Question

What would be the interaction force between two copper spheres each of mass 1 g, separated by distance 1 m, if the total electronic charge in them differed from the total charge of the nuclei by one percent.?

Answer 1

Total number of atoms in 1 g of copper spheres = $\frac{1}{63.54}\times 6.023\times {10}^{23}$

Total nuclear charge = $\frac{1}{63.54}\times 6.023\times {10}^{23}\times 1.6\times {10}^{-19}\times 29$

Now the charge on the sphere = Total nuclear charge - Total electron charge

(The total electronic charge in them differed from the total charge of the nuclei by one percent.)
=$\frac{1}{63.54}\times 6.023\times {10}^{23}\times 1.6\times {10}^{-19}\times 29\times \frac{1}{100}=4.398\times {10}^{2}C$

Force of interaction between two charged sphere

$F=\frac{1}{4\pi {ϵ}_0}\frac{{(4.398\times {10}^2)}^2}{1^2}N$

$=9\times {10}^9\times 19.348\times {10}^{4}N$

$=1.74\times {10}^{15}N$