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- Thread starter crx
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- #1

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- #2

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The derivative is

So it requires a current

Bob S

- #3

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Q = CV = e

Now insert a dielectric of relative permittivity e and thickness d and area A in the capacitor. Now

Q' = C'V

where C = ee

There is a current in the external circuit that increases the charge Q on the plates to maintain the voltage V on the capacitor.

Bob S

- #4

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Q = CV = e_{0}AV/d.

Now insert a dielectric of relative permittivity e and thickness d and area A in the capacitor. Now

Q' = C'V

where C = ee_{0}A/d

There is a current in the external circuit that increases the charge Q on the plates to maintain the voltage V on the capacitor.

Bob S

Yes,but this is because of the dielectric molecules are shielding and weakening the electric field, so the capacitor will need more charges to reach the power supply voltage, so there will be a current in the external circuit.

What i would like to know is that if a plate capacitor with no dielectric in vacuum (with a pretty large gap ), connected to a AC supply, will have a magnetic field exactly in the area between the plates where there are no moving charges, but only variable electric field...

- #5

- 4,662

- 6

From Maxwells equations,Yes,but this is because of the dielectric molecules are shielding and weakening the electric field, so the capacitor will need more charges to reach the power supply voltage, so there will be a current in the external circuit.

What i would like to know is that if a plate capacitor with no dielectric in vacuum (with a pretty large gap ), connected to a AC supply, will have a magnetic field exactly in the area between the plates where there are no moving charges, but only variable electric field...

Curl H = sigma E + e e

so a varying voltage across the capacitor creates a magnetic field, even when the conductivity sigma = 0.

Bob S

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