Is there a B field in a charging spherical capacitor?

AI Thread Summary
In a charging spherical capacitor, the growing electric field between the plates indicates a displacement current, which suggests the presence of a magnetic field. It is hypothesized that the magnetic field may be zero due to symmetry, but this could change with the introduction of an input current. If a spherically symmetric input current is assumed, it might cancel the displacement current, while straight wires supplying current would disrupt symmetry and result in a non-zero magnetic field with azimuthal symmetry. The discussion references Ampere's law and its application to both spherical and parallel plate capacitors to analyze the magnetic field. The conversation emphasizes the need for further exploration of the spherical case to clarify the existence of the magnetic field.
scoomer
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Suppose a spherical capacitor is being charged. In this case the E field between the plates is growing with time which implies a displacement current which in turn implies a B field. How would one find this B field if it does exists? I'm guessing the B field is zero because of symmetry. I have searched the web but cannot find a reference - can anyone point me to a reference?
 
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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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Thank you Charles Link for the suggestion. I will give it a try.
 
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