# Understanding dielectrics on the basis of induced charge and energy density

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.

$$\dots$$ why can we view a dielectric as plates of opposite charge for the calculation of field, but not for the calculation of energy density? because you are considering two different aspects of the situation.
• why wouldn't $\int\frac{1}{2}\epsilon_oE_{\text{net}}^2*dV$ take the internal dipoles into account? for example, if I bring multiple artificial dipoles and place them in otherwise empty space in a particular configuration, $\int\frac{1}{2}\epsilon_oE_{\text{net}}^2*dV$ gives the correct total energy. Isn't a dielectric medium just this? the only difference is that an actual dielectric medium would have dipoles much more closely packed. Commented Jul 6 at 9:21