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Let's say that I am inducing a sinusoidal electric field inside the dielectric. The dielectric is represented by the following relative permittivity:

$$\tilde{\epsilon}_r = 1+ i \ \frac {\sigma}{\omega \epsilon_0}$$

It is complex due to delay of the dielectric with the varying electric field.

How do I find the energy density at a particular instance?

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1 Answer 1

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The instantaneous electric energy at time $t$ is given by the well known formula:

$$\mathscr U_e(t) =\frac 12 \int_V d^3\mathbf x \ \mathbf D(\mathbf x,t)\mathbf{.E}(\mathbf x,t)$$ Now since you are working in the frequency domain (you are assuming a single frequency $\omega$), we can convert the above equation into one in terms of the phasor of the field. Let's consider the time average of the above energy over one period, which is what people mostly refer to as electric energy, without explicitly stating the averaging. $$U_e \equiv \langle\mathscr U_e(t) \rangle =\frac 12\int_V d^3\mathbf x \ \langle \mathbf D(\mathbf x,t)\mathbf{.E}(\mathbf x,t) \rangle$$ I claim that $\langle \mathbf D(\mathbf x,t)\mathbf{.E}(\mathbf x,t) \rangle = \frac 12 \mathscr Re \bigr (\mathbf D(\mathbf x) \mathbf {.E^*} (\mathbf x) \bigl)$. Where $\mathbf D(\mathbf x)$ and $\mathbf E(\mathbf x)$ are the phasors of the two fields, i.e. $\mathbf D(\mathbf x,t)=\mathscr Re \bigr (\mathbf D(\mathbf x) \ e^{-i\omega t} \bigl)$ and $\mathbf E(\mathbf x,t)=\mathscr Re \bigr (\mathbf E(\mathbf x) \ e^{-i\omega t} \bigl)$. To see this, let's evaluate the left hand side by expanding the vectors in components: $$\langle \mathbf D(\mathbf x,t)\mathbf{.E}(\mathbf x,t) \rangle =\sum_{i=1}^3 \langle D_i(\mathbf x,t)E_i(\mathbf x,t)\rangle = \sum_{i=1}^3 \langle \mathscr Re \bigr (D_i(\mathbf x)e^{-i\omega t} \bigl) \mathscr Re \bigr (E_i(\mathbf x)e^{-i\omega t} \bigl)\rangle= \sum_{i=1}^3 \langle |D_i||E_i| \mathscr Re \bigl(e^{-i(\omega t - \phi_i)} \bigr)\mathscr Re \bigl(e^{-i(\omega t - \theta_i)} \bigr)\rangle $$ Where I have expressed the complex variables $D_i$ and $E_i$ in terms of their magnitude and phase ($\phi_i$ and $\theta_i$). This gives: $$\langle \mathbf D(\mathbf x,t)\mathbf{.E}(\mathbf x,t) \rangle= \sum_{i=1}^3 |D_i||E_i| \langle \cos \bigl(\omega t - \phi_i \bigr)\cos \bigl(\omega t - \theta_i \bigr) \rangle = \sum_{i=1}^3 |D_i||E_i| \frac 12 \biggr( \langle \cos \bigl(2\omega t - \phi_i - \theta_i \bigr) \rangle + \langle \cos \bigl(\phi_i - \theta_i \bigr) \rangle \biggl ) = \sum_{i=1}^3 |D_i||E_i| \frac 12 \cos \bigl(\phi_i - \theta_i \bigr)= \frac 12 \mathscr Re \biggl (\sum_{i=1}^3 |D_i| e^{i\phi_i} |E_i|e^{-i \theta_i} \biggr )=\frac 12 \mathscr Re \biggl (\sum_{i=1}^3 D_i(\mathbf x) E_i^*(\mathbf x) \biggr ) $$ Thus our claim has been proven: $$\langle \mathbf D(\mathbf x,t)\mathbf{.E}(\mathbf x,t) \rangle = \frac 12 \mathscr Re \bigr (\mathbf D(\mathbf x) \mathbf {.E^*} (\mathbf x) \bigl)$$ With that in mind, substituting in our $U_e$ relation, we get: $$U_e =\frac 14 \mathscr Re\int_V d^3\mathbf x \ \mathbf D(\mathbf x)\mathbf{.E^*}(\mathbf x) $$ Which is our final equation for the total (time-averaged) electric energy. Equivalently, one can identify the energy density as: $$u_e =\frac 14 \mathscr Re \bigl( \mathbf D(\mathbf x)\mathbf{.E^*}(\mathbf x) \bigr)$$ The complex permittivity given in your question can be easily introduced in this new equation from $\mathbf D(\mathbf x) = \epsilon_0 \tilde{\epsilon}_r \mathbf E(\mathbf x)$. I will leave the leftover calculations for you to go through.

Also note that we have calculated the time averaged energy density. As mentioned above, this is usually what is meant when referring to the energy density in time-varying problems. However, if you want the complete instantaneous energy density at any given time, you will have to find the complete field (phasor) distribution using your $\tilde {\epsilon}_r$ relation (and boundary conditions, etc.), and then calculate the time domain fields and use our first equation for the instantaneous energy. This is because there is no practical equation connecting $\mathscr U_e(t)$ to the field phasors.

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