## Appendix A Derivation of Plasmon Frequencies for Finite Electrolyte System
### Surface Ionic Plasmons
\begin{array}{l}\frac{\partial}{\partial a}\frac{1}{|\mathbf{a}-\mathbf{r}_ {1}|}=\left\{\begin{array}{l}\sum\limits_{l=0}^{\infty}\frac{ld^{l-1}}{r_{1 }^{l+2}}P_{l}(cos\gamma),\;for\;a<r_{1},\\ -\sum\limits_{l=0}^{\infty}\frac{(l+1)r_{1}^{l}}{d^{l+2}}P_{l}(cos\gamma),\; for\;a>r_{1},\end{array}\right.\end{array}  \tag{11}\]  
where \(\mathbf{a}=a\frac{r}{r}\), \(cos\gamma=\frac{\mathbf{a}\cdot\mathbf{r}_{1}}{ar_{1}}\). Employing Equation (11) the last two terms in Equation (A8) can be transformed as follows,  
\[ \begin{array}{l}-\frac{\omega_{p}^{2}}{4\pi}\int\limits_{l}^{L}r^{2}dr \delta(a-r)\frac{\partial}{\partial r}\int\limits_{a}^{\infty}r_{1}^{2}dr_{1} \int d\Omega_{1}\frac{\delta\rho_{2}(\mathbf{r}_{1})}{|\mathbf{r}-\mathbf{r}_ {1}|}\\ =-\frac{\omega_{p}^{2}}{4\pi}a^{2}\int d\Omega_{1}\int\limits_{a}^{\infty}r_{1 }^{2}dr_{1}\delta\rho_{2}(\mathbf{r}_{1})\frac{\partial}{\partial a}\frac{1}{ \sqrt{a^{2}+r_{1}^{2}-2 ar_{1}cos\gamma}}\\ =-\frac{\omega_{p}^{2}}{4\pi}a^{2}\int d\Omega_{1}\int\limits_{a}^{\infty}r_{1 }^{2}dr_{1}\sigma(\Omega_{1})\delta(a+0+-r_{1})\sum\limits_{l=0}^{\infty} \frac{ld^{l-1}}{r_{1}^{l+1}}P_{l}(cos\gamma)\\ =-\frac{\omega_{p}^{2}}{4\pi}a^{2}\int d\Omega_{1}\sigma(\Omega_{1})\frac{1}{ a^{2}}\sum\limits_{l=0}^{\infty}\frac{4\pi l}{2l+1}\sum\limits_{m=-l}^{l}Y_{lm}( \Omega)Y_{lm}^{*}(\Omega_{1})\\ =-\omega_{p}^{2}a^{2}\sum\limits_{l=0}^{\infty}\sum\limits_{m=-l}^{l}\frac{l}{ l+1}Y_{lm}(\Omega)\int d\Omega_{1}\sigma(\Omega_{1})Y_{lm}^{*}(\Omega_{1}),\end{array}  \tag{12}\]  
and  
\[ \begin{array}{l}-\frac{\omega_{p}^{2}}{4\pi}\int\limits_{l}^{L}r^{2}dr \delta(a-r)\frac{\partial}{\partial r}\int\limits_{0}^{a}r_{1}^{2}dr_{1}\int d \Omega_{1}\frac{\delta\rho_{1}(\mathbf{r}_{1})}{|\mathbf{r}-\mathbf{r}_{1}|} \\ =-\frac{\omega_{p}^{2}}{4\pi}a^{2}\int d\Omega_{1}\int\limits_{0}^{a}r_{1}^{2} dr_{1}nF(\mathbf{r}_{1},t)(\mathbf{r}_{1})\frac{\partial}{\partial a}\frac{1}{ \sqrt{a^{2}+r_{1}^{2}-2 ar_{1}cos\gamma}}\\ =\frac{\omega_{p}^{2}}{4\pi}a^{2}\int d\Omega_{1}nF(\mathbf{r}_{1},t)\sum \limits_{l=0}^{\infty}\frac{(l+1)r_{1}^{l}}{d^{l+2}}P_{l}(cos\gamma)\\ =\omega_{p}^{2}n\sum\limits_{l=0}^{\infty}\frac{l+1}{2l+1}Y_{lm}(\Omega)\int \limits_{0}^{a}r_{1}^{2}dr_{1}\frac{l_{1}^{l}}{d^{l}}\sum\limits_{l_{1}=0}^{ \infty}\sum\limits_{m_{1}=-l_{1}}^{l_{1}}\\ \sum\limits_{i}A_{1 mi}j_{l_{1}}(k_{l_{1}i^{*}}r_{1})sin(\omega_{l_{1}i^{*}}t) \int d\Omega_{1}Y_{lm}^{*}(\Omega_{1})Y_{lm_{1}m_{1}}(\Omega_{1})\\ =\omega_{p}^{2}n\sum\limits_{l=0}^{\infty}\sum\limits_{m=-l}^{l}\sum\limits_{i} \frac{l+1}{2l+1}Y_{lm}(\Omega)A_{1 mi}\int\limits_{0}^{a}\frac{r_{1}^{l+2}dr_{1 }}{d^{l}}j_{l}(k_{li}r_{1})sin(\omega_{li}t).\end{array}  \tag{13}\]  
Equation (A8) attains thus the form,  
\[ \begin{array}{l}\frac{\partial^{2}\sigma(\Omega_{l})}{\partial t^{2}}=-\omega_ {p}^{2}a^{2}\sum\limits_{l=0}^{\infty}\sum\limits_{m=-l}^{l}\frac{l}{2l+1}Y_{ lm}(\Omega)\int d\Omega_{1}\sigma(\Omega_{1})Y_{lm}^{*}(\Omega_{1})\\ +\omega_{p}^{2}n\sum\limits_{l=0}^{\infty}\sum\limits_{m=-l}^{l}\sum\limits_{i} \frac{l+1}{2l+1}Y_{lm}(\Omega)A_{1 mi}\int\limits_{0}^{a}\frac{r_{1}^{l+2}dr_{1 }}{d^{l}}j_{l}(k_{li}r_{1})sin(\omega_{li}t),\end{array}  \tag{14}\]Assuming now, \(\sigma(\Omega,t)=\sum\limits_{l=0}^{\infty}\sum\limits_{m=-l}^{l}q_{lm}(t)Y_{lm}(\Omega)\) and putting it to the above equation, we obtain,  
\[ \begin{array}{l}\sum\limits_{l=0}^{\infty}\sum\limits_{m=-l}^{l}Y_{lm}(\Omega) \frac{\partial^{2}q_{lm}(t)}{\partial t^{2}}=-\sum\limits_{l=0}^{\infty}\sum \limits_{m=-l}^{l}\frac{\omega_{p}^{2}l}{2l+1}Y_{lm}(\Omega)q_{lm}(t)\\ +\omega_{p}^{2}\sum\limits_{l=1}^{\infty}\sum\limits_{m=-l}^{l}\sum\limits_{i} \frac{l+1}{2l+1}Y_{lm}(\Omega)A_{lm}\int\limits_{0}^{a}\frac{r_{l}^{l+2}dr_{l} }{d^{l+2}}j_{l}(k_{li}r_{1})sin(\omega_{li}t).\end{array}  \tag{100}\]  
From the above equation we notice that for \(l=0\) we get \(\frac{\partial^{2}q_{00}}{\partial t^{2}}=0\) and thus \(q_{00}(t)=0\) (as \(q(0)=0\) and \(\lim_{t\rightarrow\infty}q(t)<\infty\)).