Complete question:
Check the image uploaded for the direction of the electric field, as indicated in Fig. 22.1.
Answer:
(e) the charge enclosed by the cylindrical surface is closest to - 0.28 nC
Explanation:
Given;
length of the cylinder = 0.8 m
diameter of the cylinder = 0.2 m, radius, r = 0.1 m
electric field at the entering point of the cylinder, E₁ = 2000 N/C
electric field at the end point of the cylinder, E₂ = 1000 N/C
Total electric flux in the enclosed surface due to the parallel fields, is given as;
Φ = ΣE*A*cosθ
θ = 0, since the two fields are parallel to each other
Φ = ΣE*A = E₁A + E₂A
Flux through the entering point of the cylinder, E₁ is negative and positive at the end point, E₂ since no flux enter through this point.
Φ = ΣE*A = -E₁A + E₂A
= A( -E₁ + E₂)
= πr²( -E₁ + E₂)
= π(0.1)²( -2000 + 1000)
= π(0.1)² ( -1000)
= 0.03142 ( -1000)
= -31.42 Nm²/C
Also,
Q enc. = Φ * ε₀
= -31.42 * 8.85 x 10⁻¹²
= - 0.278 x 10⁻⁹ C
= - 0.28 nC
Thus, the charge enclosed by the cylindrical surface is closest to - 0.28 nC