Answer:
0.0567 N
Explanation:
q1 = 3 micro coulomb
q2 = - 2 micro coulomb
OB = 50 cm
AB = 60 cm
By using Pythagoras theorem in triangle OAB
[tex]OA^{2}=OB^{2}+AB^{2}[/tex]
[tex]OA^{2}=50^{2}+60^{2}=6100[/tex]
OA = 78.1 cm
By using the Coulomb's law, the force on 3 micro coulomb due to -2 micro coulomb is
[tex]F=\frac {Kq_{1}q_{2}}{OA^{2}}= \frac{9 \times 10^{9}}\times 3\times 10^{-6} \times 2 \times 10^{-6}{0.781^{2}}[/tex]
F = 0.0885 N
The horizontal component of force is
= F CosФ = [tex]F\times \frac{OB}{OA}[/tex]
= 0.0885 x 50 / 78.1 = 0.0567 N
