Question

Suppose a uniform electric field of 4 N/C is in the positive x direction. When a charge is placed at and fixed to the origin, the resulting electric field on the x axis at x = 2 m becomes zero. What is the magnitude of the electric field at x = 4 m on the x axis at this time?

Answer 1

E_total = 3 N / A

Step-by-step explanation:

The electric field is a vector magnitude so when adding we must use vectors, in this case as the initial field E = 4N / c goes towards the axis axis and the field created by the fixed charge (E1) is also on the axis x we can add in scalar form.

E_total = E + E₁

the expression for the field of a point charge is

E₁ = k q₁ / r²

for the point x = 2m, they do not say that the total field is zero, so the charge q1 must be negative

E_total = E -k q₁ / r₂

we substitute

0 = E - k q₁ / r²

q₁ =
`(E r^2)/(k)`

let's calculate

q₁ =
`(4 \ 2^2)/(9 \ 10^(-9))`

q₁ = 1.78 10⁻⁹ C

now we can calculate the field for position x = 4 m

E_total = 4 - 9 10⁹ 1.78 10⁻⁹ / 4²2

E_total = 3 N / A