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A conducting sphere of radius 1 cm is raised to a potential of 8000 V. What is the energy density of the electric field just outside its surface?

1. 64 * 10⁵ J/m³
2. 8 * 10³ J/m³
3. 32 J/m³
4. 2.83 J/m³

Correct answer: 2.83 J/m³

Solution

For a sphere, V = kQ/R and surface field E = kQ/R² = V/R = 8000/0.01 = 8*10⁵ V/m. The electric energy density is u = (1/2)*epsilon0*E² = 0.5 * 8.854*10⁻¹² * (8*10⁵)² = 2.83 J/m³.

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