Charge Equality Between Different Size Pith Balls

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SUMMARY

When identical pith balls of different sizes come into contact, their charges will not be equal upon separation. The charge density becomes uniform, resulting in the smaller ball containing less charge. The potential of both balls equalizes when they are connected, and the ratio of their charges is determined by the ratio of their radii. This principle applies similarly to metal balls, confirming that the charge distribution is dependent on size.

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  • Familiarity with concepts of electric potential and charge density
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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?
 
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You know, I vaguely remember doing this problem way back in freshman physics. Just a guess on my part, but I think that putting the balls in contact would make the charge density uniform. If my assumption is correct, then when you separate the balls, the smaller one would contain less charge.
 
I'm not real sure about pith, but I can answer for metal balls...

Putting the balls in contact forces their potential to be the same. The problem is easier if the balls are separated by a distance large compared to their size. Briefly connecting them with a wire would force the potential at each ball to be the same. In this case the ratio of charges would be equal to the ratio of the radii.
 
mdelisio said:
I'm not real sure about pith, but I can answer for metal balls...

Putting the balls in contact forces their potential to be the same. The problem is easier if the balls are separated by a distance large compared to their size. Briefly connecting them with a wire would force the potential at each ball to be the same. In this case the ratio of charges would be equal to the ratio of the radii.


Are you sure it wouldn't be equal to the ratio of the cube of their radii?

I'm asking because I'm also considering lesser dimensional problems. What if, instead of spheres, we were talking about disks, or straight rods? Well, maybe I should work the problem for myself and get back to you guys.
 

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