## Two positive charges q1 = q2 = 2.0 μC are located at x = 0, y = 0.30 m and x = 0, y = -0.30 m, respectively. Third point charge Q = 4.0 μC i

Question

Two positive charges q1 = q2 = 2.0 μC are located at x = 0, y = 0.30 m and x = 0, y = -0.30 m, respectively. Third point charge Q = 4.0 μC is located at x = 0.40 m, y = 0.What is the net force ((a)magnitude and (b)direction) on charge q1 exerted by the other two charges?

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6 months 2021-07-29T09:21:41+00:00 1 Answers 81 views 0

F = 0.111015 N

Explanation:

For this exercise the force is given by Coulomb’s law

F = k q₁q₂ / r₂₁²

we calculate the electric force of the other two particles on the charge q1

Charges q₁ and q₂

the distance between them is

r₁₂ = y₁ -y₂

r₁₂ = 0.30 + 0.30

r₁₂ = 0.60 m

let’s calculate

F₁₂ = 9 10⁹ 2 10⁻⁶ 2 10⁻⁶ / 0.60 2

F₁₂ = 1 10⁻¹ N

directed towards the positive side of the y-axis

Charges 1 and 3

Let’s find the distance using the Pythagorean Theorem

r₁₃ = RA [(0.40-0) 2 + (0-0.30) 2]

r₁₃ = 0.50 m

F₁₃ = 9 10⁹ 2 10⁻⁶ 4 10⁻⁶ / 0.50²

F₁₃ = 1.697 10⁻² N

The direction of this force is on the line that joins the two charges (1 and 3), let’s use trigonometry to find the components of this force

tan θ = y / x

θ = tan⁻¹ y / x

θ = tan⁻¹ 0.3 / 0.4

tea = 36.87º

The angle from the positive side of the x-axis is

θ ’= 180 – θ

θ ’= 180 – 36.87

θ ’= 143.13º

sin143.13 = F_13y / F₁₃

F_13y = F₁₃ sin 143.13

F{13y} = 1.697 10⁻² sin 143.13

F_13y = 1.0183 10⁻² N

cos 143.13 = F_13x / F₁₃

F₁₃ₓ = F₁₃ cos 143.13

F₁₃ₓ = 1.697 10⁻² cos 143.13

F₁₃ₓ = -1.357 10-2 N

Now we can find the components of the resultant force

Fx = F13x + F12x

Fx = -1,357 10-2 +0

Fx = -1.357 10-2 N

Fy = F13y + F12y

Fy = 1.0183 10-2 + ​​1 10-1

Fy = 0.110183 N

We use the Pythagorean theorem to find the modulus

F = Ra (Fx2 + Fy2)

F = RA [(1.357 10-2) 2 + 0.110183 2]

F = 0.111015 N

Let’s use trigonometry for the angles

tan tea = Fy / Fx

tea = tan-1 (0.110183 / -0.01357)