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 :

• A. $q_1=\pm 3\mu C;q_2=\mp 1\mu C$
• B. $q_1=\mp 3\mu C;q_2=\mp 2\mu C$
• C. $q_1=\pm 5\mu C;q_2=\mp 2\mu C$
• D. None of these

Answer 1

Answer: (A)

$q_1=\pm 3\mu C;q_2=\mp 1\mu C$

Let the initial charges are $q_1$ and $q_2$.

In first case , $F=\frac{k{q}_1{q}_2}{r^2}=-0.027$ (minus sign due to attraction force)

or $\frac{(9\times {10}^9)q_1{q}_2}{1^2}=-0.027\Rightarrow q_1{q}_2=-3\times {10}^{-12}..\left(1\right)$

When a conducting wire are connected them, the charge on each particle will be half of the total charge i.e $(q_1+q_2)/2$.

For repulsion, $F=\frac{(9\times {10}^9(\frac{q_1+q_2}{2}{)}^2}{1^2}=0.009$

or $(q_1+q_2{)}^2=4\times {10}^{-12}$

or $q_1+q_2=\pm 2\times {10}^{-6}...\left(2\right)$

A) From $q_1{q}_2=-3\times {10}^{-12}$ and $q_1+q_2=+2\times {10}^{-6}$, we get after solving $q_1=3\mu C$ and $q_2=-1\mu C$

B) From $q_1{q}_2=-3\times {10}^{-12}$ and $q_1+q_2=-2\times {10}^{-6}$, we get after solving $q_1=-3\mu C$ and $q_2=1\mu C$