Question

Three charges, q1 = +1.90 10-9 C, q2 = -2.91 10-9 C, and q3 = +1.10 10-9 C, are located on the x-axis at x1 = 0, x2 = 10.0 cm, and x3 = 20.0 cm. Find the resultant force on q3.

Answer 1

Given $q_{1}=1.9*10^{-9}C$

$q_{2}=-2.91*10^{-9}C$

$q_{3}=1.1*10^{-9}C$

$d_{1}=0.2m$

$d_{2}=20cm-10cm =10cm=0.1m$

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

$F_{1}=9*10^{9}*1.9*10^{-9}C*1.1*10^{-9}C/(0.2m)^{2}$

$F_{1}=9*10^{9}*1.9*10^{-9}*1.1*10^{-9}/0.2^{2}$

$F_{1}=1.881*10^{-8}/0.2^{2}$

$F_{1}=4.7025*10^{-7}N$ towards right

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

$F_{2}=9*10^{9}*2.91*10^{-9}C*1.1*10^{-9}C/(0.1m)^{2}$

$F_{2}=9*10^{9}*2.91*10^{-9}*1.1*10^{-9}/0.1^{2}$

$F_{2}=2.8809*10^{-8}/0.1^{2}$

$F_{2}=2.8809*10^{-6}N$ towards left

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Resultant Force ,$F=F_{1}-F_{2}$

$F=4.7025*10^{-7}N-2.8809*10^{-6}N$

ANSWER: $F=-2.41065*10^{-6}N$ (negative sign shows direction as towards left)

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