Three point charges, q1 , q2 , and q3 , lie along the x-axis at x = 0, x = 3.0 cm, and x = 5.0 cm, respectively. calculate the m

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Three point charges, q1 , q2 , and q3 , lie along the x-axis at x = 0, x = 3.0 cm, and x = 5.0 cm, respectively. calculate the m

agnitude and direction of the electric force on each of the three point charges when q1 = +6.0 µc, q2 = +1.5 µc, and q3 = -2.0 µc

Physics

1 answer:

Ostrovityanka [3.2K] · Answer 1 · 2026-02-24T06:50:25+03:00

Generally, the force exerted by an electric field on a point charge q located at a distance r from another charge Q can be expressed as:
F = k Qq/r²

When multiple charges are present, the resulting force is determined by vectorially adding the forces contributed by each charge.

Furthermore, charges with similar signs will repel each other, while those with opposite signs will attract. Understanding this is crucial for determining the force's direction.

We will define the direction "to the right" as the way toward increasingly positive values on the x-axis (and assign it a positive value), while "to the left" will be toward decreasing negative values on the x-axis (and we’ll assign a negative value).

Prior to analyzing each position, it is advisable to convert our variables into the right units of measurement:
q₁ = 6.0×10⁻⁶C
q₂ = 1.5×10⁻⁶C
q₃ = 2.0×10⁻⁶C
d₂ = 3×10⁻²m
d₃ = 5×10⁻²m

A) At position 1, a positive charge (q₁) experiences a repulsive force from another positive charge (q₂) directed to the left as it is pushed away, and an attractive force from the negative charge (q₃) directed to the right:
F₂₁ = 9×10⁹ · 1.5×10⁻⁶ · 6.0×10⁻⁶ / (3×10⁻²)² = 90N
F₃₁ = 9×10⁹ · 2.0×10⁻⁶ · 6.0×10⁻⁶ / (5×10⁻²)² = 43.2N
Thus, the net force on q₁ is:
F₁ = -90 + 43.2 = -46.8N (negative, indicating leftward direction)

B) For position 2, the positive charge (q₂) encounters a repulsive force from the other positive charge (q₁), which pushes it to the right, and also attracts the negative charge (q₃), also directed to the right. We anticipate F₁₂ to have the same magnitude but opposite direction compared to F₂₁ found in part A:
F₁₂ = 9×10⁹ · 6.0×10⁻⁶ · 1.5×10⁻⁶ / (3×10⁻²)² = 90N
F₃₂ = 9×10⁹ · 2.0×10⁻⁶ · 1.5×10⁻⁶ / (2×10⁻²)² = 67.5N
The total force acting on q₂ is:
F₂ = 90 + 67.5 = 157.5N (positive, hence directed to the right)

C) In position 3, we have a negative charge (q₃) that feels attracting forces from the positive charges (q₁ and q₂), both directed to the left. We can assume that F₁₃ = -F₃₁ and F₂₃ = -F₃₂:
F₁₃ = 9×10⁹ · 6.0×10⁻⁶ · 2.0×10⁻⁶ / (5×10⁻²)² = 43.2N
F₂₃ = 9×10⁹ · 1.5×10⁻⁶ · 2.0×10⁻⁶ / (2×10⁻²)² = 67.5N
Thus, the overall force experienced by q₃ will be:
F₃ = -43.2 - 67.5 = -110.7N (negative, therefore to the left)

Three point charges, q1 , q2 , and q3 , lie along the x-axis at x = 0, x = 3.0 cm, and x = 5.0 cm, respectively. calculate the m

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