Physics · Advanced university
Vacuum electric-field energy density
This asks for the electric contribution only. A magnetic field would add B²/(2μ₀).
Method and conditions
This asks for the electric contribution only. A magnetic field would add B²/(2μ₀).
uE = ε₀E²/2
Worked examples
In vacuum E=24 kV/m. Use ε₀=8.854×10⁻¹² F/m. Find electric-field energy density in μJ/m³. Round the final result to 4 decimal places if needed. Answer in μJ/m³.
Convert E to 24×10³ V/m. Compute ε₀E²/2 in J/m³, then multiply by 10⁶.
2549.952 μJ/m³
In vacuum E=97 kV/m. Use ε₀=8.854×10⁻¹² F/m. Find electric-field energy density in μJ/m³. Round the final result to 4 decimal places if needed. Answer in μJ/m³.
Convert E to 97×10³ V/m. Compute ε₀E²/2 in J/m³, then multiply by 10⁶.
41653.643 μJ/m³
Try it
In vacuum E=45 kV/m. Use ε₀=8.854×10⁻¹² F/m. Find electric-field energy density in μJ/m³. Round the final result to 4 decimal places if needed. Answer in μJ/m³.
In vacuum E=74 kV/m. Use ε₀=8.854×10⁻¹² F/m. Find electric-field energy density in μJ/m³. Round the final result to 4 decimal places if needed. Answer in μJ/m³.
Check your work
Answers and marking notes
- 8964.675 μJ/m³
Convert E to 45×10³ V/m. Compute ε₀E²/2 in J/m³, then multiply by 10⁶.
- 24242.252 μJ/m³
Convert E to 74×10³ V/m. Compute ε₀E²/2 in J/m³, then multiply by 10⁶.
Further reading and review references
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