Question

1. Charge 1 of +9 micro-coulombs is located at the origin. Charge 2 of -6 micro-coulombs is located at x = -11 cm, y = -24 cm. What is the magnitude of the electric force on charge 1 due to charge 2 in Newtons?

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2. Charge 1 of +5 micro-coulombs is located at the origin, charge 2 of -7 micro-coulombs is located at x = 0 cm, y = -11 cm, and charge 3 of +6 micro-coulombs is located at x = -17 cm, y = 0 cm. What is the direction of the total electric force on charge 1 measured counter-clockwise from the +x axis, in degrees?  

Answer 1

First find the distance between charges

$d=\sqrt{x^{2}+y^{2}}$

$d=\sqrt{(-11)^{2}+(-24)^{2}}$

$d=26.4cm=0.264m$

Given $q_{1}=9\mu C=9*10^{-6}C$

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

Force,$F=kq_{1}q_{2}/d^{2}$

$F=9*10^{9}*9*10^{-6}C*6*10^{-6}C/(0.264m)^{2}$

$F=9*10^{9}*9*10^{-6}*6*10^{-6}/0.264^{2}$

$F=0.486/0.264^{2}$

ANSWER: $F=6.973N$

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2.

Given $q_{1}=5\mu C=5*10^{-6}C$

$q_{2}=-7\mu C=-7*10^{-6}C$

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

$d_{2}=11cm=0.11m$

$d_{3}=17cm=0.17m$

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

$F_{y}=9*10^{9}*5*10^{-6}C*7*10^{-6}C/(0.11m)^{2}$

$F_{y}=9*10^{9}*5*10^{-6}*7*10^{-6}/0.11^{2}$

$F_{y}=0.315/0.11^{2}$

$F_{y}=-26.033N$(negative sign shows force is acting downwards)

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

$F_{x}=9*10^{9}*5*10^{-6}C*6*10^{-6}C/(0.17m)^{2}$

$F_{x}=9*10^{9}*5*10^{-6}*6*10^{-6}/0.17^{2}$

$F_{x}=0.27/0.17^{2}$

$F_{x}=9.343N$

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Direction

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

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

$\theta =tan^{-1}(-26.033/9.343)$

$\theta =-70.257^{\circ}$

ANSWER: $\theta =289.743^{\circ}$

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