Question

Three point charges are arranged on a line. Charge q3= +5.00 nC and is at the origin. Charge q2= -3.00 nC and is at x= +4.00 cm. Charge q1 is at x= +2.00 cm. What is q1 (magnitude and sign) if the net force on q3 is zero?

Answer 1

The distance between $q_3$ and $q_2$ is $r_{32} = 4m$, the distance between $q_3$ and $q_1$ is $r_{31} = 2m$.

According to the Coulomb's law, the force on $q_3$ from $q_2$ is:

$$F_{32} =k \dfrac{q_3q_2}{r_{32}^2}$$

where $k\approx 9\times 10^9N\cdot m^2/C^2$ is the Coulomb's constant.

The force on $q_3$ from $q_1$ is:

$$F_{31} =k \dfrac{q_3q_1}{r_{31}^2}$$

Since these forces are directed along one line, we can add them. The resultant force is then:

$$F = F_{32} + F_{31} = 0\\ k \dfrac{q_3q_1}{r_{31}^2} = -k \dfrac{q_3q_2}{r_{32}^2} \\ \dfrac{q_1}{r_{31}^2} = -\dfrac{q_2}{r_{32}^2}\\$$

Finally:

$$q_1 = -q_2\dfrac{r_{31}^2}{r_{32}^2}$$

Substituting the numbers, obtain:

$$q_1 = -(-3\times 10^{-9})\dfrac{2^2}{4^2}=+ 0.75\times 10^{-9}C = +0.75\space nC$$

Answer. $+0.75\space nC$.