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In summary, if a spherical capacitor is being charged, the E field between the plates is growing with time which implies a displacement current which in turn implies a B field. However, because the B field is zero because of symmetry, it is difficult to find this field.f

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- #2

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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.

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- #3

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

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