Question #178115

1. The force of +50.0μC at (-10.0m, 0) and +50.0μC at (10.0m, 0) on 20.0μC at (0, 0). Show solution.


2. The force of +40.0μC at (0, -3.00m) and +20.0μC at (0, 3.00m) on 50.0μC at (4.00m, 0). Show solution.


1
Expert's answer
2021-04-05T11:08:37-0400

F=F13F23=kq1q3/r132kq2q3/r232=F=F_{13}-F_{23}=k{q_1q_3}/{r_{13}^2}-kq_2q_3/r_{23}^2=


=91095010620106/102=9\cdot10^9\cdot50\cdot10^{-6}\cdot20\cdot10^{-6}/10^2-


91095010620106/102=0-9\cdot10^{-9}\cdot50\cdot 10^{-6}\cdot20\cdot10^{-6}/10^2=0 . Answer




F=F132+F232+2F13F23cos73.74°F=\sqrt{F_{13}^2+F_{23}^2+2F_{13}F_{23}\cos73.74°}



F13=kq1q3/r132=F_{13}=k{q_1q_3}/{r_{13}^2}=


=91094010650106/52=0.72 (N)=9\cdot10^9\cdot40\cdot10^{-6}\cdot50\cdot10^{-6}/5^2=0.72\ (N)



F23=kq2q3/r232=F_{23}=kq_2q_3/r_{23}^2=


=91092010650106/52=0.36 (N)=9\cdot10^9\cdot20\cdot10^{-6}\cdot50\cdot10^{-6}/5^2=0.36\ (N)



F=0.722+0.362+20.720.36cos73.74°=F=\sqrt{0.72^2+0.36^2+2\cdot0.72\cdot 0.36\cdot\cos73.74°}=



=0.89 (N)=0.89\ (N) . Answer











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