I Calculating Total Electric Charge with a Capacitor and Battery

[annamal]

With a capacitor with a dielectric with the battery on,
$E_{total} = E_0 + E_i$
$\frac{Q_t}{dC_t} = \frac{Q_0}{dC_0} + \frac{Q_i}{dC_i}$
thus,
$\frac{Q_t}{C_t} = \frac{Q_0}{C_0} + \frac{Q_i}{C_i}$
since in a battery $V_t = V_0, V_i = 0$, so either $Q_i = 0$ or $C_i = infinite$
but $Q_t = Q_i + Q_0$, which confuses me.

subscript t is total, i is the induced charge, voltage or field in the dielectric, and 0 is the vacuum.

 

[vanhees71]

I don't understand your notation. What is $C_i$? The capacitance of a usual plate capacitor is (approximately)
$C=\frac{\epsilon A}{d}$, where $\epsilon=\epsilon_0 \epsilon_r$ is the permittivity of the dielectric. The charge on the capacitor plate is
$$Q=C U = \epsilon_{r} C_{\text{vac}} U = (1+\chi) C_{\text{vac}} U,$$
where $\chi=\epsilon_r -1$ is the electrical susceptibility of the di-electric. In your notation $Q_0=C_{\text{vac}} U$ and $Q_i=\chi Q_0$.

 

[annamal]

I don't understand your notation. What is $C_i$? The capacitance of a usual plate capacitor is (approximately)
$C=\frac{\epsilon A}{d}$, where $\epsilon=\epsilon_0 \epsilon_r$ is the permittivity of the dielectric. The charge on the capacitor plate is
$$Q=C U = \epsilon_{r} C_{\text{vac}} U = (1+\chi) C_{\text{vac}} U,$$
where $\chi=\epsilon_r -1$ is the electrical susceptibility of the di-electric. In your notation $Q_0=C_{\text{vac}} U$ and $Q_i=\chi Q_0$.

$C_i$ is the capacitance of the induced dielectric

 

[annamal]

Do you see where I went wrong?

 

[alan123hk]

The equation describing a capacitor with a dielectric material is shown in post 2. I think the $C_i$ mentioned in your equation is unnecessary and rather difficult to understand. If you insist on including $C_i$ in the equation, it seems that $V_i$ should not be assumed to be 0 either.

 

[annamal]

The equation describing a capacitor with a dielectric material is shown in post 2. I think the $C_i$ mentioned in your equation is unnecessary and rather difficult to understand. If you insist on including $C_i$ in the equation, it seems that $V_i$ should not be assumed to be 0 either.

If $V_i$ is not 0 then $V_t$ does not equal $V_0$

 

[alan123hk]

If Vi is not 0 then Vt does not equal

I think the contradiction seems to be caused by you. I don't think it's actually necessary to include independent $E_i~$, $V_i~$ and $C_i~$ in the equation due to the induced charge of the dielectric material. Note that such a concept does not exist in the equations of the second post.

Or do you mean that there exists an infinitely large capacitance $C_i~$ and an infinitely small voltage $V_i~$, the product of which is exactly equal to $Q_i~$, which is the theoretical model you are trying to build?

 

[annamal]

I think the contradiction seems to be caused by you. I don't think it's actually necessary to include independent $E_i~$, $V_i~$ and $C_i~$ in the equation due to the induced charge of the dielectric material. Note that such a concept does not exist in the equations of the second post.

Or do you mean that there exists an infinitely large capacitance $C_i~$ and an infinitely small voltage $V_i~$, the product of which is exactly equal to $Q_i~$, which is the theoretical model you are trying to build?

I am saying I don't understand what $E_i$ is when the battery to the circuit is always on. According to my equations with the battery on $E_i$ has to equal 0, which confuses me as to how the induced electric field is 0 with a constant electric field.

 

[rude man]

With a capacitor with a dielectric with the battery on,
$E_{total} = E_0 + E_i$
$\frac{Q_t}{dC_t} = \frac{Q_0}{dC_0} + \frac{Q_i}{dC_i}$
thus,
$\frac{Q_t}{C_t} = \frac{Q_0}{C_0} + \frac{Q_i}{C_i}$
since in a battery $V_t = V_0, V_i = 0$, so either $Q_i = 0$ or $C_i = infinite$
but $Q_t = Q_i + Q_0$, which confuses me.

subscript t is total, i is the induced charge, voltage or field in the dielectric, and 0 is the vacuum.

