Answer:
c) The magnitud of field is 10⁷ N/C
Explanation:
An easy way to solve this exercise is by using Gauss's law
Ф = ∫ E .dA = [tex]q_{int}[/tex] / ε₀
As the balls are spherical we create a spherical Gaussian surface, in this case the electric field line for letter radii of the sphere and the scaled product is reduced to the ordinary product
E A = [tex]q_{int}[/tex] / ε₀
The area of a sphere is
A = 4π R²
[tex]q_{int}[/tex] = E 4π R² ε₀
Let's calculate the charge on the ball
[tex]q_{int}[/tex] = 1.0 10⁶ 4π 25²2 8.85 10⁻¹²
[tex]q_{int}[/tex] = 6.95 10⁻² C
Now we can calculate the taste load outside the surface of the globe R = 5 m
E = [tex]q_{int}[/tex] / A ε₀
E = [tex]q_{int}[/tex] / (4π R² ε₀)
E = 6.95 10⁻² / (4π 5² 8.85 10⁻¹²)
E = 2.5 10⁷ C / m
The magnitud of field is 10⁷ N/C