Question

Solve for the net force experienced by q3.

           

 

 

      q1           2 cm                q2                     4 cm                 q3

          . ___________________. _________________________.

      - 20 statC                           - 30 statC                                 - 10 statC

 

Answer 1

$F=F_1+F_2=k\frac{q_1q_3}{(0.02+0.04)^2}+k\frac{q_2q_3}{(0.04)^2}=$

$=9\cdot10^9\cdot \frac{20\cdot 10}{(0.02+0.04)^2}+9\cdot10^9\cdot \frac{30\cdot 10}{(0.04)^2}=2.2\cdot10^{15} (N)$ . Answer