- #1
{???}
- 57
- 7
I wrote and solved this problem but am having serious doubts about the answer I obtained.
Two point charges [itex]\pm q[/itex] move along the [itex]z[/itex]-axis with velocity [itex]\pm v[/itex]. If they are at the origin when [itex]t=0[/itex], what is the electric field magnitude a distance [itex]r[/itex] from the [itex]z[/itex]-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:
[tex]
\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2},\qquad
\mathcal{E}=-\frac{\mathrm{d}\Phi_B}{\mathrm{d}t},\qquad
\oint\vec{B}\cdot\mathrm{d}\vec{s}=\mu_0\epsilon_0\frac{\mathrm{d}\Phi_E}{\mathrm{d}t}.
[/tex]
The positions of the point charges are [itex](0,0,\pm z)[/itex] where [itex]z=vt[/itex]. We wish to determine the [itex]\vec{E}[/itex] and [itex]\vec{B}[/itex] fields a distance [itex]r[/itex] from the axis where [itex]z=0[/itex], i.e., at [itex](x,y,0)[/itex] where [itex]x^2+y^2=r^2[/itex]. The first contribution is that of two opposite point charges; then [itex]E^{(0)}=-E_z^{(0)}[/itex], where
[tex]E_z^{(0)}=\frac{q}{4\pi\epsilon_0}\frac{1}{r^2+z^2}\frac{-z}{\sqrt{r^2+z^2}}
+\frac{-q}{4\pi\epsilon_0}\frac{1}{r^2+z^2}\frac{z}{\sqrt{r^2+z^2}}
=-\frac{q}{2\pi\epsilon_0}\frac{z}{(r^2+z^2)^\frac{3}{2}}.[/tex]
Now consider the electric flux [itex]\Phi_E^{(0)}[/itex] through a circle of radius [itex]R[/itex] at [itex]z=0[/itex]. Then [itex]\Phi_E^{(0)}=\int\mathrm{d}\Phi_E^{(0)}[/itex], where we integrate over circular rings of thickness [itex]\mathrm{d}r[/itex]:
[tex]\mathrm{d}\Phi_E^{(0)}=\vec{E}^{(0)}\cdot\mathrm{d}\vec{A}=E_z^{(0)}\,\mathrm{d}A
=\left(-\frac{q}{2\pi\epsilon_0}\frac{z}{(r^2+z^2)^\frac{3}{2}}\right)(2\pi r\,\mathrm{d}r)
=-\frac{q}{2\epsilon_0}\frac{z(2r\,\mathrm{d}r)}{(r^2+z^2)^\frac{3}{2}}[/tex]
[tex]\Rightarrow\Phi_E^{(0)}=\int\mathrm{d}\Phi_E^{(0)}
=-\frac{qz}{2\epsilon_0}\int_0^R\frac{2r\,\mathrm{d}r}{(r^2+z^2)^\frac{3}{2}}
=-\frac{qz}{2\epsilon_0}\int_{z^2}^{R^2+z^2}u^{-\frac{3}{2}}\,\mathrm{d}u
=-\frac{qz}{2\epsilon_0}\big[-2u^{-\frac{1}{2}}\big]_{z^2}^{R^2+z^2}[/tex]
where we substituted [itex]u=r^2+z^2[/itex], [itex]\mathrm{d}u=2r\,\mathrm{d}r[/itex], [itex]u(0)=z^2[/itex], and [itex]u(R)=R^2+z^2[/itex]: Thus
[tex]\Phi_E^{(0)}(r)=-\frac{qz}{\epsilon_0}\left(\frac{1}{z}-\frac{1}{\sqrt{r^2+z^2}}\right)
=\frac{q}{\epsilon_0}\left(\frac{z}{\sqrt{r^2+z^2}}-1\right).[/tex]
Recall the Amp\`ere-Maxwell law,
[tex]\oint\vec{B}\cdot\mathrm{d}\vec\ell=\mu_0I_\mathrm{enc}+\mu_0\epsilon_0\frac{\mathrm{d}\Phi_E}{\mathrm{d}t}.[/tex]
By symmetry, we have that [itex]\vec{B}^{(0)}[/itex] is purely azimuthal, so [itex]\oint\vec{B}\cdot\mathrm{d}\vec\ell=(2\pi r)B_\phi^{(0)}[/itex]. Since [itex]I_\mathrm{enc}=0[/itex] (there is no current enclosed by the circular loop), we have
[tex](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]
=\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}[/tex]
[tex]=\mu_0q\left(\frac{(1)(\sqrt{r^2+z^2})-(\frac{z}{\sqrt{r^2+z^2}})(z)}{r^2+z^2}\right)v
=\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)
=\mu_0qv\frac{r^2}{(r^2+z^2)^\frac{3}{2}}[/tex]
[tex]\Rightarrow B_\phi^{(0)}=\frac{\mu_0qv}{2\pi}\frac{r}{(r^2+z^2)^\frac{3}{2}}.[/tex]
Now, [itex]B^{(0)}[/itex] itself changes with time, and a changing magnetic field induces an electric field according to Faraday's law:
[tex]\oint\vec{E}\cdot\mathrm{d}\vec\ell=-\frac{\mathrm{d}\Phi_B}{\mathrm{d}t}.[/tex]
