# Spherical Capacitor with Frequency Dependent Dielectric

• CopyOfA
In summary, the problem involves finding the electric field and electric flux density for a spherical capacitor with an inner conductor of radius ##a## and outer conductor of radius ##c##, with a dielectric filling the space between the conductors. 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)## and the task is to find the electric field and electric flux density in the regions ##a<r<b## and ##b<r<c##, as well as the free and induced surface charge densities on the surfaces ##r=a##, ##r=b##
CopyOfA
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?

That approach looks longer than I would expect.
I would follow the hint of the professor, as such a charge variable allows to calculate all variables step by step instead of getting huge systems of equations. You do not need it, but it makes the solution easier.

## 1. What is a spherical capacitor with frequency dependent dielectric?

A spherical capacitor with frequency dependent dielectric is a type of capacitor where the dielectric material, which is the insulating material between the two metal plates, has a varying dielectric constant with respect to frequency. This means that the capacitance of the capacitor will change depending on the frequency of the electric signal passing through it.

## 2. How does the capacitance of a spherical capacitor with frequency dependent dielectric change with frequency?

The capacitance of a spherical capacitor with frequency dependent dielectric will increase with increasing frequency. This is because the dielectric constant of the material also increases with frequency, allowing the capacitor to store more charge at higher frequencies.

## 3. What are some examples of materials that can be used as frequency dependent dielectrics in a spherical capacitor?

Some examples of materials that can be used as frequency dependent dielectrics in a spherical capacitor include semiconductors such as silicon, ferroelectric materials such as barium titanate, and piezoelectric materials such as quartz.

## 4. What are the applications of a spherical capacitor with frequency dependent dielectric?

A spherical capacitor with frequency dependent dielectric can be used in electronics and telecommunications, such as in filters, oscillators, and tunable resonant circuits. It can also be used in scientific research, such as in the study of high frequency phenomena and in precision measurements.

## 5. How is the behavior of a spherical capacitor with frequency dependent dielectric different from a regular capacitor?

The behavior of a spherical capacitor with frequency dependent dielectric is different from a regular capacitor in that its capacitance will vary with frequency, whereas a regular capacitor will have a constant capacitance. This makes it useful in applications where a variable capacitance is needed, such as in tuning circuits.

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