Two identical particles are charged and held at a distance of 1m form each pther. They are found to be attraction each other with a force of 0.027N . Now they are connected by a conducting wire so that charge folws between them. Whe the charge flow stop, they are found to be repelling each other with a force of 0.009N. Find the initial charge on each particle.

Answer

`pm3muC,pm1muC`

Explanation

As there is attraction (here we have assumed `Q_(1)` and `Q_(2)` in `muC`), so we get
`(9xx10^(9)xxQ_(1)Q_(2)xx10^(-12))/(1^(2))=-0.027` or `Q_(1)Q_(2)=-3`
Negative sign is due to attraction
For repulsion :
`(9xx10^(9)xx((Q_(1)+Q_(2))/(2))^(2)xx10^(-12))/(1^2)=0.009`.
or `(Q_(1)+Q_(2))=4`
or `Q_(1)+Q_(2)= +-2`
Here, we have two sets of equations.
(i) From` Q_(1)Q_(2)= -3` and `Q_(1)+Q_(2)=2` we get `3mu `C and `-1muC`.
(ii) From ` Q_(1)Q_(2)= -3` and `Q_(1)+Q_(2)= -2` we get `-3muC` and `1muC`.
Both are valid.