Respuesta :

The magnitude of the electric field at x =4m on the x axis at this time 1 N/C.

Electric field at position 4 m

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?

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