More Problems

Problem 1:

Consider only Maxwell's equations for electrostatics.
(a)  How many equations are there?
(b)  How many independent and dependent variables are there?
(c)  You want to find a unique solution in a volume V subject to given boundary conditions.  Do you need all the equations, or are they overspecified. Give an example.

Solution:

• Concepts:
  Maxwell's equations for electrostatics
• Reasoning:
  In static situations the equations for the electric field and the magnetic field decouple.
• Details of the calculation:
  (a)  ∇·E = ρ/ε₀,  ∇×E = 0, (SI units).
  There are four equations.
  ∇·E = ρ/ε₀ (Gauss' law) is a scalar equation, ∇×E = 0, is a vector equation in 3D, for example one equation for each Cartesian coordinate of ∇×E.
  (b)  There are 3 independent variables, for example the Cartesian coordinates x, y, z, and 4 dependent variables, for example the three Cartesian components of E(x,y,z) and ρ(x,y,z).
  (c)  The equations are not overspecified.  We have 4 equations and 4 unknowns.  To find a unique solution in a volume V subject to given boundary conditions we also need to know the the charge distribution inside the volume.
  ∇×E = 0 yields E = -∇Φ, but ∇²Φ = -ρ/ε₀ specifies the "initial conditions".

Problem 2:

A conducting sphere of radius R is centered at the origin.  The sphere has a net charge q.  Another point charge q is located on the z-axis at z = d > R.
(a)  Find the force they exert on each other.
(b)  Simplify the expression for F if d = 2R.  Is the force attractive or repulsive?

Solution:

• Concepts
  The method of images
• Reasoning:
  For the point charge q is located on the z axis at d, place an image charge q' = -Rq/d on the z-axis at z' = R²/d and a point charge q'' = q + Rq/d at the origin.  This will keep the surface of the sphere an equipotential surface, and the sphere will have a net charge q.
• Details of the calculation:
  (a)  F_z =  k_eqq''/d² - k_eqq'/(d -  R²/d)²
  = (k_eq²/d²)[(1 + R/d - Rd/(d - R²/d)²].
  (b)  If d = 2R, F_z = (k_eq²/d²)(3/2 - 8/9).
  The force is repulsive.

Problem 3:

Consider an insulating spherical half-shell with radius R resting on a very large (infinite) grounded conducting plane as shown in the figure.  The shell carries a surface charge density σ = σ₀cosθ.

(a)  Find the potential everywhere in the region z > 0.  (Choose your approach.)
(b)  Find the surface charge density on the conducting plane.

Solution:

• Concepts:
  Method of images, boundary value problems
• Reasoning:
  Method of images:  The potential in the region z > 0 is the same as the potential of a sphere with surface charge density σ = σ₀cosθ.  The potential of such a sphere can be found by solving a boundary value problem.
  Φ(r,θ) = ∑_n=0^∞[A_nrⁿ + B_n/rⁿ⁺¹]P_n(cosθ) is the most general solution inside and outside of the sphere, since ρ = 0 inside and outside of the sphere.  To find the specific solution we apply boundary conditions.
• Details of the calculation:
  (a)  Assume Φ₁(r,θ) = ArP₁(cosθ) = Arcosθ inside the sphere and Φ₂(r,θ) = (B/r²)P₁(cosθ) = (B/r²)cosθ outside 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.  The potential vanishes as r goes to infinity.
  Boundary conditions:
  Φ is continuous across the boundary.  Φ₁(R,θ) = Φ₂(R,θ).  A = B/R³.
  (E₂ - E₁)·n₂ = σ/ε₀, dΦ₂/dr - dΦ₁/dr|_R = -σ/ε₀.  -2(B/R³)cosθ - Acosθ = -σcosθ/ε₀.
  -2A - A = -σ/ε₀.  A = σ/3ε₀.
  Φ₁(r,θ) = (σ₀/3ε₀) r cosθ inside the sphere and Φ₂(r,θ) = (σ₀R³/3ε₀r²)cosθ outside the sphere. 
  So for our problem in the region z ≥ 0 we have Φ₁(r,θ) = (σ₀/3ε₀) z and Φ₂(r,θ) = (σ₀R³/3ε₀r²)cosθ.
  
  (b)  The electric field in region 1 at z = 0 is -dΦ₁/dz = -σ₀/3ε₀.   The surface charge density is σ = -σ₀/3.
  The electric field in region 2 is a pure dipole field of a dipole with dipole moment p = k 4πR³σ₀/3.
  E(r) = [1/(4πε₀)](1/r³)[3(p·r)r/r² - p] = [1/(4πε₀)](1/r³)[2pcosθ e_r + psinθ e_θ].
  At z = 0  E(r) = -[1/(4πε₀)](1/r³)p k.  The surface charge density is σ =  -(R³σ₀/r³)/3.