# Electric field between charging capacitor

• Aziza
In summary, the electric field between a discharging capacitor can be calculated by taking the time varying electric field and adding the time varying magnetic field.

#### Aziza

I am confused about how to find the electric field between a discharging capacitor.

For example, let's say I have a fully charged capacitor, such that there is no current flow anywhere. Then there is an electric field between the capacitor which equals 2*sigma/epsilon_0.

But now the ciruit is opened, such that current begins flowing to discharge the capacitor. The the original E-field is decreasing (but still present), causing a B field to be generated, which also causes another E field to be generated. Is the E field then the sum of these two E-fields??

This is a good question. First I would like to point out one interesting thing you seem to be missing:
The the original E-field is decreasing (but still present), causing a B field to be generated, which also causes another E field to be generated. Is the E field then the sum of these two E-fields??
Now that you are adding a second E field that was generated by the changes of the first, wouldn't that second E also change, thus generating a second B field to be generated, causing yet a third E field?Anyway, one way to analyze this problem might be to imagine an idealized circuit where an infinite parallel-plate capacitor discharges exponentially according to its RLC time constant. See here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capdis.html. If you use that sort of arithmetic to calculate the electric field in the capacitor, you will get a good approximation in most realistic situations, and the magnetic effects you're talking about probably represent small corrections.

I believe the magnetic effects only really arise in the "edge effects" that you get rid of by making the two plates of your parallel plate capacitor have infinite extent in space. I think you can see this through staring at the Maxwell equation:
$$\nabla \times \mathbf{B} = \mu_0\left(\mathbf{J} + \varepsilon_0 \frac{\partial \mathbf{E}} {\partial t} \right)$$
J will be zero in the region between the plates of the capacitor so, $$\nabla \times \mathbf{B} = \mu_0\ \varepsilon_0 \frac{\partial \mathbf{E}} {\partial t}$$
Which means that the curl is a constant inside the capacitor (since we're assuming the region between the plates of the capacitor has constant E and constant ∂E/∂t).

If that were so then the magnetic field would cancel everwhere except near the edges. One way to think about this is if you think of the curl as the circulation around a point, then, for example, the circulation at a distance r to the left of point x would cancel the circulation around the point located 2r to the left of point x.

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Jolb said:
This is a good question. First I would like to point out one interesting thing you seem to be missing:

Now that you are adding a second E field that was generated by the changes of the first, wouldn't that second E also change, thus generating a second B field to be generated, causing yet a third E field?

Anyway, one way to analyze this problem might be to imagine an idealized circuit where an infinite parallel-plate capacitor discharges exponentially according to its RLC time constant. See here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capdis.html. If you use that sort of arithmetic to calculate the electric field in the capacitor, you will get a good approximation in most realistic situations, and the magnetic effects you're talking about probably represent small corrections.

I believe the magnetic effects only really arise in the "edge effects" that you get rid of by making the two plates of your parallel plate capacitor have infinite extent in space.

thanks for the response!
Also, I just saw a Griffiths problem where he treats the E-field between charging capacitor by just saying that E(t)=σ(t)/ε, where σ is the surface charge and is increasing...so then this equation is not completely correct...but since the B field for given changing E field is small, I guess it makes sense that the correction is small

It is just one electric field with 2 components :
1st component generated by the time varying charge distribution (in the plates of the capacitor) and
2nd component generated by the time varying magnetic field.

The time varying magnetic field has also 2 components:

1st component generated by the time varying current
2nd component generated by the time varying electric field

In electromagnetism we just have 4 basic "entities": Charges, Currents(=moving charges), Electric Field, Magnetic Field. It is just that these 4 entities interact with each other (for example an electric field interacts with charges and set em in motion which creates a current which creates a magnetic field which will alter the electric field which will alter the flow of charges and so on) that makes it kind of confusing when we try to think in a linear way of cause and effect. In short the fields affect the currents and charges distributions which in turn affect the fields. Maxwell's 4 equations give us with full mathematical accuracy how these 4 entities interact with each other.

Aziza said:
there is an electric field between the capacitor which equals 2*sigma/epsilon_0.
The E field (which should be sigma/epsilon_0) is given by Gauss's law, which holds even for a discharging capacitor. The B field due to the current cannot change Gauss's law.

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clem said:
The E field (which should be sigma/epsilon_0) is given by Gauss's law, which holds even for a discharging capacitor. The B field due to the current cannot change Gauss's law.

Gauss's law give us the delusion that the E-field depends only on the charge density which is true only for the electrostatic case.

Faraday's law also give us the delusion that the E-field depends only on time varying magnetic fields but that's true only if there arent any charges

What happens is that by knowing only the divergence or only the curl of a vector field u cannot uniquely determine the field. Only if u know both div and curl u can determine it.

Gauss's law shows the E-field depends only on the charge density, which is true for all cases.

No, in the general case you need the full Maxwell equations to determine the electromagnetic field. The charge density only determines $\vec{\nabla} \cdot \vec{E}$ but not $\vec{\nabla} \times \vec{E}$!

I think that one way of looking at the modification of the E field whilst the Capacitor charge is changing could be to think in terms of a 'back emf' due to the inductance of the space between the plates.

## What is an electric field?

An electric field is a region in space where electrically charged particles experience a force. It is created by the presence of charged particles and can be either positive or negative.

## How is an electric field created between charging capacitors?

An electric field is created between charging capacitors due to the accumulation of opposite charges on each plate. This accumulation of charges creates an imbalance of electric potential, resulting in an electric field.

## What is the direction of the electric field between charging capacitors?

The direction of the electric field between charging capacitors is from the positively charged plate to the negatively charged plate. This is because opposite charges attract each other, and the electric field is directed towards the negative charges.

## How does the distance between the charging capacitors affect the electric field?

The electric field between charging capacitors is inversely proportional to the distance between the plates. This means that as the distance between the plates increases, the electric field decreases, and vice versa.

## What is the equation for calculating the electric field between charging capacitors?

The electric field between charging capacitors can be calculated using the equation E = Q/εA, where Q is the charge on the plates, ε is the permittivity of the material between the plates, and A is the area of the plates.