where the integration extends throughout the whole space unoccupied by conductors. We have by partial integration
| ∫ ∫ ∫ ( | dV | ) | ² | dx dy dz = ∫ ∫ V | dV | dy dz − ∫ ∫ ∫ V | d²V | dx dy dz, |
| dx | dx | dx² |
and two similar equations in y and z. Hence
| 1 | ∫ ∫ ∫ {( | dV | ) | ² | + ( | dV | ) | ² | + ( | dV | ) | ² | } dx dy dz = |
| 8π | dx | dy | dz |
| 1 | ∫ ∫ V | dV | dS − | 1 | ∫ ∫ ∫ V∇V dx dy dz |
| 8π | dn | 8π |
(22)
where dV/dn means differentiation along the normal, and ∇ stands for the operator d²/dx² + d²/dy² + d²/dz². Let E be the resultant electric force at any point in the field. Then bearing in mind that σ = (1/4π) dV/dn, and ρ = −(1/4π) ∇V, we have finally
| 1 | ∫ ∫ ∫ E²dv = | 1 | ∫ ∫ Vσ dS + | 1 | ∫ ∫ ∫ Vρ dv. |
| 8π | 2 | 2 |
The first term on the right hand side expresses the energy of the surface electrification of the conductors in the field, and the second the energy of volume density (if any). Accordingly the term on the left hand side gives us the whole energy in the field.
Suppose that the dielectric has a constant K, then we must multiply both sides by K and the expression for the energy per unit of volume of the field is equivalent to ½DE where D is the displacement or polarization in the dielectric.