Respuesta :
The correct answer is option C, 0.50 N to the left.
Given,
The charges,
q₁=5×10⁻⁶ C
q₂=4.0×10⁻⁶ C
q₃=10.0×10⁻⁶ C
The distance between the 1st charge and the second charge, r₁=20 cm=0.20 m
The distance between 2nd charge and the 3 rd charge, r₂=30 cm=0.30 m
As all the charges have the same sign, the nature of the force is repulsive. Thus the force applied by the q₂ is directed to the left. And the force applied by the charge q₃ is pointed to the right.
From Coulomb's law, the net electrostatic force on q₂ is given by,
[tex]\begin{gathered} F=\frac{kq_1q_2}{r^2_1}-\frac{kq_2q_3}{r^2_2} \\ =kq_2(\frac{q_1}{r^2_1}-\frac{q_3}{r^2_2}) \end{gathered}[/tex]On substituting the knwon values,
[tex]\begin{gathered} F=9\times10^9\times4\times10^{-6}(\frac{5\times10^{-6}}{0.20^2}-\frac{10\times10^{-6}}{0.30^2}) \\ =0.5\text{ N} \end{gathered}[/tex]Thus the magnitude of the net electrostatic force on the charge q₂ is 0.5 N
As the force to the left is greater than the force to the right, the net force will be directed to the left.
Therefore the correct answer is option A, 0.50 N to the left.
