Problem 12 - Advanced Engineering Mathematics

Question

A conducting sphere of radius \(c\) is grounded and placed in a uniform electric field that has intensity \(E\) in the \(z\) -direction. The potential \(u(r, \theta)\) outside the sphere is determined from the boundary-value problem \(\frac{\partial^{2} u}{\partial r^{2}}+\frac{2}{r} \frac{\partial u}{\partial r}+\frac{1}{r^{2}} \frac{\partial^{2} u}{\partial \theta^{2}}+\frac{\cot \theta}{r^{2}} \frac{\partial u}{\partial \theta}=0, \quad r>c, 0<\theta<\pi\) \(u(c, \theta)=0, \quad 0<\theta<\pi\) \(\lim _{\sim \infty} u(r, \theta)=-E z=-E r \cos \theta\) Show that $$ u(r, \theta)=-E r \cos \theta+E \frac{c^{3}}{r^{2}} \cos \theta $$

Step-by-Step Solution

Answer

The solution is \(u(r, \theta) = -E r \cos \theta + E \frac{c^3}{r^2} \cos \theta\).

Step 1: Understand the Problem

This boundary value problem involves finding the electric potential \(u(r, \theta)\) in spherical coordinates for \(r > c\). The sphere is embedded in a uniform electric field, which must be evaluated using the given partial differential equation and boundary conditions.

Step 2: Consider the General Solution Form

The general solution to Laplace's equation in spherical coordinates for azimuthal symmetry can be expressed as a sum of Legendre polynomials: \(u(r, \theta) = \sum_{l=0}^{\infty} \left( A_l r^l + \frac{B_l}{r^{l+1}} \right) P_l(\cos\theta)\). This considers the radial behavior and the angular dependency with Legendre polynomials \(P_l(\cos\theta)\).

Step 3: Apply First Boundary Condition

Use the boundary condition \(u(c, \theta) = 0\). Substitute \(r = c\) in the assumed potential form. This imposes the condition \(\sum_{l=0}^{\infty} \left( A_l c^l + \frac{B_l}{c^{l+1}} \right) P_l(\cos\theta) = 0\).

Step 4: Apply the Condition at Infinity

The behavior as \(r \to \infty\) is \(\lim_{r \to \infty} u(r, \theta) = -E r \cos \theta\). Thus, for large \(r\), the \(A_1\) term (\(A_1 r P_1(\cos\theta)\)) should match \(-E r \cos \theta\) implying \(A_1 = -E\). Terms with other \(A_l\) for \(l eq 1\) should not grow, so they must be zero.

Step 5: Solve for Remaining Constants

The condition \(u(c, \theta) = 0\) combines with \(A_1 = -E\) to solve \(\frac{B_1}{c^2} = Ec\) from \(\frac{-E c}{c^2} + \frac{B_1}{c^3} = 0\). Solving gives \(B_1 = Ec^3\).

Step 6: Form the Solution

Substituting \(A_1 = -E\) and \(B_1 = Ec^3\) into \(u(r, \theta)\), we get: \(u(r, \theta) = -E r \cos \theta + E \frac{c^3}{r^2} \cos \theta\). Each term corresponds to a distinct physical effect: the uniform field and the induced field due to grounding the sphere.

Key Concepts

Boundary Value Problem

Boundary value problems are important in physics and engineering because they define how a physical system behaves given certain constraints. In this problem, we have a conducting sphere in an electric field. We need to find the electric potential outside the sphere. The essence of a boundary value problem lies in determining a function (here, the electric potential \(u(r, \theta)\)) that satisfies a differential equation and follows specific boundary conditions. For this sphere, the boundary conditions are:

• The potential is zero on the surface of the sphere (\(u(c, \theta) = 0\)).
• The potential behaves as \(-E r \cos \theta\) far away from the sphere, indicating the influence of the uniform electric field.

Laplace's equation, which the potential satisfies, is a specific type of differential equation widely used in potential theory. Solving it requires adjusting to the boundary conditions of the problem while considering the geometry of the system.

Electric Potential in Spherical Coordinates

Electric potential in spherical coordinates provides a way to describe the potential field in a system where spherical symmetry is likely. Unlike Cartesian coordinates, spherical coordinates use radius \(r\), polar angle \(\theta\), and azimuthal angle \(\phi\) to describe points in space. This perspective is particularly convenient when dealing with spheres or point charges.In the given problem, electric potential is examined for a spherical conductor placed in a uniform electric field. The governing Laplace's equation in spherical coordinates is: \[ \frac{\partial^{2} u}{\partial r^{2}}+\frac{2}{r} \frac{\partial u}{\partial r}+\frac{1}{r^{2}} \frac{\partial^{2} u}{\partial \theta^{2}}+\frac{\cot \theta}{r^{2}} \frac{\partial u}{\partial \theta}=0 \] The solution of this equation needs to consider the radial (\(r\)) and angular (\(\theta\)) components independently. Since the sphere’s potential is zero, the potential outside must change smoothly from this boundary condition to the condition at infinity.Spherical coordinates naturally separate the radial and angular dependencies in problems of symmetry, lending themselves to using solutions like the Legendre polynomials to express potential efficiently.

Legendre Polynomials

Legendre polynomials are a set of orthogonal polynomials used to solve problems with spherical symmetry. They appear in solutions to Laplace's equation where azimuthal symmetry is present.Legendre polynomials can help express functions dependent on angles in spherical systems. The key feature is that these polynomials correspond to a family \(P_l(\cos \theta)\), indexed by \(l\), that simplifies harmonic functions.They arise naturally when solving problems like the electrostatic potential outside the conducting sphere in the given problem, leading to a series solution:\[ u(r, \theta) = \sum_{l=0}^{\infty} \left( A_l r^l + \frac{B_l}{r^{l+1}} \right) P_l(\cos\theta) \]In this problem, each term in the series accounts for how the sphere distorts the electric field.

• The coefficient \(A_1 = -E\) corrects the electric field within the sphere’s range.
• The coefficient \(B_1 = Ec^3\) captures the sphere's influence at the boundary.

These terms create a solution that balances external field effects and satisfies the boundary conditions at the sphere's surface.