Consider an infinitesimal rectangle of height [itex]\mathrm{d}z[/itex] and width [itex]\mathrm{d}r[/itex] about a point with [itex]z=0[/itex] a distance [itex]r[/itex] away from the [itex]z[/itex]-axis. Then the radial components cancel, and the lengthwise components give
[tex]\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
\frac{\mathrm{d}\Phi_B}{\mathrm{d}t}=\frac{\mathrm{d}B_\phi^{(0)}}{\mathrm{d}t}\,\mathrm{d}r\,\mathrm{d}z.[/tex]
By symmetry, [itex]E^{(1)}=E_z^{(1)}[/itex]. We now have
[tex][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
\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}.[/tex]
The right-hand side is given by
[tex]\frac{\mathrm{d}B_\phi^{(0)}}{\mathrm{d}t}
=\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}
=\frac{\mu_0qv}{2\pi}\frac{\mathrm{d}}{\mathrm{d}z}\left[\frac{r}{(r^2+z^2)^\frac{3}{2}}\right]v
=\frac{\mu_0qv^2}{2\pi}\left(-\frac{3}{2}\frac{r(2z)}{(r^2+z^2)^\frac{5}{2}}\right)[/tex]
[tex]\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)
\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)[/tex]
[tex]=\frac{\mu_0qv^2}{2\pi}\left(-\frac{3}{2}\int\frac{z\,\mathrm{d}u}{u^\frac{5}{2}}\right)
=\frac{\mu_0qv^2}{2\pi}\left(-\frac{3}{2}\right)\left(-\frac{2}{3}\frac{z}{u^\frac{3}{2}}\right)
=\frac{\mu_0qv^2}{2\pi}\frac{z}{(r^2+z^2)^\frac{3}{2}}.[/tex]
Note that [itex]1/\sqrt{\mu_0\epsilon_0}=c[/itex], so [itex]\mu_0=1/(\epsilon_0c^2)[/itex] and thus
[tex]E_z^{(1)}=\frac{v^2}{c^2}\frac{q}{2\pi\epsilon_0}\frac{z}{(r^2+z^2)^\frac{3}{2}}.[/tex]
We have obtained [itex]E_z^{(1)}[/itex] by noting that the changing electric field [itex]\vec{E}^{(0)}[/itex] induced a magnetic field [itex]\vec{B}^{(0)}[/itex], which in turn induced an electric field [itex]\vec{E}^{(1)}[/itex]. However, the procedure does not stop there; the electric field [itex]\vec{E}^{(1)}[/itex] induces a magnetic field [itex]\vec{B}^{(1)}[/itex], which induces an electric field [itex]\vec{E}^{(2)}[/itex].
To determine [itex]E^{(2)}[/itex], we replace [itex]q[/itex] with [itex]q'=-\frac{v^2}{c^2}q[/itex] so that [itex]E^{(1)}=-(q'/2\pi\epsilon_0)[z/(r^2+z^2)^\frac{3}{2}][/itex] (which is exactly the same as [itex]E^{(0)}[/itex] with [itex]q'[/itex] instead of [itex]q[/itex]). Therefore
[tex]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}}[/tex]
It is now clear that the [itex]n[/itex]th-order induced electric field is then
[tex]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}}.[/tex]
There are infinitely many contributions, which we sum via superposition:
[tex]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}}
=\frac{1}{1+v^2/c^2}\frac{q}{2\pi\epsilon_0}\frac{z}{(r^2+z^2)^\frac{3}{2}}.[/tex]
Now, I am familiar with the deep connection between relativity and electromagnetism, so when a [itex]1/(1+v^2/c^2)[/itex] showed up instead of a [itex]1/(1-v^2/c^2)[/itex], 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 [itex]\pm q[/itex] move along the [itex]z[/itex]-axis with velocity [itex]\pm v[/itex]. If they are at the origin when [itex]t=0[/itex], what is the electric field magnitude a distance [itex]r[/itex] from the [itex]z[/itex]-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:
[tex]
\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2},\qquad
\mathcal{E}=-\frac{\mathrm{d}\Phi_B}{\mathrm{d}t},\qquad
\oint\vec{B}\cdot\mathrm{d}\vec{s}=\mu_0\epsilon_0\frac{\mathrm{d}\Phi_E}{\mathrm{d}t}.
