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. The electric fields E 1 and E 2, at the ends of the cylindrical surface, have magnitudes of 2000 N/C and 1000 N/C respectively, and are directed as shown. In Fig. 22.1, the charge enclosed by the cylindrical surface is closest to:

a. 0.28 nC
b. -0.60 nC
c. 1.2 nC
d. -1.2 nC
e. -0.28 nC

Complete question:

Check the image uploaded for the direction of the electric field, as indicated in Fig. 22.1.

(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

pe = weightxheight = 2000x1000=2,000,000joules

b. false

explanation:

suppose a rock in free fall is dropped off a cliff. after falling for 5 seconds, its speed would not be 100 m/s.