# Capacitor in presence of an external electric field

Let's consider a capacitor made of a couple of parallel metal strips (suppose they are made of perfect electric conductor) as shown in the figure, which represents a little capacitor.

Suppose an external (time variant) electric field is applied in the space in which the capacitor is placed, as shown in figure.

Such an electric field may be for instance that of an incident orthogonal electromagnetic wave, as shown in the following picture.

What happens? I don't know which are cause and effects in this situation.

I'd say that:

1. The external electric field induces separation of charges in the two strips. So, plus charges on one strip and minus charges on the other.

2. The previous situation is like that happens on a capacitor when supplied by a voltage source. But, in such a case, I'd say that the voltage source provides charges to the capacitor metal plates, and then the resulting charge distribution generates the electric field. In this case, however, the electric field is already existing since it's the external electric field (the incident wave).

In other words, is the electric field between the plates the original external incident E field or is it the E field generated by the charges separated by the original incident E field?

## 3 Answers

In other words, is the electric field between the plates the original external incident E field or is it the E field generated by the charges separated by the original incident E field?

Using Superposition theorem

Electric field between the plates is due to

1.Electric field of electromagnetic wave 2.induce charges of 1st plate 3.induce charges of 2nd plate .

Net electric field between the plates $$=\vec(E)+\vec(E_1)+\vec(E_2)+\vec(E_3)+\vec(E_4)$$

But if plates are very very close to each other then we can assume that they will behave as parllel sheet of infinite length and we get net electric field between plates is $$\vec(E)$$ because between plates $$\vec(E_1)+\vec(E_2)+\vec(E_3)+\vec(E_4)=0$$

and final distribution will look like -

And from uniqueness theorem this distribution is unique(but function of time ) for a given wave and charges on conductor (=0).

But what if plates are not very close to each other ?

Then we cannot assume as parllel sheet of infinite length and hence a distorted field will be obtained in between of plates and for that we have to solve Laplace equations with suitable boundary conditions which is too complicated

Note-$$\vec(E),\vec(E_1),\vec(E_2),\vec(E_3),\vec(E_4)$$ are electric field due to electromagnetic wave, due to induce charges on 1st plate (2 surface) and due to induce charges on 2nd plates (2surface) respectively.

• Thank you for your answer. I've not understood why do the E fields of each charge distribution in each plate neglect themselves. Why don't they add themselves like in a capacitor? Mar 24, 2021 at 16:08
• @Kinka-Byo,I added More details , take a look ,and the key point for these distribution is to get net electric field zero between the surfaces of same plate because Electrostatic field is zero inside a conductor! Mar 24, 2021 at 16:30
• @Kinka-Byo because if you look closely the figure I drawn , you'll found that net charge on each plate is zero but in case of capacitor there is some finite amount of charges are present in each plates Mar 24, 2021 at 16:36
• @Kinka-Byo E1 and E2 are electric fields due to 2surfaces of 1st plate and E3,E4 are electric fields due to 2 surfaces of 2nd plate . Each plate has two surfaces (outer and inner) so total 4surfaces are there and corresponding to each surface we can assume one electric field Mar 25, 2021 at 5:52
• @Kinka-Byo , of course charges -metal plate-charges because this configuration make sure that electric field is Zero inside plate (between outer and inner surface). and linear distribution is valid only if we assume infinite plates otherwise there will be highly complicated fringing fields present due to edges Mar 25, 2021 at 6:06

The difference in external electric field is all you might expect if it is parallel at all times but not radial or orthogonal and assuming that the object does not interfere with the continuity of the field gradient.

If the external field has no significant gradient in a small distance then neither will the detected field or voltage across the plates.

The external electric field induces separation of charges in the two strips. So, plus charges on one strip and minus charges on the other.

the two electrodes are already separated – how would the missing / over charges from one travel to the other?

No, what changes (if the charge is fixed), then the voltage (i.e. the energy per charge) changes.

The previous situation is like that happens on a capacitor when supplied by a voltage source.

A voltage source fully defines the electric field between the plates: Voltage divided by distance, and the voltage source holds that voltage constant.

To keep the voltage constant, a current will have to flow reflecting that to keep voltage constant, the charge must change.

• "the two electrodes are already separated – how would the missing / over charges from one travel to the other?" We can consider two surfaces of each plate such that although net charge is zero (on each plate) but the surface of plates facing each other contain +ve and -ve charge Mar 24, 2021 at 15:15
• sure, the charges within the conductor can shift, though they won't: we're modelling this as PEC, so they are all perfectly on the edge, anyway. Mar 24, 2021 at 15:38