The real reason for a capacitor having the same amounts of + and - charges on the two plates

[feynman1]

Most textbooks say that a capacitor whether it be a single one or one in series/parallel should have equal amounts of + and – charges on both plates and that they mostly conclude the + charges attract the same amount of – charges on the other plate without giving any reason.

Now I claim that this is supported by Gauss’ law!

When a capacitor is fully charged, there’s no electric field (no current) in the wires connecting both plates of a fully charged capacitor and there can’t be any net charge on the capacitor when enclosing the whole capacitor by a Gaussian surface.

When a capacitor isn’t fully charged, there’re 2 currents in the same direction flowing to both plates though not through the interior of the capacitor. There can’t be any net charge on the capacitor when enclosing the whole capacitor by a Gaussian surface as the whole electric flux is canceled out to 0.

Do you all agree with this argument?

 

[BvU]

Are you mixing up capacitors with rechargeable batteries ?
And electric flux with electric current ?

 

[feynman1]

Are you mixing up capacitors with rechargeable batteries ?
And electric flux with electric current ?

no why

 

[BvU]

What is a capacitor that 'is fully charged' ?
What is a capacitor that 'isn't fully charged' ?

What is the difference ?

as the whole electric flux is canceled out to 0

Is that a cause or a consequence ?

 

[feynman1]

What is a capacitor that 'is fully charged' ?
What is a capacitor that 'isn't fully charged' ?

What is the difference ?

Is that a cause or a consequence ?

When connected to a battery, a capacitor is fully charged when there's no current and not fully charged when there's a current.

 

[BvU]

It's the other way around !

 

[feynman1]

It's the other way around !

Obviously you misunderstood me. When meaning there’s no current, I mean the flow of charges is finished after there was a current. When meaning there’s a current, I mean the charges aren’t balanced in magnitude yet.

 

[BvU]

Obviously you misunderstood me. When meaning there’s no current, I mean the flow of charges is finished after there was a current. When meaning there’s a current, I mean the charges aren’t balanced in magnitude yet.

Obviously. So you don't talk about capacitors but about a closed circuit that is charging a capacitor.
Did I understand correctly at least that ?
So, a connected circuit, like :

And what is it you want to do with Gauss' law ?

 

[feynman1]

Correct. Take a Gaussian surface enclosing the capacitor (both plates). There's no electric field in both wires connecting both plates. So there's no net electric flux out of the Gaussian surface. So there's no net charge in the capacitor.

 

[BvU]

There's no electric field in both wires connecting both plates. So there's no net electric flux out of the Gaussian surface

Sounds to me like "it's not raining here, so it's not raining anywhere" :
You need to prove this for the entire Gaussian surface -- not just the minimal entry and exit points of the wires.

 

[hutchphd]

If I shuffle across the carpet and pick up the circuit and set it back down on the table there will be a net charge on the capacitor (and the battery and the wires!).
There is no net charge because there is no net charge supplied.

 

[vanhees71]

The most simple example is a spherical capacitor with two spherical shells at radii ##a_1## and ##a_2##. Assuming the total charge is ##0## by symmetry you have a Coulomb potential between the plates and constant potential everywhere else:
$$\Phi(r)=\frac{Q}{4 \pi \epsilon_0 r} \quad \text{for} \quad a_1 \leq r \leq a_2.$$
Then you have
$$U=\Phi(a_1)-\Phi(a_2)=\frac{Q (a_2-a_1)}{4 \pi \epsilon_0 a_1 a_2}.$$
The charge density at ##a_1## is [EDIT: corrected missing factors ##\epsilon_0##]
$$\sigma_1=\epsilon_0 \vec{e}_r \cdot \vec{E}=\frac{Q}{4 \pi a_1^2} \; \Rightarrow Q_1=4 \pi a_1^2 \sigma_1=Q$$
and
$$\sigma_2=-\epsilon_0 \vec{e}_r \cdot \vec{E}=-\frac{Q}{4 \pi a_2^2} \; \Rightarrow Q_2=4 \pi a_2^2 \sigma_2=-Q.$$
That's the usual configuration, where you assume that the capacitor as a whole is uncharged.

Of course you can also solve the problem by putting arbitrary charges ##Q_1## and ##Q_2## on the inner and outer sphere which by symmetry spread homogeneously over the corresponding surfaces.
$$\Phi(r)=\begin{cases} Q_1/(4 \pi \epsilon_0 a_1) &\text{for} \quad r<a_1,\\
Q_1/(4 \pi \epsilon_0 r) &\text{for} \quad a_1 \leq r \leq a_2 ,\\
(Q_1+Q_2)/(4 \pi \epsilon_0 r)-\frac{Q_2}{4 \pi \epsilon_0 a_2} & \text{for} \quad r>a_2.
\end{cases}$$

 

[hutchphd]

If the capacitor is charging, ##\dfrac{\partial \rho}{\partial t}## is a positive quantity and the entire right hand side is negative.

