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 ;-).