Water bending with comb: but where do the electrons go?

[guywithdoubts]

I suppose the stream isn't being charged!

 

[berkeman]

Um, link please?

 

[davenn]

Summary:: ... and there's no clear discharge, ...

That you can see !

The charged ( energised) comb or other object being brought close to the water stream
will gain/loose electrons via whatever is supporting it, your hand/body, clamp etc

Whether the electrostatic charging of the comb etc is positive or negative depends on what it is made of.
Some materials will gain electrons during the "charging" process ( the frictional rubbing) making it overall negatively charged.
Other materials will have the electrons stripped off it and will leave it positively charged.

Electrons ( charges) will flow into or out of the comb/other object via what is supporting it till a balance of charges is achieved

Either way, the effect on the stream of water will be the same, the effect will lessen as the charge balance in the comb has evened out.

Now I haven't tried this experiment for many years, but from memory ( some one will surely correct me), I recall of the direction of
the bend in the stream of water will be different for a positively or negatively charged comb/object

 

[davenn]

Bend Water with Static Electricity - Scientific American

Electrostatically Bending Water - YouTube

Bending water - static attraction | Experiments | Naked Scientists (thenakedscientists.com)

cheers
Dave

 

[etotheipi]

Now I haven't tried this experiment for many years, but from memory ( some one will surely correct me), I recall of the direction of
the bend in the stream of water will be different for a positively or negatively charged comb/object

The water molecules are polar, so if the comb is positively charged, overall all the water molecules rotate until the side with the $\delta^-$ oxygen is facing toward the comb and the two $\delta^+$ hydrogens are facing away from the comb, i.e. the stream of water essentially becomes an 'electric dipole', and the net force on the stream is toward the comb.

If you switch the polarity of the comb, the water molecules will just overall rotate in the opposite direction [now the electronegative oxygens point away from the comb, and the electropositive hydrogens point toward the comb], the direction of the induced dipole is reversed and the net force on the stream is once again toward the comb.

So I believe whichever polarity you give the comb, i.e. whatever material in the triboelectric series you use to charge the comb, the water is always going to bend toward the comb.

Either way, the effect on the stream of water will be the same, the effect will lessen as the charge balance in the comb has evened out.

I'm not totally sure about this, if the comb is supposed to be a perfect insulator (it will never be, but as an approximation), then whatever net charge you provide it during the charging should remain on the comb and not flow off of the comb whilst it is being supported. The effect should be pretty constant.

But in reality it probably does decrease a little bit, because there are no perfect insulating materials!

 

[sophiecentaur]

The effect on the water is an example of Electric Induction and not charge transfer. It's the same mechanism that allows the charged comb to pick up neutral pieces of dust and paper. The field around the comb polarises the charges in the dust particles (as mentioned above about the water). The 'like' charges end up further from the comb than the unlike charges. Net effect is attraction. If the particles were already charged then they could be either attracted or repelled, depending on the sign.

Check this link about the Electrophorus (brilliant name) for more about electric induction. We used these in Physics in 1961.

The equivalent magnetic effect is how a magnet will always attract a non-magnet by magnetic induction (giving it 'poles) but it will repel OR attract the end of a magnet, depending on the permanent poles of the test magnet.

PS a weakly polarised magnet or weakly charged particle will still be attracted due to a strong external field.

 

[sophiecentaur]

I recall of the direction of
the bend in the stream of water will be different for a positively or negatively charged comb/object

I doubt this was what you saw. If the effect were not due to induction (no charge transfer```) it would last hardly any time at all, bearing in mind that the Capacitance of the comb would be only a few tens of pF.

 

[davenn]

I doubt this was what you saw. If the effect were not due to induction (no charge transfer```) it would last hardly any time at all, bearing in mind that the Capacitance of the comb would be only a few tens of pF.

I really don't understand that response ??
what does induction have to do with it ?

 

[davenn]

The effect on the water is an example of Electric Induction and not charge transfer.

?

I didnt say anything about charge transfer to the water stream

I'm not totally sure about this, if the comb is supposed to be a perfect insulator (it will never be, but as an approximation), then whatever net charge you provide it during the charging should remain on the comb and not flow off of the comb whilst it is being supported. The effect should be pretty constant.

of course the charge will drain away from or into whatever is supporting it

 

[etotheipi]

of course the charge will drain away from or into whatever is supporting it

A perfect insulator cannot be charged or discharged by conduction, it is why we have to resort to the triboelectric effect in order for electron transfer to occur to initially charge the comb.