[/tex]
The Attempt at a Solution
The positions of the point charges are [itex](0,0,\pm z)[/itex] where [itex]z=vt[/itex]. We wish to determine the [itex]\vec{E}[/itex] and [itex]\vec{B}[/itex] fields a distance [itex]r[/itex] from the axis where [itex]z=0[/itex], i.e., at [itex](x,y,0)[/itex] where [itex]x^2+y^2=r^2[/itex]. The first contribution is that of two opposite point charges; then [itex]E^{(0)}=-E_z^{(0)}[/itex], where
[tex]E_z^{(0)}=\frac{q}{4\pi\epsilon_0}\frac{1}{r^2+z^2}\frac{-z}{\sqrt{r^2+z^2}}
+\frac{-q}{4\pi\epsilon_0}\frac{1}{r^2+z^2}\frac{z}{\sqrt{r^2+z^2}}
=-\frac{q}{2\pi\epsilon_0}\frac{z}{(r^2+z^2)^\frac{3}{2}}.[/tex]
Now consider the electric flux [itex]\Phi_E^{(0)}[/itex] through a circle of radius [itex]R[/itex] at [itex]z=0[/itex]. Then [itex]\Phi_E^{(0)}=\int\mathrm{d}\Phi_E^{(0)}[/itex], where we integrate over circular rings of thickness [itex]\mathrm{d}r[/itex]:
[tex]\mathrm{d}\Phi_E^{(0)}=\vec{E}^{(0)}\cdot\mathrm{d}\vec{A}=E_z^{(0)}\,\mathrm{d}A
=\left(-\frac{q}{2\pi\epsilon_0}\frac{z}{(r^2+z^2)^\frac{3}{2}}\right)(2\pi r\,\mathrm{d}r)
=-\frac{q}{2\epsilon_0}\frac{z(2r\,\mathrm{d}r)}{(r^2+z^2)^\frac{3}{2}}[/tex]
[tex]\Rightarrow\Phi_E^{(0)}=\int\mathrm{d}\Phi_E^{(0)}
=-\frac{qz}{2\epsilon_0}\int_0^R\frac{2r\,\mathrm{d}r}{(r^2+z^2)^\frac{3}{2}}
=-\frac{qz}{2\epsilon_0}\int_{z^2}^{R^2+z^2}u^{-\frac{3}{2}}\,\mathrm{d}u
=-\frac{qz}{2\epsilon_0}\big[-2u^{-\frac{1}{2}}\big]_{z^2}^{R^2+z^2}[/tex]
where we substituted [itex]u=r^2+z^2[/itex], [itex]\mathrm{d}u=2r\,\mathrm{d}r[/itex], [itex]u(0)=z^2[/itex], and [itex]u(R)=R^2+z^2[/itex]: Thus
[tex]\Phi_E^{(0)}(r)=-\frac{qz}{\epsilon_0}\left(\frac{1}{z}-\frac{1}{\sqrt{r^2+z^2}}\right)
=\frac{q}{\epsilon_0}\left(\frac{z}{\sqrt{r^2+z^2}}-1\right).[/tex]
Recall the Amp\`ere-Maxwell law,
[tex]\oint\vec{B}\cdot\mathrm{d}\vec\ell=\mu_0I_\mathrm{enc}+\mu_0\epsilon_0\frac{\mathrm{d}\Phi_E}{\mathrm{d}t}.[/tex]
By symmetry, we have that [itex]\vec{B}^{(0)}[/itex] is purely azimuthal, so [itex]\oint\vec{B}\cdot\mathrm{d}\vec\ell=(2\pi r)B_\phi^{(0)}[/itex]. Since [itex]I_\mathrm{enc}=0[/itex] (there is no current enclosed by the circular loop), we have
[tex](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]
=\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}[/tex]
[tex]=\mu_0q\left(\frac{(1)(\sqrt{r^2+z^2})-(\frac{z}{\sqrt{r^2+z^2}})(z)}{r^2+z^2}\right)v
=\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)
=\mu_0qv\frac{r^2}{(r^2+z^2)^\frac{3}{2}}[/tex]
[tex]\Rightarrow B_\phi^{(0)}=\frac{\mu_0qv}{2\pi}\frac{r}{(r^2+z^2)^\frac{3}{2}}.[/tex]
Now, [itex]B^{(0)}[/itex] itself changes with time, and a changing magnetic field induces an electric field according to Faraday's law:
[tex]\oint\vec{E}\cdot\mathrm{d}\vec\ell=-\frac{\mathrm{d}\Phi_B}{\mathrm{d}t}.[/tex]
