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Question Number 10899 by Saham last updated on 01/Mar/17

Answered by sandy_suhendra last updated on 02/Mar/17

Commented by sandy_suhendra last updated on 02/Mar/17

q_1 =q_2 =6μC=6.10^(−6) C q_3 =2μC=2.10^(−6) C F_1 =((kq_1 q_3 )/R_(13) ^2_ ) = ((9.10^9 ×6.10^(−6) ×2.10^(−6) )/(0.05^2 )) = 43.2 N F_2 =F_1 = 43.2 N sinθ=(3/5) ⇒ cos2θ=1−2sin^2 θ=1−((18)/(25))=(7/(25))=0.28 F_R =(√((F_1 )^2 +(F_2 )^2 +2F_1 F_2 cos2θ)) =(√(43.2^2 +43.2^2 +2×43.2×43.2×0.28)) =(√(4,777.57)) = 69.12 N

$$\mathrm{q}_{\mathrm{1}} =\mathrm{q}_{\mathrm{2}} =\mathrm{6}\mu\mathrm{C}=\mathrm{6}.\mathrm{10}^{−\mathrm{6}} \mathrm{C} \\ $$$$\mathrm{q}_{\mathrm{3}} =\mathrm{2}\mu\mathrm{C}=\mathrm{2}.\mathrm{10}^{−\mathrm{6}} \mathrm{C} \\ $$$$\mathrm{F}_{\mathrm{1}} =\frac{\mathrm{kq}_{\mathrm{1}} \mathrm{q}_{\mathrm{3}} }{\mathrm{R}_{\mathrm{13}} ^{\mathrm{2}_{} } }\:=\:\frac{\mathrm{9}.\mathrm{10}^{\mathrm{9}} ×\mathrm{6}.\mathrm{10}^{−\mathrm{6}} ×\mathrm{2}.\mathrm{10}^{−\mathrm{6}} }{\mathrm{0}.\mathrm{05}^{\mathrm{2}} }\:=\:\mathrm{43}.\mathrm{2}\:\mathrm{N}\:\:\:\: \\ $$$$\mathrm{F}_{\mathrm{2}} =\mathrm{F}_{\mathrm{1}} =\:\mathrm{43}.\mathrm{2}\:\mathrm{N} \\ $$$$\mathrm{sin}\theta=\frac{\mathrm{3}}{\mathrm{5}}\:\Rightarrow\:\mathrm{cos2}\theta=\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \theta=\mathrm{1}−\frac{\mathrm{18}}{\mathrm{25}}=\frac{\mathrm{7}}{\mathrm{25}}=\mathrm{0}.\mathrm{28}\:\:\:\:\: \\ $$$$\mathrm{F}_{\mathrm{R}} =\sqrt{\left(\mathrm{F}_{\mathrm{1}} \right)^{\mathrm{2}} +\left(\mathrm{F}_{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{2F}_{\mathrm{1}} \mathrm{F}_{\mathrm{2}} \:\mathrm{cos2}\theta} \\ $$$$\:\:\:\:\:=\sqrt{\mathrm{43}.\mathrm{2}^{\mathrm{2}} +\mathrm{43}.\mathrm{2}^{\mathrm{2}} +\mathrm{2}×\mathrm{43}.\mathrm{2}×\mathrm{43}.\mathrm{2}×\mathrm{0}.\mathrm{28}}\:\:\:\: \\ $$$$\:\:\:\:\:=\sqrt{\mathrm{4},\mathrm{777}.\mathrm{57}}\:=\:\mathrm{69}.\mathrm{12}\:\mathrm{N} \\ $$

Commented by Saham last updated on 02/Mar/17

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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