Question

Two positive charges separated by a distance of 0.3 m repel each other by a force of 1.6 N. If the sum of two charges is $10\mu C$ then find a separate value of each charge.

Answer 1

Hint: We’ve been given two conditions a) sum of charges is 10 b) charges separated by a distance of 0.3 m repel each other by a force of 1.6 N. Using the Coulomb's law of force between two electrically charged particles i.e. $F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$ we can get the another equation and solving the two would give the answer.

Complete step by step answer:
Coulomb's law of force between two electrically charged particles states that the magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them. Mathematically, we can write
 $F = {k_e}\dfrac{{{q_1}{q_2}}}{{{r^2}}} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$ where ${k_e} = 8.988 \times {10^9}N{m^2}{C^{ - 2}}$ is called the Coulomb's constant and ${\varepsilon _0}$ is the electric constant.

Let the charges given in the question be ${Q_1},{Q_2}$respectively. Let the distance between them be r and the force between them be F. Particles are assumed to be point sized.
According to the question,
r= 0.3m
F= 1.6N
\[
  {Q_1} + {Q_2} = 10\mu C = 10 \times {10^{ - 6}}C = {10^{ - 5}}C \\
   \Rightarrow {Q_1} + {Q_2} = {10^{ - 5}}C\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_1 \\
 \]
Then, using the Coulomb's law, we can write
$
  F = {k_e}\dfrac{{{Q_1}{Q_2}}}{{{r^2}}} \\
  1.6N = 8.988 \times {10^9}N{m^2}{C^{ - 2}} \times \dfrac{{{Q_1}{Q_2}}}{{{{\left( {0.3m} \right)}^2}}} \\
   \Rightarrow {Q_1}{Q_2} = 16 \times {10^{ - 12}}{C^2}\_\_\_\_\_\_\_\_\_\_\_\_\_2 \\
 $
Now we can solve the equations 1 and 2 to find the values of charges as,
$
  {Q_1}\left( {{{10}^{ - 5}} - {Q_1}} \right) = 16 \times {10^{ - 12}} \\
   \Rightarrow {10^{ - 5}}{Q_1} - Q_1^2 = 16 \times {10^{ - 12}} \\
   \Rightarrow Q_1^2 - {10^{ - 5}}{Q_1} + 16 \times {10^{ - 12}} = 0 \\
   \Rightarrow \left( {{Q_1} - 8 \times {{10}^{ - 6}}} \right)\left( {{Q_1} - 2 \times {{10}^{ - 6}}} \right) = 0 \\
   \Rightarrow {Q_1} = 8 \times {10^{ - 6}}C,2 \times {10^{ - 6}}C \\
 $
And on substituting the value of ${Q_1}$ in equation 1 we get ${Q_2} = 2 \times {10^{ - 6}}C,8 \times {10^{ - 6}}C$.

Therefore, the charges are $2\mu C$ and $8\mu C$.

Note: We can also make a quadratic in the second charge and substitute the value then. Also, this question can be solved without changing the units from microcoulomb to coulomb as it gets cancelled out later, but it is still to always solve questions in SI units so as to avoid any error.