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An electrostatic potential (mks units) is $$ \varphi(\mathbf{r})=\frac{Z}{4 \pi \varepsilon_{0}} \cdot \frac{e^{-a r}}{r} . $$ Reconstruct the electrical charge distribution that will produce this potential. Note that \(\varphi(r)\) vanishes exponentially for large \(r\), showing that the net charge is zero. $$ \text { ANS. } \rho(r)=Z \delta(r)-\frac{Z a^{2}}{4 \pi} \frac{e^{-a r}}{r} \text { . } $$

Short Answer

Expert verified
The charge distribution is \( \rho(r) = Z \delta(r) - \frac{Z a^2}{4\pi} \frac{e^{-a r}}{r} \).

Step by step solution

01

Identify the Potential Function

Given the electrostatic potential function: \[ \varphi(\mathbf{r})=\frac{Z}{4 \pi \varepsilon_{0}} \cdot \frac{e^{-a r}}{r}.\] This potential function decreases exponentially as \( r \to \infty \). It suggests that the system is neutral with localized charge distribution.
02

Apply Poisson's Equation

To find the charge distribution \( \rho(\mathbf{r}) \) that produces the given potential, use Poisson's equation for electrostatics: \[ abla^2 \varphi(\mathbf{r}) = - \frac{\rho(\mathbf{r})}{\varepsilon_0}.\] Our task is to find \( abla^2 \varphi(\mathbf{r}) \) and then solve for \( \rho(\mathbf{r}) \).
03

Compute the Laplacian in Spherical Coordinates

Since the potential \( \varphi(r) \) is spherically symmetric, use the Laplacian in spherical coordinates:\[ abla^2 \varphi(r) = \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{d\varphi}{dr} \right).\] Substituting \( \varphi(r) = \frac{Z}{4\pi \varepsilon_0} \frac{e^{-ar}}{r} \) into the Laplacian formula gives:\[ \frac{d\varphi}{dr} = \frac{Z}{4\pi \varepsilon_0} \left( -\frac{e^{-ar} (ar + 1)}{r^2} \right).\]
04

Derive the Laplacian Expression

Continue by finding:\[ \frac{d}{dr}\left( r^2 \frac{d\varphi}{dr} \right) = \frac{d}{dr} \left( \frac{-Z \cdot e^{-ar} (ar + 1)}{4\pi \varepsilon_0} \right).\] Differentiate, noticing exponential decrease and factor contributions.
05

Solve for the Charge Distribution \(\rho(r)\)

Simplify the expression to solve for \( abla^2 \varphi \) and, thereby, \( \rho(r) \):\[ \rho(r) = \varepsilon_0 abla^2 \varphi(r) \]Converting back into the expression for \( \rho(r) \), remember the Dirac delta function contribution from central charges. Hence:\[ \rho(r) = Z \delta(r) - \frac{Z a^2}{4\pi} \frac{e^{-ar}}{r}.\] Be sure to acknowledge the properties of the Dirac delta \( \delta(r) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electrostatic potential
Electrostatic potential is a fundamental concept in electrostatics that helps describe the energy required to move a charge within an electric field. In this context, the potential \( \varphi(\mathbf{r}) = \frac{Z}{4 \pi \varepsilon_{0}} \cdot \frac{e^{-a r}}{r} \) is given in spherical coordinates. This expression represents a potential that decreases exponentially with distance \( r \) from the source, indicating localized charges.

The electrostatic potential tells us how a charge would influence the space around it. It's vital to note that this potential is influenced by the constants \( Z \), which can be related to the source charge, \( \varepsilon_0 \), the permittivity of free space, and \( a \), a decay constant.

Understanding electrostatic potential is crucial because it connects electric fields to potential energy, assisting in solving problems relating to electric forces, energy conservation, and electric field distributions.
Charge distribution
Charge distribution describes how electric charge is spread in space and is closely linked to the electrostatic potential. Using Poisson's equation, \( abla^2 \varphi(\mathbf{r}) = - \frac{\rho(\mathbf{r})}{\varepsilon_0} \), we can reconstruct the charge distribution \( \rho(r) \) from the given potential. Poisson’s equation is essential in electrostatics because it describes how charge distribution leads to electric potential, allowing us to calculate charges if we know the potential.

For our potential \( \varphi(r) \), the solution gives \( \rho(r) = Z \delta(r) - \frac{Z a^2}{4\pi} \frac{e^{-ar}}{r} \). This indicates two key components:
  • The term \( Z \delta(r) \) represents a point charge at the origin, with the Dirac delta function \( \delta(r) \) ensuring that this charge is centralized.
  • The term \(-\frac{Z a^2}{4\pi} \frac{e^{-ar}}{r} \) represents a distributed charge that decreases exponentially with distance, aligning with the potential's decay behavior.
Understanding charge distribution is pivotal as it tells us how electric fields are generated and altered within different materials and geometries.
Laplacian in spherical coordinates
The Laplacian is a mathematical operator used to compute the divergence of the gradient, which in physics frequently helps describe the behavior of scalar fields such as electric potential. In spherical coordinates, especially useful for problems with radial symmetry, the Laplacian takes a unique form.

For a spherically symmetric function \( \varphi(r) \), the Laplacian is expressed as:
  • \( abla^2 \varphi(r) = \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{d\varphi}{dr} \right) \).
This allows us to manage the derivatives more effectively by considering the geometric properties inherent to spherical coordinates.

By substituting the given potential into this formula, the Laplacian reveals how rapid changes occur with respect to \( r \). Calculating these derivatives step by step provides insight into the intricate balance of forces within the field and allows us to solve for the charge distribution effectively. Harnessing the Laplacian in spherical coordinates is vital for solving electrostatic problems with spherical geometries and understanding symmetrical physical systems.

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Most popular questions from this chapter

The one-dimensional Schrödinger wave equation for a particle in a potential field \(V=\frac{1}{2} k x^{2}\) is $$ -\frac{\hbar^{2}}{2 m} \frac{d^{2} \psi}{d x^{2}}+\frac{1}{2} k x^{2} \psi=E \psi(x) . $$ (a) Using \(\xi=a x\) and a constant \(\lambda\), we have $$ a=\left(\frac{m k}{\hbar^{2}}\right)^{1 / 4}, \quad \lambda=\frac{2 E}{\hbar}\left(\frac{m}{k}\right)^{1 / 2} $$ show that $$ \frac{d^{2} \psi(\xi)}{d \xi^{2}}+\left(\lambda-\xi^{2}\right) \psi(\xi)=0 $$ (b) Substituting $$ \psi(\xi)=y(\xi) e^{-\xi^{2} / 2} $$ show that \(y(\xi)\) satisfies the Hermite differential equation.

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The differential equation for the population of a radioactive daughter element is $$ \frac{d N_{2}(t)}{d t}=\lambda_{1} \exp \left(-\lambda_{1} t\right)-\lambda_{2} N_{2}, $$ \(\lambda_{1} \exp \left(-\lambda_{1} t\right)\) being the rate of production resulting from the decay of the parent element, \(\lambda_{1}=0.10 \mathrm{~s}^{-1}, \lambda_{2}=0.08 \mathrm{~s}^{-1}\). Integrate this \(\mathrm{ODE}\) from \(t=0\) out to \(t=40 \mathrm{~s}\) for the initial condition \(N_{2}(0)=0\). Tabulate and plot \(N_{2}(t)\) vs \(t\).

Show that Legendre's equation has regular singularities at \(x=-1,1\), and \(\infty\).

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