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I wrote and solved this problem but am having serious doubts about the answer I obtained.
Two point charges \pm q move along the z-axis with velocity \pm v. If they are at the origin when t=0, what is the electric field magnitude a distance r from the z-axis?
The only things I seemed to need for this problem are Coulomb's law, Maxwell's correction to Ampere's law, and Faraday's law:
<br /> \vec{E}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2},\qquad<br /> \mathcal{E}=-\frac{\mathrm{d}\Phi_B}{\mathrm{d}t},\qquad<br /> \oint\vec{B}\cdot\mathrm{d}\vec{s}=\mu_0\epsilon_0\frac{\mathrm{d}\Phi_E}{\mathrm{d}t}.<br />
The positions of the point charges are (0,0,\pm z) where z=vt. We wish to determine the \vec{E} and \vec{B} fields a distance r from the axis where z=0, i.e., at (x,y,0) where x^2+y^2=r^2. The first contribution is that of two opposite point charges; then E^{(0)}=-E_z^{(0)}, where
E_z^{(0)}=\frac{q}{4\pi\epsilon_0}\frac{1}{r^2+z^2}\frac{-z}{\sqrt{r^2+z^2}}<br /> +\frac{-q}{4\pi\epsilon_0}\frac{1}{r^2+z^2}\frac{z}{\sqrt{r^2+z^2}}<br /> =-\frac{q}{2\pi\epsilon_0}\frac{z}{(r^2+z^2)^\frac{3}{2}}.
Now consider the electric flux \Phi_E^{(0)} through a circle of radius R at z=0. Then \Phi_E^{(0)}=\int\mathrm{d}\Phi_E^{(0)}, where we integrate over circular rings of thickness \mathrm{d}r:
\mathrm{d}\Phi_E^{(0)}=\vec{E}^{(0)}\cdot\mathrm{d}\vec{A}=E_z^{(0)}\,\mathrm{d}A<br /> =\left(-\frac{q}{2\pi\epsilon_0}\frac{z}{(r^2+z^2)^\frac{3}{2}}\right)(2\pi r\,\mathrm{d}r)<br /> =-\frac{q}{2\epsilon_0}\frac{z(2r\,\mathrm{d}r)}{(r^2+z^2)^\frac{3}{2}}
\Rightarrow\Phi_E^{(0)}=\int\mathrm{d}\Phi_E^{(0)}<br /> =-\frac{qz}{2\epsilon_0}\int_0^R\frac{2r\,\mathrm{d}r}{(r^2+z^2)^\frac{3}{2}}<br /> =-\frac{qz}{2\epsilon_0}\int_{z^2}^{R^2+z^2}u^{-\frac{3}{2}}\,\mathrm{d}u<br /> =-\frac{qz}{2\epsilon_0}\big[-2u^{-\frac{1}{2}}\big]_{z^2}^{R^2+z^2}
where we substituted u=r^2+z^2, \mathrm{d}u=2r\,\mathrm{d}r, u(0)=z^2, and u(R)=R^2+z^2: Thus
\Phi_E^{(0)}(r)=-\frac{qz}{\epsilon_0}\left(\frac{1}{z}-\frac{1}{\sqrt{r^2+z^2}}\right)<br /> =\frac{q}{\epsilon_0}\left(\frac{z}{\sqrt{r^2+z^2}}-1\right).
Recall the Amp\`ere-Maxwell law,
\oint\vec{B}\cdot\mathrm{d}\vec\ell=\mu_0I_\mathrm{enc}+\mu_0\epsilon_0\frac{\mathrm{d}\Phi_E}{\mathrm{d}t}.
