Answer:
q2 = - 8 × [tex]10^{-6}[/tex] C
negative sign because attract together
Explanation:
given data
q1 = +3.2 µC = 3.2 × [tex]10^{-6}[/tex] C
distance r = 0.28 m
force F = 2.9 N
to find out
q2 (magnitude and sign)
solution
we know that here if 2 charge is unlike charge
than there will be electrostatic force of attraction , between them
now we apply coulomb law that is
F = [tex]\frac{q1q2}{4\pi \epsilon *r^{2}}[/tex] ............1
here we know [tex]\frac{1}{4\pi \epsilon}}[/tex] = 9 × [tex]10^{9}[/tex] Nm²/C²
so from equation 1
2.9 = 9 × [tex]10^{9}[/tex] × [tex]\frac{3.2*10^{-6}*q2}{0.28^{2}}[/tex]
q2 = [tex]\frac{2.9*0.28^2}{9*10^9*3.2*10^{-6}}[/tex]
q2 = - 8 × [tex]10^{-6}[/tex] C
