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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Answer:
a) -1.325 μC
b) 4.17μC
Explanation:
First, you need to know that charge is conserved. So, the adition of the charges, as there is no lost in charge, should always be the same. Also, after the wire is removed, both spheres will have the same charge, trying to find equilibrium. In summary:
We know both q1f and q2f must be positive, because the negative charge at the beginning was the the smaller.
The electrostatic force is equal to:
K is the Coulomb constant, equal to 9*10^9 Nm^2/C^2
Now, we are told that the electrostatic force after the wire is equal to 0.0443 N:
Originally, the force is negative because it was an attraction force, therefore, its direction was opposite to the direction of the repulsive force after the wire:
Solving the quadratic equation:
for this values q_1 wil be:
So as you can see, the negative charge will always be -1.325 μC and the positive 4.17μC