By symmetry, we have that \vec{B}^{(0)} is purely azimuthal, so \oint\vec{B}\cdot\mathrm{d}\vec\ell=(2\pi r)B_\phi^{(0)}. Since I_\mathrm{enc}=0 (there is no current enclosed by the circular loop), we have
(2\pi r)B_\phi^{(0)}=0+\mu_0\epsilon_0\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{q}{\epsilon_0}\left(\frac{z}{\sqrt{r^2+z^2}}-1\right)\right]<br /> =\mu_0\epsilon_0\left(\frac{q}{\epsilon_0}\right)\frac{\mathrm{d}}{\mathrm{d}z}\left[\frac{z}{\sqrt{r^2+z^2}}\right]\frac{\mathrm{d}z}{\mathrm{d}t}
=\mu_0q\left(\frac{(1)(\sqrt{r^2+z^2})-(\frac{z}{\sqrt{r^2+z^2}})(z)}{r^2+z^2}\right)v<br /> =\mu_0qv\left(\frac{\frac{r^2+z^2}{\sqrt{r^2+z^2}}-\frac{z^2}{\sqrt{r^2+z^2}}}{r^2+z^2}\right)<br /> =\mu_0qv\frac{r^2}{(r^2+z^2)^\frac{3}{2}}
\Rightarrow B_\phi^{(0)}=\frac{\mu_0qv}{2\pi}\frac{r}{(r^2+z^2)^\frac{3}{2}}.
Now, B^{(0)} itself changes with time, and a changing magnetic field induces an electric field according to Faraday's law:
\oint\vec{E}\cdot\mathrm{d}\vec\ell=-\frac{\mathrm{d}\Phi_B}{\mathrm{d}t}.
Consider an infinitesimal rectangle of height \mathrm{d}z and width \mathrm{d}r about a point with z=0 a distance r away from the z-axis. Then the radial components cancel, and the lengthwise components give
\oint\vec{E}\cdot\mathrm{d}\vec\ell=E_z^{(1)}(r)\,\mathrm{d}z-E_z^{(1)}(r+\mathrm{d}r)\,\mathrm{d}z,\qquad<br /> \frac{\mathrm{d}\Phi_B}{\mathrm{d}t}=\frac{\mathrm{d}B_\phi^{(0)}}{\mathrm{d}t}\,\mathrm{d}r\,\mathrm{d}z.
By symmetry, E^{(1)}=E_z^{(1)}. We now have
[E_z^{(1)}(r)-E_z^{(1)}(r+\mathrm{d}r)]\,\mathrm{d}z=-\frac{\mathrm{d}B^{(0)}}{\mathrm{d}t}\,\mathrm{d}r\,\mathrm{d}z<br /> \Rightarrow\frac{E_z^{(1)}(r+\mathrm{d}r)-E_z^{(1)}(r)}{\mathrm{d}r}=\frac{\mathrm{d}E_z^{(1)}}{\mathrm{d}r}=\frac{\mathrm{d}B_\phi^{(0)}}{\mathrm{d}t}.
The right-hand side is given by
\frac{\mathrm{d}B_\phi^{(0)}}{\mathrm{d}t}<br /> =\frac{\mathrm{d}}{\mathrm{d}z}\left[\frac{\mu_0qv}{2\pi}\frac{r}{(r^2+z^2)^\frac{3}{2}}\right]\frac{\mathrm{d}z}{\mathrm{d}t}<br /> =\frac{\mu_0qv}{2\pi}\frac{\mathrm{d}}{\mathrm{d}z}\left[\frac{r}{(r^2+z^2)^\frac{3}{2}}\right]v<br /> =\frac{\mu_0qv^2}{2\pi}\left(-\frac{3}{2}\frac{r(2z)}{(r^2+z^2)^\frac{5}{2}}\right)
\Rightarrow\frac{\mathrm{d}E_z^{(1)}}{\mathrm{d}r}=\frac{\mu_0qv^2}{2\pi}\left(-\frac{3}{2}\frac{r(2z)}{(r^2+z^2)^\frac{5}{2}}\right)<br /> \Rightarrow E_z^{(1)}=\frac{\mu_0qv^2}{2\pi}\left(-\frac{3}{2}\int\frac{2rz\,\mathrm{d}r}{(r^2+z^2)^\frac{5}{2}}\right)
=\frac{\mu_0qv^2}{2\pi}\left(-\frac{3}{2}\int\frac{z\,\mathrm{d}u}{u^\frac{5}{2}}\right)<br /> =\frac{\mu_0qv^2}{2\pi}\left(-\frac{3}{2}\right)\left(-\frac{2}{3}\frac{z}{u^\frac{3}{2}}\right)<br /> =\frac{\mu_0qv^2}{2\pi}\frac{z}{(r^2+z^2)^\frac{3}{2}}.
Note that 1/\sqrt{\mu_0\epsilon_0}=c, so \mu_0=1/(\epsilon_0c^2) and thus
E_z^{(1)}=\frac{v^2}{c^2}\frac{q}{2\pi\epsilon_0}\frac{z}{(r^2+z^2)^\frac{3}{2}}.
