A cube is bounded by x = 0 to a, y = 0 to a and z = 0 to a. The electric field is E = E₀x î, where E₀ = 4 × 10⁴ N C⁻¹ m⁻¹ and a = 2 cm. If the enclosed charge is Q × 10⁻¹⁴ C, find Q. Take ε₀ = 9 × 10⁻¹² C² N⁻¹ m⁻².
Physics · Electrostatics · Gauss law
JEE Mains JEE Advanced

Question

A cube is bounded by x = 0 to a, y = 0 to a and z = 0 to a. The electric field is E = E₀x î, where E₀ = 4 × 10⁴ N C⁻¹ m⁻¹ and a = 2 cm. If the enclosed charge is Q × 10⁻¹⁴ C, find Q. Take ε₀ = 9 × 10⁻¹² C² N⁻¹ m⁻².

Physics · Electrostatics · Gauss law

G1 Physics · Accepted Answer

Electric field: E = E₀x î Only the two faces normal to the x-axis contribute to net flux. At x = 0: E = 0 At x = a: E = E₀a Net flux: Φ = (E₀a)(a²) = E₀a³ By Gauss's law: qₑₙc = ε₀Φ = ε₀E₀a³ = (9 × 10⁻¹²)(4 × 10⁴)(2 × 10⁻²)³ = 288 × 10⁻¹⁴ C Therefore Q = 288.