# How to statically charge something

Here is an interesting question. How would you go about statically charging something using batteries( AA or AAA)? I am thinking of utilizing static electricity to lift something really light. Like a piece of paper or a thin piece of cloth etc. Any thoughts?

Simon Bridge
Homework Helper
Depends on the something ... simplest would be to attach one terminal of a battery to the thing and the other to something else. Works well for conductors - see "capacitor".

Probably need more than one battery.

You could use the battery to drive an electric motor that turns a wheel that has a cloth on it (say) and apply the turning cloth to the object you want to charge.

meBigGuy
Gold Member
seems like a good place to start is reading the wikipedia article on static electricity.
http://en.wikipedia.org/wiki/Static_electricity
Maybe you can ask more specific questions after doing that.

Sorry about the late reply I was caught up in some homework.
I understand the concept of charging a conductor but I am looking to charge non- conductors. Any resources you know of that tell which material charges which when rubbed. At the moment I am only looking to charge different types of paper and cloth but I feel that once I am able to perform my experiment successfully with these I might be able to move onto other materials.
Also thanks for the wiki article I will make sure I go over the entire thing. Any other online resources will also be helpful.

Ok so I've been doing some reading and came across the concept of electrostatic induction. What I understand for dielectrics is that if a positively charged object is brought close to them they will tend to that charged object. I looked at the comb and paper experiment. When a comb goes through hair it becomes positively charged. When this comb is brought close the paper the electrons in the paper are attracted to the comb and since the paper is very light the paper moves towards the comb. My concern now is how to charge metal positively so that it takes the place of the comb. Will simply connecting it to the positive terminal of the battery be enough?

Simon Bridge
Homework Helper
For a sufficiently powerful battery - sure.
Common static charges end up in the 1000s of volts.
Humans usually cannot detect less than 3000V.

That wikipedia link you were given has a whole section on energies.

This is why I suggested some means to step-up the voltage.
The suggestion to drive a motor for friction-charging is not a joke - look up "Van der Graaf generator".

What you have is still very open-ended:
You can refine your question by being specific - i.e. how far away do you want to attract the paper from and how big is the bit of paper?

Have a look at:
http://faculty.mint.ua.edu/~pleclair/ph102/Notes/older/electrostatics_only.pdf p72

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I want to attract the paper from about 0.5 to 1 cm, I'll see if I can go even closer. And I would say a standard 8.5" by 11" paper for now. I will expand on different materials after I have the paper down.
I am still reading and I will post more questions as they come up. I have done the calculations from your second link for my purposes. I need about 50 mN of force to lift the paper and to do this I need a charge of 23.57 nC. However I think I might double the number just to account for errors and also so I can move the paper diagonally rather than just straight up. I've read about van de Graaﬀ Generators and I think I can get my hands on the materials needed to make one. One thing that keeps appearing is the importance of humidity. If humidity is a big issue I might need to look at alternative methods. Is it possible to counter the effect of humidity while still staying below the safe limit(350 mJ)?

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Simon Bridge
Homework Helper
I would say a standard 8.5" by 11" paper for now.
Ownch ... that's almost 5 grams.

Lets say you make a two metal plates the same dimensions as the sheet and put them a distance d apart. Bias the plates wrt each other by V, and put the paper on the bottom plate.

Model: the power supply takes charges from the paper and transfers them to the top plate.
We can ask what minimum voltage is needed to lift the sheet.

The force on the sheet is the charge on it times the electric field - to lift the sheet, this force must be at least the mass of the paper times the acceleration due to gravity. I can get the electric field and transferred charge from the capacitor equations.

E=V/d, Q=ε0AV/d, condition: mg=QE

Eliminate Q and solve for voltage:

$$V=\left(\frac{mgd^3}{\epsilon_0^2A^2}\right)^{1/3}$$
Using standard values:
A=0.0616m2, d=0.01m, m=0.0045kg, g=9.8N/kg, ε0=8.9×10-12m-3kg-1s4A2
Code:
octave:67> A,d,m,g,e
A =  0.061600
d =  0.010000
m =  0.0045000
g =  9.8000
e =  8.9000e-12
octave:68> V=((m*g*d^3)/(e^2*A^2))^(1/3)
V =  5.3142e+05

Gives me V=531,000V over a 1cm separation. i.e. you will get sparks and crispy-fried paper.