..
Why is this average not zero? I believe ##\dfrac{\partial \rho}{\partial t}## is positive on one plate and negative on the other.

 

[kuruman]

..
Why is this average not zero? I believe ##\dfrac{\partial \rho}{\partial t}## is positive on one plate and negative on the other.

You are correct. I tried fixing the post, but it was beyond repair so I deleted it.

 

[feynman1]

The most simple example is a spherical capacitor with two spherical shells at radii ##a_1## and ##a_2##. Assuming the total charge is ##0## by symmetry you have a Coulomb potential between the plates and constant potential everywhere else:
$$\Phi(r)=\frac{Q}{4 \pi \epsilon_0 r} \quad \text{for} \quad a_1 \leq r \leq a_2.$$
Then you have
$$U=\Phi(a_1)-\Phi(a_2)=\frac{Q (a_2-a_1)}{4 \pi \epsilon_0 a_1 a_2}.$$
The charge density at ##a_1## is
$$\sigma_1=\vec{e}_r \cdot \vec{E}=\frac{Q}{4 \pi a_1^2} \; \Rightarrow Q_1=4 \pi a_1^2 \sigma_1=Q$$
and
$$\sigma_2=-\vec{e}_r \cdot \vec{E}=-\frac{Q}{4 \pi a_2^2} \; \Rightarrow Q_2=4 \pi a_2^2 \sigma_2=-Q.$$
That's the usual configuration, where you assume that the capacitor as a whole is uncharged.

Of course you can also solve the problem by putting arbitrary charges ##Q_1## and ##Q_2## on the inner and outer sphere which by symmetry spread homogeneously over the corresponding surfaces.
$$\Phi(r)=\begin{cases} Q_1/(4 \pi \epsilon_0 a_1) &\text{for} \quad r<a_1,\\
Q_1/(4 \pi \epsilon_0 r) &\text{for} \quad a_1 \leq r \leq a_2 ,\\
(Q_1+Q_2)/(4 \pi \epsilon_0 r)-\frac{Q_2}{4 \pi \epsilon_0 a_2} & \text{for} \quad r>a_2.
\end{cases}$$

In this spherical configuration you assume there’s no outgoing electric field onto the wires if there’s wires connected to both plates. Then that’s the same assumption I made for parallel plate capacitors: no outgoing electric field leaving the capacitor, thus no net charge within the Gaussian surface enclosing the capacitor.

 

[Delta2]

One of the crucial and vital assumptions we do in circuit theory is that the current along the same branch of a circuit is the same and independent of the position in the branch. By applying this assumption in the case of a circuit with a batter and a capacitor we can conclude that if ##q(x_1,t)=\int I(x_1,t) dt## is the charge in one plate of the capacitor positioned at ##x_1## and ##q(x_2,t)=\int I (x_2,t) dt## is the charge in the other plate of the capacitor positioned at ##x_2## then because ##I(x_2,t)=I(x_1,t)=I(t)## it will be that ##q(x_1,t)=q(x_2,t)## for any time instant t.
However this assumption of circuit theory is valid only if the wavelength of the current is big relative to the distance between the capacitor's plates (so practically valid for almost all cases up to microwave frequencies).

 

[feynman1]

One of the crucial and vital assumptions we do in circuit theory is that the current along the same branch of a circuit is the same and independent of the position in the branch. By applying this assumption in the case of a circuit with a batter and a capacitor we can conclude that if ##q(x_1,t)=\int I(x_1,t) dt## is the charge in one plate of the capacitor positioned at ##x_1## and ##q(x_2,t)=\int I (x_2,t) dt## is the charge in the other plate of the capacitor positioned at ##x_2## then because ##I(x_2,t)=I(x_1,t)=I(t)## it will be that ##q(x_1,t)=q(x_2,t)## for any time instant t.
However this assumption of circuit theory is valid only if the wavelength of the current is big relative to the distance between the capacitor's plates (so practically valid for almost all cases up to microwave frequencies).

##I(x_2,t)=I(x_1,t)## can explain a single capacitor having a 0 net charge. But for capacitors in series, how does that apply to a capacitor in the middle?

 

[Delta2]

##I(x_2,t)=I(x_1,t)## can explain a single capacitor having a 0 net charge. But for capacitors in series, how does that apply to a capacitor in the middle?

