Final answer:
The energy density in the electric field near the surface of a charged isolated metal sphere with a diameter of 12 cm and a potential of 8000 V is approximately 7.885 × 10^-3 J/m³.
Step-by-step explanation:
To calculate the energy density in the electric field near the surface of a charged isolated metal sphere, we use the concept of electric potential and electric field.
The electric field (E) near the surface of a sphere with potential (V) is given by
E = V / r,
where r is the radius of the sphere. Here, r = 12 cm / 2 = 6 cm = 0.06 m, and V = 8000 V.
So, the electric field near the surface is E = 8000 V / 0.06 m = 133333.33 V/m.
The energy density (u) in the electric field is given by
u = ½ ε₀ E²,
where ε₀ is the permittivity of free space (ε₀ = 8.85 × 10^-12 C²/N·m²).
Substituting the value of E, we get
u = ½ × 8.85 × 10^-12 C²/N·m² × (133333.33 V/m)²
u = ½ × 8.85 × 10^-12 × 1.7778 × 10^10,
u = 7.885 × 10^-3 J/m³.
The energy density in the electric field near the surface of the sphere is approximately 7.885 × 10^-3 J/m³.