# Spherical Capacitor with Frequency Dependent Dielectric

This is a long post. Sorry...

1. Homework Statement

We are given a spherical capacitor with an inner conductor of radius ##a## and outer conductor of radius ##c##. The space between the conductors is half filled (##a<r<b##) with a dielectric with permittivity ##\varepsilon\left(\omega\right)## and vacuum in the region ##b<r<c##. The potential difference between the conductors is ##V\left(t \right) = \phi\left(c,t \right) - \phi\left(a,t \right) = V_{0}\cos\left(\omega t\right)##. Find the electric field and the electric flux density in the regions ##a<r<b## and ##b<r<c##. Also find the free and induced surface charge density residing on the surfaces ##r=a##, ##r=b##, and ##r=c##.

## Homework Equations

This is a spherically symmetric problem, so the electric field and electric flux density will be functions of ##r## only. Since there is no free charge inside the sphere,
$$\nabla^2 \phi\left(r,t\right) = 0$$ Also,
1. ##\mathbf{E}\left(r,t\right) = E_{r}\left(r,t\right) = -\nabla \phi##
2. ##\mathbf{D}\left(r,t\right) = D_{r}\left(r,t\right) = \varepsilon(r,\omega)\mathbf{E}\left(r,t\right)##
The boundary conditions are:
1. Denoting the regions as 1 (##a<r<b##) and 2 (##b<r<c##), $$\phi_{2}\left(c,t\right) - \phi_{1}\left(a,t\right) = V_{0}\cos\left(\omega t\right)$$
2. At ##r=b##, $$\mathbf{D}_{1} = \mathbf{D}_2 \Longrightarrow \varepsilon\left(\omega\right)\nabla \phi_{1}\left(b,t\right) = \varepsilon_{0}\nabla \phi_{2}\left(b,t\right)$$ and $$\phi_{1}\left(b,t\right) = \phi_{2}\left(b,t\right)$$

## The Attempt at a Solution

1. At ##r=b##, the induced surface charge density of the dielectric is: $$\sigma_{P}\left(r=b\right) = \hat{r} \cdot \left(\mathbf{P}_1 - \mathbf{P}_2\right) = \hat{r} \cdot \mathbf{P}_1 = \hat{r} \cdot \varepsilon_{0}\chi_{1}\left(\omega\right)\mathbf{E}_1\left(b,t\right) = \left(\varepsilon\left(\omega\right) - \varepsilon_0\right)E_{r}\left(b,t\right)$$
2. At ##r=a##, the free surface charge density is: $$\sigma_{f}\left(a\right) = \hat{r}\cdot \mathbf{D}_{1}\left(a,t\right) = \varepsilon\left(\omega\right)E_{r}\left(a,t\right)$$
The solution to ##\nabla^2 \phi\left(r,t\right) = 0## is the Legendre polynomials, and for a sphere: $$\phi\left(r,t\right) = \begin{cases} \dfrac{-A_1}{r} + B_1 & a<r<b\\ \dfrac{-A_1}{r} + B_1 & b<r<c \end{cases}$$ Hence, in region (1), $$E_{r,1}\left(r,t\right) = -\nabla \phi_{1} = -\dfrac{A_1}{r^2}$$ and $$D_{r,1}\left(r,t\right) = \varepsilon\left(\omega\right)E_{r,1} = -\varepsilon\left(\omega\right)\dfrac{A_1}{r^2}$$ In region (2), $$E_{r,2}\left(r,t\right) = -\dfrac{A_2}{r^2}$$ and $$D_{r,2}\left(r,t\right) = -\varepsilon_{0}\dfrac{A_2}{r^2}$$ Using boundary conditions: numer (2) yields $$-\varepsilon\left(\omega\right)\dfrac{A_1}{b^2} = -\varepsilon_{0}\dfrac{A_2}{b^2} \Longrightarrow A_{2} = \dfrac{\epsilon\left(\omega\right)}{\varepsilon_{0}}A_1$$ and $$\dfrac{-A_1}{b} + B_1 = \dfrac{-A_2}{b} + B_2\\\dfrac{-A_1}{b} + B_1= \dfrac{-1}{b}\left( \dfrac{\epsilon\left(\omega\right)}{\varepsilon_{0}}A_1\right)+ B_2\\ \dfrac{A_1}{b}\left( \dfrac{\epsilon\left(\omega\right)}{\varepsilon_{0}} - 1\right) = B_2 - B_1$$ Using B.C. number (1) yields: $$V_0 \cos\left(\omega t\right) = \dfrac{-A_2}{c} + B_2 + \dfrac{A_1}{a} - B_1 = -\dfrac{1}{c}\left(\dfrac{\epsilon\left(\omega\right)}{\varepsilon_{0}}A_1\right) + \dfrac{A_1}{a} + \dfrac{A_1}{b}\left( \dfrac{\epsilon\left(\omega\right)}{\varepsilon_{0}} - 1\right)\\ V_{0}\cos\left(\omega t\right) = \left[\dfrac{\varepsilon\left(\omega\right)}{\varepsilon_0}\left(\dfrac{1}{b} - \dfrac{1}{c}\right) + \left(\dfrac{1}{a} - \dfrac{1}{b}\right)\right]A_1 = K_1 A_1$$
Hence $$A_1 = \dfrac{V_0}{K_1}\cos\left(\omega t\right)$$ $$A_2 = \dfrac{\epsilon\left(\omega\right)}{\varepsilon_{0}} \dfrac{V_0}{K_1}\cos\left(\omega t\right) = \left[\dfrac{1}{b} - \dfrac{1}{c} + \dfrac{\varepsilon_0}{\varepsilon\left(\omega\right)}\left(\dfrac{1}{b} - \dfrac{1}{a}\right)\right]^{-1} V_0 \cos\left(\omega t\right) = \dfrac{V_0}{K_2}\cos\left(\omega t\right)$$
Since we need the difference $$B_2 - B_1 = \dfrac{A_1}{b}\left( \dfrac{\epsilon\left(\omega\right)}{\varepsilon_{0}} - 1\right)$$ we can choose ##B_1 = 0## and then $$B_2 = \dfrac{A_1}{b}\left( \dfrac{\epsilon\left(\omega\right)}{\varepsilon_{0}} - 1\right) = \left(\dfrac{\varepsilon\left(\omega\right)}{\varepsilon_0} - 1\right)\dfrac{V_0}{b K1}\cos\left(\omega t\right)$$
Finally, $$\phi_{1}\left(r,t\right) = -\dfrac{V_0}{K_1 r}\cos\left(\omega t\right) = \dfrac{abc\varepsilon_0}{\varepsilon\left(\omega\right)\left(ab-ac\right) + \varepsilon_{0}\left(bc-ac\right)}\dfrac{V_0}{r}\cos\left(\omega t\right)$$
and, $$\phi_{2}\left(r,t\right) = -\dfrac{V_0}{K_2}\cos\left(\omega t\right) + \left(\dfrac{\varepsilon\left(\omega\right)}{\varepsilon_0} - 1\right)\dfrac{V_0}{b K1}\cos\left(\omega t\right) \\ \phi_{2}\left(r,t\right) = \dfrac{ac\left(r\left(\epsilon\left(\omega\right) - \varepsilon_0\right) - b\varepsilon\left(\omega\right)\right)}{\varepsilon_0\left(a c - b c\right) + \varepsilon\left(\omega\right)\left(a c - a b\right)}\dfrac{V_0}{r}\cos\left(\omega t \right)$$

