Energy Density of Fields Calculator

The total electromagnetic energy density ($u$) in a vacuum is the sum of the electric field energy density and the magnetic field energy density. Mathematically, it is expressed as:

$u = \frac{1}{2} \varepsilon_0 E^2 + \frac{1}{2 \mu_0} B^2$

Where $\varepsilon_0$ represents the permittivity of free space ($8.854 \times 10^{-12} \text{ F/m}$), $E$ is the electric field strength in volts per meter or Newtons per Coulomb, $\mu_0$ is the permeability of free space ($4\pi \times 10^{-7} \text{ H/m}$), and $B$ is the magnetic field strength in teslas. Each component contributes independently to the overall energy stored in the field.

Consider a region of space where an electric field ($E$) of $100,000 \text{ N/C}$ and a magnetic field ($B$) of $0.05 \text{ T}$ are simultaneously present. Let us calculate the total energy density.

First, calculate the electric energy density component: $\frac{1}{2} (8.854 \times 10^{-12}) \times (100,000)^2 = 0.5 \times 8.854 \times 10^{-12} \times 10,000,000,000 = 0.04427 \text{ J/m}^3$.

Next, calculate the magnetic energy density component: $\frac{1}{2 \times (4\pi \times 10^{-7})} \times (0.05)^2 = \frac{0.0025}{2.5133 \times 10^{-6}} = 994.72 \text{ J/m}^3$.

Adding both values together gives a total energy density of approximately $994.76 \text{ J/m}^3$, demonstrating that the magnetic field dominates the energy storage in this specific scenario.

Always double-check your input units before running the calculation; electric fields must be in Newtons per Coulomb (or Volts per meter) and magnetic fields must be in Teslas. Keep in mind that in many standard capacitor or inductor problems, only one field type is active, meaning the other variable can be safely set to zero. Finally, remember that dielectric materials alter the permittivity constant, so standard vacuum formulas will require adjustments if analyzing fields inside polarized matter.