Question

In a vacuum, two particles have charges of q1 and q2, where q1-+3.80 HC. They are separated by a distance of 0.205 m, and particle 1 experiences an attractive force of 3.36 N. What is the magnitude of q2? Submit Answer Incorrect. Tries 3/10 Previous Tries

Answer 1

$From\,\,Coulobm's\,\,law\,\,$

$Force\,\,F=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{d^2}$
$Where\,\, \frac{1}{4\pi\varepsilon_0}=9\times10^9\,\,\frac{N.m^2}{C^2}\,\,is\,\,Coulomb's\,\,law constant\\ \varepsilon_0\,\, is\,\,vacuum\,\, permittivity\\$

$In\,\,the\,\,given\,\,problem,\\ F=3.36\,\,N\\ q_1=+3.80\,\,\mu C=+3.80\times10^{-6}\,\,C\\ d=0.205\,\,m\\ q_2=?$

  $\therefore \,\,\,\,3.36=9\times10^9\frac{3.38\times10^{-6}q_2}{(0.205)^2}\\\\ q_2=\frac{3.36\times(0.205)^2}{(9\times10^9)\times(3.38\times10^{-6})}\,\,\,\,C\\\\ q_2=4.642\times10^{-6}\,\,C \\q_2=4.642\,\,\mu C$

$As\,\,particle\,\,1\,\,experiences\,\,an\,\,attractive\,\,force,\,\,\\ then\,\,particle\,\,2\,\,must\,\,be\,\,negative\,\,in\,\,nature.$

$\dpi{120} \therefore q_2=4.642\,\,\mu C$