Solve for the net force experienced by q3.
q1 2 cm q2 4 cm q3
. ___________________. _________________________.
- 20 statC - 30 statC - 10 statC
"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
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