More Problems

Problem 1:

A charged metal sphere with charge +Q and radius 'a' is positioned at the center of a neutral, spherical metal shell with inner radius 'b' and outer radius 'c'.

(a)  Calculate the surface charge densities σ_a, σ_b, σ_c on the inner sphere, the inner surface of the shell, and the outer surface of the shell, respectively.
(b)  Calculate the electric field E everywhere, and sketch E vs Radius.
(c)  Calculate the potential V(0) at the center of the inner sphere.  Use the usual convention of a reference point of infinity V(∞)  = 0.

• Concepts:
  Gauss' law
• Reasoning:
  The field due to a spherically symmetric charge distribution can be found from Gauss' law.
• Details of the calculation:

A small conducting ring of radius a is located in the xy-plane centered at the origin.  A positive charge Q is place on the ring.
(a)  Find the potential Φ on the z-axis at z > a and expand your expression in powers of a/z.
(b)  The potential Φ(r,θ) at a arbitrary points in space with r > a can be expanded in terms of Legendre polynomials,  Φ(r,θ) = ∑_n=0^∞[A_nrⁿ + B_n/rⁿ⁺¹]P_n(cosθ). 
Find the expansion coefficients.

Solution:

• Concepts:
  Symmetry, Φ(r,θ) = ∑_n=0^∞[A_nrⁿ + B_n/rⁿ⁺¹]P_n(cosθ) is the general solution to Laplace's equation for problems with azimuthal symmetry.
• Reasoning:
  Φ(r) = [1/(4πε₀)]∫ρ(r')dV'/|r - r'|.  We can evaluate this integral for r = rk. 
  For r > a, Φ(r) satisfies Laplace's equation, ∇²Φ(r) = 0.  The general solution is
  Φ(r,θ) = ∑_n=0^∞[A_nrⁿ + B_n/r^n-1]P_n(cosθ). 
  We can find the coefficients by comparing the general solution to the solution found by integration on the z-axis.
• Details of the calculation:
  (a)  Φ(z) = [1/(4πε₀)]Q/(a² + z²)^½.
  Expanding (a² + z²)^-½ = (1/z)(1 + a²/z²)^-½ = (1/z)(1 - ½a²/z² + ½(3/2)a⁴/z⁴ - ½(3/2)(5/2)a⁶/z⁶ + ...)
  = (1/z)(1 + ∑_n=1^∞(-1)ⁿ(a/z)²ⁿ(2n - 1)!!/2ⁿ).
  Φ(z) = [Q/(4πε₀z](1 + ∑_n=1^∞(-1)ⁿ(a/z)²ⁿ(2n - 1)!!/2ⁿ).
  (b)  For r > a  ∑_n=0^∞(B_n/rⁿ⁺¹)P_n(cosθ).  Only the even parity terms contribute and all A_n are zero since Φ = 0 at infinity. 
  We can therefore write  Φ(r,θ) = ∑_n=0^∞(B_2n/r²ⁿ⁺¹)P_2n(cos(θ)). 
  On the positive z-axis  Φ(z) = ∑_n=0^∞(B_2n/z²ⁿ⁺¹), since P_n(1) = 1 for all n.
  Equating the two expressions for Φ(z), we have
  Φ(z) = ∑_n=0^∞(B_2n/z²ⁿ⁺¹) = [Q/(4πε₀)](1/z + ∑_n=1^∞(-1)ⁿ(a²ⁿ/z²ⁿ⁺¹)(2n -1)!!/2ⁿ).
  Equating term with equal powers of z we have B₀ = Q/(4πε₀),
  B_2n≠0 = [Q/(4πε₀)](-1)ⁿa²ⁿ(2n -1)!!/2ⁿ.

Problem 4:

The z-axis is the symmetry axis of a very long cylinder of radius a, made from dielectric material of relative permittivity ε = Kε₀.  The cylinder carries a free surface charge density σ_free = σ_f0cosφ.  The electric field inside and outside the cylinder is of the form
E_in = -A₁i = -A₁cosφ (ρ/ρ) + A₁sinφ (φ/φ),
E_out = (A₂cosφ (ρ/ρ) + A₂sinφ (φ/φ))/ρ².
Use the boundary conditions conditions for E and D to find the polarization P of the cylinder in terms of K and σ_f0

Solution:

• Concepts:
  Boundary conditions for the electric field and the electric displacement
• Reasoning:
  The given E_in and E_out can satisfy the boundary conditions.  The constants A₁ and A₂ for which the boundary conditions are satisfied are unique.  Given E_in, we can find P.
• Details of the calculation:
  Boundary conditions:
  (E₂ - E₁)∙n₂ = (σ/ε₀),  (E₂ - E₁)∙t = 0,  (D₂ - D₁)∙n₂ = σ_f.
  (E_out - E_in)∙t = 0 --> A₁ = A₂/a²,  E_in = (-A₂cosφ (ρ/ρ) + A₂sinφ (φ/φ))/a².
  (D_out - D_in)∙n_out = σ_free -->  ε₀A₂/a² + εA₂/a² = σ_f0,  A₂ = σ_f0a²/(ε₀ + ε).
  E_in = -A₁i = σ_f0/(ε₀ + ε)i.
  P = ε₀χ_eE_in = ε₀(K - 1)E_in = -ε₀(K - 1)σ_f0/(ε₀ + ε)i = -[(K - 1)/(K  + 1)] σ_f0i.

  Or we can use
  (E_out - E_in)∙n_out = σ/ε₀ = (σ_free + σ_bound)/ε₀ -->  A₂ = (σ_f0 + σ_b0)a²/(2ε₀),
  where σ_bound = σ_b0cosφ.
  σ_f0/(ε₀ + ε) = (σ_f0 + σ_b0)/(2ε₀),  [(ε₀ - ε)/(ε₀ + ε)]σ_f0 = [(1 - K)/(1 + K)]σ_f0 = σ_b0.
  On the x-axis at x = a, P∙n = P = σ_b0.  Therefore
  P =  -[(K - 1)/(K + 1)]σ_f0i.