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 =

Question

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40 cm, and –60 µC at x = 60 cm, what is the magnitude of the electrostatic force on the +32-µC charge? *

1 answer:

Sav [3.1K] · Answer 1 · 2026-03-04T18:19:57+03:00

Response:

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

Clarification:

Let q₁ = +32 µC, located at x₁ = 0

q₂ = +20 µC, positioned at x₂ = 40 cm = 0.4 m

q₃ = -60 µC, placed at x₃ = 60 cm = 0.6 m

Define the force magnitude on the +32 µC charge from the +20 µC charge as F₁ (the force on q₁ due to q₂).

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

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

Define the force magnitude on the +32 µC charge from the -60 µC charge as F₂ (the 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 resultant electrostatic force on the 32 µC charge is $F_{net} = |F_{2} + F_{3}|$

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