Question

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 magnitude 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

Answer 1

In general, we know that the force of the electric field exerted on a point charge q at distance r from charge Q is:
F = k Qq/r²

If you have more than one charge, then the total force is the vectorial sum of the forces created by each charge.

Moreover, two charges with the same sign repulse each other, while two charges with opposite sign attract each other. This is fundamental to understand the direction of the force.

We will define "to the right" the direction towards increasing positive values of the x-axis (and we will assign a positive value), and "to the left" the direction towards decreasing negative values of the x-axis of x (and we will assign a negative value).

Before considering each position, it is better to transform our data into the correct units of measurements:
q₁ = 6.0×10⁻⁶C
q₂ = 1.5×10⁻⁶C
q₃ = 2.0×10⁻⁶C
d₂ = 3×10⁻²m
d₃=  5×10⁻²m

A) In position 1, we have a positive charge (q₁) on which is exerted a repulsive force by another positive charge (q₂) - which will be to the left because the charge q₁ will be pushed away- and an attractive force by a negative charge - which will be 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
The total force exerted on q₁ will be:
F₁ = -90 + 43.2 = - 46.8N (negative, then to the left)

B)In position 2, we have a positive charge (q₂) on which is exerted a repulsive force by another positive charge (q₁) - which will be to the right because the charge q₂ will be pushed away- and an attractive force by a negative charge (q₃) - which will be to the right. We expect F₁₂ to be equal in magnitude but opposite to F₂₁ found in point 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 exerted on q₂ will be:
F₂ = 90 + 67.5 = 157.5N (positive, then to the right)

C) In position 3, we have a negative charge (q₃) on which is exerted an attractive force by a positive charge (q₁) - which will be to the left - and an attractive force by another positive charge (q₂) - which will be to the left. We expect 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
The total force exerted on q₂ will be:
F₃ = -43.2 - 67.5 = -110.7N (negative, then to the left)