Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.108 n when separated by 50.0 cm, center-to-center. the spheres are then connected by a thin conducting wire. when the wire is removed, the spheres repel each other with an electrostatic force of 0.360 n. what were the initial charges on the spheres?

The initial charges of the spheres were q₁=6.7712×10⁻⁶C and q₂=-4.4350×10⁻⁶C.

Explanation:

As the spheres attract each other, the charges of the spheres are opposite.

The atracction force is given by:

F=

where:

K: Coulomb constant

q₁: charge of sphere 1

q₂: charge of sphere 2

r: distance between both charges

The electrostatic atraction force is 0.108 N so:

0.108N=-8.99×10⁹

q₁·q₂=

q₁·q₂=-3.003×10⁻¹² C²

When the wire is connected the charges are equally distributed as the spheres are identical. Hence, the final charge is of each sphere is

The repel force is 0.360 N and it is given by:

F=

F=

Then, we get a secong equation:

(q₁+q₂)²=

(q₁+q₂)=√4.004×10⁻¹¹ C²

q₁+q₂=6.3277×10⁻⁶ C

We solve the equation system:

We replace q₁ in the second equation:

The solutions are:

q₁=6.7712×10⁻⁶C

q₂=-4.4350×10⁻⁶C

answer: electrons