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Question 1 (13 marks) A charge Q is located on the top-left corner of a square. Charges of 2Q, 3Q and 4Q are located on the other corners as shown above in the Figure. Assuming each side of the square is 5cm in length, and the charge Q= 2.5 µC, determine the net force exerted on the 2Q charge due to the other charges.

User Icekomo
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1 Answer

4 votes

Answer:

F = 45 (2.4142 i ^ + 4.414 j ^)

F = 226.40 N, θ= 61.3

Step-by-step explanation:

For this exercise we will use that the forces are vectors and we will add them, the force due to being electric charges must comply, go Coulomb's law

F₁₂ = k q₁ q₂ / r₁₂²

To apply this equation to our case, they indicate that all charges are of the same sign and their value, also the charge Q is located in the upper left corner, unfortunately the diagram of the other charges is not loaded, but the most general is that is in sequence, see attached, for the sum of the forces let's add its components

x-axis (horizontal)

Fₓ = F₂₁ + F₂₄ₓ

Let's use trigonometry for the force component, as the weathering figure is a square the angle between the force and the x axis is 45

cos 45 = F₂₄ₓ / F₂₄

F₂₄ₓ = F₂₄ cos 45

y-axis (vertical)


F_(y) = F₂₃ + F_{24y}

sin 45 = F_{24y} / F₂₄

F_{24y} = F₂₄ sin 45

let's search every distance

The side of the square is worth l = 5cm = 0.05 m

we can find the diagonal with the Pythagorean theorem

d = √(l² + l²) = l √2

now we can search every force

F₂₁ = k q₁ q₂ / r₁₂²

F₂₁ = k Q 2Q / l²

F₂₁ = k 2Q² / l²

this force points in the positive x direction

F₂₃ = k q₂ q₃ / r₂₃²

F₂₃ = k 2Q 3Q / l²

this force points in the direction of the positive y

F₂₄ = k q₄q₂ / d²

F₂₄ = k 4Q 2Q / 2 l²

let's find the resultant in each x

X axis

Fₓ = k 2Q² / l² + k 8Q² / 2 l² cos 45

Fₓ = k 2Q² / l² (1 + 4/2 cos 45)

Fₓ = k 2Q²/ l² 2.4142 i ^

Axis y

F_{y} = k 6Q² / l² + k 8Q² / 2 l² sin 45

F_{y} = k 2Q² / l² (3 + 4/2 sin45)

F_{y} = k 2Q² / l² (4,414)

the resultant force

F = Fₓ i ^ + F_{y} j ^

F = k 2Q² / l² ( 2.4142 i ^ + 4.414 j ^)

let's substitute the values

F = 9 10⁹ 2 (2.5 10⁻⁶) 2 / 0.05² (2.4142 i ^ + 4.414 j ^)

F = 45 (2.4142 i ^ + 4.414 j ^)

The result we can give is this form and in the form of module and angle

let's use the Pythagorean theorem to find the modulus

F =√ (Fₓ² +
F_(y)²)

F = 45 √ (2.4142² + 4.414²)

F = 226.40 N

we use trigonometry for the angle, measured from the x-axis

tan θθ = Fy / Fx

θ = tan⁻¹ (4.414 / 2.4142)

θ= 61.3

Question 1 (13 marks) A charge Q is located on the top-left corner of a square. Charges-example-1
User John Ellis
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