Answer to Question #178210 in Physics for Shekainah

Question

A +100 μC charge is placed at the origin. A -50 μC is placed at x=2m and a +200 μC is placed at x= -4m.

What is the net electric force acting on the +100 μC charge?

Answer 1

Let $q_1 = +100\mu C = 0.1\times 10^{-3}C, q_2 = -0.05\times 10^{-3}C, q_3 = 0.2\times 10^{-3}C$

The distance between $q_1$ and $q_2$ is $r_{12} = 2m$, the distance between $q_1$ and $q_3$ is $r_{13} = 4m$.

Then, according to the Coulomb's law, the force on $q_1$ from $q_2$ is:

$$F_{12} =k \dfrac{q_1q_2}{r_{12}^2}$$

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

$$F_{12} =9\times 10^9\cdot \dfrac{0.1\times 10^{-3}\cdot (-0.05\times 10^{-3})}{2^2} = -11.25N$$

The force on $q_1$ from $q_3$ is:

$$F_{13} =k \dfrac{q_1q_3}{r_{13}^2}$$ $$F_{13} =9\times 10^9\cdot \dfrac{0.1\times 10^{-3}\cdot 0.2\times 10^{-3}}{2^2} = 45N$$

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

$$F_1 = F_{12} + F_{13} = -11.25N + 45N = 33.75N$$

And directed as shown in figure below:

Answer. 33.75 N.