Question

In the region \(-\infty

Short answer

The relationship is \(\tilde{V}(\alpha) = \frac{\tilde{\rho}(\alpha)}{\epsilon_0 (\alpha^2 + q^2)}. The potential at (x,0,0) is \(\frac{A}{\pi} \int_{-\infty}^{\infty} \frac{\sin kt}{k(k^2+q^2)} dk\).

Step-by-step solution

Understand the given charge-density wave

The charge-density wave is given by \( \rho(\textbf{r})= A \cos q x\). This wave only varies along the x-direction.

Expressing charge-density \(\rho(\textbf{r})\) in integral form

The wave \( \rho(\textbf{r})\) is represented as \(\rho(\textbf{r})=\frac{e^{i q x}}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}\tilde{\rho}(\alpha) e^{i x z} d \alpha\).

Identifying the potential \(V(\textbf{r})\)

The potential \(V(\textbf{r})\) is given by \(V(\textbf{r})=\frac{e^{i q x}}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}\tilde{V}(\alpha) e^{i x z} d \alpha\).

Relation between \(\tilde{V}(\alpha)\) and \(\tilde{\rho}(\alpha)\)

Using Poisson's equation in Fourier space, the potential and charge density are related by: \(abla^2 V = - \frac{\rho}{\epsilon_0}\).In Fourier space, \(\-k^2 \tilde{V}(k) = -\frac{\tilde{\rho}(k)}{\epsilon_0}\), where \(k = \sqrt{\alpha^2+\beta^2+\gamma^2} = \sqrt{\alpha^2+q^2}\), leading to \(\tilde{V}(\alpha) = \frac{\tilde{\rho}(\alpha)}{\epsilon_0 (\alpha^2 + q^2)}\).

Substituting the given \(\tilde{\rho}(\alpha)\) into the potential expression

The charge density given is \(\rho(\mathbf{r})= A \cos q x\), which in Fourier space (\( \tilde{\rho}(\alpha)\)) gives, mathematically, \(\frac{e^{i q x}}{\sqrt{2 \pi}} \tilde{\rho}(\alpha) e^{i x z}\). Substitution into Fourier transformed Poisson eq. gives: \(\tilde{V}(\alpha)=\frac{1}{\epsilon_0(\alpha^2+q^2)}\tilde{\rho}(\alpha)\).

Integrate over \(\alpha\)

Expressing potential at the point (x,0,0), substituting boundary (-t to t) limits for non-zero terms simplify: \(. Finally: V(x,0,0)= \frac{A}{\pi} \int_{-\infty}^{\infty} \frac{\sin (kt)}{k(k^2+q^2)} dk\).

Key concepts

Poisson's equation

Poisson's equation is a fundamental part of electrostatics. It relates the electric potential to the charge density in a region. The equation is given by: \[ abla^2 V = - \frac{\rho}{\text{ε}_0} \] where \(V\) is the electric potential, \(ρ\) is the charge density, and \(ε_0\) is the permittivity of free space. In simple terms, it means that if you know the charge distribution, you can find the resulting electric potential.

• The Laplacian operator \( abla^2 \), appears in the equation. It measures the second spatial derivatives of the potential.
• The term \( - \frac{\rho}{\text{ε}_0} \) signifies the source term; it shows how charges affect the potential field.

Fourier transform

The Fourier transform is a mathematical tool that transforms a function of time (or space) into a function of frequency (or wave number). It is very useful in solving differential equations, including Poisson's equation. The general form of the Fourier transform of a function \(f(x)\) is: \[ \tilde{f}(\beta) = \frac{1}{\text{π}} \bigint_{- \text{∞}}^{\text{∞}} f(x) e^{-ix\beta} dx \]Using the Fourier transform, Poisson's equation in the original (spatial) domain \( abla^2 V = - \frac{\rho}{\text{ε}_0} \) can be converted into the frequency domain: \[ - k^2 \tilde{V}(k) = - \frac{\tilde{\rho}(k)}{\text{ω}_0} \]This approach simplifies many problems, as it turns complicated differential equations into algebraic equations.

Electric potential

The electric potential \(V\) is a scalar quantity representing the electric potential energy per unit charge at a point in space. To find the electric potential caused by a charge distribution, you can use Poisson's equation: \[ abla^2 V = - \frac{\rho}{\text{ε}_0} \]In our specific problem, we are dealing with a charge-density wave: \[ \rho(\textbf{r}) = A \text{ cos} qx \]By using Poisson's equation in Fourier space, we relate the transformed potential and charge density: \[ \tilde{V}(\beta) = \frac{\tilde{\rho}(\beta)}{\text{ω}_0 (\beta^2 + q^2)} \]

• This relationship can be reverted back into the spatial domain to get our electric potential.
• The final result for the potential at point \((x, 0, 0)\) after integrating and simplifying is: \[ V(x, 0, 0) = \frac{A}{ \text{π}} \bigint_{-\text{∞}}^{\text{∞}} \frac{ \text{sin}(kt)}{k(k^2 + q^2)} dk \]