Magnetic field and displacement current

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A parallel plate capacitor in a vacuum can indeed generate a magnetic field due to a variable electric field, even without moving electrical charges. The discussion highlights that when a voltage is applied, the relationship between charge (Q), capacitance (C), and voltage (V) leads to a displacement current that can create a magnetic field. Inserting a dielectric affects the electric field and requires additional charge to maintain voltage, resulting in a current in the external circuit. Theoretical backing from Maxwell's equations supports the idea that a changing electric field is sufficient to produce a magnetic field. Thus, a capacitor connected to an AC supply can exhibit a magnetic field in the absence of moving charges.
crx
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Is there really a magnetic field around a capacitors (parallel plate) in vacuum dielectric fed by a variable voltage? Is there an experiment that can prove that we don't need actually a moving electrical charge to create a magnetic field, but a variable electric field in vacuum its enough?
 
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The relation between charge Q and voltage V on a capacitor C is
Q = CV
The derivative is
dQ/dt = I = C dV/dt

So it requires a current I to produce a variable voltage dV/dt on a capacitor. So there is a moving electrical charge around the capacitor.
Bob S
 
Here is an example of the displacement current creating a real current. Take an air capacitor C with gap d and area A. Put and maintain a voltage V on it. So the charge
Q = CV = e0AV/d.
Now insert a dielectric of relative permittivity e and thickness d and area A in the capacitor. Now
Q' = C'V
where C = ee0A/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
 
Bob S said:
Here is an example of the displacement current creating a real current. Take an air capacitor C with gap d and area A. Put and maintain a voltage V on it. So the charge
Q = CV = e0AV/d.
Now insert a dielectric of relative permittivity e and thickness d and area A in the capacitor. Now
Q' = C'V
where C = ee0A/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...
 
crx said:
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...
From Maxwells equations,
Curl H = sigma E + e e0 dE/dt = e e0/d dV/dt,
so a varying voltage across the capacitor creates a magnetic field, even when the conductivity sigma = 0.
Bob S
 
I'm working through something and want to make sure I understand the physics. In a system with three wave components at 120° phase separation, the total energy calculation depends on how we treat them: If coherent (add amplitudes first, then square): E = (A₁ + A₂ + A₃)² = 0 If independent (square each, then add): E = A₁² + A₂² + A₃² = 3/2 = constant In three-phase electrical systems, we treat the phases as independent — total power is sum of individual powers. In light interference...

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