Simply, why is the energy density in a dielectric medium = $\frac{1}{2} K \epsilon_o E^2$?
For a simple case such as that of a capacitor with a dielectric medium inside it, to find the magnitude of induced charge, the net electric field can viewed as: $\vec{\text{E}}_{\text{net}} = \vec{\text{E}}_{\text{o}} + \vec{\text{E}}_{\text{p}}$ where $\vec{\text{E}}_{\text{o}}$ is the electric field due to charges on the capacitor plate, and $\vec{\text{E}}_{\text{p}}$ is the electric field due to the oppositely induced charges on the surface of the dielectric.
Further, this can be written as $E_{\text{net}} = E_{\text{o}} - E_{\text{p}}$.
Also, by the definition of a dielectric: $\vec{\text{E}}_{\text{net}} = \frac{\vec{\text{E}}_{\text{o}}}{K}$
the previous two equations imply $E_{\text{p}} = E_{\text{o}}(1-\frac{1}{K})$
Since $E_{\text{o}} = \frac{Q}{A\epsilon_o}$ and $E_{\text{p}} = \frac{Q_p}{A\epsilon_o}$, where $Q$ is the charge on capacitor plate and $Q_p$ is the induced charge, we can imply $$Q_p = Q(1-\frac{1}{K})$$
$E_{\text{p}} = \frac{Q_p}{A\epsilon_o}$, as used above, actually means we have substituted the dielectric medium with two plates of $Q_p$ and $-Q_p$ charge at the two surfaces of the dielectric. This is why the denominator shows $\epsilon_o$ rather than $K\epsilon_o$.
Now my question is, why can't the energy density be evaluated following this logic of substituting a dielectric with two plates of induced charge?
If I do so, $energy density = \frac{1}{2}\epsilon_oE_{\text{net}}^2$ and not $\frac{1}{2}K\epsilon_oE_{\text{net}}^2$, so what is the fundamental difference between two plates holding charge $Q_p$ & $-Q_p$, and a dielectric medium? I imagined a dielectric is just a medium that can produce two plates of opposite charge of a magnitude that is related very specifically to the field it is subjected to.
The proof for $\frac{1}{2}K\epsilon_oE_{\text{net}}^2$ given in Dr. HC Verma's, Concepts of Physics is using a capacitor with dielectric K, which will have energy = $\frac{1}{2}CV^2$, where C is $\frac{K\epsilon_oA}{d}$. (The proof for $\frac{1}{2}CV^2$ was calculating work done in pulling apart two oppositely charged plates from zero separation to separation = d.)
$$u = \frac{U}{Ad} = \frac{\frac{1}{2}\frac{K\epsilon_oA}{d}V^2}{Ad} = \frac{1}{2}K\epsilon_oE_{\text{net}}^2$$
To conclude, my question is, why can we view a dielectric as plates of opposite charge for the calculation of field, but not for the calculation of energy density? Besides, I have always learned $\int\frac{1}{2}\epsilon_oE_{\text{net}}^2*dV$ includes all potential energy which manifests in the form of interaction energy, self-energy, etc. If we have already considered the contribution of the dielectric in the net field, why again must we multiply $\epsilon_o$ with K? However, intuitively it does feel correct that making a field in a dielectric would require more energy than making the same field in air.
