# Capacitor Electric Field in a Dielectric?

In summary: The electrostatic field inside the battery is in opposition to that in the wires and capacitor. There is no net E field in the battery, assuming zero internal resistance.

Hi,

I was reading the capacitor and its operation and it is written there that when the DC is applied to the capacitor it becomes open circuit. The main mechanism behind this phenomena is explained as below:-

"When DC voltage is applied the charge started accumulating on the plates of the capacitor. The plate which is connected with the positive terminal of the battery has positive charge accumulation and the plate which is connected with the negative terminal of the battery has negative charge accumulation. With this charge accumulation the electric field exist in the dielectric that actually opposes the voltage applied from the source"

My Question is that since the accumulation of the charge is according to the terminals of the battery so therefore the direction of Electric field should also be like that than how does they oppose each other and prevent the voltage to stop.

Correct me Please if I am wrong.

Many Thanks for your valuable time and Help.

Hello,
The electric field inside the dielectric is as you say, it does not oppose the electric field from the source, it is rather in the same direction as the electric field from the source.

It is the electric field inside the connecting wire that opposes the electric field from the voltage source inside the connecting wire. To understand this , imagine the connecting wire from the positive terminal of the source to the positive plate of the capacitor. The positive terminal of the source creates an electric field inside the connecting wire that attracts the electrons from the connecting wire to the positive terminal of the source. But also the positive plate of capacitor creates an electric field that also attracts the electrons to the positive plate of the capacitor. These two electric fields are opposite.

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Hi,

I was reading the capacitor and its operation and it is written there that when the DC is applied to the capacitor it becomes open circuit. The main mechanism behind this phenomena is explained as below:-

"When DC voltage is applied the charge started accumulating on the plates of the capacitor. The plate which is connected with the positive terminal of the battery has positive charge accumulation and the plate which is connected with the negative terminal of the battery has negative charge accumulation. With this charge accumulation the electric field exist in the dielectric that actually opposes the voltage applied from the source"

My Question is that since the accumulation of the charge is according to the terminals of the battery so therefore the direction of Electric field should also be like that than how does they oppose each other and prevent the voltage to stop.

Correct me Please if I am wrong.

Many Thanks for your valuable time and Help.
There are two electric fields inside the battery: one generated by the battery's emf acting to move charge through the wires to the capacitor plates, and the other electrostatic which opposes the flow of charge inside the battery. They are of equal magnitude and opposing direction assuming zero internal resistance.

Outside the battery there are only electrostatic fields. This is true for the wires (which however will have a very low E field due to their high conductance), and the space between the plates. Current flow and the electrostatic E fields in the wires and between the plates are all in the same direction. The electrostatic field inside the battery is in opposition to that in the wires and capacitor. There is no net E field in the battery, assuming zero internal resistance.

rude man said:
There are two electric fields inside the battery: one generated by the battery's emf acting to move charge through the wires to the capacitor plates, and the other electrostatic which opposes the flow of charge inside the battery. They are of equal magnitude and opposing direction assuming zero internal resistance.
Allow me to disagree about what the E-field from the EMF does. It acts to move charges through the battery and opposes the Electrostatic field from the poles of the battery inside the battery. Outside the battery the E-field from the EMF is zero, it is the electrostatic field from the poles that (initiates the process of creating surface charges in the connecting wires ) and acts to move charges through the wires to the capacitor plates.

Outside the battery there are only electrostatic fields. This is true for the wires (which however will have a very low E field due to their high conductance), and the space between the plates. Current flow and the electrostatic E fields in the wires and between the plates are all in the same direction.
Agreed up to here
The electrostatic field inside the battery is in opposition to that in the wires and capacitor.
Hmm, not sure what you had in mind when you wrote that sentence. I can certainly give examples where the electrostatic field inside the wires or inside the capacitor are in the same direction and trend with the electrostatic field inside the battery.

If what you meant is that the EMF-field inside the battery can be in opposite direction of the electrostatic field inside the capacitor and /or the wires then I agree to that.

There is no net E field in the battery, assuming zero internal resistance.
Agreed.

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EVERYTHINGISNOTHING
Delta² said:
Allow me to disagree about what the E-field from the EMF does. It acts to move charges through the battery and opposes the Electrostatic field from the poles of the battery inside the battery. Outside the battery the E-field from the EMF is zero, it is the electrostatic field from the poles that (initiates the process of creating surface charges in the connecting wires ) and acts to move charges through the wires to the capacitor plates.
I don't disagree with this but it's the emf that propels charge thru the entire circuit. No emf, no current, no electrostatic field ...
Agreed up to here
Hmm, not sure what you had in mind when you wrote that sentence. I can certainly give examples where the electrostatic field inside the wires or inside the capacitor are in the same direction and trend with the electrostatic field inside the battery.
If that were so then the circulation of the electrostatic field around the loop would not be zero. Or so says Kirchhoff. Therefore, the electrostatic field in the battery is is in opposition to that in the wires and in the capacitor. Would like to see some of those examples you cite.
If what you meant is that the EMF-field inside the battery can be in opposite direction of the electrostatic field inside the capacitor and /or the wires then I agree to that.
I did not say that; it's wrong. The emf field, the current and the electrostatic field in the wires and the capacitor are all in the same direction. (In that sense the emf is pushing the current in its direction.) The only differing one is the electrostatic field in the battery.

