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Spherical Capacitor Discharging Through Radial Resistor

  • #1

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

  • #2
nrqed
Science Advisor
Homework Helper
Gold Member
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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.
 
  • #3
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.
 

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