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Recently, I did an experiment with salt and electric current. I set up an open electric current in a used battery holder where the missing part of the circuit is two metal plates. I then fill this open space with table salt. Then, I close the circuit by putting a 3V battery inside the battery holder. Judging by the rise in the circuit's temperature, I think there was a current flowing through the circuit. I then turn the whole battery holder upside down and the salt stayed there instead of falling down. Once I removed the battery, the salt fell down onto the table. I repeated the experiment 3 times with the same result.

I don't know what is the force that was keeping the salt in place? Can somebody please help me. Thank you. enter image description here

P/s: Sorry for my terrible drawing, it's my first time using a website to sketch a circuit (or anything).

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    $\begingroup$ Would you mind adding a sketch of the setup? $\endgroup$
    – Steeven
    Commented Oct 21, 2019 at 12:12
  • $\begingroup$ I suspect that your table salt isn't perfectly dry. $\endgroup$
    – PM 2Ring
    Commented Oct 21, 2019 at 12:28
  • $\begingroup$ I suspected salt is not dry as well but what I don't understand is what force kept them from falling off from the circuit when electric current went through them. $\endgroup$
    – random
    Commented Oct 21, 2019 at 12:31
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    $\begingroup$ Very interesting experiment. You should repeat it with dry salt (2 h @ 150 C). Moist salt conduct electricity, dry does not. +1 from me. $\endgroup$
    – Gert
    Commented Oct 21, 2019 at 14:23
  • $\begingroup$ How big is the salt chamber? $\endgroup$
    – Keith McClary
    Commented Oct 22, 2019 at 3:41

1 Answer 1

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It sounds like the electric field (perhaps added by the current flow) turned the salt crystals into electric dipoles, which were then attracted to each other and to the charged plates.

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  • $\begingroup$ Can you be a bit more detail, please? $\endgroup$
    – random
    Commented Oct 21, 2019 at 21:57
  • $\begingroup$ This might work with much higher voltages. $\endgroup$
    – Keith McClary
    Commented Oct 22, 2019 at 4:18

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