Final answer:
To calculate the electric flux density at 3 m from the center of two concentric spheres with given charge densities we sum the total charges from both spheres and apply Gauss's Law which relates the enclosed charge to the electric flux density at a given point outside concentric spheres.
Step-by-step explanation:
The question involves calculating the electric flux density at a point 3 m from the center of two concentric spheres with given charge densities. According to Gauss's Law, the electric flux Φ through a closed surface is equal to the enclosed charge Q divided by the permittivity of free space ε₀; mathematically, Φ = Q/ε₀. Since the observation point is outside both spheres, the charge on each sphere can be treated as if it were concentrated at the center. The combined charge from both spheres is the sum of the charge on each sphere, which is the product of charge density and the surface area (4πr²) for each sphere. the total charge by the smaller sphere (radius 1 m) is Q₁ = σ₁ * 4πr₁². Substituting, we get Q₁ = 8 nC/m² * 4π * (1 m)² = 8 nC/m² * 4π nC. The total charge by the larger sphere (radius 2 m) is Q₂ = σ₂ * 4πr₂², giving us Q₂ = -6 mC/m² * 4π * (2 m)² = -6 mC/m² * 16π mC.
Adding both charges Q₁ + Q₂ gives the total charge enclosed. To find the electric flux density D at a distance of 3 m from the center, we use the relation D = Q_enclosed/(4πr²). The radius here is the distance from the center to the observation point, which is 3 m. Substituting the known values, we get the electric flux density D = (Q₁ + Q₂) / (4π * (3 m)²).