Question

each experiences an attractive force of 3.7 N. The spheres are then touched together and moved back to a distance r apart. Find the magnitude of the new force on each sphere.

Answer 1

Complete Question;

Two otherwise identical, small conducting spheres have charges +6.0 µC and −2.0 µC. When placed a distance r apart,each experiences an attractive force of 3.7 N. The spheres are then touched together and moved back to a distance r apart. Find the magnitude of the new force on each sphere.

Answer:

F= 1.23 N

Explanation:

The total charge on the spheres, is + 6 μC + (-2 μC) = +4 μC.

The magnitude of  the attractive force between them, according Coulomb's Law (assuming both spheres can be treated as point charges), can be expressed as follows:

 $F =\frac{k*6e-6*2e-6}{r^{2}} = 3.7N$ (1)

Now, if we put in contact both spheres, due to the universal principle of conservation of charge, total charge on both spheres must remain the same.

Due to both spheres are conductors, this excess charge will be evenly spread out on their surfaces in contact each other, which means that each sphere, after being separated, will have a charge of +2 μC.

So, the general expression for the magnitude of the force between them, for these new values of charges, can be expressed as follows:

$F = \frac{k*2e-6*2e-6}{r^{2}}$ (2)

Dividing both sides in (1) and (2), we get:  

$F =\frac{k*6e-6*2e-6}{k*2e-6*2e-6} = \frac{3.7N}{Fx}$

Simplifying common terms, and rearranging, we can solve for Fx, as follows:

Fx = 3.7N / 3 = 1.23 N (the  force is repulsive).