In reality there are no perfect insulators, but I just tried the effect now with a comb and didn't notice any change in the effect over time. That suggests that the charge I put on the comb, as far as I could tell, remained on the comb during the whole "experiment", and didn't 'drain away'. Otherwise we would not get the effect!

what does induction have to do with it ?

The comb induces effectively a dipole in the stream of water, due to preferential alignment of the water molecules in response to the static electric field produced by the comb! The result that the water attracts the comb no matter the charge of the comb indicates that this is indeed an effect pertaining to induction!

 

[sophiecentaur]

I really don't understand that response ??
what does induction have to do with it ?

That's the proper term for it. (We were taught it in school and its in all those textbooks.) It's not to be confused with an emf induced in a moving conductor. The charge separation is 'induced' by the applied field. Needless to say, in the environment of a laboratory sink, the humidity will cause leakage but the loss of charge is not due to the induced charge displacement on the water stream.

The force, due to the charge imbalance is small, which is why it only works for a narrow free falling stream of water or small items like dust. If you look at that link on the electrophorus you'll see one other example that can give you spark after spark without having any effect on the surface charge of the dielectric one its base. The Whimshurst Machine is another more impressive use of the effect.

 

[tech99]

This experiment is used as a demonstration of the polar nature of water. But does it work with a non polar fluid? The comb can attract bits of paper and they are not polar.

 

[etotheipi]

This experiment is used as a demonstration of the polar nature of water. But does it work with a non polar fluid? The comb can attract bits of paper and they are not polar.

the bits of paper contain atoms, and the individual atoms can be polarised! That is, an external static electric field can induce a polarisation in the electron clouds around each of the the atoms, and once again cause the paper to act like an electric dipole.

 

[tech99]

I suppose that water will have an induced charge greater than, say, paper, as its permittivity is high. The force depends on this charge, according to Coulomb.

 

[sophiecentaur]

I suppose that water will have an induced charge greater than, say, paper, as its permittivity is high. The force depends on this charge, according to Coulomb.

It would also imply that metal dust could work even better. The snag might be that metal. being high density, would need greater force on it to produce the same visible effect.

 

[dRic2]

Is it because the comb polarizes water or because it simply attracts the positive ions that may be present (K+, Na+) in the water and not balanced by their counterparts (negative ions)? Anyone has distilled water to test it ?

 

[sophiecentaur]

Is it because the comb polarizes water or because it simply attracts the positive ions that may be present (K+, Na+) in the water and not balanced by their counterparts (negative ions)? Anyone has distilled water to test it ?

If the water’s in drops then they’re just polarized. If it’s in a continuous stream from the tap, there will be more displacement of the electrons (one way or another). That means more Capacitance, if you look at it that way.
That would be an easy experiment to do with a comb and tap, to see if the displacement changes for stream vs drops.

 

[Orthoceras]

The Kelvin water dropper shows that a positive comb will charge the stream negatively (positive ions a repelled, negative ions are attracted). If the stream forms droplets near the comb, they will be charged negatively. I don't see why the force due to polarization is commonly believed to be stronger than the electrostatic force.

 

[sophiecentaur]

I don't see why the force due to polarization is commonly believed to be stronger than the electrostatic force.

I can't see that, either; is it really "commonly believed"? When there's merely polarisation, there are two different charges, relatively close to each other and the two forces will be in opposition in any field. The net effect can only be non-zero if the field is not uniform (from a point or line charge, for instance).
The main difference between the two mechanisms is that polarisation is easy to arrange whereas putting a net charge on an object is much less easy to achieve.
(They are both "electrostatic forces".)

 

[sophiecentaur]

The Kelvin water dropper shows that a positive comb will charge the stream negatively (positive ions a repelled, negative ions are attracted). If the stream forms droplets near the comb, they will be charged negatively.

