0
$\begingroup$

I have come across many hand dryers that attempt to dry your hands really fast after you wash them. Here are two of them:

XLERATOR http://www.exceldryer.com/

Dyson Airblade http://www.dysonairblade.com/homepage.asp

So I guess I have a ridiculously high standard cause I think even these are too slow. Would it be possible to create a large static electric field to attract the water molecules off of your hand? Can someone offer some ideas that would remove the water off of your hand using some type of electric field generated by charged plates or something?

|cite|improve this question
$\endgroup$
6
$\begingroup$

Water molecules don’t carry an electric charge (and if they do, you don’t want them on your hands…). The dipole moment of water molecules can only be used to rotate them in space, not to move them. Additionally, the forces that apply to water molecules on your hands also apply to water molecules in your hands. So even if you somehow managed to apply a sensible force on these water molecules, this would get rather uncomfortable. The same problem arises if you attempt to heat them up by means of electric resonance (similar to a microwave).

I therefore doubt that it would be possible to build a device based on electric fields rather than moving air, that removes water molecules from the surface of your skin.

However, it might be possible to vaporise the water on your hands using strong infrared lamps. This might lead to other problems, though, such as the focusing of infrared radiation on small areas of the skin by water drops.

|cite|improve this answer
$\endgroup$
  • 2
    $\begingroup$ Just one question... As water is an electric dipole, Of course it could be bent using charged insulators so that the dipoles align in the direction of external field thereby attracting water molecules towards them. What do you say about that? $\endgroup$ – Waffle's Crazy Peanut Oct 11 '12 at 18:46
  • $\begingroup$ Your external field would have to be strong enough to break the water molecule apart (so that the positively charged part and the negatively charged part can move independently). However, if you break the water molecule, the atoms will most likely regain their respective electrons and will be uncharged (again). This will change if the electric field is non-uniform and dynamic, but you would have to be very clever to move water molecules around with such a field. $\endgroup$ – Claudius Oct 11 '12 at 18:51
  • 1
    $\begingroup$ And this exactly builds on the non-uniformity of the electrical field :) The force from the negative partial charges towards the positively-charged rod exceeds the force away from the rod due to positive partial charges since there is no dedicated ‘negative pole’, the field lines of the electrical field are non-uniform (roughly speaking, they point towards the rod from all directions, which helps cancelling out the effect experienced by the positive partial charges). If you took a plate condensator (uniform field), the effect described in the link would not appear. $\endgroup$ – Claudius Oct 11 '12 at 19:05
  • $\begingroup$ Furthermore, the argument that each force on the water on your hand also applies to your hands themselves still stands. I would be rather uncomfortable holding my hand close to such a device. $\endgroup$ – Claudius Oct 11 '12 at 19:16
  • $\begingroup$ Maintaining the field shouldn’t be a problem, putting the negative pole ‘somewhere far away’ should make it possible to maintain it without the need for static charges. $\endgroup$ – Claudius Oct 11 '12 at 19:22
0
$\begingroup$

I think it is possible, as water is attracted towards regions with higher electric field.However, the static charge on the electrodes creating the field would give you shock every time you touch them, so it does not seem to be too comfortable/safe solution.

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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