Answer to Question #95903 in Electric Circuits for Dawson

Question #95903
Four charges,
q1 = +150 µC, q2 = +47 µC, q3 = −134 µC, and q4 = +22 µC,
are fixed at the corners of a 4 m by 5 m rectangle, as illustrated in the figure below. What are the magnitude (in N) and the direction (in degrees counterclockwise from the +x-axis) of the net force acting on q1? (Assume the x-axis extends from q1 to the right.)
1
Expert's answer
2019-10-07T10:46:51-0400

Force betwwn two particales q1 and q2 is given by:q_1\ and\ q_2\ is\ given\ by:\\

F=kq1q2r2F=\frac{kq_1q_2}{r^2}

Let q1 be A, q2 be B, q3 be C and q4 be Dq_1\ be\ A,\ q_2\ be \ B,\ q_3\ be\ C\ and\ q_4\ be\ D


Force on A due to B :

FAB=9×109×150×47×101242=3.97 N along x axisSimilarly,F_{AB}=\frac{9\times10^9\times150\times47\times10^{-12}}{4^2}=3.97\ N\ along\ -x\ axis\\Similarly,\\

Force on Adue to C :

FAC=4.4412 NF_{AC}=4.4412\ N

Direction of this force is : θ1=tan154=\theta_1=tan^{-1}\frac{5}{4}= 51.3° clockwise of x axis51.3\degree\ clockwise\ of\ x\ axis


Similarly,

Force on A due to D :

FAD=1.188 N along y axisF_{AD}=1.188\ N\ along\ y\ axis


resultant of FAB and FAD  is:F_{AB}\ and\ F_{AD}\ \ is:

FABD=3.972+1.1882=4.144 N Direction of this resultant is:F_{ABD}=\sqrt{3.97^2+1.188^2}=4.144\ N\ \\Direction\ of\ this\ resultant\ is:

θ2=tan13.971.188=73.34° anticlockwise of y axis\theta_2=tan^{-1}\frac{3.97}{1.188}=73.34\degree\ anticlockwise\ of\ y\ axis


Now net force on A is given by:

Fnet=4.1442+4.44122+2×4.144×4.4412cos214.64°=2.57 NF_{net}=\sqrt{4.144^2+4.4412^2+2\times4.144\times4.4412\cos214.64\degree}=2.57\ N


Direction of this net force is given by ;

θ=121.33° clockwise of FABDOr,((90+73.34)121.33)=42.01° anticlockwise of x axis\theta=121.33\degree\ clockwise\ of\ F_{ABD}\\Or,\\((90+73.34)-121.33)=42.01\degree\ anticlockwise\ of\ x\ axis


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