For me to help you need to define your terms.

 

[rude man]

But:
there is only one E field and that is E = V/d. V is the battery emf. It makes no difference if there is a dielectric present or not, assuming the dielectric fully fills the interplate space.
EDIT:
If you remove the dielectric, $Q_{free}$ changes but E does not. $Q_{free}$ is the plate charge.

 

[vanhees71]

What happens physically is quite simple: You have the two plates of the capacitor charged with a charge $Q$ and $-Q$. The corresponding electric field polarizes the dielectric medium, i.e., it shifts the bound charges within this medium a bit from their equilibrium position without an external field. This field opposes the external field. That's why the relative permittivity $\epsilon_r>1$, which means you need a larger charge $Q$ to get the same electric-potential difference between the plates than for the same capacitor without a dielectric (vacuum) between the plates.

For a homogeneous isotropic dieelectric the electrostatic equations simply read
$$\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{E}=\frac{1}{\epsilon} \rho,$$
where $\epsilon=\epsilon_r \epsilon_0>\epsilon_0$ is the permittivity of the dielectric. The solutions are of course the same as in vacuum only with this changed permittivity. The upshot is that the capacity with dielectric is larger by a factor of $\epsilon_r$ compared to the capacity without the dielectric.

 

[alan123hk]

I am saying I don't understand what Ei is when the battery to the circuit is always on. According to my equations with the battery on Ei has to equal 0, which confuses me as to how the induced electric field is 0 with a constant electric field.

IMHO, it's hard to understand what you're thinking from the equation, because I don't think you explain in enough detail what each term in the equation means and where exactly it is in the capacitor. So actually from your first equation I'm not sure what you mean.

So I think another way might be better. I recommend you to refer to the link below, which has a clear and detailed explanation of how dielectric capacitors work. Notably, the electric field is higher in the gap between the metal plate and the dielectric surface. But due to the extremely narrow width of this gap, its effect can be ignored when deriving the equation.

https://www.feynmanlectures.caltech.edu/II_10.html

 

[annamal]

With a capacitor with a dielectric with the battery on,
$E_{total} = E_0 + E_i$
$\frac{Q_t}{dC_t} = \frac{Q_0}{dC_0} + \frac{Q_i}{dC_i}$
thus,
$\frac{Q_t}{C_t} = \frac{Q_0}{C_0} + \frac{Q_i}{C_i}$
since in a battery $V_t = V_0, V_i = 0$, so either $Q_i = 0$ or $C_i = infinite$
but $Q_t = Q_i + Q_0$, which confuses me.

subscript t is total, i is the induced charge, voltage or field in the dielectric, and 0 is the vacuum.

$E_0$ is the electric field of the capacitor plates with no dielectric in between. $E_i$ is the induced electric field due to the induced charge on the dielectric.
$E_{total} = E_0 + E_i$
It follows that
$V_{total} = V_0 + V_i$ where
$V_0$ is the voltage of the capacitor plates with no dielectric in between. $V_i$ is the induced voltage due to the induced charge on the dielectric.
With a battery constantly on, $V_{total} = V_0$ therefore $V_i = 0$. Since $V_i = \frac{Q}{C}$, either Q = 0 or C = infinite? Is that correct?

 

[rude man]

$E_0$ is the electric field of the capacitor plates with no dielectric in between. $E_i$ is the induced electric field due to the induced charge on the dielectric.
$E_{total} = E_0 + E_i$
It follows that
$V_{total} = V_0 + V_i$ where
$V_0$ is the voltage of the capacitor plates with no dielectric in between. $V_i$ is the induced voltage due to the induced charge on the dielectric.
With a battery constantly on, $V_{total} = V_0$ therefore $V_i = 0$. Since $V_i = \frac{Q}{C}$, either Q = 0 or C = infinite? Is that correct?

Why do you add E_0 + E_i ? There is only one electric field as I said. It's V/d.

I see no meaning to your $V_i$. There is only one voltage and that is the voltage between the plates, again irrespective of presence or absence of dielectric.

 

[annamal]

Why do you add E_0 + E_i ? There is only one electric field as I said. It's V/d.

I see no meaning to your $V_i$. There is only one voltage and that is the voltage between the plates, again irrespective of presence or absence of dielectric.

