The net force on the middle charge due to the four other charges is zero.
The given parameters:
- q₁ = -2 nC
- q₂ = 2 nC
- q₃ = 2 nC
- q₄ = -2 nC
- middle charge, q₀ = 1 nC
What is the force between two charges?
- The attractive or repulsive force between two charges is determined by Coulomb's law.
The magnitude of the force on the middle charge due to the four other charges is calculated as follows;
[tex]F_0{net} = F_{01} + F_{02} + F_{03} + F_{04}\\\\F_0{net} = \frac{k\times q_0(-q_1)}{r^2} + \frac{k\times q_0(q_2)}{r^2} + \frac{k\times q_0(q_3)}{r^2} + \frac{k\times q_0(-q_4)}{r^2}\\\\F_0{net} = - \frac{k\times q_0q_1}{r^2} + \frac{k\times q_0q_2}{r^2} + \frac{k\times q_0q_3}{r^2} - \frac{k\times q_0q_4}{r^2}\\\\F_0{net} = \frac{kq_0}{r^2} (-q_1 + q_2 + q_3 - q_4)\\\\recall, \ |q_1| = q_2 = |q_3| = |q_4| = 2 \ nC\\\\F_0{net} = \frac{kq_0}{r^2} (-2 + 2+ 2-2)\\\\F_0{net} = \frac{kq_0}{r^2}(0)\\\\[/tex]
[tex]F_0{net} = 0[/tex]
Thus, the net force on the middle charge due to the four other charges is zero.
Learn more about net force on charges here: https://brainly.com/question/14361879
