The electric field between two infinite parallel conducting plates with uniform surface charge densities +σ and −σ is σ/ε₀, so n=1.

Concept and Derivation

Consider two identical, infinite, plane conducting plates kept parallel to each other and separated by a distance d, with surface charge densities +σ and −σ on the facing surfaces.

For a single infinite plane sheet of charge (conductor or insulator) with surface charge density σ, Gauss’s law gives a uniform electric field of magnitude E = σ/(2ε₀) on each side of the sheet, directed away from the sheet if it is positively charged and towards the sheet if it is negatively charged.

Now superpose fields due to both plates:

• Between the plates, the field due to the positively charged plate points from +σ towards −σ, while the field due to the negatively charged plate also points from +σ to −σ, so the magnitudes add.
• Hence E_between = σ/(2ε₀) + σ/(2ε₀) = σ/ε₀.

Outside the plates, the two fields are equal in magnitude and opposite in direction, so they cancel and the net electric field is zero.

Explanation of Standard Options

For such multiple-choice questions, typical options for the field between the plates are:

• 0: Incorrect, because fields from the two plates add (same direction) between them, not cancel.
• σ/(2ε₀): Incorrect, this is the field due to a single infinite sheet, not the combined field of two oppositely charged plates.
• σ/ε₀: Correct, since each plate contributes σ/(2ε₀) and they add between the plates.
• 2σ/ε₀: Incorrect, it would correspond to incorrectly doubling the already-added result or misusing the single-sheet formula.

Thus the correct value of n is 1, and the electric field between the plates is uniform, equal to σ/ε₀, and directed from the positively charged plate towards the negatively charged plate.