Acharge of 8.0 pc is distributed uniformly on a spherical surface (radius = 2.0 cm), and a second charge of â3.0 pc is distributed uniformly on a concentric spherical surface (radius = 4.0 cm). determine the magnitude of the electric field 5.0 cm from the center of the two surfaces.

According to Gauss' law, the electric field outside a spherical surface uniformly charged is equal to the electric field if the whole charge were concentrated at the center of the sphere.

Therefore, when you are outside two spheres, the electric field will be the overlapping of the two electric fields:
E(r > r₂ > r₁) = k · q₁/r² + k  · q₂/r² = k · (q₁ + q₂) / r²
where:
k = 9×10⁹ N·m²/C²

We have to transform our data into the correct units of measurement:
q₁ = 8.0 pC = 8.0×10⁻¹² C
q₂ = 3.0 pC = 3.0×10⁻¹² C
r = 5.0 cm = 0.05 m

Now, we can apply the formula:
E(r) = k · (q₁ + q₂) / r²
      = 9×10⁹ · (8.0×10⁻¹² + 3.0×10⁻¹²) / (0.05)²
      = 39.6 N/C

Hence, the magnitude of the electric field 5.0 cm from the center of the two surfaces is E = 39.6 N/C