Two particles are located on the x axis of a Cartesian coordinate system Particle 1 carries a charge of 3.0 nC and is at x 40 mm and particle 2 carries a charge of 3.0 nC and is at x 40 mm Particle 3 which carries a charge of 9.0 μC is located on the positive y axis 90 mm from the origin What is the magnitude of the vector sum of the electric forces exerted on particle 3?

What is the direction angle of the vector sum of the electric forces exerted on particle 3 measured counterclockwise from the positive x axis?

The easiest way to solve this problem is to find the force on each axis, the two particles on the x-axis have the same charge and are one at x = 40 mm and the other at x = -40mm,

X axis

Fₓ = F₁₃ₓ - F₂₃ₓ = 0

Axis y

[tex]F_{y}[/tex] = [tex]F_{13y}[/tex] + [tex]F_{23y}[/tex]

The modulus of this force is the same

Fy = 2 [tex]F_{13y}[/tex]

Let's reduce the magnitudes to the SI system

q₁ = 3.0 nC = 3.0 10⁻⁹ C

x₁ = 40 mm = 0.040 m

q₂ = 3.0 10⁻⁹ C

x₂ = -0.040 m

q₃ = 9.0 10⁻⁹ C

y₃ = 0.090 m

Let's look for the electric force

= k q₁ q₃ / r₁₃²

r₁₃ = √ ((0.04-0)² + (0-0.09)²)

r₁₃ = 0.09849 m

= 8.99 10⁹ 3 10⁻⁶ 9 10⁻⁶ / 0.09849

[tex]F_{13}[/tex] = 2.552 10¹ N = 25.52 N

To find the component of this force we will use trigonometry

tan θ = y / x

θ = tan⁻¹ 90/40

θ = 66⁰

Now we can find the vertical components of this force

sin 66 = [tex]F_{13y}[/tex] / [tex]F_{13}[/tex]

= sin66

= 25.52 sin66

= 22.8 N

The total vertical force is

= 2 22.8

= 45.6 N

We can find the magnitude of the total force using the Pythagorean theorem

F² = Fₓ² + ²

F² = ²

F = 45.6 N

The angle can be found using trigonometry

Tan θ = / Fₓ

θ = 90

onyebuchinnaji onyebuchinnaji · Answer 2 · 2022-01-14T12:52:32+01:00

The vector sum of the electric forces exerted on particle 3 is 0.05 N.

The direction of the angle of the vector is 66⁰.

The given parameters:

Charge of particle 1 = 3 nC
Charge of particle 2 = 3 nC
Position of particle 1 = 40 mm x-axis
Position of particle 2 = 40 mm x - axis
Charge of particle 3 = 9 μC
Position of particle 3 = 90 mm y - axis

The distance between particle 1, 2 and 3 is calculated as follows;

[tex]r = \sqrt{90^2 + 40^2} \\\\r = 98.49 \ mm[/tex]

The particle 1 and 2 are in the same position, the total charge = 3 nC + 3nC = 6 nC

The vector sum of the electric forces exerted on particle 3 is calculated as follows;

[tex]F = \frac{kq\times q_3}{r^2} \\\\F = \frac{9\times 10^9 \times 6 \times 10^{-9} \times 9 \times 10^{-6}}{(98.49 \times 10^{-3})^2} \\\\F = 0.05 \ N[/tex]

The direction of the angle of the vector is calculated as follows;

[tex]\theta = tan^{-1} (\frac{90}{40} )\\\\\theta = 66^0[/tex]

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