Question

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1. A non-uniform electric field given by E = 5.0x i + 3.0 j (in SI units) pierces the Gaussian cube as shown in the figure. What is the electric flux through:

a.the right face

b.the left face, and

c.the top face of the cube?

Answer 1

The flux through the right face is φ = 60 N·m²/C and through the left face is  φ = - 20 N·m²/C and through the top is   φ = 12 N·m²/C

Explanation:

E = 5.0 x i + 3.0 j

• x = 1 m from the origin, for the left face
• x = 3 m from the origin, for the right face
• x = 0 at the top face which is in upward direction

Solution:

Flux through the right face.

E = 5.0 x i + 3.0 j

Substitute the value of x = 3 m (for the right face)

E = 5.0(3) i + 3.0 j

E = 15 i + 3 j

φ = ∫ E·d A

φ = ∫ (15 i + 3 j) · d A i

φ = ∫ 15 d A

φ = 15 ∫d A

The integral of d A is A

φ = 15 A

The are of the cube is A = (2)^2 = 4 m^2

φ = 15 (4 m^2)

φ = 60 N·m²/C

Now flux for the left face:

E = 5.0 x i + 3.0 j

E = 5.0(1) i + 3.0 j

E = 5 i + 3 j

φ = ∫ E·d A

φ = ∫ (5 i + 3 j) · d A -i

φ = ∫ - 5 d A = - 5 ∫d A   = - 5 A

φ = - 5 (4 m^2)

φ = - 20 N·m²/C

Flux at the top face of the cube:

E = 5.0 x i + 3.0 j

φ = ∫ E·d A

φ = ∫ (5.0 x i + 3.0 j) · d A j

φ = ∫ 3.0 d A

φ = 3 ∫ d A   = 3 A  = 3 (4 m²)

φ = 12 N·m²/C