We have obtained E_z^{(1)} by noting that the changing electric field \vec{E}^{(0)} induced a magnetic field \vec{B}^{(0)}, which in turn induced an electric field \vec{E}^{(1)}. However, the procedure does not stop there; the electric field \vec{E}^{(1)} induces a magnetic field \vec{B}^{(1)}, which induces an electric field \vec{E}^{(2)}.
To determine E^{(2)}, we replace q with q'=-\frac{v^2}{c^2}q so that E^{(1)}=-(q'/2\pi\epsilon_0)[z/(r^2+z^2)^\frac{3}{2}] (which is exactly the same as E^{(0)} with q' instead of q). Therefore
E_z^{(2)}=-\left(\frac{v^2}{c^2}\right)^2\frac{q}{2\pi\epsilon_0}\frac{z}{(r^2+z^2)^\frac{3}{2}}
It is now clear that the nth-order induced electric field is then
E_z^{(n)}=-\left(-\frac{v^2}{c^2}\right)^n\frac{q}{2\pi\epsilon_0}\frac{z}{(r^2+z^2)^\frac{3}{2}}.
There are infinitely many contributions, which we sum via superposition:
E=-E_z=-\sum_{n=0}^\infty E_z^{(n)}=\sum_{n=0}^\infty\left[\left(-\frac{v^2}{c^2}\right)^n\right]\frac{q}{2\pi\epsilon_0}\frac{z}{(r^2+z^2)^\frac{3}{2}}<br /> =\frac{1}{1+v^2/c^2}\frac{q}{2\pi\epsilon_0}\frac{z}{(r^2+z^2)^\frac{3}{2}}.
Now, I am familiar with the deep connection between relativity and electromagnetism, so when a 1/(1+v^2/c^2) showed up instead of a 1/(1-v^2/c^2), I started wondering if I did something wrong. I have checked my signs and directions of induced fields five times, and they do appear to be correct. I'm starting to think that I don't want to use Coulomb's law for point charges moving at relativistic speeds. But then what do I use?
Homework Statement
Two point charges \pm q move along the z-axis with velocity \pm v. If they are at the origin when t=0, what is the electric field magnitude a distance r from the z-axis?
Homework Equations
The only things I seemed to need for this problem are Coulomb's law, Maxwell's correction to Ampere's law, and Faraday's law:
<br /> \vec{E}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2},\qquad<br /> \mathcal{E}=-\frac{\mathrm{d}\Phi_B}{\mathrm{d}t},\qquad<br /> \oint\vec{B}\cdot\mathrm{d}\vec{s}=\mu_0\epsilon_0\frac{\mathrm{d}\Phi_E}{\mathrm{d}t}.<br />
The Attempt at a Solution
The positions of the point charges are (0,0,\pm z) where z=vt. We wish to determine the \vec{E} and \vec{B} fields a distance r from the axis where z=0, i.e., at (x,y,0) where x^2+y^2=r^2. The first contribution is that of two opposite point charges; then E^{(0)}=-E_z^{(0)}, where
E_z^{(0)}=\frac{q}{4\pi\epsilon_0}\frac{1}{r^2+z^2}\frac{-z}{\sqrt{r^2+z^2}}<br /> +\frac{-q}{4\pi\epsilon_0}\frac{1}{r^2+z^2}\frac{z}{\sqrt{r^2+z^2}}<br /> =-\frac{q}{2\pi\epsilon_0}\frac{z}{(r^2+z^2)^\frac{3}{2}}.
