Question

Three point charges are positioned on the x axis. If the charges and corresponding positions are +32 µC at x = 0, +20 µC at x = 40 cm, and –60 µC at x = 60 cm, what is the magnitude of the electrostatic force on the +32-µC charge? *

Answer 1

Answer:

Magnitude of the electrostatic force on the +32 µC charge, $F_{net} = 12 N$

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)

$F_{2} = \frac{kq_{1}q_{2} }{x_{2}^2 }$

$F_{2} = \frac{9 * 10^{9} * 32 * 10^{-6} * 20 * 10^{-6} }{0.4^2 }\\F_{2} = 36 N$

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)

$F_{3} = \frac{kq_{1}q_{3} }{x_{3}^2 }$

$F_{3} = -\frac{9 * 10^{9} * 32 * 10^{-6} * 60 * 10^{-6} }{0.6^2 }\\F_{3} =-48 N$

The electrostatic force on the 32 µC charge, $F_{net} = |F_{2} + F_{3}|$

$F_{net} =| 36 + (-48)| \\F_{net} =|- 12 N| \\ F_{net} = 12 N$