There is a lot of confusion out there regarding the two different kinds of E fields; perhaps you might want to read my recent Insight publication regarding MIT's Dr. Lewin's conundrum: https://www.physicsforums.com/insights/a-new-interpretation-of-dr-walter-lewins-paradox/

Hold on I make a figure I have in mind with a simple circuit with a resistor capacitor and a battery. Do you know some free available software that I could use or I ll just do it using Windows Paint !??

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As you can see the E-field inside the capacitor and the electrostatic E-field inside the battery are in the same trend and direction...

I think what confuses you is that the E-stat field inside the battery opposes the current flow, while the E_capacitor is in the same trend with current flow, however both of those fields have the same direction and trend.

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Delta² said:
View attachment 228756 Hold on I make a figure I have in mind with a simple circuit with a resistor capacitor and a battery. Do you know some free available software that I could use or I ll just do it using Windows Paint !??
As you can see the E-field inside the capacitor and the electrostatic E-field inside the battery are in the same trend and direction...
No they're not; they are in opposition to each other. But if the E field in the battery is meant to be the electrostatic field you have just agreed with me since your two E fields do oppose each other and the circulation is zero.

I thought you said you agreed that the net E field in the battery is zero - what happened? Fact is that the emf E field in the battery points - to +, giving zero net E field in the battery since net E field = Estatic + Eemf, added algebraically.

If we take as positive direction of the loop the counter clock wise direction, then E_statbat can be taken as negative while E_capacitor as positive when calculating the ##\int Edl## around the loop. I don't know if that's what you mean when you say they oppose each other...

But if we view them purely as vectors, it is clear that E_statbat and E_capacitor have same direction and trend...

Delta² said:
If we take as positive direction of the loop the counter clock wise direction, then E_statbat can be taken as negative while E_capacitor as positive when calculating the ##\int Edl## around the loop. I don't know if that's what you mean when you say they oppose each other...
That is exactly what I mean.
But if we view them purely as vectors, it is clear that E_statbat and E_capacitor have same direction and trend...
It is not clear at all.

OK, you are the battery, a swimmer swimming in the direction of your battery E field in a fast-moving river. The fast-moving current is in the direction of the capacitor E field. Are you swimming in the direction ("trend") of the current or against it?

rude man said:
It is not clear at all.
don't understand how you can say its not clear at all, if you view the two vectors just as I draw them in the image and forget for the moment that we talk about currents, emfs, electric circuits, just view purely the two vectors, what can you say?
OK, you are the battery, a swimmer swimming in the direction of your battery E field in a fast-moving river. The fast-moving current is in the direction of the capacitor E field. Are you swimming in the direction ("trend") of the current or against it?

Well, the current (officially speaking the current density) changes direction and trend through out the loop. Inside the battery the current has opposite trend than (the hypothetical current or displacement current) inside the capacitor. BUT that's one of the reasons current , officially speaking , is not a vector quantity in physics therefore it doesn't have direction and trend, current density has direction and trend.

I think all this boils down to the fact that when you say "oppose" or "in same direction" you refer with the relation to what happens to the current flow (or how we evaluate the ##\int Edl##) , while I refer to the definitions of direction and trend viewing them purely as vectors. So for you the two fields in the image are opposite because the field inside the battery opposes current flow, while the field inside the capacitor is at the same trend with current flow.

Delta² said:
... while I refer to the definitions of direction and trend viewing them purely as vectors. "
OK, purely as vectors, using conventional polar coordinates (r,θ) assuming the path is circular:

Eemf = +Eemf θ (counter-clockwise)
E
statbatt = -Estatbatt θ (clockwise)
Ecap = +Ecap θ (counter-clockwise)

Vectors are in bold. θ is the unit vector in the +θ direction (counter-clockwise).
Estatbatt and Ecap have opposite signs! They point in opposite directions!

rude man said:
OK, purely as vectors, using conventional polar coordinates (r,θ) assuming the path is circular:

Eemf = +Eemf θ (counter-clockwise)
Estatbatt = -Estatbatt θ (clockwise)
Ecap = +Ecap θ (counter-clockwise)

Vectors are in bold. θ is the unit vector in the +θ direction (counter-clockwise).
Estatbatt and Ecap have opposite signs! They point in opposite directions!
Interesting you switch the problem to circular path and polar coordinates, when it is obvious that for rectangular path and cartesian coordinate the vectors have same direction.

I think your second equation is wrong or at least incomplete i would say cause the unit vectors in polar coordinate systems are variable vectors. Both ##\vec{r}## and ##\vec{\theta}## depend on the polar coordinates ##(r,\theta)##. If we assume the emf source to be at ##(1,\frac{\pi}{2})## and the capacitor at ##(1,\frac{3\pi}{2})## then it is ##\vec{\theta}(1,\frac{\pi}{2})=-\vec{\theta}(1,\frac{3\pi}{2})##