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Assuming that we have two plates providing very strong electric fields. Two wires are held close to them to induce charges in the wires. enter image description here


Assuming that an empty capacitor is charged with the charges on the other side of wires and then disconnected. If this process is repeated by empty capacitors multiple times so that potential difference between wires get close to zero, what will happen to charge formation in the wires? (There are images at the bottom of this post of what I imagine)

  1. Do charges close to the electric fields get less dense? Do they move at all (images 2 to 7)?
  2. If the charges close to the electric field do not move, what will happen on wire edges? How can this be connected to the rule of zero electric field in a conductor?
  3. If the charges close to the electric field get less dense, is it possible to shortcut the two wires and recharge the wire edges (images 8-11)?
  4. If it is possible, then will the charge density be less than initial step (not discharged by capacitors) or it'll be the same as initial step?

I did a very simple test but I found this logic to be false, however, my electric field surely was not strong enough to maybe fill the wire edge to make charges stack farther than wire edge. What I inferred was that electric charges stick so hard close to the electric field and there is no charge built up in the wires to be shortcut. Or at least shortcut does not work even when I get sure that there is no more charge to store on an empty capacitor by the wires.

  1. Assuming that the electric field is very strong and wire be narrow, will all charges stack on the edge or they may stack a bit farther to the electric field because of high charge density on wire edge (image 1)? What about when potential of wires are consumed (image 7)?

enter image description here

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  • $\begingroup$ You could attach plates to the ends of the wires with surface areas much larger than that of the wires. Then the charges will spread over them instead of the wires. This might make the question easier to answer. $\endgroup$ Jan 27, 2020 at 16:05
  • $\begingroup$ My intention is to know if it is possible to fill the edge of conductor, like wire, with so many charges that the charges are forced to stay significantly farther to the electric field. A metal plate has so much space for charges that even a huge electric field can't fill it with charges. Now, if two sharp conductors are close to strong opposite electric fileds, if they get shortcut, will charges move because of the high density of charges on the tips? The intention of removing the left side charges is to make sure that they do not involve when there is a shortcut. $\endgroup$ Jan 28, 2020 at 8:22

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enter image description here

While charges will accumulate on the edge near the charged plate, there is no cause for charges to accumulate on the other side of the wire. So I think positive charges will evenly distribute throughout the wire.

I think repeating connecting and disconnecting to an empty capacitor is essentially equivalent to short-circuiting or grounding (in the case both wires are initially 0V respect to the ground) because an empty capacitor will pass currents before it is fully charged and repeating this process will pass all currents as long as there is a voltage difference.

From the above two considerations, I say that the procedure shown in slides 2-7 (connect & disconnect empty capacitor) and 8-9 (short-circuit in the middle of the wire) in your slides are essentially equivalent. So I need to consider only a few states.

The attached figure is the three states I considered.

  1. When there is no wire, the voltage gradually decreases as the distance from a charged plate increases.

  2. When there is a wire, the voltage inside the wire drops sharply at the edge due to the concentration of negative charges at the edge. After that point, the voltage remains slightly positive due to the excess positive charges remained in the wire.

  3. When the wire is grounded, excess positive charges are taken away to the ground, but the negative charges at the edge remain there because they are strongly attracted to a charged plate.

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  • $\begingroup$ I think your argument is more logical than mine. Actually, left side of the wire has less density and lose density not the right side close to the electric field. However, I believe that there would be a very small change close to electric field because the left side charges attract the right side charges, therefore, after grounding, the left side charges should get even closer to the electric field. $\endgroup$ Jan 28, 2020 at 8:13

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