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(Electric) Scalar and vector potential

  • Thread starter lailola
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Homework Statement



In the problem, the electric scalar and vector potentials are,
[itex]\phi=0, \vec{A}=A_0 e^{i(k_1 x-2k_2y-wt)}\vec{u_y}[/itex]

I have to find E, B and S.

Then, I have to calculate [itex]\phi '[/itex] that satisfies [itex]div\vec{A}+\frac{\partial \phi '}{\partial t}=0[/itex] Then calculate E and B.

Is it possible to find [itex]\vec{A}'[/itex] and [itex]\phi'[/itex] that satisfy the previous equation and produce the same E and B as [itex]\vec{A}[/itex] and [itex]\phi[/itex]?

Homework Equations


[itex]\vec{E}=-grad\phi-\frac{\partial \vec{A}}{c\partial t}=0[/itex]
[itex]\vec{B}=rot(\vec{A})[/itex]


The Attempt at a Solution



Using the equations I find:

[itex]\vec{E}=A_0wi/c e^{i(k_1x-2k_2y-wt)}\vec{u_y}[/itex]

[itex]\vec{B}=A_0ik_1 e^{i(k_1x-2k_2y-wt)}\vec{u_k}[/itex]

[itex]\vec{S}=\frac{c}{4\pi} \vec{E}x\vec{B}=1/(4\pi) A_0^2wk_1sin^2(k_1x-2k_2y-wt)\vec{u_x}[/itex]

For the next part I find,

[itex]\phi ' = - 2K_2 c A_0/(w) e^{i(k_1x-2k_2y-wt)}+ constant(x,y)[/itex]

Then, I calculate E and B as before.

I don't know how to answer the last part. Any idea?

Thank you.
 
Last edited:

Answers and Replies

  • #2
vela
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Homework Statement



In the problem, the electric scalar and vector potentials are,
[itex]\phi=0, \vec{A}=A_0 e^{i(k_1 x-2k_2y-wt)}[/itex]
Your expression for the vector potential isn't a vector.

I have to find E, B and S.

Then, I have to calculate [itex]\phi '[/itex] that satisfies [itex]div\vec{A}+\frac{\partial \phi '}{\partial t}=0[/itex] Then calculate E and B.

Is it possible to find [itex]\vec{A}[/itex] and [itex]\phi'[/itex] that satisfy the previous equation and produce the same E and B as [itex]\vec{A}[/itex] and [itex]\phi[/itex]?

Homework Equations


[itex]\vec{E}=-grad\phi-\frac{\partial \vec{A}}{c\partial t}=0[/itex]
[itex]\vec{B}=rot(\vec{A})[/itex]


The Attempt at a Solution



Using the equations I find:

[itex]\vec{E}=A_0w/c e^{i(k_1x-2k_2y-wt)}\vec{u_y}[/itex]

[itex]\vec{B}=A_0ik_1 e^{i(k_1x-2k_2y-wt)}\vec{u_k}[/itex]

[itex]\vec{S}=\frac{c}{4\pi} \vec{E}x\vec{B}=1/(4\pi) A_0^2wk_1sin^2(k_1x-2k_2y-wt)\vec{u_x}[/itex]

For the next part I find,

[itex]\phi ' = - 2K_2 c A_0/(w) e^{i(k_1x-2k_2y-wt)}+ constant(x,y)[/itex]

Then, I calculate E and B as before.

I don't know how to answer the last part. Any idea?

Thank you.
 
  • #3
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Your expression for the vector potential isn't a vector.
Presumably [itex]A_0[/itex] is itself a vector, which makes that definition perfectly valid.

I don't know how to answer the last part. Any idea?

Thank you.
Imagine taking your current definitions for [itex]A[/itex] and [itex]\phi[/itex] (I'm renaming your [itex]\phi'[/itex] to [itex]\phi[/itex] for simplicity, and so that I can reuse the symbol [itex]\phi'[/itex] below), and adding new quantities to them. So something like:
[tex]A \rightarrow A + A'\\
\phi \rightarrow \phi + \phi'[/tex]

Now plug those definitions into your previous equations, and see if you can use them to come up with some constraints on the forms that [itex]A'[/itex] and [itex]\phi'[/itex] would have to take, in order to cause [itex]E[/itex] and [itex]B[/itex] to still come out the same.
 
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  • #4
vela
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Presumably [itex]A_0[/itex] is itself a vector, which makes that definition perfectly valid.
I don't think so. A0 appears in the expressions for the electric and magnetic fields along with unit vectors. It's pretty clear that the OP meant for A0 to denote a scalar.
 
Last edited:
  • #5
368
12
Hmm, you're right. Guess I didn't look close enough at those E/B solutions.

lailola, can you clarify what the original problem is, and how you came up with your definitions for E and B?
 
  • #6
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I don't think so. A0 appears in the expressions for the electric and magnetic fields along with unit vectors. It's pretty clear that the OP meant for A0 to denote a scalar.
I forgot to write the vector in the expression of the vector potential. I've written it.
 
  • #7
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Presumably [itex]A_0[/itex] is itself a vector, which makes that definition perfectly valid.



Imagine taking your current definitions for [itex]A[/itex] and [itex]\phi[/itex] (I'm renaming your [itex]\phi'[/itex] to [itex]\phi[/itex] for simplicity, and so that I can reuse the symbol [itex]\phi'[/itex] below), and adding new quantities to them. So something like:
[tex]A \rightarrow A + A'\\
\phi \rightarrow \phi + \phi'[/tex]

Now plug those definitions into your previous equations, and see if you can use them to come up with some constraints on the forms that [itex]A'[/itex] and [itex]\phi'[/itex] would have to take, in order to cause [itex]E[/itex] and [itex]B[/itex] to still come out the same.
Ok, thank you!
 

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