Capacitor and electrostatic field

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Main Question or Discussion Point

Is there any electrostatic field around the leads of a charged capacitor? Let's take just the negative one. If I take a piece of tissue and put close to that terminal it will attract or repel the paper? And if not, why?
 

Answers and Replies

kuruman
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Sure. Suppose the capacitor is connected to a battery. Don't forget that the plates and the connecting conducting wires are equipotentials all the way to the battery terminals. There will be a positive and negative charge distribution over the entire two equipotential surfaces. Thus, there will be electric field lines and hence electric forces. Of course, most of the charge will be distributed on the inner surface of the capacitor plates which means that the electric field is by far strongest between the plates. Whether the leads will be able to attract or repel a piece of paper depends on the paper, if it's charged and with what kind of charge.
 
berkeman
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If I take a piece of tissue and put close to that terminal it will attract or repel the paper?
Just to add to the good reply by @kuruman -- The force would depend on the amount of charge on the tissue and the capacitance between the tissue and the lead. That capacitance depends on the areas of the tissue and the lead, and the separation between them.

So unless there is a lot of charge on the tissue, the force on it will be very small. The area of the lead is small, so the capacitance with the tissue is also very small.

To get good electrostatic force, you want to have a lot of capacitance between the two charged objects (a lot of area and close spacing). Hope that helps. :smile:
 
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Yes, this help. I think I have to charge that capacitor to high voltage to see some effect, that's why I cannot do an experiment. But how can I calculate something? Voltage vs. field strength at 1mm from the lead for example. The capacity of capacitor is also important? And if I connect the lead to a larger metal plate, aluminum foil for example? It can increase the effect?
 
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Of course, most of the charge will be distributed on the inner surface of the capacitor plates which means that the electric field is by far strongest between the plates.
Are you sure the inner field doesn't hold down all of e- and p+ force?
 
kuruman
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Are you sure the inner field doesn't hold down all of e- and p+ force?
I don't understand what you mean by this.
 
berkeman
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Are you sure the inner field doesn't hold down all of e- and p+ force?
The two capacitor plates have the bulk of the electric field energy between them, yes, but the two leads outside the capacitor still have a small parasitic capacitance between them, and they have the same voltage Vc between them that the plates inside the capacitor have.
 
berkeman
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But how can I calculate something? Voltage vs. field strength at 1mm from the lead for example.
If you have 1V between two objects that are 1mm apart, the electric field (E-field) between them will be 1V/0.001m = 1000 V/m.
 
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I don't understand what you mean by this.
I mean: why field strength stronger between the plates? Electron and proton are point charges, field is like a bubble around them. Half of the strength between the plates is on each terminal, isn't it?
 
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If you have 1V between two objects that are 1mm apart, the electric field (E-field) between them will be 1V/0.001m = 1000 V/m.
And how much is that? Can you compare with something?
 
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What is the spacing of the plates compared to the spacing of the leads? Are you familiar with the formula for capacitance in terms of plate area and spacing?
Uh, OK, I understand. But what if I connect that lead to a larger plate, an aluminum foil for example?
 
kuruman
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Uh, OK, I understand. But what if I connect that lead to a larger plate, an aluminum foil for example?
Then you will have a capacitor with a new capacitance that will depend on the size, shape and position of the foil. Capacitance is a quantity that depends only on the geometry of the two conductors.
 
sophiecentaur
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Are you sure the inner field doesn't hold down all of e- and p+ force?
The Potential Difference between the terminals and between the plates is the same. The FIELD (Volts per metre) is a lot higher between the plates because the spacing is very small.
I think what you are getting at here is that the charges on the leads are less than the charges on the plates, which is fair enough. One way to look at this is to consider the capacitor as two separate capacitors in parallel. The big one (the plates) is a few μF and it is in parallel with the small one (the leads) of about 1pF. The charges are shared accordingly.
 

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