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Parallel plate sheets, field in regions

Short answer

By superposition, the electric fields of the sheets cancel outside the plates and add inside. Hence E = 0 for y<0 and y>b. Between the plates (0<y<b) E = (ρ_s/ε₀) a_y, uniform and directed from the positive to the negative plate.

Find the electric field intensity in all three regions due to an infinite sheet parallel plate capacitor having surface charge density ρ_s C/m² and −ρ_s C/m² and placed at y = 0 and y = b respectively.

The field of one infinite sheet of charge ρ_s is E = (ρ_s/2ε₀) a_n, where a_n is the unit normal pointing away from the sheet (for positive ρ_s). Superpose the fields of the two sheets.

   Region I      |   Region II     |   Region III
   (y < 0)       |  (0 < y < b)    |   (y > b)
                 |                 |
               +ρ_s              −ρ_s
              (y = 0)           (y = b)
                 |   ──→ a_y       |

Field of each sheet

  • Sheet +ρ_s at y = 0: E₁ = +(ρ_s/2ε₀) a_y for y > 0, and −(ρ_s/2ε₀) a_y for y < 0.
  • Sheet −ρ_s at y = b: the field points towards a negative sheet. So E₂ = +(ρ_s/2ε₀) a_y for y < b, and E₂ = −(ρ_s/2ε₀) a_y for y > b.

Total field

RegionE₁E₂E = E₁ + E₂
y < 0−(ρ_s/2ε₀) a_y+(ρ_s/2ε₀) a_y0
0 < y < b+(ρ_s/2ε₀) a_y+(ρ_s/2ε₀) a_y(ρ_s/ε₀) a_y
y > b+(ρ_s/2ε₀) a_y−(ρ_s/2ε₀) a_y0

Answer: E = 0 for y < 0, E = (ρ_s/ε₀) a_y V/m for 0 < y < b (from the positive to the negative plate), and E = 0 for y > b. The field is confined between the plates and is uniform there.

✓ Answer checked by Claude Opus 5.5 (high effort)

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