Answer : [tex]F = 3.5\times10^{3}\ N[/tex]
Explanation :
Given that
Radius of sphere [tex]r = 5.90\times 10^{-15}\ m[/tex]
The distance between the centers of the two spheres is
[tex]r = 2\times 5.90\times 10^{-15}\ m[/tex]
The charge of the sphere [tex]q = 46\times1.6\times10^{-19} C[/tex]
The magnitude of the repulsive force between the charges pushing them a part is
Using coulomb law
[tex]F = \dfrac {kq_{1}q_{2}}{r^{2}}[/tex]
[tex]F = \dfrac{9\times10^{9}\times (46\times1.6\times10^{-19})^{2}C^{2}}{2\times(5.90\times10^{-15})^{2}\ m^{2}}[/tex]
[tex]F = 3501.3\ N[/tex]
[tex]F = 3.5\times10^{3}\ N[/tex]
Hence, this is the required solution.
