Electric field due to an infinite line of charge with linear density λ at distance r from the line is:

Electric field due to an infinite line of charge with linear density λ at distance r from the line is:

• A. E = λ /(2π ε0 r), radial outward for λ>0
• B. E = λ /(4π ε0 r)
• C. E = (λ r)/(2π ε0)
• D. E = λ /(2π ε0 r^2)

Answer: (A)

A central idea here is symmetry and Gauss’s law for an infinite line of charge. The field from such a line depends only on the distance r from the line and points radially outward (for positive linear density). Take a cylindrical Gaussian surface with radius r and length L coaxial with the line. The electric field is perpendicular to the axis and has the same magnitude at every point on the curved surface, so the flux through the sides is E(2πrL). The ends contribute nothing because E is perpendicular to them. The charge enclosed is λL. Gauss’s law gives E(2πrL) = (λL)/ε0, so E = λ/(2π ε0 r). The direction is outward if λ is positive. Other expressions don’t fit the cylindrical symmetry: 1/(4π ε0 r) is the field of a point charge, (λr)/(2π ε0) grows with r, and 1/(2π ε0 r^2) decays too quickly and isn’t consistent with the line’s symmetry.

A central idea here is symmetry and Gauss’s law for an infinite line of charge. The field from such a line depends only on the distance r from the line and points radially outward (for positive linear density).

Take a cylindrical Gaussian surface with radius r and length L coaxial with the line. The electric field is perpendicular to the axis and has the same magnitude at every point on the curved surface, so the flux through the sides is E(2πrL). The ends contribute nothing because E is perpendicular to them. The charge enclosed is λL. Gauss’s law gives E(2πrL) = (λL)/ε0, so E = λ/(2π ε0 r). The direction is outward if λ is positive.

Other expressions don’t fit the cylindrical symmetry: 1/(4π ε0 r) is the field of a point charge, (λr)/(2π ε0) grows with r, and 1/(2π ε0 r^2) decays too quickly and isn’t consistent with the line’s symmetry.