Question

A point charge of 3×10^-6 is 12cm distance.second point charge -1.5×10^-6c calculate the magnitude of the force on each charge

Answer 1

Answer:

2.81N

Explanation:

F=kq₁q₂/r²

$=\frac{9*10^{9}*3*10^{-6}*(-1.5)*10^{-6}}{12^{2}*10^{-4}}\\=\frac{-40.5*10^{-3}}{144*10^{-4}}$

=-0.281*10

=2.81N

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