Consider an infinitesimal rectangle of height [itex]\mathrm{d}z[/itex] and width [itex]\mathrm{d}r[/itex] about a point with [itex]z=0[/itex] a distance [itex]r[/itex] away from the [itex]z[/itex]-axis. Then the radial components cancel, and the lengthwise components give
[tex]\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
\frac{\mathrm{d}\Phi_B}{\mathrm{d}t}=\frac{\mathrm{d}B_\phi^{(0)}}{\mathrm{d}t}\,\mathrm{d}r\,\mathrm{d}z.[/tex]
By symmetry, [itex]E^{(1)}=E_z^{(1)}[/itex]. We now have
[tex][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
\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}.[/tex]
The right-hand side is given by
[tex]\frac{\mathrm{d}B_\phi^{(0)}}{\mathrm{d}t}
=\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}
=\frac{\mu_0qv}{2\pi}\frac{\mathrm{d}}{\mathrm{d}z}\left[\frac{r}{(r^2+z^2)^\frac{3}{2}}\right]v
=\frac{\mu_0qv^2}{2\pi}\left(-\frac{3}{2}\frac{r(2z)}{(r^2+z^2)^\frac{5}{2}}\right)[/tex]
[tex]\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)
\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)[/tex]
[tex]=\frac{\mu_0qv^2}{2\pi}\left(-\frac{3}{2}\int\frac{z\,\mathrm{d}u}{u^\frac{5}{2}}\right)
=\frac{\mu_0qv^2}{2\pi}\left(-\frac{3}{2}\right)\left(-\frac{2}{3}\frac{z}{u^\frac{3}{2}}\right)
=\frac{\mu_0qv^2}{2\pi}\frac{z}{(r^2+z^2)^\frac{3}{2}}.[/tex]
Note that [itex]1/\sqrt{\mu_0\epsilon_0}=c[/itex], so [itex]\mu_0=1/(\epsilon_0c^2)[/itex] and thus
[tex]E_z^{(1)}=\frac{v^2}{c^2}\frac{q}{2\pi\epsilon_0}\frac{z}{(r^2+z^2)^\frac{3}{2}}.[/tex]
We have obtained [itex]E_z^{(1)}[/itex] by noting that the changing electric field [itex]\vec{E}^{(0)}[/itex] induced a magnetic field [itex]\vec{B}^{(0)}[/itex], which in turn induced an electric field [itex]\vec{E}^{(1)}[/itex]. However, the procedure does not stop there; the electric field [itex]\vec{E}^{(1)}[/itex] induces a magnetic field [itex]\vec{B}^{(1)}[/itex], which induces an electric field [itex]\vec{E}^{(2)}[/itex].
To determine [itex]E^{(2)}[/itex], we replace [itex]q[/itex] with [itex]q'=-\frac{v^2}{c^2}q[/itex] so that [itex]E^{(1)}=-(q'/2\pi\epsilon_0)[z/(r^2+z^2)^\frac{3}{2}][/itex] (which is exactly the same as [itex]E^{(0)}[/itex] with [itex]q'[/itex] instead of [itex]q[/itex]). Therefore
[tex]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}}[/tex]
It is now clear that the [itex]n[/itex]th-order induced electric field is then
[tex]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}}.[/tex]
There are infinitely many contributions, which we sum via superposition:
[tex]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}}
=\frac{1}{1+v^2/c^2}\frac{q}{2\pi\epsilon_0}\frac{z}{(r^2+z^2)^\frac{3}{2}}.[/tex]
Now, I am familiar with the deep connection between relativity and electromagnetism, so when a [itex]1/(1+v^2/c^2)[/itex] showed up instead of a [itex]1/(1-v^2/c^2)[/itex], 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?