Using these results,
\begin{align*} E_{r}\left(r,t\right) &= \begin{cases} -\dfrac{V_0}{K r^2}\cos\left(\omega t\right) & a<r<b\\ \mbox{}\\ -\dfrac{\varepsilon\left(\omega\right)}{\varepsilon_0}\dfrac{V_0}{Kr^2}\cos\left(\omega t\right), & b<r<c \end{cases}\\ D_{r}\left(r,t\right) &= \begin{cases} -\varepsilon\left(\omega\right)\dfrac{V_0}{K r^2}\cos\left(\omega t\right) & a<r<b\\ \mbox{}\\ -\varepsilon\left(\omega\right)\dfrac{V_0}{Kr^2}\cos\left(\omega t\right), & b<r<c \end{cases} \end{align*}

\begin{align*} \sigma_f\left(a\right) &= -\varepsilon\left(\omega\right)\dfrac{V_0}{K a^2}\cos\left(\omega t\right)\\ \sigma_P\left(b\right) &= \left(\varepsilon_0 - \varepsilon\left(\omega\right)\right)\dfrac{V_0}{K b^2}\cos\left(\omega t\right) \end{align*}

The surface charge density at ##r=c## will be equal and opposite that at ##r=a##.

Using these quantities to calculate the capacitance yields:

\begin{align*} C = \dfrac{Q}{\Delta V} = \dfrac{\sigma_f A_S}{V_0 \cos\left(\omega t\right)} = \dfrac{4\pi abc \varepsilon\left(\omega\right)\varepsilon_0}{\varepsilon_{0}\left(bc -ac\right) + \varepsilon\left(\omega\right)\left(ab-ac\right)} \end{align*}

My professor gave an instruction at the end of the problem saying: "To simplify notation you may introduce as an intermediate quantity the charge on the outer sphere provided you eventually give an expression for it in terms of the voltage." I don't know if I am making some wrong assumptions or why I would need this intermediate term as I already found the surface charge densities. Any suggestions?