Question

Two positive charges distant $0.1 m$ apart, repel each other with a force of $18 N$. If the sum of the two charges is $9 \mu C$ , then calculate their separate values.

Answer 1

Hint: In the given question, the separation between the charges is given. The force between the two forces is also given. We will find the product of the charges. We have the sum of the charges. To get the values of the charges, we will find the difference of the charges. We can easily obtain the values of both the charges.

Complete step-by-step solution:
Given: the separation between two charges is $0.1 m$.
$q + Q = 9 \mu C$
$F = 18 N$
The formula for force between two charges in air is given by:
$F= \dfrac{1}{4 \pi \epsilon_{o}} \dfrac{qQ}{r^{2}}$
Where, q and Q are the charges.
r is the distance between two charges.
$\epsilon_{o}$ is the permittivity in air.
$K = \dfrac{1}{4 \pi \epsilon_{o}} = 9 \times 10^{9} $
We find the product of the charges using the above formula.
$\therefore$ put the values of F, r and K.
$18=9 \times 10^{9} \times \dfrac{qQ}{0.1^{2}}$
$\implies qQ = 20 \times 10^{-12} C $
It is given $q + Q = 9 \times 10^{-6} C$
Using this formula,
$\left(q-Q \right)^{2} = \left(q+Q \right)^{2} – 4Qq$
$\implies \left(q-Q \right)^{2} = \left(9 \times 10^{-6} \right)^{2} – 4 \times 20 \times 10^{-12} $
$\implies \left(q-Q \right)^{2} = 81 \times 10^{-12} – 4 \times 20 \times 10^{-12} $
$\implies q – Q = 1 \times 10^{-6} C$
After solving, we get $q = 5 \mu C\ or\ 4 \mu C$
$Q = 5 \mu C\ or\ 4 \mu C$
Note: Coulomb's law is a trial law of physics that quantifies the amount of force between two stationary charged particles. The force between charged bodies at rest is called Coulomb force. Coulomb’s Law can only be applied in those cases where the inverse square law is obeyed. It is challenging to implement Coulomb’s law where charges are in uncertain shape because we cannot define the distance among the charges. The law cannot be utilized directly to measure the charge on the giant planets.