Question

A charge \(+2 q\) is situated at the origin and charges of \(-q\) are situated at distances \(\pm a\) from it along the polar axis. By relating it to the generating function for the Legendre polynomials, show that the electrostatic potential \(\Phi\) at a point \((r, \theta, \phi)\) with \(r>a\) is given by $$ \Phi(r, \theta, \phi)=\frac{2 q}{4 \pi \epsilon_{0} r} \sum_{s=1}^{\infty}\left(\frac{a}{r}\right)^{2 s} P_{2 s}(\cos \theta). $$

Short answer

Using the generating function for Legendre polynomials, the potential \( \Phi \) at \( (r, \theta, \phi) \) for \( r > a \) is \[ \Phi(r, \theta, \phi) = \frac{2q}{4\pi \epsilon_0 r} \sum_{s=1}^{\infty} \left(\frac{a}{r}\right)^{2s} P_{2s}(\cos \theta) \]

Step-by-step solution

- Understand the setup and potential equation

Consider the electrostatic potential \( \Phi\ \) due to a charge distribution. We have a charge \( +2q \) at the origin (0, 0, 0) and charges of \( -q \) positioned at \( (0, 0, a) \) and \( (0, 0, -a) \). The goal is to express the potential \( \Phi \) at a point \( (r, \theta, \phi) \) for \( r > a \).

- Potential due to individual charges

The potential at \( (r, \theta, \phi) \) due to a point charge is given by \[\Phi = \frac{Q}{4 \pi \epsilon_0 R}\]where \(\Q\ \) is the charge and \(\R\ \) is the distance from the charge to the point.

- Distance approximation for \( r > a \)

For \( r > a \), the distances from the charges \( -q \) to the point \( (r, \theta, \phi) \) can be approximated using the binomial expansion:\[ R = \sqrt{r^2 + a^2 - 2ar\cos\theta} \approx r \left(1 - \frac{a}{r} \cos\theta\right) \]This will help us simplify further calculations.

- Potential due to combined charges

The total potential at \( (r, \theta, \phi) \) is the sum of the potentials due to each charge:\[ \Phi = \frac{2q}{4\pi \epsilon_0 r} + \frac{-q}{4\pi \epsilon_0 \sqrt{r^2 - 2ar\cos\theta + a^2}} + \frac{-q}{4\pi \epsilon_0 \sqrt{r^2 + 2ar\cos\theta + a^2}} \]

- Simplification using binomial expansion

Using the earlier approximation for \( r > a \) and expanding using the binomial theorem, the potential can be expressed as:\[ \Phi = \frac{2q}{4\pi \epsilon_0 r} \left[ 1 - \frac{1}{2} \cdot \left( \frac{a}{r} \cos \theta \right)^2 - \frac{1}{2} \cdot \left( -\frac{a}{r} \cos\theta \right) \right] \]

- Expression in terms of Legendre polynomials

Recognize that the resulting series can be expressed in terms of Legendre polynomials \( P_{2s}(\cos \theta) \). Specifically, \[ \Phi = \frac{2 q}{4\pi \epsilon_0 r} \sum_{s=1}^{\infty} \left(\frac{a}{r}\right) ^{2s} P_{2s} (\cos \theta) \]

Key concepts

Electrostatic Potential

Electrostatic potential is a fundamental concept in electromagnetism. It represents the potential energy per unit charge at a point in space due to the presence of electric charges. In our exercise, we need to determine the electrostatic potential at a point \((r, \theta, \phi)\) due to a given charge distribution.An important point to remember is that the potential due to a single point charge \(Q\) at a distance \(R\) from it is given by: \[\Phi = \frac{Q}{4 \pi \epsilon_0 R}\] Here, \( \epsilon_0 \) is the permittivity of free space. The superposition principle tells us that if we have multiple point charges, the total potential at a point is simply the sum of the potentials due to individual charges. So, in our problem, we consider the potentials due to \(+2q\) at the origin and \(-q\) at positions \( (0,0,a) \) and \( (0,0,-a) \).

Binomial Expansion

In mathematical analysis, binomial expansion is a useful tool for approximating expressions. It allows us to simplify complex square roots and other functions.In this exercise, binomial expansion is crucial when dealing with distances for \(r > a\). To approximate the distance \(R\) from the charges \(-q\) to a point \((r, \theta, \phi)\), we use:\[ R = \sqrt{r^2 + a^2 - 2ar \cos \theta} \approx r \left(1 - \frac{a}{r} \cos \theta \right) \]By expanding and simplifying using binomial series, we get terms in powers of \(\frac{a}{r}\). These approximations allow further simplification when summing the potentials. We get a more manageable series form like \[ \Phi = \frac{2q}{4\pi \epsilon_0 r} \left[ 1 - \frac{1}{2} \left( \frac{a}{r} \cos\theta \right)^2 \] for easier calculation and to relate it to Legendre polynomials later.

Charge Distribution

Charge distribution refers to how electric charge is spread out in space. For our problem, the charge distribution consists of a \(+2q\) charge at the origin and two \(-q\) charges positioned symmetrically at \( (0,0,a) \) and \( (0,0,-a) \).This specific arrangement affects the resulting electrostatic potential due to the interplay of these charges. The charges create respective potentials that sum up, following the principle of superposition.
This symmetrical arrangement allows us to take advantage of Legendre polynomials, which are orthogonal polynomials arising in solutions of Laplace's equation in spherical coordinates. For this problem, it simplifies the series representation of the potential \( \Phi \) which can be expressed as:
\[ \Phi = \frac{2 q}{4\pi \epsilon_0 r} \sum_{s=1}^{\infty} \left(\frac{a}{r}\right)^{2s} P_{2s} (\cos \theta) \]Here, \( P_{2s}(\cos \theta) \) are Legendre polynomials of even order, reflecting the symmetrical distribution of our charges along the polar axis.