Step 1: Start from the work done to charge the capacitor.
When charge is transferred to build up a field, the total electrostatic energy stored is \(U = \dfrac{Q^2}{2C}\), where \(Q = CV\) is the final charge.
Step 2: Express charge through the surface charge density.
For a parallel plate capacitor the charge is \(Q = \sigma A\), where \(\sigma\) is the surface charge density and \(A\) the plate area. The field between the plates is related to the charge density by \(E = \dfrac{\sigma}{\varepsilon_0}\), so \(\sigma = \varepsilon_0 E\) and hence \(Q = \varepsilon_0 E A\).
Step 3: Write capacitance and combine.
Using \(C = \dfrac{\varepsilon_0 A}{d}\),
\(U = \dfrac{Q^2}{2C} = \dfrac{(\varepsilon_0 E A)^2}{2\,(\varepsilon_0 A/d)} = \dfrac{\varepsilon_0^2 E^2 A^2 d}{2\,\varepsilon_0 A} = \dfrac{1}{2}\varepsilon_0 E^2 A d.\)
Step 4: Interpret as energy per unit volume.
The field fills the region of volume \(A d\) between the plates. Dividing,
\(u = \dfrac{U}{A d} = \dfrac{1}{2}\varepsilon_0 E^2.\)
Step 5: Conclusion.
This confirms the field itself stores energy with density
\[\boxed{u = \frac{1}{2}\varepsilon_0 E^2}\]