Electrostatics Part 1

According to Coulomb's law, the force of attraction or repulsion is directly proportional to the product of distance between them and inversely proportional to the square of the distance between them.$F = K \frac{q_1 q_2}{r^2}$Here we have, $bo q_1$ is the charge of the first mass, $bo q_2$ is the charge of the second mass, $bo r$ is the distance and $K$ is the proportionally constant. Given: Charge of first sphere $q_1 = 2.1\,\mathrm{nc}$and Charge of second sphere $q_2 = -0.1\,\mathrm{NC}$ when they connect with each other the charge becomes, $q = \frac{q_2 - q_1}{2}$$\Rightarrow q = \frac{2.1-0.1}{2}$$\Rightarrow q = 1\,\mathrm{nC}$ σ According to Coulomb's law, we have;$F = \frac{1}{4\pi E_o} \frac{q_1 q_2}{r^2}$Now, on putting the given values we have;$F = 9 \times 10^9 \times \frac{1 \times 10^{-9} \times 1 \times 10^{-9}}{0.5^2}$$F = 9 \times 10^9 \times \frac{10^{-18}}{0.25}$$\Rightarrow F = 36 \times 10^{-9}\,\mathrm{N}$$\;\text{ Hence, }\; F = 36 \times 10^{-9}\,\mathrm{N}$