Capacitor Charge Calculation with Dielectric Insertion

[Callix]

Homework Statement

A capacitor connected to a battery initially holds a charge of +q on its positive plate and -q on its negative plate. The electric field between the plates is initially E. A dielectric material is then inserted that polarizes in such a way as to produce an electric field of -0.30E (where the - sign indicates that this field opposites that of E). Determine the new charge stored on the positive plate.

Homework Equations

Eo-E=Ef
E=kq^2/r^2

The Attempt at a Solution

I did my work on Word using the equation tool just to make the notation format easier to read.

Is my logic correct here?

Any help would be greatly appreciated! :)

 

[ehild]

Homework Statement

A capacitor connected to a battery initially holds a charge of +q on its positive plate and -q on its negative plate. The electric field between the plates is initially E. A dielectric material is then inserted that polarizes in such a way as to produce an electric field of -0.30E (where the - sign indicates that this field opposites that of E). Determine the new charge stored on the positive plate.

Homework Equations

Eo-E=Ef
E=kq^2/r^2

The Attempt at a Solution

I did my work on Word using the equation tool just to make the notation format easier to read.

Is my logic correct here?

Any help would be greatly appreciated! :)

The equation in red is the Coulomb force between two equal point charges. It is not electric field, and it is irrelevant to this problem.

 

[Callix]

The equation in red is the Coulomb force between two equal point charges. It is not electric field, and it is irrelevant to this problem.

Oh duh, it's just q from the source, not q^2

Could we solve it using the equation for capacitance?

 

[ehild]

What i

Oh duh, it's just q from the source, not q^2

Could we solve it using the equation for capacitance?

What do you mean?

 

[Callix]

What i

What do you mean?

Well since it's capacitor maybe we can use C=Q/V=Q/Ed?

 

[ehild]

Well since it's capacitor maybe we can use C=Q/V=Q/Ed?

That is correct. But the problem is not clear. Is the capacitor connected to the battery during the whole experiment?

 

[Callix]

That is correct. But the problem is not clear. Is the capacitor connected to the battery during the whole experiment?

Yes.

 

[ehild]

Yes.

Does the electric field change between the plates then when the dielectric material is inserted?

 

[Callix]

Does the electric field change between the plates then when the dielectric material is inserted?

The field is the same between the empty space, but is different inside the dielectric, correct?

 

[ehild]

No, the field is voltage over the distance between the plates, neither of them changes.

 

[Callix]

No, the field is voltage over the distance between the plates, neither of them changes.

Okay, so where can I go from there?

 

[ehild]

You need to determine the new charge on the plates. Have you learned about the electric displacement? https://www.physicsforums.com/threads/what-is-the-electric-displacement-field.763099/
How is it related to the surface charge density?
What does it mean that the dielectrics is polarized?

 

[Callix]

You need to determine the new charge on the plates. Have you learned about the electric displacement? https://www.physicsforums.com/threads/what-is-the-electric-displacement-field.763099/
How is it related to the surface charge density?
What does it mean that the dielectrics is polarized?

Well, E=σ/ε where σ=Q/A. So would the new charge be Q=EεA?

 

[Callix]

Can anyone else help me with this? =/

 

[ehild]

Yes, the new charge is Q'=EεA. But you do not know ε.
The field polarizes the dielectric and dipole chains appear. The ends of the chains neutralize some of the surface charges. How much do they neutralize in this case? It was given, that the dipoles would make a field of 0.3 E, opposite to the original.

 

[Callix]

Yes, the new charge is Q'=EεA. But you do not know ε.
The field polarizes the dielectric and dipole chains appear. The ends of the chains neutralize some of the surface charges. How much do they neutralize in this case? It was given, that the dipoles would make a field of 0.3 E, opposite to the original.

Does this image depict what you're describing correctly?

 

[Callix]

deleted

 

[ehild]

Yes.

 

[Callix]

Yes.

Okay, so then would the new value for ε be the dielectric constant?
So κ=Eo/E = E/0.7E=1.4286?

 

[ehild]

The new electric field is the same as the initial one was, U/d. See Post #10.

 

[Callix]

The new electric field is the same as the initial one was, U/d. See Post #10.

Okay, got it. So voltage is the same because of the presence of the battery and obviously separation does not change, thus field is constant. So how can I find the new charge if we don't know ε?

 

[NascentOxygen]

Your sketch in #16 suggests you are assuming that the added material occupies less than the full gap between the plates? However, this fraction doesn't seem to be specified.

 

[ehild]

You know that E=E₁-E₂, as your picture shows. E₂=0.3E. The new surface charge corresponds to the field E₁, σ'=ε₀E₁. So what is σ' in terms of σ?

 

[Callix]

Your sketch in #16 suggests you are assuming that the added material occupies less than the full gap between the plates? However, this fraction doesn't seem to be specified.

That's true. I made the assumption because all of the examples that our professor has mentioned has been when there is some separation between the two. But the question never suggests that there is any.

You know that E=E₁-E₂, as your picture shows. E₂=0.3E. The new surface charge corresponds to the field E₁, σ'=ε₀E₁. So what is σ' in terms of σ?

σ=q/A

 

[ehild]

I asked the relation between the new and the initial surface charges.

 

[Callix]

I asked the relation between the new and the initial surface charges.

Oh okay, well if
σ=ε₀E₁
σ'=0.7σ=0.7ε₀E₁

Or would it be by a factor of 0.3?

 

[ehild]

Oh okay, well if
σ=ε₀E₁
σ'=0.7σ=0.7ε₀E₁

Or would it be by a factor of 0.3?

What do you think, is the new charge more or less than the initial one? Why we use dielectrics in the capacitors? To increase their ability to store more charge.
Look at your picture. The actual E is E = E1-E2, and E2=0.3 E.

 

[Callix]

What do you think, is the new charge more or less than the initial one? Why we use dielectrics in the capacitors? To increase their ability to store more charge.
Look at your picture. The actual E is E = E1-E2, and E2=0.3 E.

Right, so then it would be σ'=0.7σ=0.7ε₀E₁.

 

[ehild]

Is the new charge less than the initial one?

 

[Callix]

Is the new charge less than the initial one?

Well logically I would suppose not then, since C=q/V and we already know V is constant because of the battery, so q must change. Adding dielectrics help increase C, so that means q increases as well.

So before, were you saying that E=E1-(-E2)? Or E=E1+(-E2), because it seems like you're hinting that it's the first, which in that case it would be by a factor of 1.7.

 

[ehild]

Well logically I would suppose not then, since C=q/V and we already know V is constant because of the battery, so q must change. Adding dielectrics help increase C, so that means q increases as well.

So before, were you saying that E=E1-(-E2)? Or E=E1+(-E2), because it seems like you're hinting that it's the first, which in that case it would be by a factor of 1.7.

It is the first, E1=E+E2, but E2 is the field of the dipoles, which is 0.3E, if the E2 means the magnitude of the dipole field.

 

[Callix]

It is the first, E1=E+E2, but E2 is the field of the dipoles, which is 0.3E, if the E2 means the magnitude of the dipole field.

Right, so then it would increase by a factor of 1.7, so
σ'=1.7σ=1.7ε₀E₁.
And that is the surface charge density on the dielectric?

 

[ehild]

Right, so then it would increase by a factor of 1.7, so
σ'=1.7σ=1.7ε₀E₁.
And that is the surface charge density on the dielectric?

Why 1.7? It is wrong.