I understand how the displacement current work in dielectric such as in parallel plate capacitors but don't know how they would work in a vacuum where there is no dielectric present. Can someone please explain.
Displacement current generation in dielectric and in vacuum? How is it different
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The end is the net result. The blue and red dots in the middle of the second diagram "cancel out", so to speak. This results in the third diagram being an accurate model of the real physical situation in the second diagram.
Think of the device with zero anode current and both plates at 0V and the the device has been allowed to run a while:
If the inequality holds:
\$ \textrm{If }I_{annode} = 0 \textrm{ then } I_{emmission}+I_{return} =0 \$
this only makes sense if you put a direction on the emission and return currents so one opposes the other. This represents the electrons jumping off the hot cathode, hanging around in between the plates (or bouncing off of one and back to the other), but the net charge on the cathode is going to be zero.
In circuit you have to have define a current with a direction and a polarity. So in this case I'm going to define all currents as cathode to anode as positive and use the actual electron flow moving in the direction as positive (which is reversed from circuit current nomenclature). But this means that in this operation \$ I_{emission} \$ and \$I_{return} \$ are have opposing signs on their values
\$I_{emmission} + I_{return} =0 \$
Now lets suppose that we increase the voltage on the anode to a positive value: (Upper diagram). Electrons are heated off the plate (\$ I_{emission} \$) but not all of them come back because some of them hit the anodes metal (its now positive) and flow out of the diode.
\$I_{emmission}+I_{return} = I_{anode} \$
Since there are always electrons bouncing back to the cathode \$ I_{return} \$ is always negative \$ I_{return}<0\$ (in the way I've defined things). With the author including the inequality \$ I_{return}<0\$, it means that the return current is defined as going from cathode to anode and is always negative.
The important takeaway is that with a negative return current you can't have more anode current than emission current. \$I_{emmission}> I_{anode} \$ And that means you'll have to keep the cathode sufficiently hot.
Diagrams from and further reading here
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Best Answer
The displacement current only refers to the current flowing in the wires of the capacitor when it is being charged or discharged. Since the charges cannot jump from one plate to the other, the charges are instead displaced, building up on one plate and depleting the other, hence the term 'displacement current'. This charge displacement results in an increasing electric field in the direction of the current flow. The increasing electric field generates a circulating magnetic field that would have the same magnitude as if there was actually a current carrying wire there. Since electric and magnetic fields can be present in free space, this effect can be seen in a vacuum.