# Spherical Capacitor Discharging Through Radial Resistor

## Homework Statement

A spherical capacitor has internal radius $a$ and external radius $b$. At time $t = 0$, the charge of the
capacitor is $Q_0$ Then the two shells are connected by a resistor in the radial direction of resistance $R$. Find the Poynting vector and the energy dissipation in a small volume around the resistor and compare it to the time variation of the electrostatic energy.

## Homework Equations

\begin{aligned} \nabla \times \vec{\mathbf{B}} - \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \mu_0 \vec{\mathbf{j}} \\ \nabla \cdot \vec{\mathbf{E}} & = \frac{\rho}{\epsilon_0} \\ \nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\ \nabla \cdot \vec{\mathbf{B}} & = 0 \end{aligned}

## The Attempt at a Solution

When the resistor is connected to the shells the system becomes an RC circuit, so the charge on the capacitor is now $Q(t) = Q_0 \exp{(-t/\tau)}$ with $\tau =RC = \frac{4\pi R}{\frac{1}{a} -\frac{1}{b}}$. This results in a flow of current but is the first source of confusion for me. Does the current flow through the resistor connecting the shells? If so then we have a radial current $\vec{\mathbf{j}} =-\frac{dQ}{dt} \vec{\hat{r}}$. The electric field between the shells is also changing as the charge decreases so there is a displacement current determined by $\vec{\mathbf{E(t)}} = \frac{Q(t)}{r^2 \epsilon_0} \vec{\hat{r}}$. Using these I could find the magnetic field around the resistor from Maxwell's equations and then Poynting's vector by taking a cross product and then integrate for a surface surrounding the resistor to compare against the time derivative of the electrostatic energy $U = \frac{Q(t)^2}{2C}$. But before proceeding further I want to make sure that I'm understanding the physics here properly.

## Answers and Replies

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## Homework Statement

A spherical capacitor has internal radius $a$ and external radius $b$. At time $t = 0$, the charge of the
capacitor is $Q_0$ Then the two shells are connected by a resistor in the radial direction of resistance $R$. Find the Poynting vector and the energy dissipation in a small volume around the resistor and compare it to the time variation of the electrostatic energy.

## Homework Equations

\begin{aligned} \nabla \times \vec{\mathbf{B}} - \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \mu_0 \vec{\mathbf{j}} \\ \nabla \cdot \vec{\mathbf{E}} & = \frac{\rho}{\epsilon_0} \\ \nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\ \nabla \cdot \vec{\mathbf{B}} & = 0 \end{aligned}

## The Attempt at a Solution

When the resistor is connected to the shells the system becomes an RC circuit, so the charge on the capacitor is now $Q(t) = Q_0 \exp{(-t/\tau)}$ with $\tau =RC = \frac{4\pi R}{\frac{1}{a} -\frac{1}{b}}$.
Are you using units where $\epsilon_0=1$? If not, you are missing some factors there.

To answer your question, yes, the current will be flowing through the resistor.

So then the rest of the solution would proceed `as I've outlined? Use the expression for curl in spherical coordinates, find which component of B is non-zero and then turn the crank to find Poynting's vector and compare with the time derivative of the electrostatic energy? Regarding units I'm missing $\mu_0$ and $\epsilon_0$ in the Ampere-Maxwell law.