3
$\begingroup$

if you take a volume of saltwater, and charge it to a high voltage, will the water molecules repel each other and climb the walls of the container slightly, or does the fact that water is polar prevent that? Are there any conductive fluids that will repel themselves and climb the container walls when charged?

Edit: I guess this would be a glass container with water on the inside and a metal plating around the outside. The water would be charged to a high voltage relative to the plating.

$\endgroup$
5
  • $\begingroup$ What do you mean "charge"? Water by itself is not a capacitor, you can't "charge" it. $\endgroup$
    – ACuriousMind
    Commented Sep 17, 2015 at 17:10
  • $\begingroup$ High voltage relative to what? The walls of the container? Voltage is always defined in a relative way, so you need to say relative to what. $\endgroup$ Commented Sep 17, 2015 at 17:16
  • $\begingroup$ As it is a conductor in equilibrium the surplus charges will be located close to the surface and therefore not pushing the bulk in any direction but lower the uniform pressure, if the repelling force is stronger than the cohesion forces (i.e. the negative electric surface tension stronger than the intrinsic surface tension) I guess a charged droplet would split apart, I guess you could estimate the sufficient charge for this quite easily (and from this the sufficient voltage (as a spherical droplet is a capacitor), which I guess will be ridiculously high). $\endgroup$ Commented Sep 17, 2015 at 17:43
  • $\begingroup$ In the scenario in the edit (v2) the fluid would rise along the walls, even if you only charged the plate. Do you know the experiment, where a dielectric fluid is sucked into a capacitor against gravity? (But there the reduction of the electric field energy due to the polarization of the dielectric is the driving force). $\endgroup$ Commented Sep 17, 2015 at 18:46
  • $\begingroup$ Thanks Sebastian! I'll see if I can find that experiment! $\endgroup$ Commented Sep 17, 2015 at 19:13

1 Answer 1

2
$\begingroup$

There are a lot of neat effects when you use high voltages with water. See this video for an example. In this case, the voltage is applied across two different beakers. Watch carefully and you can see that the water does rise up the slope of the beaker with a high enough voltage. One important detail is that this water has to be very pure; ionic impurities such as in salt water would cancel the effect. This is because the ions act as mobile charges, making the water conductive. What would happen is that the mobile charges would crowd around the oppositely charged electrodes, creating an opposite voltage, and cancelling out any effect. This is the same theory as to why you get current when you put a voltage across a metal conductor.

Why does this effect occur? Across large voltages, it's believed that a larger portion of the water ionizes itself into OH- and H+. These are the charges that are providing the electrostatic attraction when the water connects. Why exactly it forms a water bridge is still somewhat mysterious. Evidently, the charges aren't completely mobile. Otherwise, after the water bridge connects once, the charges would transfer and the water bridge would break. This video provides a possible explanation to what happens.

Your question also touches on an idea called Debye screening. When you have ions dissolved in water, the surrounding polar molecules tends to effectively reduce the electric field around the ions. This image somewhat illustrates this idea.

electric double layer

If your fluid had a net charge, rather than climbing up the walls of the container, the charged molecules would just tend to orient themselves near the outer surfaces of the fluid. The repulsive force wouldn't be strong enough to defy gravity. There are however, surface tension effects that make water drops on a surface form into beads rather than sitting flat on a surface, or climb up glass surfaces. This is an entirely different effect however, and is more an interaction between the polar water and a surface.capillary action

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.