Answer to Question #126622 in Molecular Physics | Thermodynamics for P

Question

Two objects are identical and small enough that their sizes can be ignored relative to the distance between them, which is 0.278 m. In a vacuum, each object carries a different charge, and they attract each other with a force of 1.62 N. The objects are brought into contact, so the net charge is shared equally, and then they are returned to their initial positions. Now it is found that the objects repel one another with a force whose magnitude is equal to that of the initial attractive force. What is the initial charge on each object, the answer to part (a) being the one with the greater (and positive) value?

Answer 1

Initially, WLOG let $q_1\geq q_2$ ,thus magnitude is

$$F=k\frac{q_1q_2}{r^2}\implies q_1q_2=\frac{r^2F}{k}$$

Suppose $q$ charge is shared ,thus $q=\frac{q_1-q_2}{2}$ ,hence charge on each object are $\frac{q_1+q_2}{2}$ ,Therefore

$$F=k\frac{(\frac{q_1+q_2}{2})^2}{r^2}\implies q_1+q_2=\pm\sqrt{\dfrac{4r^2F}{k}}=\pm2\sqrt{q_1q_2}\\ \implies (\sqrt{q_1}\mp\sqrt{q_2})^2=0\implies q_1=q_2$$

Hence,

$$q_1=\sqrt{\frac{r^2F}{k}}=2.31\mu C\\q_2=-2.31\mu C$$