Answer:
E = 5.65 x 10¹⁰ N/C
Explanation:
First we need to find the total charge on the sphere. So, we use the following formula for that purpose:
[tex]q = \sigma V\\[/tex]
where,
q = total charge on sphere
V = Volume of Sphere = [tex]\frac{4}{3} \pi r^3 = \frac{4}{3} \pi (0.08\ m)^3 = 0.335\ m^3[/tex]
σ = volume charge density = 1.5 C/m³
Therefore,
[tex]q = (0.335\ m^3)(1.5\ C/m^3) \\q = 0.502 C[/tex]
Now, we use the following formula to find the electric field due to this charged sphere:
[tex]E = \frac{kq}{r^2}[/tex]
where.
E = Electric Field Magnitude = ?
k = Coulomb's Constant = 9 x 10⁹ N.m²/C²
r = radius of sphere = 8 cm = 0.08 m
Therefore,
[tex]E = \frac{(9\ x\ 10^9\ Nm^2/C^2)(0.502 C)}{(0.08\ m)^2}\\\\[/tex]
E = 5.65 x 10¹⁰ N/C
