How Close Must a Charged Pen Be to Pick Up Aluminium Foil?

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SUMMARY

The discussion focuses on the physics problem of determining how close a charged plastic pen must be to a piece of aluminum foil to pick it up. The pen, charged to 10 nanoC, generates an electric field described by the formula E = 1/(4 pi epsilon) * Q/d^2. The user attempts to model the aluminum foil as a capacitor and applies relevant equations, including the electric field of a disk and force equations, to solve the problem. The user seeks validation of their approach and calculations, indicating a need for clarification on the application of these concepts.

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Homework Statement


A clear plastic pen of length 14.5 cm is rubbed with wool. The charged plastic pen is held over a small uncharged disk shaped piece of aluminium foil of diameter 0.5cm. How close you must move the pen to the foil in order to pick up the foil?



The Attempt at a Solution



Ok I thought I would let the disk be a diople and let the pen be a point charge but nothing worked. Please explain to me how I should do this question.

Thank you very much
 
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You migth want to study the forum rules. You MUST attempt the problem, and tells us what relations and formulas you know.
 
This is what I did?

Ok this is what I did but don't know whether it is right or not:
E = 1/(4 pi epslon) * Q/d^2 is the electric field of the pen

I said that the aluminium foil works like a capacitor therefore:
(qA)/epsilon - 1/(4 pi epslon) * Q/d^2 = 0 also Q=10 nanoC the charge of the pen

then I can figure out the charge of the foil on one side I think(I hope this is right)

Electricfield of disk = ((Q/A)*(s/R))/2epsilon where s=thickness of the disk
also m=dV is the formula for density, but is V= pi r^2 s
then if the above is right we know F = mg
F = ((Q/A)*(s/R))/2epsilon m = D pi r^2 s
therefore:
((Q/A)*(s/R))/2epsilon = D pi r^2 s g

But is this even right, I really need you guys to help me please HELP me !
 

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