Electric force between objects that are in brief contact

AI Thread Summary
When two small charged objects are brought into brief contact, they share their charge before being separated. The initial charges of -3.00 x10^-6 and -1.50 x10^-6 lead to a redistribution of charge, resulting in a new charge on each object. Once separated to a distance of 0.500m, the electric force between them can be calculated using the formula F=KQ1Q2/r^2. The correct electric force after contact is 1.82 x10^1 N, confirming that the objects are likely conductors. Understanding charge redistribution is crucial for accurate calculations in such scenarios.
Jabababa
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Homework Statement



Two small charged objects of the same mass and volume are brought into brief contact with each other. If the initial charges on the two objects are -3.00 x10^-6 and -1.50 x10^-6, what is the electric force between the objects when they are separated to a distance of 0.500m after contact?

The answer is 1.82 x10^1 N

Homework Equations



F=KQ1Q2/r^2


The Attempt at a Solution



I thought it is just plug in the numbers but i get a different answer. So is there anything that goes on when they are in contact?

F= (9.0x10^9)(-3.00x10^-6)(-1.50x10^-6)/(0.500)^2
=0.162N
 
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The important thing here is that they are brought into contact first.
What changes does that cause?
 
im not really sure...

umm their initial distance? Or it will have to be something to do with their charge (transfer of electro maybe)?
 
I think the objects you have are conductors?? What happens in this case?? And anyway are you sure the result is not 1.82x10^(-1)?? (you put 10^1)...
 
I think the objects you have are conductors??
I think you have to assume they're conductors, since it gives the right answer :)

And anyway are you sure the result is not 1.82x10^(-1)??
Yes - this is the answer I got.

Jabababa - When you put two unequally charged conductors in contact, what happens to the charges on them?
 
ap123 said:
I think you have to assume they're conductors, since it gives the right answer :)

Just as I thought :smile:
 
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