- #1
Decimal
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- Homework Statement
- A uniformly charged infinite sheet in the x-y plane starts moving with velocity ##v## in the x direction at time ##t=0##. The spatial charge distribution is then ##\rho(\vec{r}) = \sigma_0 \delta(z)##. The current density for ##t>0## is then ##\vec{J} = \sigma_0 v \delta(z) \hat{x}##.
Calculate the retarded vector potential a distance ##z## above the moving sheet for ##t>0##.
- Relevant Equations
- ##\vec{A} = \frac{\mu_0}{4\pi} \int \frac{\vec{J}(\vec{r}', t_r)}{|\vec{r}-\vec{r}'|} d^3r'. ## Here ##t_r## is the retarded time and ##r## the distance to the observation point
Hello,
I start by applying the integral for the vector potential ##\vec{A}## using cylindrical coordinates. I define ##r## as the distance to the ##z##-axis. This gives me the following integral,$$\vec{A} = \frac{\mu_0}{4\pi} \sigma_0 v 2 \pi \hat{x} \int_0^{\sqrt{(ct)^2-z^2}} \frac{r}{\sqrt{r^2+z^2}} dr.$$ Here I have adjusted my integration boundary such that only the current which actually influences the observation point influences the potential. Obviously this boundary increases with time.
The above integral gives the following result, $$\vec{A} = \frac{\mu_0 \sigma_0 v }{2} ct \ U(t-\frac{|z|}{c}) \hat{x}.$$ Here I have defined ##U(x)## as the unit step function, since ##\vec{A} = 0## if ##t<\frac{|z|}{c}##. Now I am not sure if this is a correct result. On one hand it implies that the potential increases to infinity as ##t \rightarrow \infty##, instead of reducing to the static result. I know that in the static case using the above integral for infinite current distributions doesn't converge anyway, but I don't see a different way of finding the retarded potential for this set-up.
Also with this this vector potential the resulting magnetic field is 0 almost everywhere, except for maybe some delta functions that originate from the step function. This is because ##\vec{A}## is mostly rotationless. This also seems weird to me, and again does not correspond with the static case.
So I feel like I am either starting with the wrong formula entirely or missing some important part of the problem. Any help would be very much appreciated!
I start by applying the integral for the vector potential ##\vec{A}## using cylindrical coordinates. I define ##r## as the distance to the ##z##-axis. This gives me the following integral,$$\vec{A} = \frac{\mu_0}{4\pi} \sigma_0 v 2 \pi \hat{x} \int_0^{\sqrt{(ct)^2-z^2}} \frac{r}{\sqrt{r^2+z^2}} dr.$$ Here I have adjusted my integration boundary such that only the current which actually influences the observation point influences the potential. Obviously this boundary increases with time.
The above integral gives the following result, $$\vec{A} = \frac{\mu_0 \sigma_0 v }{2} ct \ U(t-\frac{|z|}{c}) \hat{x}.$$ Here I have defined ##U(x)## as the unit step function, since ##\vec{A} = 0## if ##t<\frac{|z|}{c}##. Now I am not sure if this is a correct result. On one hand it implies that the potential increases to infinity as ##t \rightarrow \infty##, instead of reducing to the static result. I know that in the static case using the above integral for infinite current distributions doesn't converge anyway, but I don't see a different way of finding the retarded potential for this set-up.
Also with this this vector potential the resulting magnetic field is 0 almost everywhere, except for maybe some delta functions that originate from the step function. This is because ##\vec{A}## is mostly rotationless. This also seems weird to me, and again does not correspond with the static case.
So I feel like I am either starting with the wrong formula entirely or missing some important part of the problem. Any help would be very much appreciated!
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