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Electromagnetic field acting on a conducting infinite plate

  • Thread starter papercace
  • Start date
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1. Homework Statement
Consider an electromagnetic field in an empty space in the region ##0 \leq z \leq a## with the following non-zero components:
$$E_x = -B_0\frac{\omega a}{\pi}\sin\left(\frac{\pi z}{a}\right)\sin\left( ky-\omega t\right)\\
B_z = B_0\frac{ka}{\pi}\sin\left(\frac{\pi z}{a}\right)\sin\left( ky-\omega t\right)\\
B_y= B_0\cos\left(\frac{\pi z}{a}\right)\cos\left( ky-\omega t\right)
$$
Determine the condition for which this field satisfies Maxwell's equations. Assume that the fields are zero for z<0 and that there is a perfectly conducting plate in the z=0 plane and determine the surface charge density and surface current density on the plate.

2. Homework Equations
Maxwell's equations:
$$I. \quad \nabla \cdot \mathbf{E}=\frac{\rho}{\epsilon_0}\\
II. \quad \nabla \cdot \mathbf{B}=0\\
III. \quad \nabla \times \mathbf{E}=-\frac{\partial\mathbf{B}}{\partial t}\\
IV. \quad \nabla \times \mathbf{B}=\mu_0 \mathbf{J}+\mu_0\epsilon_0\frac{\partial \mathbf{E}}{\partial t}
$$

3. The Attempt at a Solution
It's the second part I have problems with. If you use MI (Maxwell I) you get ##\mathbf{E}=0 \Rightarrow \rho=0##. That's all fine and dandy, but if you use MIV you get
$$\nabla \times \mathbf{B}=\mu_0 \mathbf{J} = \left(\frac{k^2a}{\pi}+\frac{\pi}{a}\right)B_0\cos(ky-\omega t) \hat{\mathbf x}\\
\left(\frac{\partial \mathbf{E}}{\partial t}=0 \quad \text{at} \quad z=0\right).$$
So we've got a current density but no charge density and no electric field at ##z=0## even though I think there should be some, especially since we have a changing B-field which should induce an E-field. Something is obviously wrong with my thought process.
 

TSny

Homework Helper
Gold Member
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2,658
For the first part of the problem, did you find that there must exist a nonzero current density ##\mathbf{J}(y, z, t)## for the region ##z>0##?

It's the second part I have problems with. If you use MI (Maxwell I) you get ##\mathbf{E}=0 \Rightarrow \rho=0##. That's all fine and dandy, but if you use MIV you get
$$\nabla \times \mathbf{B}=\mu_0 \mathbf{J} = \left(\frac{k^2a}{\pi}+\frac{\pi}{a}\right)B_0\cos(ky-\omega t) \hat{\mathbf x}
$$
Shouldn't the right side of the equation contain ##\sin(\frac{\pi z}{a})## which would go to zero for ##z = 0##?

So we've got a current density but no charge density and no electric field at ##z=0## even though I think there should be some, especially since we have a changing B-field which should induce an E-field. Something is obviously wrong with my thought process.
Maxwell's equation MIII says that at a point where B is changing with time, the curl of E must be nonzero at that point. But, E itself does not have to be nonzero at that point. Consider the example of a uniform magnetic field in a circular region that is changing with time. At the center of the region the E field remains zero even though there is a changing magnetic field at the center.
See http://sdsu-physics.org/physics180/physics196/images_196/31_Electricfield1.jpg and note that E = 0 at r = 0.

To answer the second part of the problem, you will need to know the boundary conditions on E and B at the surface of a perfect conductor. See equations (1300) to (1303) here
https://farside.ph.utexas.edu/teaching/jk1/lectures/node112.html
 

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