Suppose I take a vacuum tube and accelerate electrons in it by electric fields then collide it on a copper plate.

Will the electrons then go inside it and generate an electric current by putting a positively charged panel on the opposite end to create a potential difference?

If yes how can I extract the electrons from the positive plate back into the vacuum tube?

Thanks! Also you can see a diagram I made for what I am trying to say,

enter image description here

  • $\begingroup$ Once you have passed some current through the tube, a corresponding number of electrons is on the anode, and there is no reason/need to extract and send back. No charge on anode. You should revert the current and heat the anode to go back with difficulties (improper anode material) $\endgroup$ – jaromrax Mar 24 '15 at 8:16

It will help if you study this diagram of what a vacuum tube is


If a cathode is heated, it is found that electrons from the cathode become increasingly active and as the temperature increases they can actually leave the cathode and enter the surrounding space.

When an electron leaves the cathode it leaves behind a positive charge, equal but opposite to that of the electron. In fact there are many millions of electrons leaving the cathode. As unlike charges attract, this means that there is a force pulling the electrons back to the cathode. Unless there are any further influences the electrons would stay in the vicinity of the cathode, leaving the cathode as a result of the energy given to them as a result of the temperature, but being pulled back by the positive charge on the cathode.

The electrons are produce by heating a filament. Electrons are bound in atoms and even facing a vacuum stay there, unless kicked out, which is what heating a filament does. If you place the tube in an electric circuit, if the filament is not heated, no current will flow through because there is nothing to conduct electricity in the vacuum. Extracting electrons from the filament allows a current to form and close the circuit when they hit the other plate. In this form to get electrons again out of the second plate one would have to somehow heat the second plate.

Better study a bit the link given above.

  • $\begingroup$ This is not what i am saying did u see the diagram link? Please see it. The wires are connected outside the tube for a load. $\endgroup$ – Bhavesh Mar 24 '15 at 5:57
  • $\begingroup$ Yes, I saw your diagram, if you have the whole outside the vacuum electrons will be absorbed by the glass interface, and if any remain, by the air. It is only in vacuum that one can have an electron cloud ejected through heating to turn into a beam. If you are transferring them with wires your diagram is similar as the one i posted above , if the two plates above are not a power supply. In that case no current will flow because there will be no positive attracting the electrons to the anode. $\endgroup$ – anna v Mar 24 '15 at 6:15
  • $\begingroup$ Heating the filament will keep ejecting electrons and there may be some transient currents outside , charging a capacitor, which your two plates are. But the electron cloud will with equal probability be attracted back to where it came, since the filament is turning positive by the electrons leaving. $\endgroup$ – anna v Mar 24 '15 at 6:28
  • $\begingroup$ But if i positively charge the anode then will there be a current? $\endgroup$ – Bhavesh Mar 24 '15 at 8:20
  • $\begingroup$ Yes, but it will be like charging a capacitor , because the dc electric circuit will not close, it will reach a saturation. $\endgroup$ – anna v Mar 24 '15 at 9:10

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