Capacitor and dielectrics help

[vortexbaby]

This is a conceptial problem. A slab of dielectric, with a dielectric constant K between an isolated parallel-plate capacitor, is removed. The two plates of the capacitor have charges Q and -Q, which do not change when the slab is removed. Is the energy stored in the capacitor more, less, or the same after the dielectric is removed? If it is more, what is the source of the extra energy? If less, where does the lost energy go?

Now I know that the Potential Difference (V) increases when the slab is removed, plus the capacitance (C) decreases. This means that the energy stored in the capacitor should increase, but where does this extra energy come from? I've worked on this problem for so long that I've confused myself.

Vortexbaby

 

[lalbatros]

Let's assume the capacitor is disconnected from any external circuit.

Removing the dielectric would indeed increase the energy stored in the capacitor.
This means that you need to bring energy in order to remove the dielectric: the dielectric is attracted by the capacitor. This is clear from the forces involved: the (opposite) surface charges on the dielectric are attracted by the plates. You need to work against these forces to remove the dielectric. The funny thing is that this work is stored in the capacitor!

On the other side, the attraction force between the two plates will increase since the "shielding" by the dielectric has be removed.

The details of this process might be a little bit more complicated, like the exact shape of the electric field lines when the dielectric is removed. This shows the "magic" of the energy theorem that allows us to quantitfy this attraction force between the plates and the slab is a very easy way.

Have a look at this site for applications:

www.nasatech.com/Briefs/Apr05/NPO30747.html​

Would that also play a role sometimes in adhesion? Any idea?

Michel

 

[Andrew Mason]

This is a conceptial problem. A slab of dielectric, with a dielectric constant K between an isolated parallel-plate capacitor, is removed. The two plates of the capacitor have charges Q and -Q, which do not change when the slab is removed. Is the energy stored in the capacitor more, less, or the same after the dielectric is removed? If it is more, what is the source of the extra energy? If less, where does the lost energy go?

Now I know that the Potential Difference (V) increases when the slab is removed, plus the capacitance (C) decreases. This means that the energy stored in the capacitor should increase, but where does this extra energy come from? I've worked on this problem for so long that I've confused myself.

Since the potential difference (energy per unit charge) increases, the total energy increases. It can only come from the energy used in removing the dielectric. What forces are acting on the dielectric?

AM

 

[Andrew Mason]

Double post

This appears to be a https://www.physicsforums.com/showthread.php?t=126216"

AM

 

[ZapperZ]

Thread has been merged.

Zz.