Two point charges (q1 = -4.8μC and q2 = 6.7 μC) are fixed along the x-axis, separated by a distance d = 7.6 cm. Point P is locat
1 answer:
Answer:
1) Eₓ = 8.99 10³3 (6.7 /(d-0.076)² - 4.8 / d²) ,
2) E_{y}= 1,708 10⁴/d² N /C
3) Eₓ = 8.99 10³ (-0.7/d² + 6.7 / (d-0.076)²)
4) magnitud is double and maintains the same angle
5) There is no change that can be made.
Explanation:
The electric field is a vector quantity with formula
E = k q / r²
Where k is the Coulomb constant with value 8.99 10⁹ N m² / C², q is the load and r is the distance from the load to the point of interest.
The total field is
E = E1 + E2
E = k q₁ / r₁² + k q₂ / r₂²
Since the field is a vector quantity we will look for each component
1) x axis
Let's look for the distance for simplicity, suppose the origin of the coordinate system is at point 1
x₁ = x- x₀
x = d - 0 = d
Point 2
x₂ = d- 0.076
Eₓ = E₁ₓ + E₂ₓ
Eₓ = k q₁ / x₁² + k q₂ / x₂²
Eₓ = 8.99 10⁹ (- 4.8 10⁻⁶/ d² + 6.7 10⁻⁶ / (d-0.076)² )
Eₓ = 8.99 10³3 (6.7 /(d-0.076)² - 4.8 / d²)
.) Y axis
Distance
The points are on the x-axis so the y coordinate of the charges is zero
y₁ = y-y₀
y₁ = d-0
y₂ = d-0
= E_{1y} + E_{2y}
E_{y} = k q₁ / y₁² + k q₂ / y₂²
E_{y} = 8.99 10⁹ (-4.8 10⁻⁶ / d² + 6.7 10⁻⁶ / d²)
E_{y} = 8.99 10³ (1.9 / d²)
E_{y}= 1,708 10⁴/d² N /C
For explicit value of the electric field we must know the coordinates of point P
3) A third load is placed at a point d = 7.6 cm, on the y-axis, we calculate the field on the x-axis
Distance
x₃ = d - 0 = d
Eₓ = E₁ₓ + E₂ₓ + E₃ₓ
Eₓ = k (q₁ / x₁² + q₂ / x₂² + q₃ / x₃²)
Eₓ = 8.99 10⁹ (- 4.8 / d² + 6.7 / (d-0.076)² + 3.1 / d²) 10⁻⁶
Eₓ = 8.99 10³ (-0.7/d² + 6.7 / (d-0.076)²)
4) The electric field at point P is
E₀ = RA (Eₓ² + ²)
With the new charges
Eₓ = 2 Eₓ₀
E_{y} = 2 E_{yo}
E = √ (4Exo + 4E_{yo})
E = 2 E₀
As the two components increase the same amount the angle does not change
The answer: magnitud is double and maintains the same angle
5) For the field to be zero at point P the two components must be zero
Y Axis
Ey = 8.99 10⁹ (-4.8 10⁻⁶/d² + 6.7 10⁻⁶ / d²)
For this amount to be zero you must increase q₁ by keeping your sign
X axis
Ex = 8.99 10⁹ (- 4.8 / d² + 6.7 / (d-0.076)² + 3.1 / d²) 10⁻⁶
For this component you must also increase q1 without changing sign
But we have a problem, the magnitude of the change is different on each axis and, so there is no way to take the point / p to a zero field
The serious answer: There is no change that can be made.
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