I Calculating Total Electric Charge with a Capacitor and Battery

[hutchphd]

But in my experience it (and its cousins) causes inordinate confusion. I do not really understand why

 

[annamal]

and you get the same field as without dielectric (always connected to the same battery).

So the induced field on the dielectric is 0 in the case it is always connected to the battery?

 

[nasu]

So the induced field on the dielectric is 0 in the case it is always connected to the battery?

No, not at all. I did not say this and does not follow from what I said. How can you draw this conclusion from my post?

 

[hutchphd]

So the induced field on the dielectric is 0 in the case it is always connected to the battery?

WTF is the "induced field"...it is not a thing. The presence of the battery means the voltage (potential difference) is held constant. Because the geometry doesn't change that means the E field cannot change. When there is dielectric the charge in fact then must change. Finis.

 

[annamal]

WTF is the "induced field"...it is not a thing. The presence of the battery means the voltage (potential difference) is held constant. Because the geometry doesn't change that means the E field cannot change. When there is dielectric the charge in fact then must change. Finis.

See this: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/dielec.html
And
here:
https://opentextbc.ca/universityphysicsv2openstax/chapter/molecular-model-of-a-dielectric/

they state $E_{total} = E_0 + E_i$ where $E_i$ is the electric field induced in the dielectric

 

[annamal]

No, not at all. I did not say this and does not follow from what I said. How can you draw this conclusion from my post?

Ok, you're not answering my question do you agree that $E_{total} = E_0 + E_i$ where $E_i$ is the electric field induced in the dielectric
If so, with a battery constantly on, do you agree that $E_{total} = E_0$? If so, that means $E_i$ =0. Are we on the same page here?

 

[nasu]

Ok, you're not answering my question do you agree that $E_{total} = E_0 + E_i$ where $E_i$ is the electric field induced in the dielectric
If so, with a battery constantly on, do you agree that $E_{total} = E_0$? If so, that means $E_i$ =0. Are we on the same page here?

I am afraid that you are not reading what I write or you don't pay attention.

I already told you that if by E_o you mean the field in absence of dielectric, then your formula is nonsense. I did not use the word "nonsense" but this was the meaning of the first part of my explanation.

In the second part I told you what makes sense: to add te field of the free charges and the field of the dielectric polarization. None of them is zero.

So, one more time:
In absence of dielectric, you have only the field of free charges: E₁=V/d
With dielectric you have the field of the free charges (which is not E₁ but a larger value, E₁') and the field of the dielectric, E_d.
The total field in the case with dielectric is $\vec{E}_2 = \vec{E}_1'+\vec{E}_d$ where the directions of the two vectors on the right hand side are opposite so you actualy subtract the magnitudes.
None of the magnitudes E₁, E₂, E₁' or E_d is zero.
But because the capacitor is connected to the same battery we have E₁=E₂=V/d where V is the potential difference provided by the battery and d is the distance between the plates of the capacitor.

 

[annamal]

I am afraid that you are not reading what I write or you don't pay attention.

I already told you that if by E_o you mean the field in absence of dielectric, then your formula is nonsense. I did not use the word "nonsense" but this was the meaning of the first part of my explanation.

In the second part I told you what makes sense: to add te field of the free charges and the field of the dielectric polarization. None of them is zero.

So, one more time:
In absence of dielectric, you have only the field of free charges: E₁=V/d
With dielectric you have the field of the free charges (which is not E₁ but a larger value, E₁') and the field of the dielectric, E_d.
The total field in the case with dielectric is $\vec{E}_2 = \vec{E}_1'+\vec{E}_d$ where the directions of the two vectors on the right hand side are opposite so you actualy subtract the magnitudes.
None of the magnitudes E₁, E₂, E₁' or E_d is zero.
But because the capacitor is connected to the same battery we have E₁=E₂=V/d where V is the potential difference provided by the battery and d is the distance between the plates of the capacitor.

Ok in the last part you say E₁=E₂=V/d, since $\vec{E}_2 = \vec{E}_1'+\vec{E}_d$, would $\vec{E}_d = 0$?

 

[alan123hk]

@annamal

I'm getting more and more confused.

I think the crux of the problem is that you should think about interpreting $E_0$ as the electric field created by the free charge on the capacitor plate $Q_0$, not the electric field of a capacitor without a dielectric.

Doing so makes the equation $E_{total} = E_0 + E_i~$ reasonable and universally applicable, including different cases with and without batteries connected, etc.

Simply put, it is $E_{total}$ = $E_0$ (generated by free charge on the capacitor plate) + $E_i$ (generated by induced charge on the dielectric surface).

This may help reduce misunderstandings and promote consensus.

 

[artis]

@annamal Changes in E field have their highest speed in vacuum (speed of light c) and a lower speed in any other medium. Also E field is not blocked in vacuum (as there is nothing in the way) but is always obstructed in other mediums. Molecules in solids and liquids and gasses normally have no net E field but in the presence of an E field they become polarized, this polarization always is such that it cancels/opposes the source/applied E field.
But none of the fields are zero, the source E field from the plates has a positive value and the opposing field from the dielectric has a negative value if compared with the source field , if looked from the perspective of the dielectric then the field of the dielectric has also a positive value but it's polarity is opposite to that of the plates.
I do not see how any of the fields can be zero.

The question then simply becomes what has to change to keep things in balance with the dielectric opposing the plate E field and the answer is that for a capacitor that has a voltage/current source connected more charge needs to be supplied.

(b) Now think about the same situation, but the empty capacitor being first connected to the battery and then you disconnect it and bring in the dielectric afterwards. What stays now constant and what changes? How can the total field be understood as the superposition of the field of the "free charges" and the field due to the "response of the medium", i.e., due to its polarization?

What do you think is the answer @annamal , the dielectric obstructs the field just like in the battery case but unlike the battery case what is missing?

 

[nasu]

Ok in the last part you say E₁=E₂=V/d, since $\vec{E}_2 = \vec{E}_1'+\vec{E}_d$, would $\vec{E}_d = 0$?

Why would Ed equal to zero? Are you actually reading the posts or just the last line? I just said that E1, E1', E2 and Ed are all non-zero. How do you draw the conclusion that Ed is zero? The fact that E1=E2 implies that E1 =E1'-Ed (in terms of magnitudes).As E1 is not the same as E1' there is no way to conclude that Ed is zero.

 

[alan123hk]

@annamal You seem to mention several times that $E_i$ generated by the induced charge $Q_i$ on the surfaces of the dielectric becomes zero, but $E_i$ is not equal to zero in the presence of a dielectric. Maybe the following diagram can help you understand.

In this example, the relative permittivity $(\epsilon_r)~$is clearly equal to $\frac {E_0} {E_t}=\frac{Q_0}{Q_t}=\frac {5}{2}=2.5$

 

[berkeman]

Thread closed for a bit for Moderation...

Update, after a Mentor discussion, the thread will remain closed. Thank you everyone for trying to help the OP.