Three charges are placed at the corners of a square


Asked by maham237 @ in Physics viewed by 251 People


Consider 4 charges placed at the corners of a square with side 1.25m as shown. What are the magnitude and direction of the electrostatic force on Q1 resulring from other three charges?
(Note: Please put your final answer in 4 decimal places.). pls answer:)​

Answered by maham237 @



Answer:

F = 2,8289 i ^ + 1,0909 j ^) 10⁻² N

F = 3.0226 10⁻² N
, θ = 21.16º

Explanation:


For this exercise we use Coulomb's law

F = k q₁q₂ / r₁₂²


We also use that the force is a vector magnitude, so we must calculate each component of the force
, see the adjoint for the direction of the vectors

X axis

Fₓ = -F₁₄ + F₁₃ₓ


Y axis

F_{y} = F₁₂ -F_{13y}

let's look for the expression for each force

where the side of the square is a = 1.25 m


F₁₂ = k Q₁Q₂ / a²

F₁₄ = k Q₁Q₄ / a²


the distance between 1 and 3 is

d = √(a² + a²) = a √2

F₁₃ = k Q₁Q₃ / d²


let's use trigonometry to find the components

cos 45 = F₁₃ₓ / F₁₃

F₁₃ₓ = F₁₃ cos 45

F₁₃ₓ = k Q₁Q₃ / 2a²

sin 45 = F_{13y} / F₁₃

F_{13y} = F₁₃ sin 45

F_{13y} = k Q₁Q₃ / 2a² sin 45

Taking all terms, we substitute in the force for each axis

X axis

Fₓ = - k Q₁Q₄ / a² + k Q₁Q₃ / 2a₂ cos 45

Fₓ = k Q₁ / a² ( -Q₄ + Q₃ /2 cos 45)

Fₓ = 9 10⁹ 1.5 10⁻⁶ / 1.25² (- 4.5 10⁻⁶ + 3.5/2 cos 45 10⁻⁶)

Fₓ = 8.64 10³ (3.2626 10⁻⁶)

Fₓ = 2.8189 10⁻² N

Y axis

F_{y} = k Q₁Q₂ / a² - k Q₁Q₃ /2a² sin 45

F_{y} = k Q₁ / a² (Q₂ - Q₃ /2 sin45)

F_{y} = 9 10⁹ 1.5 10⁻⁶/ 1.25² (2.5 10⁻⁶ - 3.5/2 sin 45 10⁻⁶)

F_{y} = 8.64 10³ (1.26256 10⁻⁶)

F_{y} = 1.0909 10⁻² N



The result can be given in two ways

1) F = Fₓ i ^ + F_{y} j ^

F = 2,8289 i ^ + 1,0909 j ^) 10⁻² N

2) in the form of a module and an angle, for which we use the Pythagorean theorem and trigonometry

F = √ (Fₓ² + F_{y}²)

F = 10⁻² √ (2,8189² + 1,0909²)

F = 3.0226 10⁻² N

tan θ = F_{y} / Fx

θ = tan⁻¹ (F_{y} / Fₓ)

θ = tan⁻¹ (1.0909 / 2.8189)

θ = 21.16º


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