Question

Three point charges are placed on the x-axis. A charge of + 2.0 * mu * C is placed at the origin - 2.0mu * C to the right at x = 50 cm , and +4.0 at the 100 cm mark. What are the magnitude and direction of the electrostatic force that acts on the charge at the origin?

Answer 1

The magnitude of the electrostatic force that acts on the charge at the origin is given by the Coulomb's law:

$$F_0 =\left| k \dfrac{q_0q_1}{r_{01}^2} + k \dfrac{q_0q_2}{r_{02}^2} \right|$$

where $k = 9\times 10^9N\cdot m^2/C^2$, $q_0 = 2\times 10^{-6}C$ is the charge placed at the origin, $q_1 = -2\times 10^{-6}C$ is the charge placed at x=50cm, $q_2 = 4\times 10^{-6}C$ is the charge placed at x=100cm, $r_{01} = 50cm = 0.5m$ is the distance between 0 and 1 charges and $r_{02} = 100cm = 1m$ is the distance between 0 and 2 charges. Thus, obtain:

$$F_0 = \left| 9\times 10^9\cdot \left( \dfrac{2\times 10^{-6}\cdot (-2\times 10^{-6})}{0.5^2} + \dfrac{2\times 10^{-6}\cdot 4\times 10^{-6}}{1^2} \right) \right| =72\times10^3 N$$

As far as the interation force with the charge 1 is stronger, then this force will be directed in positive x direction.

Answer. $72\times10^3 N$, positive x.