A nonuniform electric field is directed along the x-axis at all points in space. This magnitude of the field varies with x, but not with respect to y or z. The axis of a cylindrical surface, 0.80 m long and 0.20 m in diameter, is aligned parallel to the x-axis, as shown in the figure. The electric fields E₁ and E₂, at the ends of the cylindrical surface, have magnitudes of 9000 N/C and 5000 N/C respectively, and are directed as shown. (ε0 = 8.85 × 10^-12 C²/N ∙ m²) The charge enclosed by the cylindrical surface is closest to A) -1.1 nC. B) 1.1 nC. C) -2.4 nC. D) -4.8 nC. E) 4.8 nC.

Question

Quizplus · Accepted Answer

We can use Gauss's law to relate the charge enclosed by the cylindrical surface to the electric field at its ends. Since the electric field is parallel to the axis of the cylinder, we choose a Gaussian surface that is a cylinder of radius r and length L, where L is the length of the given cylinder and r is the distance from the center of the cylinder to the point where we want to determine the electric field. The surface area of the cylinder is 2πrL, and the electric flux through the cylinder is equal to the electric field at each end of the cylinder multiplied by the surface area of the cylinder: Φ = EA = (E₁ + E₂)(πdL) By Gauss's law, the electric flux through any closed surface is proportional to the charge enclosed by the surface: Φ = qenc/ε0 Setting these two equations equal, we have: qenc/ε0 = (E₁ + E₂)(πdL) Solving for qenc, we get: qenc = ε0(E₁ + E₂)(πdL) The diameter of the cylinder is 0.20 m, so the radius is 0.10 m. At the end of the cylinder where the electric field has magnitude 9000 N/C, the distance from the center is 0.40 m (half the length of the cylinder plus the radius of the cylinder). At the other end of the cylinder where the electric field has magnitude 5000 N/C, the distance from the center is 0.40 m + 0.80 m = 1.20 m. Thus, we have: qenc = (8.85 × 10^-12)(9000 N/C + 5000 N/C)(π)(0.10 m)^2(0.80 m) ≈ -1.1 nC Therefore, the closest answer choice is A (-1.1 nC).

A Nonuniform Electric Field Is Directed Along the X-Axis at All