Q 1 = 2 × 1 0 − 6 C Q_1=2\times10^{-6}\space C Q 1 = 2 × 1 0 − 6 C
Q 2 = 4 × 1 0 − 6 C Q_2=4\times10^{-6}\space C Q 2 = 4 × 1 0 − 6 C
Q 3 = − 3 × 1 0 − 6 C Q_3=-3\times10^{-6}\space C Q 3 = − 3 × 1 0 − 6 C
F 13 = k Q 1 Q 3 ( 15 × 1 0 − 2 ) 2 = 9 × 1 0 9 ( 2 × 1 0 − 6 ) ( 3 × 1 0 − 6 ) ( 15 × 1 0 − 2 ) 2 F_{13}=k\dfrac{Q_1Q_3}{(15\times10^{-2})^2}=9\times10^9\dfrac{(2\times10^{-6})(3\times10^{-6})}{(15\times10^{-2})^2} F 13 = k ( 15 × 1 0 − 2 ) 2 Q 1 Q 3 = 9 × 1 0 9 ( 15 × 1 0 − 2 ) 2 ( 2 × 1 0 − 6 ) ( 3 × 1 0 − 6 )
F 13 = 2.4 N F_{13}=2.4\space N F 13 = 2.4 N
F 23 = k Q 2 Q 3 ( 18.02 × 1 0 − 2 ) 2 = 9 × 1 0 9 ( 4 × 1 0 − 6 ) ( 3 × 1 0 − 6 ) ( 18.02 × 1 0 − 2 ) 2 F_{23}=k\dfrac{Q_2Q_3}{(18.02\times10^{-2})^2}=9\times10^9\dfrac{(4\times10^{-6})(3\times10^{-6})}{(18.02\times10^{-2})^2} F 23 = k ( 18.02 × 1 0 − 2 ) 2 Q 2 Q 3 = 9 × 1 0 9 ( 18.02 × 1 0 − 2 ) 2 ( 4 × 1 0 − 6 ) ( 3 × 1 0 − 6 )
F 23 = 3.32 N F_{23}=3.32\space N F 23 = 3.32 N
Angle between F 23 F_{23} F 23 and F 13 F_{13} F 13
tan θ = 10 15 \tan\theta=\dfrac{10}{15} tan θ = 15 10
θ = 33.69 ° \theta=33.69\degree θ = 33.69°
Resultant force,
F = F 13 2 + F 23 2 − 2 F 13 F 23 cos θ F=\sqrt{F_{13}^2+F_{23}^2-2F_{13}F_{23}\cos\theta} F = F 13 2 + F 23 2 − 2 F 13 F 23 cos θ
F = 1.876 N F=1.876\space N F = 1.876 N
Angle made by resultant force with x axis
ϕ = 180 ° − θ 2 = 163.155 ° \phi=180\degree-\dfrac{\theta}{2}=163.155\degree ϕ = 180° − 2 θ = 163.155°
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