# Prospects for the Development of High Energy Density Dielectric Capacitors
## 3 Analysis of Dielectric Cells Using Nanoparticles Embedded in a Host Polymer
### Polymer Composites Using High Dielectric Constant Particles  
In this section of the paper, the case of a polymer composite using high dielectric constant particles is analyzed. In practice, ceramic particles like BaTiO\({}_{3}\) are used. These particles are not conductors, but charge and energy are stored in the particles due to polarization of the molecules in the applied electric field. The same approach as used in the previous analyzes will be used in this analysis.  
The first step will be to analysis the response of a spherical particle in the field E\({}_{0}\). This a classic problem in electrostatics and the solution is shown in most text books [44]. A key aspect of obtaining the solution is the boundary condition (b.c.) at the interface between the particle and the host polymer. The b.c. is  
\[({\rm D}_{2}-{\rm D}_{1}).\ n=0\ {\rm at}\ r=r_{\rm p}\ {\rm for\ dielectrics}\]  
D is a vector and is equal \(\epsilon\)E. Hence  
\[\epsilon_{2}\ {\rm E}_{2n}=\epsilon_{1}\ {\rm E}_{1n}\ {\rm at}\ r=r_{\rm p},\ n\ {\rm is\ the\ normal\ component\ of}\ {\rm E}\]  
Applying the b.c. to the general solution for the potential function U inside the particle  
\[{\rm U}={\rm A}_{2}\ r\ {\rm cos}\theta,\ \theta\ {\rm is\ the\ angle\ between}\ r\ {\rm and}\ {\rm E}_{0}\]  
One finds  
\[{\rm A}_{2}=-3\ {\rm E}_{0}/(\epsilon_{2}/\epsilon_{1}+2),\ {\rm U}=-{\rm A}_{2}\ r\ {\rm cos}\theta\]  
And the electric field E\({}_{2}\) inside the particle is constant and in the direction of E\({}_{0}\).  
\[{\rm E}_{2}=-{\rm A}_{2}=3\ {\rm E}_{0}/(\epsilon_{2}/\epsilon_{1}+2),\ {\rm define}\ {\rm e }_{\rm f}=3/(\epsilon_{2}/\epsilon_{1}+2)\]  
Hence, the electric field inside the particle is greatly reduced.  
Now we can analyze the effect of the particles on the effective dielectric constant \(\epsilon_{3}\) of the composite. The capacitance C\({}_{\rm p}\) of the particle is given by  
\[{\rm C}_{\rm p}=\epsilon_{2}\ \pi\ {\rm r}_{\rm p}{}^{2}\]  
And  
\[{\rm C}_{\rm ptot}={\rm C}_{\rm p}{\rm N}_{\rm p}=\epsilon_{2}\ \pi\ {\rm r}_{\rm p}{}^{2}\ ( \dot{\rm f}_{\rm v}\ A\delta/4/3\pi{\rm r}_{\rm p}{}^{3})\]  
The total charge stored by polarization of the particles is  
\[{\rm Q}_{\rm tot}={\rm C}_{\rm ptot}\ \Delta{\rm V}=\epsilon_{2}\ ({\rm f}_{\rm v}\ A \delta/4/3{\rm r}_{\rm p}{}^{3})({\rm r}_{\rm p}{\rm e}_{\rm f}{\rm E}_{0})\]The charge of the composite polymer is given by  
\[\text{Q}_{\text{polym}}=\varepsilon_{3}\text{ A}/\delta\text{ V}\]  
Hence \(\varepsilon_{3}/\varepsilon_{2}=3/8\left(3/(\varepsilon_{2}/\varepsilon_{1}+2) \right)\text{f}_{\text{v}}\ \delta/\tau_{\text{P}}\).  
Consider the example: \(\text{f}_{\text{v}}=0.2\), \(\delta=2\mu\), \(\tau_{\text{P}}=2\) nm, \(\varepsilon_{2}=3000\), \(\varepsilon_{1}=8\), \(\text{e}_{\text{f}}=0.008\).  
For this case, \(\varepsilon_{3}/\varepsilon_{2}=0.015\) and \(\varepsilon_{3}=45\).  
If the high dielectric constant particles and conducting particles were used to form a polymer composite, the effects of the two types of particles would be multiple. The dielectric constant of the composite is projected to be  
\[\varepsilon_{\text{compos}}/\varepsilon_{\text{host}}=(\varepsilon_{3}/ \varepsilon_{\text{host}})(\varepsilon_{\text{condpt}}/\varepsilon_{3})\]