No you add the electric fields.
See this: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/dielec.html
And
here:
https://opentextbc.ca/universityphysicsv2openstax/chapter/molecular-model-of-a-dielectric/

 

[rude man]

No you add the electric fields.
See this: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/dielec.html
And
here:
https://opentextbc.ca/universityphysicsv2openstax/chapter/molecular-model-of-a-dielectric/

That's if there is no battery connected. You specified a battery connected.
Horse of different color.
As long as a battery is connected the E field is the same whether your airgap is filled with air or with a dielectric.

 

[alan123hk]

@annamal Thanks for the further explanation and reference link. This will make communication easier and reduce misunderstandings.

Now what I'm confused about is that since you used the method shown in the link to split the electric field in the dielectric capacitor into two parts, $E_o$ and $E_i~$, then you can't say that $V_i$ is equal to 0 because it means that $E_i$ is equal to 0, nor can you say that $E_{total} = E_o +E_i~$, since $E_o$ represents the electric field created only by the charges on the two metal plates of the capacitor in the absence of a dielectric. But after adding the dielectric, the free charge on the two metal plates also increases, which means that the electric field generated by the free charge on the two metal plates also increases.

Also note that the total or effective electric field is always the same as mentioned in post #16

 

[artis]

@annamal the easiest way to remember that a dielectric due to it's polarization in the presence of an E field opposes the applied E field (the polarization tries to neutralize the applied field) is to know that whenever there is a dielectric instead of vacuum, the capacitance rating goes up with nothing else changed.

more capacitance per same surface area means more charge per same surface area , if more charge is packed into the same area that means the E field increases in strength.
The other way to achieve this would be to increase the capacitor charging voltage or as in your example with battery it would be to add another battery in series.

Is this a school subject or are you just trying to get a better understanding of how capacitors work?

 

[nasu]

If the capacitor is connected to the battery the increased charge density results in keeping the field at the same value as without dielectric and not in an increase of the field.

 

[annamal]

@annamal Thanks for the further explanation and reference link. This will make communication easier and reduce misunderstandings.

Now what I'm confused about is that since you used the method shown in the link to split the electric field in the dielectric capacitor into two parts, $E_o$ and $E_i~$, then you can't say that $V_i$ is equal to 0 because it means that $E_i$ is equal to 0, nor can you say that $E_{total} = E_o +E_i~$, since $E_o$ represents the electric field created only by the charges on the two metal plates of the capacitor in the absence of a dielectric. But after adding the dielectric, the free charge on the two metal plates also increases, which means that the electric field generated by the free charge on the two metal plates also increases.

Also note that the total or effective electric field is always the same as mentioned in post #16

But you can say $E_{total} = E_o +E_i$ when the battery is not constant...

 

[rude man]

@annamal

more capacitance per same surface area means more charge per same surface area , if more charge is packed into the same area that means the E field increases in strength./quote

If you move a dielectric into a vacuum with the battery connected that statement is incorrect.

Yes the charge density on the plates increases but the E field stays the same. Again, assuming the gap is fully filled by the dielectric.

But maybe that is not what you meant to say?

 

[hutchphd]

more capacitance per same surface area means more charge per same surface area , if more charge is packed into the same area that means the E field increases in strength.

I too am nonplussed by this statement. In the usual model the dielectric fills the space between conductors and the charges (real and polarization) are limited to the surface. So this does not make sense. More free charge is required to maintain the (line integral of) the E field $\therefore$the capacity is increased.

 

[alan123hk]

@annamal First I want to say that there are now two slightly different ways of interpreting dielectric capacitance, as shown below, and they are of course essentially equivalent.

I prefer this method
https://www.feynmanlectures.caltech.edu/II_10.html

The electric field is divided into two parts - $E_0$ due to the free charge $Q_0$ on the capacitor plates and $E_i$ due to the induced charge $Q_i$ on the surfaces of the dielectric
https://opentextbc.ca/universityphysicsv2openstax/chapter/molecular-model-of-a-dielectric/

But note that the $E_0~$ mentioned in the link you provided is created by the free charge $Q_0$ on the capacitor plates, not the electric field when there is no dielectric between the capacitor plates as you said. I think they are completely different situations.

E0 is the electric field of the capacitor plates with no dielectric in between

But you can say Etotal=Eo+Ei when the battery is not constant...

Sorry, I don't understand very well, can you explain a little more?

