Changes in capacitor after dielectric inserted

• Krushnaraj Pandya
In summary: 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...something else.
Krushnaraj Pandya
Gold Member

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.

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
Delta² said:
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

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

Delta² said:
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?

Delta² said:
That's strange, does your textbook agrees that the electric field remains unchanged?
yes it does.

Krushnaraj Pandya said:
yes it does.
but it explicitly mentions F being K^2F later

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

Delta² said:
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

Krushnaraj Pandya said:
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.

Delta² said:
It must have an error regarding the force. Cant understand it otherwise.
Alright. Thank you very much.

Delta² said:
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##

Delta² said:
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

Krushnaraj Pandya said:
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

1. How does inserting a dielectric affect the capacitance of a capacitor?

Inserting a dielectric material between the plates of a capacitor increases the capacitance, or the ability of the capacitor to store electric charge. This is because the dielectric material has a higher permittivity than air, allowing more electric flux to pass through the capacitor.

2. What is the relationship between the dielectric constant and the capacitance of a capacitor?

The capacitance of a capacitor is directly proportional to the dielectric constant of the material between its plates. This means that as the dielectric constant increases, the capacitance also increases.

3. How does the presence of a dielectric affect the electric field and voltage in a capacitor?

The presence of a dielectric material between the plates of a capacitor reduces the electric field and voltage across the plates. This is because the dielectric material reduces the amount of electric flux that can pass through the capacitor, thereby decreasing the electric field and voltage.

4. Can inserting a dielectric change the energy stored in a capacitor?

Yes, inserting a dielectric material can change the energy stored in a capacitor. The energy stored in a capacitor is directly proportional to the capacitance, so as the capacitance increases due to the presence of a dielectric, the energy stored also increases.

5. How does the type of dielectric material used affect the capacitance of a capacitor?

The type of dielectric material used can significantly impact the capacitance of a capacitor. Different materials have different dielectric constants, which directly affect the capacitance. Some materials, such as ceramic or paper, have higher dielectric constants than others, resulting in a larger increase in capacitance when inserted between the plates of a capacitor.

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