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I have a vector potential given by:

$\mathbf{A}(x,t) = \mathbf{e}_{y}\frac{1}{2} e^{-(x-ct)^{2}/{4a^{2}}}$

Now, the question is "Determine the E and B under the condition that the scalar potential vanishes $V = 0$."

But I'm not quite sure what it means when $V=0$ ?

As far as I can see, the B-field is given by:

$\mathbf{B}=\nabla \times \mathbf{A}$

And then I have that:

$\mathbf{E} = -\nabla V$

So is it just this straightforward ? That I find the B-field from A, and since $V = 0$, the E-field is zero, or am I doing it wrong ?

Thanks in advance :)

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2 Answers 2

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The relation $\mathbf{E} = -\nabla V$ holds only in the absence of vector potential, otherwise the electric field changes to $$ \mathbf{E} = -\nabla V-\frac{\partial\mathbf{A}}{\partial t}. $$ The reason for this is that when you introduce vector potential by $\mathbf{B} = \nabla\times\mathbf{A}$, Faraday's law reads $$ \nabla\times\mathbf{A}+\frac{\partial}{\partial t}(\nabla\times\mathbf{A}) = \nabla\times\left(\mathbf{E}+\frac{\partial\mathbf{A}}{\partial t}\right) = 0. $$ This can be solved generally by putting the bracket equal to a gradient of a scalar function $-\nabla V$ which gives the result for electric field in terms of both scalar and vector potentials given above.

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    $\begingroup$ So actually, since $V=0$ the $\mathbf{E}$-field is just the negative derivative of $\mathbf{A}$ with respect to $t$ ? $\endgroup$ Commented May 26, 2013 at 16:13
  • $\begingroup$ @DenverDang Exactly :) $\endgroup$ Commented May 26, 2013 at 16:14
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The expression for the electric field includes derivatives of the vector potential with respect to time.

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