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Coco12
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Homework Statement
A single isolated proton is fixed on a surface. Where must another proton be located in relation to the first in order that the electrostatic force of repulsion would just support its weight?
2)Two identical objects have charges from +6.0*10^-6 and -2.0*10^-6, respectively. When placed a distance d apart, their force of attraction is 2,0N. If the objects are touched together, then they moved to a distance of separation of 2d, what will be the new force between them?
Homework Equations
F=kq1q2/d^2
The Attempt at a Solution
1) the electrostatic force would be zero? And then how would u solve for the distance??
2) I thought the inverse square law would mean that the new force would be 1/4 of its original (2.0) however the ans is .17