Question

Two identical particles are charged and held at a distance of $1m$ from each other. They are found to be attracting each other with a force of $0.027N$. Now, they are connected by a conducting wire so that charge flows between them. When the charge flow stops, they are found to be repelling each other with a force of $0.009N$. Find the initial charge on each particle:
$\begin{aligned} & A){{q}_{1}}=\pm 3\mu C;{{q}_{2}}=\mp 1\mu C \\\ & B){{q}_{1}}=\mp 3\mu C;{{q}_{2}}=\pm 2\mu C \\\ & C){{q}_{1}}=\pm 5\mu C;{{q}_{2}}=\mp 2\mu C \\\ & D)\text{ }None\text{ }of\text{ }these \\\ \end{aligned}$

Answer 1

Hint: We know that the force between two charges is given by Coulomb's law. We can write the equations for initial and final cases using Coulomb's law. As the charge flows between the two charges, the final charge is average of the initial charge.

Complete step-by-step answer:
Let us consider that the charges on the two particles are ${{Q}_{1}}$ and ${{Q}_{2}}$ respectively. If the force acting between these charges are taken to be $F$ and the distance between the two particles is taken to be $d$, then by Coulomb’s law, we can write,
$F=\dfrac{k{{Q}_{1}}{{Q}_{2}}}{{{d}^{2}}}$
In the question, it has been mentioned that initially, the force between the particles is $0.027N$. Therefore, we can write,
$\begin{aligned} & 0.027=\dfrac{k{{Q}_{1}}{{Q}_{2}}}{{{(1)}^{2}}} \\\ & \Rightarrow k{{Q}_{1}}{{Q}_{2}}=0.027\to \text{ equation(1)} \\\ \end{aligned}$
Again, when the two particles are connected by a conducting wire, the charges on the two particles will become equal. If $q$is taken to be the new charge of the particles, then,
$q=\dfrac{{{Q}_{1}}+{{Q}_{2}}}{2}$
Hence the Coulomb force $F'$will be
$F'=\dfrac{k{{(\dfrac{{{Q}_{1}}+{{Q}_{2}}}{2})}^{2}}}{{{d}^{2}}}$
So, according to the question, we can write,
$\begin{aligned} & 0.009=\dfrac{k{{(\dfrac{{{Q}_{1}}+{{Q}_{2}}}{2})}^{2}}}{{{(1)}^{2}}} \\\ & \Rightarrow k{{(\dfrac{{{Q}_{1}}+{{Q}_{2}}}{2})}^{2}}=0.009\to \text{ equation(2)} \\\ \end{aligned}$
After simplifying the above equations, and substituting the given values, we find that,
${{Q}_{1}}{{Q}_{2}}=-3\times {{10}^{-12}}\text{( - because the charges are opposite)}$
$\begin{aligned} & {{({{Q}_{1}}+{{Q}_{2}})}^{2}}=4\times {{10}^{-12}} \\\ & \Rightarrow ({{Q}_{1}}+{{Q}_{2}})=\pm 2\times {{10}^{-6}} \\\ \end{aligned}$
So, from the above equations, we can find that, the answer to the given question is
$g$ ${{Q}_{1}}=\pm 3\mu C\text{ }and\text{ }{{Q}_{2}}=\mp 1\mu C\text{ }$
So, as per the question, if ${{Q}_{1}}={{q}_{1}}\text{and }{{Q}_{2}}={{q}_{2}}$, then the answer is option
$A){{q}_{1}}=\pm 3\mu C;{{q}_{2}}=\mp 1\mu C$

Note: Students must be careful while using the Coulomb’s law, because many times, there is a mistake in taking the denominator. Students need to remember that according to this law, the force is always inversely proportional to the ‘square of the distance’ between the charges and not the ‘distance’.