Answer:
E = 1.16 x 10^4 N/C
Step by step explanation:
The magnitude of the electric field (E) at a distance (r) from a point charge (q) is given by Coulomb's Law:
E = k*q/r^2
where k is Coulomb's constant (9.0 x 10^9 N*m^2/C^2).
In this case, we have a charge q = 8.6 μC (microCoulombs) located at a distance r = 4.0 m. So, plugging in the values:
E = (9.0 x 10^9 N*m^2/C^2) * (8.6 x 10^-6 C) / (4.0 m)^2
E = 1.16 x 10^4 N/C
Therefore, the magnitude of the electric field 4.0 m away from the charge is 1.16 x 10^4 N/C, directed radially outward from the charge (since the charge is positive).