Question

Electrostatic charge is distributed in a sphere of radius \(R\) centred on the origin. Determine the form of the resultant potential \(\phi(\mathbf{r})\) at distances much greater than \(R\), as follows. (a) express in the form of an integral over all space the solution of $$ \nabla^{2} \phi=-\frac{\rho(\mathbf{r})}{\epsilon_{0}} $$ (b) show that, for \(r \gg r^{\prime}\), $$ \left|\mathbf{r}-\mathbf{r}^{\prime}\right|=r-\frac{\mathbf{r} \cdot \mathbf{r}^{\prime}}{r}+\mathrm{O}\left(\frac{1}{r}\right) $$ (c) use results (a) and (b) to show that \(\phi(\mathbf{r})\) has the form $$ \phi(\mathbf{r})=\frac{M}{r}+\frac{\mathbf{d} \cdot \mathbf{r}}{r^{3}}+\mathrm{O}\left(\frac{1}{r^{3}}\right) $$ Find expressions for \(M\) and \(\mathbf{d}\), and identify them physically.

Short answer

The potential at large distances is \( \phi(\mathbf{r}) \approx \frac{M}{r} + \frac{\mathbf{d} \cdot \mathbf{r}}{r^3} \) where \( M \) is the total charge and \( \mathbf{d} \) is the dipole moment.

Step-by-step solution

Define the Poisson Equation

Given the Poisson equation \(abla^{2} \phi = -\frac{\rho(\mathbf{r})}{\epsilon_{0}}\), we need to express the solution as an integral over all space.

Solution for Poisson Equation

The Green's function solution for the Poisson equation is \[ \phi(\textbf{r}) = \frac{1}{4\pi\epsilon_{0}} \int \frac{\rho(\textbf{r}')} {|\textbf{r} - \textbf{r}'|} d^3r' \]

Approximation for Large Distances (r >> r')

When \(r \gg r'\), we can approximate the denominator using the binomial expansion: \[ |\mathbf{r} - \mathbf{r}'| = r - \frac{\mathbf{r} \cdot \mathbf{r}'}{r} + \mathrm{O}\left(\frac{1}{r}\right) \]

Integrate and Identify Terms

Substitute the approximate form of \( |\mathbf{r} - \mathbf{r}'| \) into the integral from Step 2. Then separate the terms at leading order in \( \frac{1}{r} \) to find the form of the potential:

Identify the Terms M and d

For the leading term in \( \frac{1}{r} \), we get \[ M = \frac{1}{\epsilon_{0}} \int \rho(\mathbf{r}') d^3r' \] which is the total charge.For the leading term in \( \frac{\mathbf{r} \cdot \mathbf{r}'}{r^3} \), we get \[ \mathbf{d} = \frac{1}{\epsilon_{0}} \int \rho(\mathbf{r}') \mathbf{r}' d^3r' \] which is the dipole moment.

Key concepts

Poisson Equation

The Poisson equation is fundamental in electrostatics. It relates the electrostatic potential \( abla^2 \phi \) to the charge density \( \rho(\mathbf{r}) \). The equation is given by: \[ abla^2 \phi = -\frac{\rho(\mathbf{r})}{\epsilon_0} \] Here, \( \epsilon_0 \) is the permittivity of free space. The Poisson equation tells us how charges within a distribution generate an electrostatic potential \( \phi \). Finding a solution involves integrating the charge density over all space using a helpful mathematical tool called the Green's function. This leads us to an integral form of the potential at any point in the space around the charges.

Green's Function

The Green's function is a method used to solve differential equations like the Poisson equation. It simplifies the process of integrating over charge distributions. For the Poisson equation, the Green's function solution is: \[ \phi(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \int \frac{\rho(\mathbf{r}')} {|\mathbf{r} - \mathbf{r}'|} d^3r' \] This integral evaluates the potential \( \phi \) at any point \( \mathbf{r} \) by summing the contributions from all charge elements \( \rho(\mathbf{r}') \) spread across the entire space. The factor \( |\mathbf{r} - \mathbf{r}'| \) is the distance between the observation point and each charge element. This distance plays a critical role in determining how strongly a given charge element affects the potential at \( \mathbf{r} \).

Binomial Expansion

Binomial expansion is a mathematical technique used to approximate expressions in physics. When the distance between the observation point \( r \) and charge element point \( r' \) is large \( (r \gg r') \), we can use the binomial expansion to simplify the expression for \( |\mathbf{r} - \mathbf{r}'| \): \[ |\mathbf{r} - \mathbf{r}'| = r - \frac{\mathbf{r} \cdot \mathbf{r}'}{r} + \mathrm{O}\left(\frac{1}{r}\right) \] This approximation helps in breaking down complex integrals into more manageable terms. It shows that for large distances, the denominator can be expanded into a series where the terms decrease in significance based on higher powers of \( \frac{1}{r} \).

Total Charge

The total charge, denoted by \(M\), within a given volume is calculated by integrating the charge density \(\rho(\mathbf{r}')\) over that volume: \[ M = \frac{1}{\epsilon_0} \int \rho(\mathbf{r'}) d^3r' \] This term emerges when we integrate the leading order term in \( \frac{1}{r} \) from the binomial approximation. Physically, \(M\) represents the sum of all charges present in the distribution. In the context of the resulting potential \( \phi(\mathbf{r}) \), it shows the primary contribution to the potential at large distances is mostly dependent on this total charge.

Dipole Moment

The dipole moment, denoted by \( \mathbf{d} \), measures the separation of positive and negative charges within a distribution. It's calculated by multiplying each charge element by its position \( \mathbf{r}' \) and integrating over the entire space: \[ \mathbf{d} = \frac{1}{\epsilon_0} \int \rho(\mathbf{r}') \mathbf{r}' d^3r' \] The dipole moment indicates the strength and direction of the dipole formed by the separation. In the electrostatic potential \( \phi(\mathbf{r}) \), the term involving \( \mathbf{d} \) contributes to the potential with a dependence on \( \frac{1}{r^3} \). This makes it significant at shorter distances compared to the total charge term.