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Sorry if this question doesnt belong here. I have no clue of what I'm about to ask.

I want to know, if possible, how to induce negative/positive charge on objects inside recipients? What I want to achieve is this:

Say plastic recipient A have a liquid and some metal objects inside.

We also have recipient B with the same caracteristics.

Now I want to have positive charge on everything inside recipient A and negative charge on everything inside recipient B without dropping a cable inside each recipient.

I already have positive and negative charge being supplied on different cables. I want to know how to induce the charge wirelessly.

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Easy to answer. In general a recipient has a dielectric which may be sensible to electrostatic discharge with a variable coefficients for, let's say plastic, wood, brick or glass. IF its dielectric is low yes you can easily send an electron wind wich ionizes particles inside the chamber without breaking the containment. At the point the ionization from the material inside the vessel reaches a threshold of charge two things may happen. a change in the state of matter of the material, or a breaking in the dielectric into the conductivity zone both cases depending the ionization threshold you're dealing with.

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  • \$\begingroup\$ Hey, thanks for your answer, didnt understand much of it tho. I have edited my question is your answer still valid? \$\endgroup\$ – Victor Tello Nov 18 '18 at 22:24
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To obtain charge separation, as you do in the plates of a capacitor, you need current flow. You cannot achieve current flow without some sort of conductor and a driving potential.

The conductor does not have to be metal though, you could ionize gas and provide a conductive path that way. That’s what electrostatic (ESD) guns do.

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  • \$\begingroup\$ If by charge separation you mean positive in A and negative on B, I already have that trough a machine. I want to know how to induce those charges wirelessly. \$\endgroup\$ – Victor Tello Nov 19 '18 at 0:53
  • \$\begingroup\$ @VictorTello i believe I was clear in my answer. You need a conductor, even if that conductor is not a “wire”. \$\endgroup\$ – Edgar Brown Nov 19 '18 at 0:54

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