Question

Three point charges are located along the x-axis: q1 = +5.0 μC is at 20 cm, q2 = +4.0 μC is at 0 cm, and q3 = +10.0 μC is at –30 cm. What is the net electrostatic force on q2? 0.50 N to the left 500 N to the right 0.50 N to the right 8.5 N to the left

Answer 1

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,

`\begin{gathered} F=(kq_1q_2)/(r^2_1)-(kq_2q_3)/(r^2_2) \\ =kq_2((q_1)/(r^2_1)-(q_3)/(r^2_2)) \end{gathered}`

On substituting the knwon values,

`\begin{gathered} F=9*10^9*4*10^(-6)((5*10^(-6))/(0.20^2)-(10*10^(-6))/(0.30^2)) \\ =0.5\text{ N} \end{gathered}`

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.