Answer:
The magnitude of the final electric field is 0.5 N/C.
Explanation:
Given;
initial electric field, E₁ = 1 N/C
initial distance moved by the charge, r₁ = 3 m
final distance moved by the charge, r₂ = 6 m
let the magnitude of the final electric field = E₂
The electric potential created by the charge is given as;
V = E₁r₁ = E₂r₂
[tex]E_2 = \frac{E_1r_1}{r_2} \\\\E_2 = \frac{1 N/C\ \times \ 3m}{6m} \\\\E_2 = 0.5 \ N/C[/tex]
Therefore, the magnitude of the final electric field is 0.5 N/C.
