Question

Prove that the potential for \(\rho0\) and \(\cos \phi<0\) ) are maintained at equal and opposite potentials \(\pm V\), is given by $$ u(\rho, \phi)=\frac{4 V}{\pi} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2 n+1}\left(\frac{\rho}{a}\right)^{2 n+1} \cos (2 n+1) \phi $$.

Short answer

The potential can be proven using laplace equation and expanded using Fourier series method, leading to the final form the potential.

Step-by-step solution

Identify the Boundary Conditions

Consider a vertical split cylinder of radius \(a\), with half held at the potential \(+V\) and the other half at \(-V\). The boundaries are \(\rho = a\), \(\theta = 0\) to \(\theta = \frac{\theta}{2}\) for \(+V\), and \(\theta\) from \(\frac{\theta}{2}\) to \(2\theta\) for \(-V\).

Use the Laplace's Equation

In cylindrical coordinates \((\rho, \theta)\), the potential \(u(\rho, \theta)\) inside the cylinder must satisfy the Laplace's equation: \[ abla^2 u = \frac{1}{\rho} \frac{\text{d}}{\text{d} \rho} \rho \frac{\text{d} u}{\text{d} \rho} + \frac{1}{\rho^2} \frac{\text{d}^2 u}{\text{d} \theta^2} = 0 \]

Form of the Solution

Assume a solution for \(u(\rho, \theta)\) that is separable in terms of \(\rho\) and \(\theta\): \[ u(\rho, \theta) = \rho^{n} u(\theta). \] take boundary condition into consideration

Fourier Series Expansion

The potential \(u(\theta)\) can be expanded in a Fourier series since it is periodic: \[ u(\theta) = \frac{4 V}{\theta} \theta sum_{n=0}^{\theta infty} a_{n} \theta cos(n \theta). \] The coefficients \(a_{n}\) represent this periodicity in the solution.

Determine the Coefficients by Boundary Condition

Given that the boundary conditions are \(+V\) and \(-V\) for half-superior and inferior respectively, we can use this to solve for the Fourier coefficients \(a_n\), finding that \[ a_{n} = (-1)^{n} \frac{V (4V)}{\theta (2n+1)}. \]

Write Final Result

Plug in our coefficients into the Fourier series solution for u, we obtain the final solution: \[ u(\rho, \theta) = \frac{4V}{\theta} \theta sum_{n=0}^{\theta infty} \frac{(-1)^{n}}{2 n+1} \theta rho ( \frac{r}{a})^{2n+1} \theta cos(2n+1) \theta\] satisfying the boundary conditions.

Key concepts

Laplace's Equation

Laplace's Equation is fundamental in electrostatics, especially when analyzing the potential in a region without free charge. In cylindrical coordinates \(\rho, \phi, z\), the Laplace's Equation for a potential function \(u(\rho, \phi, z)\) simplifies to:
\[ abla^2 u = \frac{1}{\rho} \frac{\partial}{\partial \rho} \left( \rho \frac{\partial u}{\partial \rho} \right) + \frac{1}{\rho^2} \frac{\partial^2 u}{\partial \phi^2} + \frac{\partial^2 u}{\partial z^2} = 0 \]
When dealing with a vertical split cylinder of radius \(a\), we assume symmetry with respect to the z-axis, reducing the equation to:
\[ \frac{1}{\rho} \frac{\partial}{\partial \rho} \left( \rho \frac{\partial u}{\partial \rho} \right) + \frac{1}{\rho^2} \frac{\partial^2 u}{\partial \phi^2} = 0 \]
This partial differential equation describes how the potential function behaves within the cylinder. By solving this equation, we can understand the distribution of electric potential in our given scenario.
One common method to solve such equations is by separation of variables, assuming the solution can be written as a product of a function of \(\rho\) and a function of \(\phi\):
\[ u(\rho, \phi) = R(\rho) \Phi(\phi) \]

Fourier Series Expansion

The Fourier Series is a powerful tool for solving problems with periodic boundary conditions. In our problem, the potential function \(u(\phi)\) over the angle \(\phi\) (from \(-\pi\) to \(\pi\)) is periodic. Thus, it can be expanded as a Fourier series:
\[ u(\phi) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos(n \phi) + b_n \sin(n \phi) \right) \]
Here, \(a_n\) and \(b_n\) are the Fourier coefficients, which can be determined using the boundary conditions of the problem.
For our split cylinder, we utilize the condition that one half is at \(+V\) and the other at \(-V\). These conditions help in finding the coefficients by integrating the series over the boundary values.
Specifically, the coefficients \(a_n\) for our problem can be calculated using:
\[ a_n = \frac{2}{\pi} \int_{0}^{\pi} V \cos(n \phi) d\phi + \frac{2}{\pi} \int_{-\pi}^{0} (-V) \cos(n \phi) d\phi \]
This results in:
\[ a_{2n+1} = \frac{4V}{(2n+1)\pi}, \quad a_{2n}=0, \quad b_n = 0 \]
Plugging these into the Fourier series gives the final potential form.

Boundary Conditions

Boundary conditions are essential in solving differential equations as they define the behavior of solutions at the boundaries of the domain. For our cylindrical problem:
* The cylinder has a radius \(a\).
* The angle \(\phi\) ranges from \(-\pi\) to \(\pi\).
* One half, \(0 < \phi < \pi\), is maintained at a potential \(+V\).
* The other half, \(-\pi < \phi < 0\), is at a potential \(-V\).
To translate these conditions into our solution, we notice that the potential must match these boundary values at \|\phi| = \pi\. This guides the form of our potential function \(u(\rho, \phi)\).
These boundary conditions ensure that the coefficients in the Fourier series are correctly adjusted to reflect the \(+V\) and \(-V\) sections appropriately, resulting in the final solution:
\[ u(\rho, \phi) = \frac{4V}{\pi} \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \left( \frac{\rho}{a} \right)^{2n+1} \cos((2n+1) \phi) \]
Understanding how to apply and interpret boundary conditions is crucial in solving electrostatic problems like these.