- #1

CopyOfA

- 35

- 1

**This is a long post. Sorry...**

1. Homework Statement

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,

- ##\mathbf{E}\left(r,t\right) = E_{r}\left(r,t\right) = -\nabla \phi##
- ##\mathbf{D}\left(r,t\right) = D_{r}\left(r,t\right) = \varepsilon(r,\omega)\mathbf{E}\left(r,t\right)##

- Denoting the regions as 1 (##a<r<b##) and 2 (##b<r<c##), $$\phi_{2}\left(c,t\right) - \phi_{1}\left(a,t\right) = V_{0}\cos\left(\omega t\right)$$
- 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

- At ##r=b##, the induced surface charge density of the dielectric is: $$\sigma_{P}\left(r=b\right) = \hat{r} \cdot \left(\mathbf{P}_1 - \mathbf{P}_2\right) = \hat{r} \cdot \mathbf{P}_1 = \hat{r} \cdot \varepsilon_{0}\chi_{1}\left(\omega\right)\mathbf{E}_1\left(b,t\right) = \left(\varepsilon\left(\omega\right) - \varepsilon_0\right)E_{r}\left(b,t\right)$$
- At ##r=a##, the free surface charge density is: $$ \sigma_{f}\left(a\right) = \hat{r}\cdot \mathbf{D}_{1}\left(a,t\right) = \varepsilon\left(\omega\right)E_{r}\left(a,t\right)$$

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*}$$

Then using the equations for free and induced surface charge density,

$$\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?