Question

A(n)-2.0 pC charge is located at (-0.80 m, 0 m) and a(n) 5.0 pC charge is located at (0.40 m, 0 m). Find the net force on a(n) 8.0 uC charge located at (0 m, 1.0 m) 13.magnitude 0.3315 N 0.2935 N 0.2778 N 0.3010 N 0.2812 N 0.3016 N C. 14. direction 127.8° 143.3° 101.6 156.20 116.8° 137.6° C.

Answer 1

$q_{1}=-2\mu C=2*10^{-6}C$

$q_{2}=5\mu C=5*10^{-6}C$

$q_{3}=8\mu C=8*10^{-6}C$

$d_{1}=\sqrt{0.8^{2}+1^{2}}$

${\color{Red} d_{1}=1.281m}$

$d_{2}=\sqrt{0.4^{2}+1^{2}}$

${\color{Red} d_{2}=1.077m}$

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$F_{1}=kq_{1}q_{3}/d_{1}^{2}$

$F_{1}=9*10^{9}*2*10^{-6}C*8*10^{-6}C/(1.281m)^{2}$

$F_{1}=9*10^{9}*2*10^{-6}*8*10^{-6}/1.281^{2}$

$F_{1}=0.144/1.281^{2}$

$F_{1}=0.08775N$

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$F_{2}=kq_{2}q_{3}/d_{2}^{2}$

$F_{2}=9*10^{9}*5*10^{-6}C*8*10^{-6}C/(1.077m)^{2}$

$F_{2}=9*10^{9}*5*10^{-6}*8*10^{-6}/1.077^{2}$

$F_{2}=0.36/1.077^{2}$

$F_{2}=0.31N$

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$F_{x}=-F_{2}cos\theta _{2}-F_{1}sin\theta _{1}$

$F_{x}=-0.31*0.4/1.077-0.08775*0.8/1.281$

${\color{Red} F_{x}=-0.1699N}$

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$F_{y}=F_{2}sin\theta _{2}-F_{1}cos\theta_{1}$

$F_{y}=0.31*1/1.077-0.08775*1/1.281$

${\color{Red} F_{y}=0.2193N}$

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$F=\sqrt{F_{x}^{2}+F_{y}^{2}}$

$F=\sqrt{(-0.1699)^{2}+(0.2193)^{2}}$

$F=0.2774N$

ANSWER :Option c. 0.2778 N

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Direction

$tan\theta =F_{y}/F_{x}$

$\theta =tan^{-1}(F_{y}/F_{x})$

$\theta =tan^{-1}(0.2193/-0.1699)$

$\theta =-52.23^{\circ}$ (2nd quadrant)

$\theta =180^{\circ}-52.23^{\circ}=127.766^{\circ}$

ANSWER: Option a.127.8 deg

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