What is the electric field just outside each side of an infinite plane sheet with surface density σ?

What is the electric field just outside each side of an infinite plane sheet with surface density σ?

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

Answer: (C)

The key idea is that an infinite plane of charge produces a uniform electric field perpendicular to the plane on both sides, with the same magnitude on each side due to symmetry. To find that magnitude, use Gauss’s law with a pillbox that straddles the plane. The field is perpendicular to the plane on both sides, and by symmetry its magnitude is the same just above and just below the sheet. The outward flux through the two faces of the pillbox is E A on the top face plus E A on the bottom face, giving 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). Direction depends on the sign of σ: if σ is positive, the field points away from the sheet on both sides; if σ is negative, it points toward the sheet on both sides. The magnitude is σ/(2 ε0) on each side, which is why the answer lists that value without a specified direction. The alternative values would correspond to different scenarios, such as a conducting sheet, where the field just outside would be σ/ε0, not σ/(2ε0).

The key idea is that an infinite plane of charge produces a uniform electric field perpendicular to the plane on both sides, with the same magnitude on each side due to symmetry. To find that magnitude, use Gauss’s law with a pillbox that straddles the plane. The field is perpendicular to the plane on both sides, and by symmetry its magnitude is the same just above and just below the sheet. The outward flux through the two faces of the pillbox is E A on the top face plus E A on the bottom face, giving 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).

Direction depends on the sign of σ: if σ is positive, the field points away from the sheet on both sides; if σ is negative, it points toward the sheet on both sides. The magnitude is σ/(2 ε0) on each side, which is why the answer lists that value without a specified direction. The alternative values would correspond to different scenarios, such as a conducting sheet, where the field just outside would be σ/ε0, not σ/(2ε0).