Now consider the electric flux \Phi_E^{(0)} through a circle of radius R at z=0. Then \Phi_E^{(0)}=\int\mathrm{d}\Phi_E^{(0)}, where we integrate over circular rings of thickness \mathrm{d}r:
\mathrm{d}\Phi_E^{(0)}=\vec{E}^{(0)}\cdot\mathrm{d}\vec{A}=E_z^{(0)}\,\mathrm{d}A<br /> =\left(-\frac{q}{2\pi\epsilon_0}\frac{z}{(r^2+z^2)^\frac{3}{2}}\right)(2\pi r\,\mathrm{d}r)<br /> =-\frac{q}{2\epsilon_0}\frac{z(2r\,\mathrm{d}r)}{(r^2+z^2)^\frac{3}{2}}
\Rightarrow\Phi_E^{(0)}=\int\mathrm{d}\Phi_E^{(0)}<br /> =-\frac{qz}{2\epsilon_0}\int_0^R\frac{2r\,\mathrm{d}r}{(r^2+z^2)^\frac{3}{2}}<br /> =-\frac{qz}{2\epsilon_0}\int_{z^2}^{R^2+z^2}u^{-\frac{3}{2}}\,\mathrm{d}u<br /> =-\frac{qz}{2\epsilon_0}\big[-2u^{-\frac{1}{2}}\big]_{z^2}^{R^2+z^2}
where we substituted u=r^2+z^2, \mathrm{d}u=2r\,\mathrm{d}r, u(0)=z^2, and u(R)=R^2+z^2: Thus
\Phi_E^{(0)}(r)=-\frac{qz}{\epsilon_0}\left(\frac{1}{z}-\frac{1}{\sqrt{r^2+z^2}}\right)<br /> =\frac{q}{\epsilon_0}\left(\frac{z}{\sqrt{r^2+z^2}}-1\right).
Recall the Amp\`ere-Maxwell law,
\oint\vec{B}\cdot\mathrm{d}\vec\ell=\mu_0I_\mathrm{enc}+\mu_0\epsilon_0\frac{\mathrm{d}\Phi_E}{\mathrm{d}t}.
By symmetry, we have that \vec{B}^{(0)} is purely azimuthal, so \oint\vec{B}\cdot\mathrm{d}\vec\ell=(2\pi r)B_\phi^{(0)}. Since I_\mathrm{enc}=0 (there is no current enclosed by the circular loop), we have
(2\pi r)B_\phi^{(0)}=0+\mu_0\epsilon_0\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{q}{\epsilon_0}\left(\frac{z}{\sqrt{r^2+z^2}}-1\right)\right]<br /> =\mu_0\epsilon_0\left(\frac{q}{\epsilon_0}\right)\frac{\mathrm{d}}{\mathrm{d}z}\left[\frac{z}{\sqrt{r^2+z^2}}\right]\frac{\mathrm{d}z}{\mathrm{d}t}
=\mu_0q\left(\frac{(1)(\sqrt{r^2+z^2})-(\frac{z}{\sqrt{r^2+z^2}})(z)}{r^2+z^2}\right)v<br /> =\mu_0qv\left(\frac{\frac{r^2+z^2}{\sqrt{r^2+z^2}}-\frac{z^2}{\sqrt{r^2+z^2}}}{r^2+z^2}\right)<br /> =\mu_0qv\frac{r^2}{(r^2+z^2)^\frac{3}{2}}
\Rightarrow B_\phi^{(0)}=\frac{\mu_0qv}{2\pi}\frac{r}{(r^2+z^2)^\frac{3}{2}}.
Now, B^{(0)} itself changes with time, and a changing magnetic field induces an electric field according to Faraday's law:
\oint\vec{E}\cdot\mathrm{d}\vec\ell=-\frac{\mathrm{d}\Phi_B}{\mathrm{d}t}.
Consider an infinitesimal rectangle of height \mathrm{d}z and width \mathrm{d}r about a point with z=0 a distance r away from the z-axis. Then the radial components cancel, and the lengthwise components give
\oint\vec{E}\cdot\mathrm{d}\vec\ell=E_z^{(1)}(r)\,\mathrm{d}z-E_z^{(1)}(r+\mathrm{d}r)\,\mathrm{d}z,\qquad<br /> \frac{\mathrm{d}\Phi_B}{\mathrm{d}t}=\frac{\mathrm{d}B_\phi^{(0)}}{\mathrm{d}t}\,\mathrm{d}r\,\mathrm{d}z.
By symmetry, E^{(1)}=E_z^{(1)}. We now have
[E_z^{(1)}(r)-E_z^{(1)}(r+\mathrm{d}r)]\,\mathrm{d}z=-\frac{\mathrm{d}B^{(0)}}{\mathrm{d}t}\,\mathrm{d}r\,\mathrm{d}z<br /> \Rightarrow\frac{E_z^{(1)}(r+\mathrm{d}r)-E_z^{(1)}(r)}{\mathrm{d}r}=\frac{\mathrm{d}E_z^{(1)}}{\mathrm{d}r}=\frac{\mathrm{d}B_\phi^{(0)}}{\mathrm{d}t}.
