Question

Solve for the net force experienced by q2. (Show your Solutions)

 

 

  q1.    20 µC

  

 

          0.3 m

 

                             

  q2    -15 µC       

                

 

         0.25 m

 

 

 q3 .  -10 µC

Answer 1

According to the Coloumb's law, the force experienced by $q_2$ from $q_1$ 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 $q_2$ from $q_3$ is:

$$F_{12} = k\dfrac{|q_2||q_3|}{r_{23}^2}$$

where $r_{23} = 0.25m$ is the distance between $q_2$ and $q_3 = -10\times10^{-6}C$. Thus, obtain:

$$F_{13} = 9\times 10^9\cdot \dfrac{20\times 10^{-6}\cdot 10\times 10^{-6}}{0.25^2} =28.8N$$

Since $q_1$ has the opposite sign with $q_2$ and $q_3$ have the same sign with $q_2$ , both forces are directed in the same direction (toward $q_1$). Thus, according to the superposition principle, the net force will be:

$$F_1 = F_{12} + F_{23} = 30N+28.8N = 58.8N$$

Answer. 58.8 N.