The electric force on q3 due to q1
F 13 = k q 1 q 3 ( 0.5 ) 2 F_{13}=k\frac{q_1 q_3}{(0.5)^2} F 13 = k ( 0.5 ) 2 q 1 q 3 F 13 = 9 ⋅ 1 0 9 6 ⋅ 1 0 − 6 ⋅ 12 ⋅ 1 0 − 6 ( 0.5 ) 2 = 2.592 ⋅ 1 0 − 6 ( N ) F_{13}=9\cdot 10^{9}\frac{6\cdot 10^{-6}\cdot 12\cdot 10^{-6}}{(0.5)^2}=2.592\cdot 10^{-6} (N) F 13 = 9 ⋅ 1 0 9 ( 0.5 ) 2 6 ⋅ 1 0 − 6 ⋅ 12 ⋅ 1 0 − 6 = 2.592 ⋅ 1 0 − 6 ( N )
The electric force on q3 due to q1
F 13 = k q 2 q 3 ( 0.4 ) 2 F_{13}=k\frac{q_2 q_3}{(0.4)^2} F 13 = k ( 0.4 ) 2 q 2 q 3 F 13 = 9 ⋅ 1 0 9 9 ⋅ 1 0 − 6 ⋅ 12 ⋅ 1 0 − 6 ( 0.4 ) 2 = 6.075 ⋅ 1 0 − 6 ( N ) F_{13}=9\cdot 10^{9}\frac{9\cdot 10^{-6}\cdot 12\cdot 10^{-6}}{(0.4)^2}=6.075\cdot 10^{-6} (N) F 13 = 9 ⋅ 1 0 9 ( 0.4 ) 2 9 ⋅ 1 0 − 6 ⋅ 12 ⋅ 1 0 − 6 = 6.075 ⋅ 1 0 − 6 ( N ) The net force on q3
F N = F 13 2 + F 23 2 + 2 F 13 F 23 c o s ( α ) F_N=\sqrt{F_{13}^2+F_{23}^2+2F_{13}F_{23}cos(\alpha)} F N = F 13 2 + F 23 2 + 2 F 13 F 23 cos ( α )
c o s ( α ) = 3 5 cos(\alpha)=\frac{3}{5} cos ( α ) = 5 3
F N = ( 2.592 ⋅ 1 0 − 6 ) 2 + ( 6.075 ⋅ 1 0 − 6 ) 2 + 2 ⋅ 2.592 ⋅ 1 0 − 6 ⋅ 6.075 ⋅ 1 0 − 6 ⋅ 3 5 F_N=\sqrt{(2.592\cdot 10^{-6})^2+(6.075\cdot 10^{-6})^2+2\cdot 2.592\cdot 10^{-6}\cdot 6.075\cdot 10^{-6}\cdot\frac{3}{5}} F N = ( 2.592 ⋅ 1 0 − 6 ) 2 + ( 6.075 ⋅ 1 0 − 6 ) 2 + 2 ⋅ 2.592 ⋅ 1 0 − 6 ⋅ 6.075 ⋅ 1 0 − 6 ⋅ 5 3
F N = 7.907 ⋅ 1 0 − 6 ( N ) F_N=7.907\cdot 10^{-6} (N) F N = 7.907 ⋅ 1 0 − 6 ( N )
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