Energy stored in capacitor w/ dielectric

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SUMMARY

The discussion centers on the energy dynamics of a parallel plate capacitor when a dielectric material is inserted. Initially, the energy stored in the capacitor is given by E1 = Q²/2C, and upon inserting a dielectric with dielectric constant K, the energy becomes E2 = Q²/2KC, where E2 is less than E1. The difference in energy is attributed to the work done in attracting the dielectric into the capacitor, which is a result of the electric field between the plates. The force required to insert the dielectric is significant, as it performs work on the dielectric material, leading to a reduction in the stored energy of the capacitor.

PREREQUISITES
  • Understanding of capacitor fundamentals, including capacitance and potential difference.
  • Knowledge of dielectric materials and their properties, specifically dielectric constant (K).
  • Familiarity with energy equations related to capacitors, such as E = Q²/2C.
  • Basic concepts of electric fields and forces acting on dielectric materials in capacitors.
NEXT STEPS
  • Explore the relationship between electric field strength and dielectric insertion in capacitors.
  • Investigate the mathematical derivation of force exerted on a dielectric in a capacitor.
  • Learn about energy conservation principles in electrical systems involving capacitors and dielectrics.
  • Examine the effects of connecting a capacitor to a voltage source during dielectric insertion.
USEFUL FOR

Students and professionals in electrical engineering, physicists studying electromagnetism, and anyone interested in the practical applications of capacitors and dielectrics in electronic circuits.

kbar1
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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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How much force does it take to insert the dielectric?
 
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.
 
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.
 
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?
 
That is my guess also, but I don't know.
 
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.
 
Guess that clears it. Thanks all.
 
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.
 

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