Magnetism, Displacement Currents and Parallel Plate Capacitors

In summary, the conversation discusses a problem involving a charging capacitor using artificial and realistic models. Part (a) focuses on finding the electric field between the plates, with the current being constant and the plates being separated by a small distance. Part (b) involves finding the displacement current through a circle and using it to calculate the magnetic field at a distance from the axis. In part (c), a cylindrical surface is used to calculate the enclosed current. Equations and rules are discussed throughout the conversation to aid in solving the problem.
  • #1
TFM
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Homework Statement



The preceding problem was an artificial model for the charging capacitor, designed to avoid complications associated with the current spreading out over the surface of the plates. For a more realistic model, imagine thin wires that connect to the centres of the plates (see figure (a)). Again, the current I is constant, the radius of the capacitor is a, and the separation of the plates is w << a. Assume that the current flows out over the plates in a such a way that the surface charge is uniform, at any given time, and is zero at time t = 0.

(a)

Find the electric field between the plates, as a function of t.

I=c dV/dt∴Idt=Cdv





(b)

Find the displacement current through a circle of radius s in the plane midway between the plates. Using this circle as your "Amperian loop", and the flat surface that spans it, find the magnetic field at a distance s from the axis.

(c)

Repair part b, but this time uses the cylindrical surface in figure (b), which is open at the right and extends to the left through the plate and terminates outside the capacitor. Notice that the displacement current through this surface is zero, and there are two contributions to the enclosed current.


Homework Equations



Not sure

The Attempt at a Solution



I'm trying to do part (a), but I am not sure of an equation which has time in it. So far, I have
got:

[tex] I = c\frac{dv}{dt} [/tex]

Thus

[tex] Idt = Cdv [/tex]

But I am not sure if this is right.

Is it? and if so, what might be some useful equations/rules to use?

Thanks in advanced,

TFM
 

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  • #2
I have done some more for part a):

[tex] \oint_S E \cdot da = \frac{Q}{\epsilon_0} [/tex]

For a capacitor, Q = CV

from the equatuion I posted in the first post,

[tex] C = I \frac{dt}{dv} [/tex]

putting together:

[tex] \oint_S E \cdot da = \frac{vI\frac{dt}{dv}}{\epsilon_0} [/tex]

Gives:

[tex] EA = \frac{vI\frac{dt}{dv}}{\epsilon_0} [/tex]

[tex] A = \pi w^2 [/tex]

giving E:

[tex] E = \frac{vI\frac{dt}{dv}}{\epsilon_0 \pi w^2 } [/tex]

Does this look right?

TFM
 
  • #3


I would like to clarify that the problem statement provided is incomplete and lacks necessary information to provide a complete response. It is important to have a clear understanding of the problem and all the given parameters before attempting to solve it.

That being said, I will provide general information on the concepts mentioned in the problem, which may help in solving it.

Magnetism is a fundamental force of nature that is caused by moving electric charges. It is characterized by the presence of a magnetic field, which exerts a force on other moving charges.

Displacement currents, on the other hand, are a type of electric current that is produced by changing electric fields. They are important in the study of electromagnetism and are a key component in Maxwell's equations.

Parallel plate capacitors are a type of circuit component that store electrical energy in the form of an electric field between two conductive plates. The capacitance of a parallel plate capacitor is dependent on the area of the plates, the distance between them, and the properties of the material between the plates.

To solve part (a) of the problem, you will need to use the relationship between current and voltage in a capacitor, which is given by I = C(dV/dt), where I is the current, C is the capacitance, and dV/dt is the rate of change of voltage. You will also need to use the fact that the surface charge on the plates is uniform and zero at t = 0.

For part (b), you will need to use Ampere's law, which relates the magnetic field around a closed loop to the current passing through the loop. You will also need to use the concept of displacement current to find the total current passing through the loop.

Lastly, for part (c), you will need to consider the contributions of the two different currents passing through the cylindrical surface and use Gauss's law to relate them to the electric field.

I hope this helps you in solving the problem. However, it is important to have a complete understanding of the problem and all the given parameters before attempting to solve it.
 

FAQ: Magnetism, Displacement Currents and Parallel Plate Capacitors

1. What is magnetism?

Magnetism is a fundamental force of nature that is responsible for the attraction and repulsion of certain materials, such as iron and steel. It is caused by the movement of electric charges, either through the flow of electricity or the spin of electrons in an atom.

2. What is displacement current?

Displacement current is a concept in electromagnetism that describes the flow of electric charge in a region where there is no actual flow of charge. It was first introduced by James Clerk Maxwell to explain the behavior of electromagnetic waves.

3. How is magnetism related to displacement current?

Magnetism is related to displacement current because both are part of the electromagnetic force. Changes in electric fields can create magnetic fields, and changes in magnetic fields can create electric fields. Displacement current is what allows electromagnetic waves, including light, to propagate through space.

4. What is a parallel plate capacitor?

A parallel plate capacitor is a device used to store electric charge. It consists of two conducting plates separated by an insulating material, or dielectric. When a voltage is applied to the plates, electric charge accumulates on each plate, creating an electric field between them.

5. How does a parallel plate capacitor work?

A parallel plate capacitor works by storing electric charge on its plates. When a voltage is applied, electrons are pushed onto one plate and pulled from the other, creating an electric field between the plates. The strength of the electric field is directly proportional to the voltage applied and inversely proportional to the distance between the plates.

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