This method just splits the electric field into two parts, but of course the net or total electric field is always equal to the supply voltage divided by the distance between the two capacitor plates. Or if there is no external voltage, it's just the voltage generated by the stored charge divided by the distance between the two plates. what does this have to do with the form of the supply voltage?

 

[annamal]

@annamal First I want to say that there are now two slightly different ways of interpreting dielectric capacitance, as shown below, and they are of course essentially equivalent.

I prefer this method
https://www.feynmanlectures.caltech.edu/II_10.html

The electric field is divided into two parts - $E_0$ due to the free charge $Q_0$ on the capacitor plates and $E_i$ due to the induced charge $Q_i$ on the surfaces of the dielectric
https://opentextbc.ca/universityphysicsv2openstax/chapter/molecular-model-of-a-dielectric/

But note that the $E_0~$ mentioned in the link you provided is created by the free charge $Q_0$ on the capacitor plates, not the electric field when there is no dielectric between the capacitor plates as you said. I think they are completely different situations.

Sorry, I don't understand very well, can you explain a little more?

This method just splits the electric field into two parts, but of course the net or total electric field is always equal to the supply voltage divided by the distance between the two capacitor plates. Or if there is no external voltage, it's just the voltage generated by the stored charge divided by the distance between the two plates. what does this have to do with the form of the supply voltage?

In post #16:

That's if there is no battery connected. You specified a battery connected.
Horse of different color.
As long as a battery is connected the E field is the same whether your airgap is filled with air or with a dielectric.

I am saying does $E_{total} = E_o +E_i$ with a constant voltage source from the battery. If so, $E_i$ will = 0 when the battery is constant since $E_{total} = E_o$

 

[artis]

If you move a dielectric into a vacuum with the battery connected that statement is incorrect.

Yes the charge density on the plates increases but the E field stays the same. Again, assuming the gap is fully filled by the dielectric.

But maybe that is not what you meant to say?

I too am nonplussed by this statement. In the usual model the dielectric fills the space between conductors and the charges (real and polarization) are limited to the surface. So this does not make sense. More free charge is required to maintain the (line integral of) the E field $\therefore$the capacity is increased.

You are both correct, I apologize I should have been clearer with my wording, at one point I had a dielectric capacitor in mind at another a vacuum one. I should have added that for a dielectric E field stays the same while charge density increases and for a vacuum increased charge density (say added another battery in series) does result in stronger E field.

 

[nasu]

In post #16:

I am saying does $E_{total} = E_o +E_i$ with a constant voltage source from the battery. If so, $E_i$ will = 0 when the battery is constant since $E_{total} = E_o$

It seems that you are adding the field between the plates without dielectric with the field in the presence of the dielectric. With the condition that the capacitor is connected to the same voltage in both cases.
Even though this makes sense mathematically, calling this "total field" does not make sense physically. These two fields do not exist at the same time so adding them has no physical meaning, you don't get a net or total field. Actually, as already discussed, the field is the same in both cases.
What you do is like adding your weight when you were 5 years old with your current weight and call this "total weight of Anna". Just that in this case the two values you add are not the same, I suppose.

You may have seen another kind of addition used in this situation: you add the field produced by the free charge on the plates with the field produced by the polarization of the dielectric. These two field do exist at the same time and their sum is the actual net field between the plates. The field of the free charge is larger in the presence of dielectric (more charge is transferred from the battery) but the field of the dielectric is in opposite direction so it cancells the effect of the extra charge and you get the same field as without dielectric (always connected to the same battery).

 

[alan123hk]

You may have seen another kind of addition used in this situation: you add the field produced by the free charge on the plates with the field produced by the polarization of the dielectric. These two field do exist at the same time and their sum is the actual net field between the plates. The field of the free charge is larger in the presence of dielectric (more charge is transferred from the battery) but the field of the dielectric is in opposite direction so it cancells the effect of the extra charge and you get the same field as without dielectric (always connected to the same battery).

I think the above statement is well said, it describes the situation completely and accurately.

 

[vanhees71]

The upshot of the whole discussion seems to be: It may be illuminating to solve for the electrostatic field between the plates for two situations:

(a) Consider the capacitor connected to the battery first being empty and then you bring in the dielectric. What stays constant? 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?

(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?

 

[nasu]

The discution of the two situations it is a quite standard (homework) problem in the introductory physics.

 

[vanhees71]

This entire thread is about a standard homework problem in introductory physics ;-).

 

[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.