Question

Solve for the net force experienced by q1. (Show your solution)

 

 

     q1           0.3 m           q2                          0.25  m                q3

          . ___________________. _________________________.

        20 µC                            - 15 µC                                        - 10 µC

 

Answer 1

According to the Coloumb's law, the force experienced by $q_1$ from $q_2$ is:

$$F_{12} = k\dfrac{|q_1||q_2|}{r_{12}^2}$$

where $k = 9\times 10^9 \space N\cdot m^2/C^2$ is the Coloumb's constant, $r_{12} = 0.3m$ is the distance between $q_1 = 20\times 10^{-6}C$ and $q_2 = -15\times10^{-6}C$. Thus, obtain:

$$F_{12} = 9\times 10^9\cdot \dfrac{20\times 10^{-6}\cdot 15\times 10^{-6}}{0.3^2} =30N$$

Similarly, the the force experienced by $q_1$ from $q_3$ is:

$$F_{12} = k\dfrac{|q_1||q_3|}{r_{13}^2}$$

where $r_{13} = 0.3m+0.25m = 0.55m$ is the distance between $q_1$ and $q_3 = -10\times10^{-6}C$. Thus, obtain:

$$F_{13} = 9\times 10^9\cdot \dfrac{20\times 10^{-6}\cdot 10\times 10^{-6}}{0.55^2} \approx 5.95N$$

Since $q_2$ and $q_3$ have the same sign, both forces are directed in the same direction. Thus, according to the superposition principle, the net force will be:

$$F_1 = F_{12} + F_{13} = 35.95N$$

Answer. 35.95 N.