An infinite charged sheet of surface density +σ and an infinite line charge of linear density +λ are parallel and separated by 5 m. Points P and Q are at perpendicular distances 3/π m and 4/π m from the line charge toward the sheet. If Eₚ/Eq = 4/a and |λ| = 2|σ|, find a.
Physics · Electrostatics · electric flux and gauss law
JEE Mains 13 April 2023 Shift 1 Morning Shift

Question

An infinite charged sheet of surface density +σ and an infinite line charge of linear density +λ are parallel and separated by 5 m. Points P and Q are at perpendicular distances 3/π m and 4/π m from the line charge toward the sheet. If Eₚ/Eq = 4/a and |λ| = 2|σ|, find a.

Physics · Electrostatics · electric flux and gauss law

JEE Mains 13 April 2023 Shift 1 Morning Shift

G1 Physics · Accepted Answer

Electric field of the sheet: Eₛ = σ/(2ε₀) Electric field of the line: Eₗ = λ/(2πε₀r) The fields oppose each other at P and Q. Using λ = 2σ: At P, r = 3/π: Eₚ = λ/[2πε₀(3/π)] - σ/(2ε₀) = 2σ/(6ε₀) - σ/(2ε₀) Taking magnitude: |Eₚ| = σ/(6ε₀) At Q, r = 4/π: Eq = 2σ/(8ε₀) - σ/(2ε₀) Magnitude: |Eq| = σ/(4ε₀) Therefore, Eₚ/Eq = (1/6)/(1/4) = 4/6 Given Eₚ/Eq = 4/a: a = 6 Final value: 6