Question

one charge of +5.0μC is placed in air at exactly x=0, and a second charge +7.0C at x=100cm. Where can a third be placed so as to experience zero net force due to the other two

Answer 1

The electric forces on the third charge $q_3$ due to charges $q_1$ and $q_2$ must be balanced:

$$F_{13}=F_{23},$$ $$\dfrac{kq_1q_3}{d^2}=\dfrac{kq_2q_3}{(r-d)^2},$$ $$\dfrac{q_1}{d^2}=\dfrac{q_2}{(r-d)^2},$$ $$q_1(r-d)^2=q_2d^2,$$ $$\dfrac{q_1}{q_2}=\dfrac{d^2}{(r-d)^2},$$ $$\sqrt{\dfrac{q_1}{q_2}}=\dfrac{d}{(r-d)},$$ $$d=\dfrac{r\sqrt{\dfrac{q_1}{q_2}}}{(1+\sqrt{\dfrac{q_1}{q_2}})},$$ $$d=\dfrac{1.0\ m\cdot\sqrt{\dfrac{5.0\cdot10^{-6}\ C}{7.0\cdot10^{-6}\ C}}}{(1+\sqrt{\dfrac{5.0\cdot10^{-6}\ C}{7.0\cdot10^{-6}\ C}})}=0.46\ m.$$