Question

The electrostatic force on a small sphere of charge $4\mu C$ due to another small sphere of charge $8\mu C$ in air is $0.2N$. What will be the distance between the two spheres?

Answer 1

Hint: The electrostatic force between two small spheres is equivalent to the force between two charged particles kept at the respective centers of the two spheres and having charge same as the total charge on the sphere. Then, we can use the coulomb’s force of attraction formula for the two charged particles.
Formula used:
$F=K\dfrac{{{Q}_{1}}{{Q}_{2}}}{{{r}^{2}}}$

Complete answer:
We will solve this problem by considering the two spheres as two charged particles at the centre of the two spheres and having a charge equal to the total charge on the respective spheres.
The Coulomb’s force $F$ between two charged particles of charges ${{Q}_{1}}$ and ${{Q}_{2}}$ and separated by a distance $r$ is given by
$F=K\dfrac{{{Q}_{1}}{{Q}_{2}}}{{{r}^{2}}}$ --(1)
Where $K=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}$ is the universal electric constant.
Now, let us analyze the question.
The first charged sphere is equivalent to a point charge of charge ${{Q}_{1}}=4\mu C=4\times {{10}^{-6}}C$ $\left( \because 1\mu C={{10}^{-6}}C \right)$
The second charged sphere is equivalent to a point charge of charge ${{Q}_{2}}=8\mu C=8\times {{10}^{-6}}C$ $\left( \because 1\mu C={{10}^{-6}}C \right)$
Let the distance between the two spheres centers (that is the distance between the equivalent charged particles) be $r$.
The coulomb force between the two forces is $F=2N$.
Therefore, using (1), we get
$2=9\times {{10}^{9}}\dfrac{\left( 4\times {{10}^{-6}} \right)\left( 8\times {{10}^{-6}} \right)}{{{r}^{2}}}$
$\therefore {{r}^{2}}=\dfrac{\left( 9\times {{10}^{9}} \right)\left( 4\times {{10}^{-6}} \right)\left( 8\times {{10}^{-6}} \right)}{2}=\dfrac{288\times {{10}^{-3}}}{2}=144\times {{10}^{-3}}=144\times 10\times {{10}^{-4}}$
Square rooting both sides we get
$\therefore \sqrt{{{r}^{2}}}=\sqrt{144\times 10\times {{10}^{-4}}}$
$\therefore r=12\sqrt{10}\times {{10}^{-2}}m=37.95\times {{10}^{-2}}m=37.95cm$ $\left( \because {{10}^{-2}}m=1cm \right)$
Therefore, the distance between the two spheres is $37.95cm$.

Note:
Students must note that the problem is specified for the medium being air. If the question was to be solved for some other medium, then the electric constant would have changed in the coulomb’s force equation. Then the electric constant would have had to be divided by the dielectric constant of the medium to get the electric constant for that medium. If the student would have proceeded with the same formula and values in a different medium, then he or she would have arrived at a completely wrong answer.