Question

What must be the distance in meters between point charge q1=36.6 uc and q2=-66.5 uc for the electrostatic force between them to have a magnitude of 5.55 N?

Answer 1

Given $q_{1}=3.66*10^{-5}C$ and $q_{2}=-6.65*10^{-5}C$

Use Formula $F=kq_{1}q_{2}/d^{2}$

Given $F=5.55N$

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

$5.55N=9*10^{9}*3.66*10^{-5}C*6.65*10^{-5}C/d^{2}$

$5.55=21.9051/d^{2}$

$d^{2}=21.9051/5.55$

$d=1.98667m$

ANSWER: $d=2m$

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