- #1

- 18

- 4

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- I
- Thread starter scoomer
- Start date

- #1

- 18

- 4

- #2

- 4,934

- 2,250

For the case of a parallel plate capacitor, the magnetic field outside of the parallel plates is actually the same as the magnetic field from the wires (assumed long and infinite) that are carrying the current ## I ## that is supplying the current to the capacitor. ## \\ ## I think in the solution to this problem you need to assume some kind of input current to feed the center capacitor. If you assume a spherically symmetric input current, I think this input current might precisely cancel the ## \epsilon_o \frac{d \vec{E}}{dt} ##. The alternative is to have straight wires supplying the current. This will disrupt the spherical symmetry of the problem, and I believe the result will be a non-zero ## \vec{B} ## . It will simply have azimuthal symmetry. ## \\ ## To elaborate on the above for the parallel plate capacitor: ## E=\frac{\sigma}{\epsilon_o} =\frac{Q}{\epsilon_o A} ##. ## \\ ## Ampere's law with just the displacement current term (for a loop outside the capacitor whose plane passes between the capacitor plates) gives: ## \oint \vec{B} \cdot dl=\mu_o \epsilon_o \int \dot{E} \, dA=\mu_o \epsilon_o \dot{E}A ##. ## \\ ## This gives ## \oint \vec{B} \cdot dl=\mu_o \epsilon_o \frac{\dot{Q}}{\epsilon_o A} A=\mu_o \dot{Q}=\mu_o I ##. ## \\ ## We see the ## \mu_o \epsilon_o \dot{E} ## term is exactly what is needed (with displacement current ## I_D=\epsilon_o \dot{E} ##) for Ampere's law to work with the plane passing between the capacitor plates, because the magnetic field ## \vec{B} ## from the wires will be continuous. ## \\ ## For the spherical case, see the 2nd paragragh above.

Last edited:

- #3

- 18

- 4

Thank you Charles Link for the suggestion. I will give it a try.

Share:

- Replies
- 8

- Views
- 1K