The right-hand side is given by
\frac{\mathrm{d}B_\phi^{(0)}}{\mathrm{d}t}<br /> =\frac{\mathrm{d}}{\mathrm{d}z}\left[\frac{\mu_0qv}{2\pi}\frac{r}{(r^2+z^2)^\frac{3}{2}}\right]\frac{\mathrm{d}z}{\mathrm{d}t}<br /> =\frac{\mu_0qv}{2\pi}\frac{\mathrm{d}}{\mathrm{d}z}\left[\frac{r}{(r^2+z^2)^\frac{3}{2}}\right]v<br /> =\frac{\mu_0qv^2}{2\pi}\left(-\frac{3}{2}\frac{r(2z)}{(r^2+z^2)^\frac{5}{2}}\right)
\Rightarrow\frac{\mathrm{d}E_z^{(1)}}{\mathrm{d}r}=\frac{\mu_0qv^2}{2\pi}\left(-\frac{3}{2}\frac{r(2z)}{(r^2+z^2)^\frac{5}{2}}\right)<br /> \Rightarrow E_z^{(1)}=\frac{\mu_0qv^2}{2\pi}\left(-\frac{3}{2}\int\frac{2rz\,\mathrm{d}r}{(r^2+z^2)^\frac{5}{2}}\right)
=\frac{\mu_0qv^2}{2\pi}\left(-\frac{3}{2}\int\frac{z\,\mathrm{d}u}{u^\frac{5}{2}}\right)<br /> =\frac{\mu_0qv^2}{2\pi}\left(-\frac{3}{2}\right)\left(-\frac{2}{3}\frac{z}{u^\frac{3}{2}}\right)<br /> =\frac{\mu_0qv^2}{2\pi}\frac{z}{(r^2+z^2)^\frac{3}{2}}.
Note that 1/\sqrt{\mu_0\epsilon_0}=c, so \mu_0=1/(\epsilon_0c^2) and thus
E_z^{(1)}=\frac{v^2}{c^2}\frac{q}{2\pi\epsilon_0}\frac{z}{(r^2+z^2)^\frac{3}{2}}.
We have obtained E_z^{(1)} by noting that the changing electric field \vec{E}^{(0)} induced a magnetic field \vec{B}^{(0)}, which in turn induced an electric field \vec{E}^{(1)}. However, the procedure does not stop there; the electric field \vec{E}^{(1)} induces a magnetic field \vec{B}^{(1)}, which induces an electric field \vec{E}^{(2)}.
To determine E^{(2)}, we replace q with q'=-\frac{v^2}{c^2}q so that E^{(1)}=-(q'/2\pi\epsilon_0)[z/(r^2+z^2)^\frac{3}{2}] (which is exactly the same as E^{(0)} with q' instead of q). Therefore
E_z^{(2)}=-\left(\frac{v^2}{c^2}\right)^2\frac{q}{2\pi\epsilon_0}\frac{z}{(r^2+z^2)^\frac{3}{2}}
It is now clear that the nth-order induced electric field is then
E_z^{(n)}=-\left(-\frac{v^2}{c^2}\right)^n\frac{q}{2\pi\epsilon_0}\frac{z}{(r^2+z^2)^\frac{3}{2}}.
There are infinitely many contributions, which we sum via superposition:
E=-E_z=-\sum_{n=0}^\infty E_z^{(n)}=\sum_{n=0}^\infty\left[\left(-\frac{v^2}{c^2}\right)^n\right]\frac{q}{2\pi\epsilon_0}\frac{z}{(r^2+z^2)^\frac{3}{2}}<br /> =\frac{1}{1+v^2/c^2}\frac{q}{2\pi\epsilon_0}\frac{z}{(r^2+z^2)^\frac{3}{2}}.
Now, I am familiar with the deep connection between relativity and electromagnetism, so when a 1/(1+v^2/c^2) showed up instead of a 1/(1-v^2/c^2), I started wondering if I did something wrong. I have checked my signs and directions of induced fields five times, and they do appear to be correct. I'm starting to think that I don't want to use Coulomb's law for point charges moving at relativistic speeds. But then what do I use?