Changes in capacitor after dielectric inserted

[Krushnaraj Pandya]

Homework Statement

A parallel plate capacitor (capacitance C) is charged with a battery of emf V volts. A dielectric slab of dielectric constant K is placed between the plates to fully occupy the space. The battery remains connected.
What are the changes in-1)C 2)Q (charge on capacitor) 3)E(E. field) 4)V(potential difference) 5)U (energy of capacitor) 6)F(Force of attraction between the plates). Symbols have usual meanings

Homework Equations

All applicable to capacitors

The Attempt at a Solution

I know that when a dielectric is inserted C2=CK (by adding 2, I'm indicating the new value). Since battery is still connected, V remains V(?). Here comes my first confusion, since E2=E/K, shouldn't V change as well. But if we assume for now that V remains V, then Q must have become KQ to make capacitance KC, either that or V became V/K..which one is happening here?
also, I don't know how to approach the 6th question ,as to how the force will change because of dielectric. I'd appreciate some help.

 

[Delta2]

When we have a battery in a circuit the voltage difference between the two ends of the battery is always (no matter what happens to the rest of the circuit) $V'=V-Ir$ where $V$ the emf of the battery and $r$ the internal resistance of the battery , and $I$ the current that flows through the battery. In our case we consider steady states or equilibrium states where $I=0$, so the voltage difference between the two ends of the battery is simply $V'=V$.

 

[Krushnaraj Pandya]

When we have a battery in a circuit the voltage difference between the two ends of the battery is always (no matter what happens to the rest of the circuit) $V'=V-Ir$ where $V$ the emf of the battery and $r$ the internal resistance of the battery , and $I$ the current that flows through the battery. In our case we consider steady states or equilibrium states where $I=0$, so the voltage difference between the two ends of the battery is simply $V'=V$.

Alright, so voltage definitely remains constant...
I calculated that if charge changes to KQ, then Electric field also remains unchanged.
And potential energy becomes KU.
The Force of attraction is Q*E, E=Q/2Aepsilon. Since charge changes to KQ, Force changes to K^2(F).
Thank you very much for your help

 

[Delta2]

I think you did a mistake regarding the Force, the force changes to $F'=Q'E'=Q'E=kQE=kF$

 

[Krushnaraj Pandya]

I think you did a mistake regarding the Force, the force changes to $F'=Q'E'=Q'E=kQE=kF$

I see my mistake, I forgot to factor in K in the denominator for E, but my textbook says the answer is K^2(F) which seems puzzling now

 

[Delta2]

That's strange, does your textbook agrees that the electric field remains unchanged?

 

[Krushnaraj Pandya]

That's strange, does your textbook agrees that the electric field remains unchanged?

yes it does.

 

[Krushnaraj Pandya]

yes it does.

but it explicitly mentions F being K^2F later

 

[Delta2]

That's strange.. Does it also agree that the new charge is Q'=kQ?

 

[Krushnaraj Pandya]

That's strange.. Does it also agree that the new charge is Q'=kQ?

It agrees to the following, new charge is KQ, new Capacitance is KC, V remains same, E remains same, U becomes kU. And finally, that F becomes (K^2)F

 

[Delta2]

It agrees to the following, new charge is KQ, new Capacitance is KC, V remains same, E remains same, U becomes kU. And finally, that F becomes (K^2)F

It must have an error regarding the force. Cant understand it otherwise.

 

[Krushnaraj Pandya]

It must have an error regarding the force. Cant understand it otherwise.

Alright. Thank you very much.

 

[Krushnaraj Pandya]

I found this link though.
https://physics.stackexchange.com/q...ates-of-a-capacitor-with-dielectric-a-battery

 

[Krushnaraj Pandya]

It must have an error regarding the force. Cant understand it otherwise.

its definitely not an error, but I don't see a concrete evaluation in the above link, just a correction in the comments that it is proportional to k^2
perhaps @CWatters knows

 

[Delta2]

I think where my post #4 goes wrong is that force is not $Q'E'$ when we insert the dielectric. Intuitively I can understand why it will be$k^2F$, because we have $kQ$ charge in the positive plate and $-kQ$charge in the negative so the coulomb force (if those charges where point charges) would be $-k^2Q^2/r^2$

 

[Krushnaraj Pandya]

force is not Q′E′

I can follow your intuition, but its still not a rigorous way to see it.
I found the following statement-
The reason is that the field acting on the capacitor plate is entirely due to the other capacitor plate; the field due to dielectric is zero outside the dielectric

This statement implies that the K we factored in the denominator is just an expression for E inside the dielectric, but the F due to the plate on the other plate is still due to E, not E/K

 

[Delta2]

I can follow your intuition, but its still not a rigorous way to see it.
I found the following statement-
The reason is that the field acting on the capacitor plate is entirely due to the other capacitor plate; the field due to dielectric is zero outside the dielectric

This statement implies that the K we factored in the denominator is just an expression for E inside the dielectric, but the F due to the plate on the other plate is still due to E, not E/K

Yes I guess that's the catch, that the field on the surface of the plate is not equal to the field inside the dielectric.

 

[Krushnaraj Pandya]

Thanks for the help :D