Question #197483

Three charges Q1= 2 microcoulomb (0,0)cm, Q2= 4 microcoulomb (0,10)cm and Q3= -3 microcoulomb (15,0)cm acts on one number another. Find the magnitude of the total electric force and the angle that the total electric force makes with the positive x-axis


1
Expert's answer
2021-05-25T10:11:40-0400

Q1=2×106 CQ_1=2\times10^{-6}\space C

Q2=4×106 CQ_2=4\times10^{-6}\space C

Q3=3×106 CQ_3=-3\times10^{-6}\space C


F13=kQ1Q3(15×102)2=9×109(2×106)(3×106)(15×102)2F_{13}=k\dfrac{Q_1Q_3}{(15\times10^{-2})^2}=9\times10^9\dfrac{(2\times10^{-6})(3\times10^{-6})}{(15\times10^{-2})^2}

F13=2.4 NF_{13}=2.4\space N


F23=kQ2Q3(18.02×102)2=9×109(4×106)(3×106)(18.02×102)2F_{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}

F23=3.32 NF_{23}=3.32\space N


Angle between F23F_{23} and F13F_{13}

tanθ=1015\tan\theta=\dfrac{10}{15}

θ=33.69°\theta=33.69\degree


Resultant force,

F=F132+F2322F13F23cosθF=\sqrt{F_{13}^2+F_{23}^2-2F_{13}F_{23}\cos\theta}

F=1.876 NF=1.876\space N


Angle made by resultant force with x axis

ϕ=180°θ2=163.155°\phi=180\degree-\dfrac{\theta}{2}=163.155\degree


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