Question

The electrostatic force on a small sphere of charge $0.4\mu C$ due to another small sphere of charge $- 0.8\mu C$ in air is ${\text{0}}{\text{.2N}}$.
(a) What is the distance between the two spheres?
(b) What is the force on the second sphere due to the first?

Answer 1

This question is based upon coulombs' law according to which like charges repel and opposite charges attract with a force directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.

Formula used:
${\text{F = }}\dfrac{{{q_1}{q_2}}}{{4\pi {\varepsilon _ \circ }{r^2}}}$
where $F$ is electrostatic force, ${q_1}\& {q_2}$ are the magnitudes of charges on both spheres, $r$ is the distance between the charges and ${\varepsilon _ \circ }$ is the permittivity of free space and the value of $\dfrac{1}{{4\pi {\varepsilon _ \circ }}} = 9 \times {10^9}N{m^2}{C^{ - 2}}$.

Complete answer:
Given that,
Electrostatic force on the first small sphere, ${\text{F = 0}}{\text{.2N}}$
Charge on first small sphere, ${q_1} = 0.4\mu C = 0.4 \times {10^{ - 6}}C$
Charge on second small sphere, ${q_2} = - 0.8\mu C = - 0.8 \times {10^{ - 6}}C$
Electrostatic force between the spheres is given by:
${\text{F = }}\dfrac{{{q_1}{q_2}}}{{4\pi {\varepsilon _ \circ }{r^2}}}$
$\therefore {r^2}{\text{ = }}\dfrac{{{q_1}{q_2}}}{{4\pi {\varepsilon _ \circ }F}}$
Putting the values we get:
​$\Rightarrow {r^2}{\text{ = }}\dfrac{{0.4 \times {{10}^{ - 6}} \times 0.8 \times {{10}^{ - 6}} \times 9 \times {{10}^9}}}{{0.2}} \\\ \Rightarrow {r^2}{\text{ = 144}} \times {\text{1}}{{\text{0}}^{ - 4}} \\\ \Rightarrow r = \sqrt {144 \times {{10}^{ - 4}}} \\\ \Rightarrow r = 12 \times {10^{ - 2}} \\\ \therefore r = 0.12 \\\$
(a) Hence the distance between the two spheres is $0.12$ meter.
(b) And the electrostatic force on the second small sphere due to the first is also the same i.e. $0.2{\text{N}}$ attractive, it will be in the opposite direction according to Newton's third law of motion.

Note:
As we know that electrostatic force is given by the formula ${\text{F = }}\dfrac{{{q_1}{q_2}}}{{4\pi {\varepsilon _ \circ }{r^2}}}$ here we substituted the mathematical values and calculated the distance between the two spheres as $0.12$ meters and the electrostatic force on the second sphere due to first will be equal and opposite as Newton's third law states that if we consider two charges then both the charges will exert an equal force on each other regardless of the magnitude of charge.