The Kelvin Dropper works entirely differently. It only works with a flow of drops; a continuous stream will not work. At no point are the drops charged until they actually fall into a bucket, at which point they share some of the accumulated charge. What happens is that the system starts with a small difference in charge (at noise level) and the drops that fall through, say the negative ring. The drop will become polarised because it is between a + ring and - bucket. Some of the gravitational potential energy of the falling drop does work to move some charge to the - bucket. A drop falling through the + ring will become polarised and some of its energy moves - charge to the - bucket. So each drop that falls through the + ring will cause some extra charge in the + bucket and each drop that falls through the - ring will increase the charge in the - bucket. The charges on the buckets (equal and opposite) will increase until the spark gap fires.

 

[Orthoceras]

The Kelvin water dropper ... only works with a flow of drops; a continuous stream will not work.

Right

At no point are the drops charged until they actually fall into a bucket,

That is your hypothesis. However, wikipedia says "When a drop breaks off the end of the left-hand stream, the drop carries a negative charge with it"

Your hypothesis that the left stream and the right stream are neutral (everywhere uncharged), implies that a kelvin water dropper could be fed by two separate water bottles, insulated from each other. However, in reality the streams must be electrically connected. Above the height where the streams form droplets, the water is a continuous conductor from left to right, and the ions are the charge carriers. The situation is somewhat similar to electrostatic induction in an electroscope, if the table is approached with a positive balloon, the table becomes negatively charged, and the leaves become positively charged. If a leaf would fall off, it would still be positive.

For the same reason droplets in the kelvin water dropper are charged, due to induction by the so called "induction rings".

 

[dRic2]

If the water’s in drops then they’re just polarized. If it’s in a continuous stream from the tap, there will be more displacement of the electrons (one way or another). That means more Capacitance, if you look at it that way.

Sorry I'm not following. I'm referring to the picture from post #3. It's clearly a stream. Why are you talking about drops ?

 

[sophiecentaur]

If the stream forms droplets near the comb, they will be charged negatively.

(This is about the original 'comb' topic) There must be two things at work. For individual drops, any charge that they could get will be small, especially if they form at the tap (far from the comb) and the polarisation within the drop will produce less net force on the way past the comb. A continuous stream goes close to the comb and the 'polarisation' can involve 'the other' charges moving all the way to earth. Also, the experiment would look different if very pure water were used because it would only work by polarisation. This can probably be regarded as a difference in capacitance, which is much higher between the comb and a stream than between the comb and a drop.

I'm referring to the picture from post #3.

The post you are querying refers to the Kelvin Dropper, which (the name suggests it) works on individual drops.

That is your hypothesis.

True; you are right and I now see it was false. The 'polarisation' takes place in the top tank. You could use the mains water supply too and the mean potential would be zero. And then the GPE from the falling charged drops allows a large increase in Electric Potential across the + and - structures because the two rings have increasingly high (magnitude) potentials. I think the dimensions of the apparatus need to be about right for the best results. The field between the dropper and the rings needs to be high so that the charge that's shifted per drop is high enough to carry a useful current into the buckets.

 

[Orthoceras]

Maybe a nice kitchen experiment, using the fact that attraction due to polarisation is necessarily attractive: hold a negatively charged comb next tot the point where the stream forms droplets. The droplets will be positive. Examine if a positively charged nylon object repels the falling droplets. If it does, the repelling force on the net charge of the water is stronger than the attractive force due to polarisation. (Which then might imply that the comb bent the stream due to a net charge of the water, instead of polarisation.)

 

[sophiecentaur]

Maybe a nice kitchen experiment, using the fact that attraction due to polarisation is necessarily attractive: hold a negatively charged comb next tot the point where the stream forms droplets. The droplets will be positive. Examine if a positively charged nylon object repels the falling droplets. If it does, the repelling force on the net charge of the water is stronger than the attractive force due to polarisation. (Which then might imply that the comb bent the stream due to a net charge of the water, instead of polarisation.)

To make that worth while, you would need to suggest where the charges would come from. If the charge on the comb appears to reduce with time, you would need a control experiment with the comb being just left there and, perhaps trying other objects and a range a of separations.

 

[dRic2]

(This is about the original 'comb' topic) There must be two things at work. For individual drops, any charge that they could get will be small, especially if they form at the tap (far from the comb) and the polarisation within the drop will produce less net force on the way past the comb. A continuous stream goes close to the comb and the 'polarisation' can involve 'the other' charges moving all the way to earth.

