Answer:
Magnitude of the electrostatic force on the +32 µC charge, [tex]F_{net} = 12 N[/tex]
Explanation:
Let q₁ = +32 µC, x₁ = 0
q₂ = +20 µC, x₂ = 40 cm = 0.4 m
q₃ = -60 µC, x₃ = 60 cm = 0.6 m
Let magnitude of the electrostatic force on the +32 µC charge due to the + 20 µC charge = F₁ (i.e force on q₁ due to q₂)
[tex]F_{2} = \frac{kq_{1}q_{2} }{x_{2}^2 }[/tex]
[tex]F_{2} = \frac{9 * 10^{9} * 32 * 10^{-6} * 20 * 10^{-6} }{0.4^2 }\\F_{2} = 36 N[/tex]
Let magnitude of the electrostatic force on the +32 µC charge due to the -60 µC charge = F₂ (i.e force on q₁ due to q₃)
[tex]F_{3} = \frac{kq_{1}q_{3} }{x_{3}^2 }[/tex]
[tex]F_{3} = -\frac{9 * 10^{9} * 32 * 10^{-6} * 60 * 10^{-6} }{0.6^2 }\\F_{3} =-48 N[/tex]
The electrostatic force on the 32 µC charge, [tex]F_{net} = |F_{2} + F_{3}|[/tex]
[tex]F_{net} =| 36 + (-48)| \\F_{net} =|- 12 N| \\ F_{net} = 12 N[/tex]
