Electric field due to an infinite plane sheet of charge with surface density σ. Which expression describes the field on either side of the sheet?

Electric field due to an infinite plane sheet of charge with surface density σ. Which expression describes the field on either side of the sheet?

• A. E = σ /(ε0) on each side
• B. E = σ /(2ε0) on each side, directed away from plane for σ>0
• C. E = σ /(2ε0) on one side only
• D. E = 2σ /(ε0)

Answer: (B)

The field from an infinite plane sheet of charge is uniform and points perpendicular to the plane on both sides. The symmetry means the magnitude cannot depend on distance, and the direction must be away from the sheet if the surface charge is positive (and toward the sheet if negative). Using Gauss’s law with a pillbox that straddles the sheet, the top and bottom faces are the only surfaces with flux. If the field has the same magnitude on both sides, the flux through the top is E A and through the bottom is E A (the directions align with the outward normals), giving a total flux of 2 E A. The enclosed charge is σ A, so Gauss’s law gives 2 E A = σ A / ε0, hence E = σ / (2 ε0). This magnitude is the same on both sides, and the direction is away from the plane for positive σ (toward the plane for negative σ). So the field on either side has magnitude σ/(2ε0) and points away from the plane when σ > 0. The other expressions either overcount the field on each side or fail to account for the field’s presence on both sides.

The field from an infinite plane sheet of charge is uniform and points perpendicular to the plane on both sides. The symmetry means the magnitude cannot depend on distance, and the direction must be away from the sheet if the surface charge is positive (and toward the sheet if negative).

Using Gauss’s law with a pillbox that straddles the sheet, the top and bottom faces are the only surfaces with flux. If the field has the same magnitude on both sides, the flux through the top is E A and through the bottom is E A (the directions align with the outward normals), giving a total flux of 2 E A. The enclosed charge is σ A, so Gauss’s law gives 2 E A = σ A / ε0, hence E = σ / (2 ε0). This magnitude is the same on both sides, and the direction is away from the plane for positive σ (toward the plane for negative σ).

So the field on either side has magnitude σ/(2ε0) and points away from the plane when σ > 0. The other expressions either overcount the field on each side or fail to account for the field’s presence on both sides.