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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

Answer :

onyebuchinnaji

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, [tex]\phi =\frac{Q_{enclosed}}{\epsilon_o}[/tex]

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

${teks-lihat-gambar} onyebuchinnaji
Cricetus

The charge enclosed by the cylindrical surface will be "-0.28 nC".

Given:

Diameter,

  • d = 0.20 m

then,

Radius,

  • r = 0.10 m

Electric field at first end,

  • [tex]E_1 = 2000 \ N/C[/tex]

Electric field at second end,

  • [tex]E_2 = 1000 \ N/C[/tex]

The flux through the end caps will be:

→ [tex]\varphi = (E_2-E_1) A[/tex]

      [tex]= -1000(\pi r^2)[/tex]

      [tex]= -1000(\pi(0.10)^2)[/tex]

With the help of Gauss law,

→ [tex]\varphi = \frac{q}{\varepsilon_0}[/tex]

→ [tex]q = \varphi \varepsilon_0[/tex]

     [tex]= -1000(\pi (0.10)^2)(8.85\times 10^{-12})[/tex]

     [tex]= -0.28 \ nC[/tex]

Thus the answer above i.e., "option e" is correct.

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