# Energy stored in capacitor w/ dielectric

1. Feb 10, 2013

### kbar1

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?

2. Feb 10, 2013

### Staff: Mentor

How much force does it take to insert the dielectric?

3. Feb 10, 2013

### kbar1

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. Feb 10, 2013

### Staff: Mentor

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. Feb 10, 2013

### kbar1

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. Feb 10, 2013

### Staff: Mentor

That is my guess also, but I don't know.

7. Feb 11, 2013

### Jano L.

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. Feb 12, 2013

### kbar1

Guess that clears it. Thanks all.

9. Feb 12, 2013

### sophiecentaur

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 realises that EM energy will be radiated due to the AC in the system, so the oscillations will always die down, in the end.

10. Jan 2, 2014

### kartik khanna

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: Jan 2, 2014
11. Jan 2, 2014

### Staff: Mentor

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.