Answer:
the correct answer is C, E’= 4E
Explanation:
In this exercise you are asked to calculate the electric field at a given point
E = [tex]k \frac{q}{r^2}[/tex]
indicates that the field is E for r = 2m
E = [tex]\frac{ k q}{4}[/tex] (1)
the field is requested for a distance r = 1 m
E ’= k \frac{q}{r'^2}
E ’= k q / 1
from equation 1
4E = k q
we substitute
E’= 4E
so the correct answer is C