The energy density of an electric field refers to the amount of energy stored per unit volume in the electric field. To derive the expression for the energy density of an electric field, we start with the electric potential energy stored in a system of charges.
Consider a region of space containing a continuous charge distribution with charge density ρ (charge per unit volume). The electric potential energy U associated with this charge distribution within a volume V is given by:
U = ∫(1/2) ε₀ E² dV
Where:
ε₀ is the vacuum permittivity (electric constant), a fundamental constant of nature.
E is the electric field intensity at a given point.
The integral is taken over the volume V.
Now, to obtain the energy density u of the electric field, we need to express the volume V in terms of the electric field E. The relation between electric field, charge density, and permittivity is given by Gauss's law:
∇ · E = ρ / ε₀
Integrating this equation over the volume V using the divergence theorem:
∫(∇ · E) dV = ∫(ρ / ε₀) dV
Applying the divergence theorem:
∫E · dA = Q / ε₀
Where:
Q is the total charge within the volume V.
The integral on the left is taken over the closed surface enclosing the volume V.
Now, divide both sides of the equation by V to get the average electric field magnitude within the volume:
(E_avg) ∫dA = Q / (ε₀ * V)
Recognize that Q/V is the charge density ρ, and the left side integral is the total surface area A of the enclosing surface:
E_avg * A = ρ / ε₀
Solve for E_avg:
E_avg = ρ / (ε₀ * A)
Substitute this expression for E_avg into the electric potential energy equation:
U = ∫(1/2) ε₀ (E_avg)² dV
U = (1/2) ε₀ ∫(ρ / (ε₀ * A))² dV
Since A * dV is the volume element dV', we can express A in terms of dV':
A = dV' / n
Where n is the unit vector normal to the enclosing surface.
Substitute this into the energy equation:
U = (1/2) ε₀ ∫(ρ / (ε₀ * (dV' / n)))² dV
U = (1/2) ε₀ ∫(ρ² / (ε₀² * dV'²)) dV
Now, integrate over the volume V:
U = (1/