Find the electrostatic potential outside a conducting sphere of radius \(a\) placed in an originally uniform electric field, and maintained at zero potential. Hint: Let the original field \(\mathbf{E}\) be in the negative \(z\) direction so that \(\mathbf{E}=-E_{0} \mathbf{k} .\) Then since \(\mathbf{E}=-\nabla \Phi,\) where \(\Phi\) is the potential, we have \(\Phi=E_{0} z=E_{0} r \cos \theta\) (Verify this!) for the original potential. You then want a solution of Laplace's equation \(\nabla^{2} u=0\) which is zero at \(r=a\) and becomes \(u \sim \Phi\) for large \(r\) (that is, far away from the sphere). Select the solutions of Laplace's equation in spherical coordinates which have the right \(\theta\) and \(\phi\) dependence (there are just two such solutions) and find the combination which reduces to zero for \(r=a\).

Step by step solution

01

Identify the Problem Context

We need to find the electrostatic potential outside a conducting sphere of radius \(a\), placed in a uniform electric field along the negative \(\text{z}\) direction and maintained at zero potential.

02

Express Original Potential

Given the uniform electric field, the potential \(\text{Φ}\) in the absence of the sphere is \(\text{Φ} = E_0 z = E_0 r \cos \theta\).

03

Form Laplace's Equation Setup

The potential \(u\) must satisfy Laplace's equation \(\abla^2 u = 0\), with the boundary conditions \(u = 0\) at \(r = a\) and asymptotically matching \(\text{Φ}\) as \(r \to \infty\).

04

General Solution to Laplace's Equation

In spherical coordinates \(r, \theta, \phi\), the general solution for \(u\) due to spherical symmetry, considering only \(\theta\) and \(r\) dependence, takes the form \[u(r, \theta) = \sum_{l=0}^{\infty} \(A_l r^l + \frac{B_l}{r^{l+1}}\) P_l(\cos \theta)\], where \(P_l\) are the Legendre polynomials.

05

Select Relevant Terms

Since we want a solution that behaves like \(E_0 r \cos \theta\) at large distances, we retain the terms for \(l = 1\). This gives two terms: \(\text{A_1 r P_1(cos \theta)}\) and \(\text{B/r^2 P_1(cos \theta)}\), where \(P_1(\cos \theta) = \cos \theta\). So, the potential is \() u(r, \theta) = (A_1 r + \frac{B_1}{r^2}) \cos \theta.\).

06

Apply Boundary Conditions

First, let’s match the boundary condition at \(r = a\): \[u(a, \theta) = (A_1 a + \frac{B_1 }{a^2}) \cos \theta.\]Since \(u = 0\) on the sphere's surface, \(\text{A_1} \text{a + B_1/a^2 = 0}\), solving gives \[B_1 = -A_1 a^3.\]

07

Match Asymptotic Behavior

At large distances (far from the sphere), \(u \to E_0 r \cos \theta \). Set \((\text{A_1})\) to match \(E_0\) coefficient, \(A_1 = -E_0\). Thus, \[u(r, \theta) = -E_0 (r - \frac{a^3}{r^2} ) \cos \theta.\]

08

Write the Potential

Combining all terms, the final electrostatic potential outside the conducting sphere is: \[\Phi (r, \theta) = -E_0 (r - \frac{a^3}{r^2}) \cos \theta.\]

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Key Concepts

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

Laplace's Equation

Laplace's equation is a second-order partial differential equation widely used in physics, particularly in electromagnetism. The equation is given as \ \( abla^{2} u = 0 \). This equation appears when describing the potential field in regions where there is no charge density. Solving Laplace's equation helps us find the electrostatic potential in the area surrounding a charged object.
For this problem, the region outside the sphere is charge-free, making Laplace's equation the correct tool for finding the potential. We apply specific boundary conditions to find a unique solution to this equation.

Spherical Coordinates

Spherical coordinates \( (r, \theta, \phi) \) are ideal for problems involving spheres or spherical symmetry. In these coordinates, the position of a point is given by:

• \( r \) (the radial distance from the origin),
• \( \theta \) (the polar angle measured from the positive z-axis),
• \( \phi \) (the azimuthal angle measured in the xy-plane from the positive x-axis).

To solve Laplace's equation for this problem, we use spherical coordinates. Given the symmetry about the z-axis, the potential \( u \) depends on \( r \) and \( \theta \) but not on \( \phi \). Therefore, our task simplifies to handling the variables \( r \) and \( \theta \) only.

Boundary Conditions

Boundary conditions are crucial for solving differential equations like Laplace's equation. They describe the behavior of the potential at certain points or surfaces. For the given problem, two boundary conditions are used:

• The potential must be zero at \( r = a \), the surface of the conducting sphere.
• The potential should asymptotically match the original potential \( \Phi = E_0 r \cos\theta \) as \( r \to \infty \).

These conditions guide us in selecting and combining solutions to Laplace's equation. By applying these constraints, we find a solution that fits the physical setup of the problem.

Legendre Polynomials

Legendre polynomials \( P_l(\cos \theta) \) are solutions to Legendre's differential equation and appear in problems with spherical symmetry. They form a complete set of orthogonal functions. In the exercise, the general solution to Laplace's equation is expressed as a series involving these polynomials: \ \[ u(r, \theta) = \sum_{l=0}^{\infty} \left( A_l r^l + \frac{B_l}{r^{l+1}} \right) P_l(\cos \theta) \]
We use Legendre polynomials to represent the angular part, \( P_1(\cos \theta) = \cos \theta \) since it matches the original potential \( \Phi = E_0 r \cos \theta \). This polynomial helps simplify the solutions and apply the necessary boundary conditions effectively.