Unless I messed something up of course ;)

That would make one spectacular show but would not achieve my goal. I can't check your work because I only know the basics about electricity and electronics as I am studying mechanical engineering. What if the separation is reduced to 0; the positively charge plate or any other object(like the comb) touching the piece of paper? Another question, if the previous situation is possible can I generate the necessary voltage by rubbing nylon with vinyl (triboelectric effect)?

dlgoff
Gold Member
Ownch ... that's almost 5 grams.

Lets say you make a two metal plates the same dimensions as the sheet and put them a distance d apart. Bias the plates wrt each other by V, and put the paper on the bottom plate.

Model: the power supply takes charges from the paper and transfers them to the top plate.
We can ask what minimum voltage is needed to lift the sheet.

The force on the sheet is the charge on it times the electric field - to lift the sheet, this force must be at least the mass of the paper times the acceleration due to gravity. I can get the electric field and transferred charge from the capacitor equations.

E=V/d, Q=ε0AV/d, condition: mg=QE

Eliminate Q and solve for voltage:

$$V=\left(\frac{mgd^3}{\epsilon_0^2A^2}\right)^{1/3}$$
Using standard values:
A=0.0616m2, d=0.01m, m=0.0045kg, g=9.8N/kg, ε0=8.9×10-12m-3kg-1s4A2
Code:
octave:67> A,d,m,g,e
A =  0.061600
d =  0.010000
m =  0.0045000
g =  9.8000
e =  8.9000e-12
octave:68> V=((m*g*d^3)/(e^2*A^2))^(1/3)
V =  5.3142e+05

Gives me V=531,000V over a 1cm separation. i.e. you will get sparks and crispy-fried paper.

Unless I messed something up of course ;)

mg=QE=(ε0AV/d)(V/d)=(ε0AV2/d2)

so,

$V=\left(\frac{mgd^2}{\epsilon_0A}\right)^{1/2}$

Unless too many beers messed something up 2.99 kV using the last equation unless I messed up. Seems more reasonable the 531 kV.

Simon Bridge
Homework Helper
Oh I wonder where I got the cube from??
That is a lot better yes - what I get from doing the algebra by cut and paste.
I should have checked :( but that's the beauty of displaying the working.

That's still 16.2μC of charge moved compared with the 23.57 nC from post #7.

What if the separation is reduced to 0;
For the setup described: charge flows from one plate to the other through the paper.
You mean if you have inductive attraction... say we move the paper so it no longer touches the bottom plate, and it is closer to the top than to the bottom?

The electric field is uniform in the setup so the force does not depend on how close the paper is to the plates.

What you want is a non-uniform field - as in the comb example:
... the positively charge plate or any other object(like the comb) touching the piece of paper?
Harder to calculate.

Another question, if the previous situation is possible can I generate the necessary voltage by rubbing nylon with vinyl (triboelectric effect)?
I have no idea.
Of course there's a much more achievable figure now and kilovolts are typical.
You see why the earlier suggestion of using the battery to drive a motor is actually sensible?

You should be able to investigate on your own from here.
Enjoy.

I talked to a professor and what he suggested was to go with the rubbing idea. At this point its mostly experimentation. Thank you for the help, I've learned quite a bit about static electricity. Hopefully I can make this work.

dlgoff
Gold Member
If you get a chance, give us an up-date on your project. Have fun. :)

I'll see what I can do. This is one of the solutions for a class project I am working on. So it is entirely possible I use another method. However this will be written in my report as a potential solution. Testing begins in about 2 weeks so I'll post the results here if I get a chance to fully test this solution out.

Simon Bridge