I really don't understand your point here, since all three (or more) capacitors are in series they belong to the same branch so by the assumption of circuit theory ##I(x_1,t)=I(x_2,t)=I(x_3,t)## and hence all the charges are equal (as it is expected for capacitors in series , they have the same charge but not necessarily the same voltage).

 

[feynman1]

I really don't understand your point here, since all three (or more) capacitors are in series they belong to the same branch so by the assumption of circuit theory ##I(x_1,t)=I(x_2,t)=I(x_3,t)## and hence all the charges are equal (as it is expected for capacitors in series , they have the same charge but not necessarily the same voltage).

Assume there are 2 capacitors in series. How to determine the charges on the 2 plates in the middle (isolated island)? Why does the left plate of the 1st capacitor have to have +q while the right plate of the 1st capacitor has -q?

 

[Delta2]

oh I see I should have been careful with the signs in my original post, it is ##q(x_1)=\int I(x_1,t)dt## if we assume that the current has such direction as to meet first the plate position at x_1 and ##q(x_2)=-\int I(x_2,t) dt##. This is because if the current "arrives" at one plate, this means that the current "leaves" from the other plate.

 

[vanhees71]

It's of course different when currents flow. Also see #15, where the argument was given that in the static situation with the capacitor connected to a voltage source you always have the configuration with ##+Q## on one plate and ##-Q## on the other such that there's no field outside the capacitor and thus no current flowing in the wires connecting the capacitor with the battery.

If a current flows, you can often use the quasistationary approximations and get to Kirchhoff's theory for compact circuits.

 

[feynman1]

It's of course different when currents flow. Also see #15, where the argument was given that in the static situation with the capacitor connected to a voltage source you always have the configuration with ##+Q## on one plate and ##-Q## on the other such that there's no field outside the capacitor and thus no current flowing in the wires connecting the capacitor with the battery.

If a current flows, you can often use the quasistationary approximations and get to Kirchhoff's theory for compact circuits.

Even when currents flow, there's still no electric field in ideal wires, so still no field outside the capacitor.

 

[sophiecentaur]

It's the other way around !

I don't understand that. Could you clear it up for me please? What would be the current value if a capacitor is fully charged?

 

[feynman1]

I don't understand that. Could you clear it up for me please? What would be the current value if a capacitor is fully charged?

no current when fully charged (when still connected to a battery).

 

[sophiecentaur]

no current when fully charged (when still connected to a battery).

Exactly. You seemed to be saying that is wrong in your post ("other way round"). But, no matter; we are in agreement about the Physics, which is what counts.

 

[feynman1]

Exactly. You seemed to be saying that is wrong in your post ("other way round"). But, no matter; we are in agreement about the Physics, which is what counts.

'the other way around' wasnt what i said

 

[sophiecentaur]

They were your words in the post. As there was no quoted context I made up my own mind what you were referring to. (And so could someone else.)?
but as we agree about the Physics there’s no harm done.

 

[BvU]

As the bishop said to the actress: "It was me"

Trying to show f1 that it's one and the same medal

 

[feynman1]

So everyone agrees on my original post?

 

[Delta2]

So everyone agrees on my original post?

No I don't agree, it is not due to Gauss's law but due to the assumption we make in circuit theory as I explained in post #16. In real world scenario and for the time dependent case the wires are not ideal, the electric field inside the wires will not be zero, neither it will be equal in the connecting wires at the two sides , the current will not be equal at the two sides, and hence the charges on the capacitor are approximately equal (and opposite) (depends on the wavelength of current in comparison with the distance between the two sides) but not exactly equal.

 

[feynman1]

No I don't agree, it is not due to Gauss's law but due to the assumption we make in circuit theory as I explained in post #16. In real world scenario and for the time dependent case the wires are not ideal, the electric field inside the wires will not be zero, neither it will be equal in the connecting wires at the two sides , the current will not be equal at the two sides, and hence the charges on the capacitor are approximately equal (and opposite) but not exactly equal.

Do you agree for ideal wires?

 

[Delta2]

Do you agree for ideal wires?

Well yes.

 

[feynman1]

Well yes.

Then why isn't the Gauss' law argument in any textbooks?

 

[Delta2]

Then why isn't the Gauss' law argument in any textbooks?

Because many textbooks don't study a subject so deeply. It will induce uneccesary complications it will be anti pedagogical.

 

[hutchphd]

So everyone agrees on my original post?

Everyone agrees that Gauss's Law is correct.
The "real reason" there is no net charge on the capacitor is that charge is conserved and no gremlins put extra charge in the circuit. Of course it is supported by Gauss's Law because there is no extra charge on the circuit and Gauss was and is correct.

I "really do" agree that it is much ado about nothing.
.