Transferring static charge from a Wimshurst machine to a foil pipe

[manicmind]

TL;DR Summary

Tried a static electricity experiment and it didn’t do what I expected. Static charge from Wimshurst machine tried transfer to foil pipe. Foil pipe impact on water none, directly from machine was strong.

Hi everyone.

I am an engineer by trade (don’t hold that against me!) but I was trying an experiment for the latest of my crazy inventions and am missing some key logic in static electricity which I was hoping you could identify.

Equipment
1. Wimshurst machine (like https://en.m.wikipedia.org/wiki/Wimshurst_machine)
2. A plastic rod with copper tape wrapped around it and insulated with electrical tape for all but two electrical wires protruding. Copper about 10cm in length and wrapped diameter 25mm but tape so thin and flexible.

What was I trying to do?
Use the Wimshurst machine to statically charge the copper tape (ie rod) and then demonstrate it held the charge by deflecting water.

I showed that the charged Wimshurst machine deflected water if it was poured near one of the spherical arm ends.
I connected one the wire of the copper tape to the sphere, charged the machine by the same number of spins used before and then disconnected the wire connecting the copper tape. Putting the copper tape (insulated) by the falling water had no effect. Putting the still charged wimshurst machine in the same position deflected the water.

Sorry its likely a very simple point about static charge capacity or something that I am missing but its been driving me crazy as I look online to the point of giving up on it so would really appreciate a pointer :)

Thanks.

 

[kuruman]

Can you post a picture of your contraption? To what did you connect the two protruding wires and why? Have you eliminated sharp edges and the corona discharge they produce?

 

[manicmind]

Can you post a picture of your contraption? To what did you connect the two protruding wires and why? Have you eliminated sharp edges and the corona discharge they produce?

Hi Kuruman. Thanks for the interest.

Simple as that for what I am trying to do atm.
So there are sharp points ie the end of the wires.
The idea was that by charging the Wimshurst machine that the copper tape (under the red electrical tape) would have the same charge as the dome it was connected with.

 

[kuruman]

Hi Kuruman. Thanks for the interest.
View attachment 293116
Simple as that for what I am trying to do atm.
So there are sharp points ie the end of the wires.
The idea was that by charging the Wimshurst machine that the copper tape (under the red electrical tape) would have the same charge as the dome it was connected with.

Thanks for posting the picture. Your arrangement looks as I imagined it from your description. First of all, a clarification of concepts. You do not expect the copper tape to have the same charge as the dome it is connected to. You expect the tape to be at the same high (relative to the ground) electric potential as the dome. Furthermore, you want it to remain at high potential when you move it close to the falling water stream which is presumably at ground potential.

My question to you is why you felt it is necessary to have the second wire coming out of the right side of the tape in the picture? It looks like that second wire is touching the ground. If that is the case, then any charge that accumulates on the tape will flow straight into the ground. It's like trying to fill a glass with a hole in its bottom. Plug the hole and the glass will get full. Here, lose the second wire and the copper tape will keep its charge because you will no longer have a completed circuit in which current flows from high electric potential to low.

 

[manicmind]

Thanks again Kuruman.

I think you just spotted my error. I started with I need charges to influence the electric field and push the water away. Went through a phase of trying to use voltage multipliers to pump electrons and create the charge. I think at that point my mind became clouded with I need voltage (forgetting charge).

So if I understand correctly; I need a high concentration of charged particles (I'm using electrons) to repel the water. This will require a high voltage but only to get the concentration of electrons. Ie if I could wave a wand and hold the electrons in place the voltage would be irrelevant to deflecting the water.

With that line of thinking it is that the Wimshurst machine has layden jars (capacitors) to hold charge whereas the pipe/copper will only hold a fraction of the electrons as it is no different to a wire. I could use two layers of copper separated by electrical tape each connected to a separate wimhurst dome to test this I suspect? Though I imagine the thickness of tape might make it a bad capacitor. Perhaps I could put an actual capacitor on but unsure how weak it would then become across the copper area.

As for the other wire. Its just a bad photo; it was hovering in the air. But i recognise the point about Corona discharge. It is there atm to allow continuity checks across the copper to ensure I didn’t break the circuit somewhere.

I really appreciate the time you have given to responding!

 

[kuruman]

I think you need to understand the basics behind what you are trying to do. The Wimshurst machine separates positive from negative charges and stores the separated charges in the two Leyden jars are essentially two capacitors in parallel. They are cross-connected, meaning that the central conducting rod of one is connected to the cylindrical conductor of the other. Thus, when the Leyden jars are charged, the two central conducting rods have a high electrostatic potential difference between them. One of the rods has a huge excess of electrons (negative charge) and the other a huge deficit of electrons (negative charge). If you leave the rods alone, they will ideally remain that way indefinitely.

Now what happens if you connect another conductor to one of the rods? In your case, the conductor is the red wire, the copper tape and the second wire coming out the other end. The key thing to bear in mind is that

​

The electrostatic potential is the same at any point in a conductor.​

This means that before connecting the two, the rod is at some very high potential and the tape at zero potential. After you connect them, you form a single conductor and you have to have the potential everywhere the same. In other words, you need to raise the electrostatic potential of the tape to match that of the rod. This is accomplished by charge flowing from the rod into the tape until there is no potential difference between the two. This happens very quickly, almost at the speed of light.

So now you have a charged tape that you can wave near the stream of water. Think of it as an extension of the charged rod. Try it.

Notes:
1. As I mentioned earlier, lose the second wire coming out the other end of the tape.
2. The plastic tape around the foil isn't doing anything except prevent you from touching the foil when it's charged. If you want to test for continuity, you can leave a bare spot and use an ohmmeter to do just that.
3. I don't know what the long black thing attached to the left handle is supposed to do. It looks like trouble.
4. A general rule for electrostatics experiment is not to try them when the humidity is high.