Spherical Capacitor Discharging Through Radial Resistor

In summary: Then I also need to worry about the energy density of the fields which is something like ##\frac{1}{2} \left( \epsilon_0 E^2 +\frac{1}{\mu_0} B^2 \right)##.In summary, in this scenario, a spherical capacitor with internal radius a and external radius b is connected by a resistor in the radial direction with resistance R. The charge on the capacitor at time t=0 is Q0. Using the RC circuit model, the charge on the capacitor can be described by Q(t) = Q0exp(-t/tau), where tau is the time constant given by RC = 4piR/(1/a - 1/b). The
  • #1
GL_Black_Hole
21
0

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.
 
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  • #2
GL_Black_Hole said:

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.
 

1. What is a spherical capacitor?

A spherical capacitor is a type of capacitor that consists of two concentric spherical conductors, separated by a dielectric material. It is used to store electrical energy and has a capacitance that depends on the size and distance between the two conductors.

2. How does a spherical capacitor discharge through a radial resistor?

When a spherical capacitor is connected to a circuit with a radial resistor, the stored electrical energy in the capacitor begins to flow through the resistor. This causes a decrease in voltage and current over time until the capacitor is fully discharged.

3. What is the difference between a spherical capacitor and a cylindrical capacitor?

The main difference between a spherical and cylindrical capacitor is the shape of their conductors. A spherical capacitor has two spherical conductors while a cylindrical capacitor has two parallel cylindrical conductors. The capacitance of a spherical capacitor also depends on the distance between the two conductors, while the capacitance of a cylindrical capacitor depends on the length and radius of the two conductors.

4. How does the capacitance of a spherical capacitor affect the discharging process?

The capacitance of a spherical capacitor determines the amount of electrical energy that can be stored. A higher capacitance means that a larger amount of energy needs to be dissipated through the radial resistor during the discharging process. Therefore, a higher capacitance will result in a longer discharging time.

5. What factors can affect the discharging time of a spherical capacitor?

The discharging time of a spherical capacitor can be affected by factors such as the capacitance, resistance of the radial resistor, and the initial voltage of the capacitor. The discharging time can also be influenced by the type of dielectric material used and the temperature of the capacitor and resistor.

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