The magnitude of the electric field at x =4m on the x axis at this time 1 N/C.
<h3>Electric field at position 4 m</h3>
Electric field at a given distance is calculated as follows;
E = kq/r²
E₂ = (9 x 10⁹ x q)/(2²)
E₂ = 2.25 x 10⁹q
E₂ + E₀ = 0
2.25 x 10⁹q + 4 = 0
2.25 x 10⁹q = - 4
q = -4 / (2.25 x 10⁹)
q = -1.78 x 10⁻⁹
E₄ = (9 x 10⁹ x (-1.78 x 10⁻⁹) ) / (4²)
E₄ = - 1 N/C
|E₄| = 1 N/C
Thus, the magnitude of the electric field at x =4m on the x axis at this time 1 N/C.
The complete question is below:
Suppose a uniform electric field of 4N/C is in the positive x direction. When a charge is placed and at a fixed to the origin, the resulting electric field on the x axis at x =2m becomes zero. What is the magnitude of the electric field at x =4m on the x axis at this time?
Learn more about electric field here: brainly.com/question/14372859
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To develop this problem we will apply Newton's laws regarding gravitational forces, both in space and on earth. From finding this relationship, leaving the variable of the dust mass open, we will find the relationship of the forces between the two surfaces. Our values are,



Distance between the center of lead sphere to dust particle



Gravitational force of the sphere on the dust particle:



Weight of the dust particle


Ratio of F and W:


Therefore the ratio is 