Potential V=100xz/(x²+4), charge on conductor

Given V = 100xz/(x² + 4) volts in free space. D = ε₀E = −ε₀∇V.

(a) D at z = 0

∂V/∂x = 100z (4 − x²)/(x²+4)²        →  0 at z = 0
∂V/∂y = 0
∂V/∂z = 100x/(x²+4)

E = −∇V = −100z(4−x²)/(x²+4)² a_x − 100x/(x²+4) a_z
At z = 0:  E = −100x/(x²+4) a_z  V/m

D(z=0) = ε₀E = −100ε₀ x/(x²+4) a_z  C/m²
       = −8.854×10⁻¹⁰ x/(x²+4) a_z

(b) z = 0 is an equipotential surface

Put z = 0 in V: V = 100x(0)/(x²+4) = 0 for every x and y. The potential is the same (0 V) everywhere on the plane, so it is an equipotential surface. This agrees with (a): E has no component tangential to the plane (E_x = E_y = 0 at z = 0).

(c) Charge on the conductor, 0 < x < 2, −3 < y < 0

Take the conductor as the plane z = 0 with the field in the region z > 0, so the outward normal is a_n = +a_z. Then ρ_s = D·a_n:

ρ_s = −100ε₀ x/(x²+4)  C/m²

Q = ∫(y=−3→0) ∫(x=0→2) ρ_s dx dy
  = −100ε₀ (3) ∫₀² x/(x²+4) dx
  = −300ε₀ · ½ ln(x²+4) |₀²
  = −150ε₀ ln(8/4)
  = −150 × 8.854×10⁻¹² × 0.6931
  = −9.21×10⁻¹⁰ C

Answer: Q ≈ −0.92 nC.