Two identical particles are charged and held at a distance of $1 m$ from each other. They are found to be attracting each other with a force of $0.027 N$ . 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 $0.009 N$ . Find the initial charge on each particle :

Question

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Answer 1

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 $\frac{( 9 \times 1 0 ^{9} ) q _{1} q _{2}}{1 ^{2}} = - 0.027 \Rightarrow q_{1} q_{2} = - 3 \times 1 0^{- 12} . . (1)$
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, $F = \frac{( 9 \times 1 0 ^{9} ( \frac{q _{1} + q _{2}}{2} ) ^{2}}{1 ^{2}} = 0.009$
or $(q_{1} + q_{2})^{2} = 4 \times 1 0^{- 12}$
or $q_{1} + q_{2} = \pm 2 \times 1 0^{- 6} . . . (2)$
A) From $q_{1} q_{2} = - 3 \times 1 0^{- 12}$ and $q_{1} + q_{2} = + 2 \times 1 0^{- 6}$ , we get after solving $q_{1} = 3 μ C$ and $q_{2} = - 1 μ C$
B) From $q_{1} q_{2} = - 3 \times 1 0^{- 12}$ and $q_{1} + q_{2} = - 2 \times 1 0^{- 6}$ , we get after solving $q_{1} = - 3 μ C$ and $q_{2} = 1 μ C$

Answer 2

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 1 0 ^{9} ) q _{1} q _{2}}{1 ^{2}} = - 0.027 \Rightarrow q_{1} q_{2} = - 3 \times 1 0^{- 12} . . (1)

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 1 0 ^{9} ( \frac{q _{1} + q _{2}}{2} ) ^{2}}{1 ^{2}} = 0.009

or

(q_{1} + q_{2})^{2} = 4 \times 1 0^{- 12}

or

q_{1} + q_{2} = \pm 2 \times 1 0^{- 6} . . . (2)

A) From

q_{1} q_{2} = - 3 \times 1 0^{- 12}

and

q_{1} + q_{2} = + 2 \times 1 0^{- 6}

, we get after solving

q_{1} = 3 μ C

and

q_{2} = - 1 μ C

B) From

q_{1} q_{2} = - 3 \times 1 0^{- 12}

and

q_{1} + q_{2} = - 2 \times 1 0^{- 6}

, we get after solving

q_{1} = - 3 μ C

and

q_{2} = 1 μ C