and then disconnected and isolated. Energy stored E1= Q2/2C.

Now if a material of appropriate dimensions and dielectric constant K is fully inserted between the plates, energy stored E2= Q2/2KC.

E2 < E1.

My question is: where did the "missing" energy go?
 
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  • #2
How much force does it take to insert the dielectric?
 
  • #3
DaleSpam said:
How much force does it take to insert the dielectric?

Pardon me if I'm wrong (I'm new to this topic), but does the force applied to push the dielectric matter? The only thing nagging me is that the energy stored in the two cases is different, and I'd like to know where the difference went.
 
  • #4
kbar1 said:
Pardon me if I'm wrong (I'm new to this topic), but does the force applied to push the dielectric matter?
Certainly it matters. It can do work on the system or allow the system to do work on the environment.

I don't know the answer to your question, but that is where I would look first.
 
  • #5
Never thought about it. Does the electric field set up between the plates oppose the insertion of the dielectric?

Wild guess on my part: The energy is used to attract the dielectric, because the capacitor system with it has lesser energy (i.e. more preferable) than the capacitor without the dielectric. Comment?
 
  • #6
That is my guess also, but I don't know.
 
  • #7
When you approach the capacitor with dielectric material, the capacitor will in fact attract the material. This pulling force performs work on the material and can be extracted or dissipated. The final energy of the condenser is lower by this amount of extracted/dissipated work.
 
  • #8
Guess that clears it. Thanks all.
 
  • #9
Note: To add a bit of understanding (or possible confusion!)
If the capacitor happens to be disconnected from the charging source / battery, the PD across it will reduce as the dielectric is inserted. If it is connected to the source then a current will flow during the insertion because the PD will be held constant by the battery.

It may be of interest to consider what exactly happens to the energy in both of these cases. If the capacitor is disconnected and the arrangement is frictionless (and no other energy losses - zero internal resistance in the battery, etc.), the dielectric will be pulled into the middle but then its KE will carry it out the other side and it will oscillate for ever, back and forth.
But there will be a smart 'someone' who realizes that EM energy will be radiated due to the AC in the system, so the oscillations will always die down, in the end.
 
  • #10
Jano L. said:
When you approach the capacitor with dielectric material, the capacitor will in fact attract the material. This pulling force performs work on the material and can be extracted or dissipated. The final energy of the condenser is lower by this amount of extracted/dissipated work.

sice the formula of force exerted on a dielectric is {εb(v^2)(k-1)}/d

sice the capacitor is isolated so v=0 hence force is zero
 
Last edited:
  • #11
Just because the capacitor is isolated doesn't mean that V=0; I don't know what would lead you to believe that. Per the setup V is nonzero and there is a force.
 

1. What is a capacitor?

A capacitor is an electronic component that is used to store electrical energy. It consists of two conductive plates separated by a dielectric material.

2. How does a capacitor store energy?

A capacitor stores energy by creating an electric field between its two plates when it is connected to a voltage source. This electric field allows for the accumulation of electrical charge on the plates, which results in the storage of energy.

3. What is the role of a dielectric in a capacitor?

A dielectric is an insulating material that is placed between the two plates of a capacitor. Its main function is to increase the capacitance of the capacitor, which allows for a higher amount of energy to be stored.

4. How does the presence of a dielectric affect the energy stored in a capacitor?

The presence of a dielectric increases the capacitance of a capacitor, which in turn increases the amount of energy that can be stored. This is because the dielectric material reduces the electric field strength between the plates, allowing for more charge to be stored on the plates.

5. What factors affect the amount of energy stored in a capacitor with a dielectric?

The amount of energy stored in a capacitor with a dielectric is affected by several factors, including the capacitance of the capacitor, the voltage applied to it, and the type of dielectric material used. Additionally, the distance between the plates and the surface area of the plates also play a role in determining the energy stored in the capacitor.

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