Question

Two charged spheres are 20cm apart and exert an attractive force of 6x10^-9 N on each other. What will be the force of attraction when the spheres are moved to 10cm apart?

Answer 1

The force of attraction between 2 charged spheres can be explained by Coulomb's law,
It states the force of attraction is directly proportional to the magnitudes of the charges and inversely proportional to the square of the distance between the charges.

/$F = \frac{k q_{1} q_{2} }{ r^{2} }$
where F - force of attraction/repulsion
q₁ and q₂ - charges of the 2 spheres 
k - Coulomb's law constant
r - distance between the spheres

In the question given, the charges of the spheres remain constant in both instances, only distance changes. Therefore (kq₁q₂) = c which is a constant 
then F = c / r²
first instance 
6 x 10⁻⁹ N = c/ (20 cm)² ---1)
F = c/(10 cm)² --- 2)
2) / 1)
$\frac{F}{6* 10^{-9} } = \frac{400}{100}$
F = 6 x 10⁻⁹ x 4
F = 2.4 x 10⁻⁸ N