Ah so you are saying that if there is a difference in the curvature between stream and drops, then I'm sure that the bending is linked to the polarization of water molecules (because polarization is an estensive quantity so it will be less for drops since they have a smaller volume). Right ?

 

[sophiecentaur]

the bending is linked to the polarization of water molecules

by implication, perhaps. But I would say that only applies to pure water. Once there are ions, it would be the drops themselves that would become polarised so it's difficult to be sure what's at work here. But the total charge will depend on the field and the dielectric constant. The charge per unit mass falling past the comb would be difficult to control and compare unless some additive could allow bubbles to form, or not, for the same flow rate.
I was thinking that the total displaced charge could be much higher if the repelled charge could go to earth.
There are perhaps too many variables in a practical experiment to come to any proper conclusion.

 

[dRic2]

There are perhaps too many variables in a practical experiment to come to any proper conclusion.

That was my point. I don't think this a very clear demonstration of the polarization of water molecules. Dissolved ions could play a role.

 

[sophiecentaur]

That was my point. I don't think this a very clear demonstration of the polarization of water molecules. Dissolved ions could play a role.

I've always assumed they would. I doubt that many of the demo's we've seen involve deionised water. I guess a similar experiment could be done with falling ballbearings or metal powder, which would be an extreme case.

 

[dRic2]

okok, I misunderstood your posts then. Sorry

 

[sophiecentaur]

okok, I misunderstood your posts then. Sorry

I'd say that the term polarisation applies whenever charges are displaced from an equilibrium state. My comments are mostly applicable for all situations with that.

There's such a lot of EM that doesn't involve 'actual contact' and flowing charges for energy transfer.

 

[shjacks45]

Summary:: The effect isn't permanent and there's no clear discharge, so where do the electrons go after bending the stream of water?

I suppose the stream isn't being charged!

Usually rub comb with silk or combing dry hair, so that's where the electrons went. Old rubber combs had a positive charge. The reason the water stream bends is due to charges being induced by the comb on the water stream.

 

[shjacks45]

Usually rub comb with silk or combing dry hair, so that's where the electrons went. Old rubber combs had a positive charge. The reason the water stream bends is due to charges being induced by the comb on the water stream.

Duh, (unless it is exceedingly pure) water conducts electricity.

 

[sophiecentaur]

Duh, (unless it is exceedingly pure) water conducts electricity.

Whether the water is pure or has ions in it, charges will still be displaced and a drop or stream will be 'polarised'.

For impure water, the resistivity is also a factor and the dielectric constant is frequency sensitive (not relevant here). The capacitance of the drop is the equivalent of an ideal capacitor with a resistor in parallel

 

[DrDu]

Once I saw a nice experiment where the experimentator used a high voltage power supply and compared the effect with and without an electric shield on the outlet of the water pipe, so that drops could form in field free space. The deflection of the drops was gone once the shield was employed which lead the experimentators to the conclusion that it is due to influence when the water droplets are formed and not due to the dielectric properties of the water drops. With a jet of water instead of droplets, influence should be even more important. Nevertheless, I tried to estimate the relative size of the two effects. When a water droplet of radius r forms at height l and transversal distance d < l over a point charge q, the influenced charge is of order of magnitude $qr^2/l^2$. The relevant maximal force will act on it when it is at the same height as the charge, so that the distance is d. Then the force will be of magnitude $F_i \sim qr^2/(l^2 d^2)$. On the other hand, the dielectric constant of the water droplet (epsilon=81) is very large. If we set it to infinity, the droplet behaves like a metallic sphere whose dipole moment can easily be calculated. The maximal force on the drop due to the charge-induced dipole moment interaction is $F_d \sim q r^3/d^5$. So their ratio is $F_d/F_i \sim r l^2/d^3$. In experiments where $d \approx l$, the dipole force is smaller due to $$r \ll d\approx l$$. This seems to be the case, when a rubber globe is used as charge. In cases, where a charged rubber rod is used, d is considerably smaller than l, and both effects may be important. It would be interesting to do this experiment with an AC high voltage charge source of few kHz, as the deflection of charged drops should vanish in the AC field, while the interaction with the induced dipoles should remain unchanged.