Problem 1:

A sphere of radius R has a spherically symmetric charge density given by ρ(r) = ρ₀( 1 - r/R).
(a)  Find the total charge Q in the sphere in terms of ρ₀ and R.
(b)  Find the electric field E both inside and outside the charge distribution using Gauss's law.
(c)  Compute the electrostatic energy stored in the configuration.

Solution:

• Concepts:
  Gauss' law
• Reasoning:
  The field due to a spherically symmetric charge distribution can be found from Gauss' law.
• Details of the calculation:
  (a)  Q = 4πρ₀∫₀^a r²dr(1 - r/R) = 4πρ₀[(R³/3) - (a³/4)] = 4πρ₀R³/12 = πρ₀R³/3.
  
  (b)  r > R :  E = (1/(4πε₀))Q/r², radially outward for positive ρ₀.
  E = ρ₀R³/(12ε₀r²) e_r.
  Φ = (1/(4πε₀))Q/r = (1/ε₀)ρ₀R³/(12r).
  r < R:  E = (1/(4πε₀)) Q_inside/r², radially outward for positive ρ₀.
  Q_inside = Q = 4πρ₀∫₀^r r'²dr'(1  -r'/R)  = 4πρ₀[(r³/3) - (r⁴/(4R))].
  E = (ρ₀r/ε₀)[(1/3) - (r/(4R))] e_r.
  Φ = Φ(R) + ∫_r^RE(r)dr = Φ(R) + (ρ₀/ε₀)∫_r^R[(r/3) - (r²/(4R))]dr
  = Φ(R) + (ρ₀/ε₀)[(r²/6) - (r³/(12R))]_r^R
  = (ρ₀/ε₀)[R²/12 + R²/6 - R²/12  - r²/6 + r³/(12R)]
  = (ρ₀/ε₀)[R²/6  - r²/6 + r³/(12R)].
  
  (c) U = (ε₀/2)∫_{all_space} E·E dV
  U = (ρ₀²/2ε₀)∫₀^r[(r²/9) - (r³/(6R)) + (r⁴/(16R²))]4πr²dr
  + (ρ₀²R⁶/288ε₀)∫_a^∞(1/r⁴)4πr²dr
  U =(ρ₀²R⁵/ε₀)*13π/630.= (ρ₀²R⁵/ε₀)*0.0648.
  or
  U = (1/2)∫₀^Rρ(r)Φ(r)dV = (4π/2)∫_r^Rρ(r)Φ(r)r²dr
  = (2πρ₀²/ε₀)∫₀^R(1 - r/R)[R²/6  - r²/6 + r³/(12R)]r²dr
  = (ρ₀²R⁵/ε₀)*0.0648.

Problem 2:

A sphere of radius R has a uniform volume charge density ρ in addition to a surface chare density σ = σ₀sin²θ.  Find the potential Φ(r) everywhere outside of the sphere.

Hint:
P₀(x) = 1
P₁(x) = x
P₂(x) = (3x² - 1)/2
P₃(x) = (5x³ - 3x)/2

Solution:

• Concepts:
  The principle of superposition, boundary value problems, azimuthal symmetry
• Reasoning:
  The potential outside the sphere is the sum of the potentials of a uniformly charged sphere and the potential of a sphere with surface charge density σ = σ₀sin²θ.  We set the potential at infinity equal to zero.
  The potential of the uniformly charged sphere for r > R is
  Φ₁(r,θ) = (4πε₀)^-1(4πρR³/3)/r = ρR³/(3ε₀r).
  The most general solution inside and outside of a sphere with given surface charge density and zro volume charge density is
  Φ₂(r,θ) = ∑_n=0^∞[A_nrⁿ + B_n/rⁿ⁺¹]P_n(cosθ).
  To find the specific solution we apply boundary conditions.
• Details of the calculation:
  σ(R,θ) = σ₀ sin²θ = σ₀(1 - cos²θ) = (2σ₀/3)(P₀ - P₂(cosθ)).
  Assume
  Φ₂(r,θ) = B₀/r + (B₂/r³)P₂(cosθ) outside the sphere,
  Φ_{2_in}(r,θ) = A₀ + (A₂r²)P₂(cosθ) inside the sphere.
  The symmetry of the situation suggests this form of the solution.  If we can satisfy the boundary conditions, the uniqueness theorem guaranties that we have found the only solution.
  Boundary conditions:
  Φ₂(R,θ) = Φ_{2_in}(R,θ)  --> A₀ = B₀/R,  A₂ = B₂/R⁵.
  (E₂ - E₁)·n₂ = σ/ε₀.
  -∂Φ₂/∂r|_R + ∂Φ_{2_in}/∂r|_R = σ/ε₀.
  B₀/R² + (3B₂/R⁴)P₂(cosθ) + (2A₂R)P₂(cosθ) = (2σ₀/(3ε₀))(1 - P₂(cosθ)).
  B₀/R² = 2σ₀/(3ε₀),  (3B₂/R⁴) + (2A₂R) = -2σ₀/(3ε₀).
  B₀ = 2σ₀R²/(3ε₀),  B₂  = -2σ₀R⁴/(15ε₀).
  Φ₂(r,θ) =  2σ₀R²/(3ε₀r) - (2σ₀R⁴/(15ε₀r³))P₂(cosθ).
  The potential everywhere outside of the the sphere is
  Φ(r,θ) = Φ₁(r,θ) + Φ₂(r,θ) = (ρR³+ 2σ₀R²)/(3ε₀r) - (2σ₀R⁴/(15ε₀r³))P₂(cosθ)
  = (ρR³+ 2σ₀R²)/(3ε₀r) + (σ₀R⁴/(15ε₀r³)) - (σ₀R⁴/(5ε₀r³))cos²(θ).
  This can be rewritten as
  Φ(r,θ) = (ρR³+ 2σ₀R²)/(3ε₀r) - (2σ₀R⁴/(15ε₀r³)) + (σ₀R⁴/(5ε₀r³))sin²(θ).

Problem 3:

An infinitely long charged wire of radius r₀ is parallel to an infinite conducting plane, at a distance h (h >> r₀) from the surface.  The potential difference between the wire and the surface is ∆V.  Derive a formula for the magnitude of the electric field just above the surface below the wire.

Solution:

• Concepts:
  Gauss' law, method of images
• Reasoning:
  The electric field outside an infinitely long straight wire with charge per unit length λ points radially away from the wire and has magnitude E = λ/( 2πε₀r). (Gauss' law)
  The electric field and potential due to a single wire with charge per unit length λ a distance h from an infinite grounded conducting plane is found using the method of images.
• Details of the calculation:
  Assume the wire has a line charge density λ.  Let the conducting plane be the xy-plane and let the wire be parallel to the x-axis at z = h.   The electric field in the region z > 0 is a superposition of the field due to the wire and the field due to an image wire with line charge density -λ, parallel to the x-axis at z = -h.  The same holds for the potential.
  For two point at radial distances a and b from the center of a wire with line charge density λ,
  we have V(b) - V(a) = -[λ/( 2πε₀)]ln(b/a).
  For our problem we have for y = 0 and z between 0 and h - r ₀ that
  E = [-λ/(2πε₀(h + z)) - λ/(2πε₀(h - z))]k = -λ/(πε₀)[h/(h² - z²)]k.
  At z = 0 E = -λ/(πε₀h) k.
  We use ∆V = [V(r₀) - V(h)]_{real wire} + [V(2h - r₀) - V(h)]_{image wire} to find λ.
  ∆V = -[λ/(2πε₀)]ln(r₀/h) + [λ/(2πε₀)]ln((2h - r₀)/h) = [λ/(2πε₀)]ln((2h - r₀)/r₀).
  λ = 2πε₀∆V/ln((2h - r₀)/r₀).
  E = 2V/[h*ln((2h - r₀)/r₀)].

Problem 4:

A uniform dielectric round plate has radius R and thickness d (R >> d).   It is uniformly polarized with the polarization P parallel to the plate.  Find the electric field generated by the polarization at the center position of the plate.

Solution:

• Concepts:
  Bound charge density, electric field due to a continuous charge distribution
• Reasoning:
  The bound surface charge density is σ = P·n = Pcosθ.  Since R >> d we can treat it as a line charge density λ = σd = Pdcosθ.
• Details of the calculation:
  Choose the coordinate system so that the polarization P points in the z-direction.
  Symmetry dictates that the electric field at the center only has a z-component.
  E_z = -4k_e∫₀^π/2Rdθ λ(θ) cosθ/R² = -(4k_ePd/R)∫₀^π/2dθ  cos²θ = -πk_ePd/R = -Pd/(4ε₀R).