Coulomb's Law: Does Charge Depend on Size?

daewoo
Messages
25
Reaction score
0

Homework Statement



Suppose you let identical pith balls come in contact to make q1=q2 would the charges be equal if the pith balls were of different size?

Homework Equations



F=k(q1)(q2)/r^2

The Attempt at a Solution



I would think they wouldn't be, because doesn't the charge require the radii?
 
Physics news on Phys.org
Force requires the distance, which the radius plays into, like your equation shows.

Charge is charge, though. It just means the charge density is bigger or smaller.
 
Charge is pretty much locked in place on a pith ball (but not on a conductor).
Very little charge would be transferred between two pith balls in contact.
q1 and q2 could take on any value, independent of the radii.
 
To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
Back
Top