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Chapter 16: Problem 62

(a) Find the electric flux through each side of a cube of edge length \(a\) in a uniform electric field of magnitude \(E\) The field direction is perpendicular to two of the faces. (b) What is the total flux through the cube?

Short Answer

Expert verified
Answer: The total electric flux through the cube is 0.

Step by step solution

01

Identify Given Information

: We are given: - Edge length of the cube (a) - A uniform electric field (E) perpendicular to two of the cube's faces
02

Calculate Flux through each face

: We will now calculate the flux through each face of the cube. Since there are three possible angles between the field lines and the faces, we will calculate the flux for each angle separately. 1. For the two faces perpendicular to the electric field, the angle between the electric field and the normal vector to the face (\(\theta\)) is 0 degrees. 2. For the two faces parallel to the electric field, the angle between the electric field and the normal vector to the face (\(\theta\)) is 90 degrees. 3. For the two faces perpendicular to the field but on the opposite sides, the angle between the electric field and the normal vector to the face (\(\theta\)) is 180 degrees.
03

Flux through faces perpendicular to the electric field (0 degrees)

: For the faces perpendicular to the electric field, we can use the electric flux formula: \(\Phi = EAcos(\theta)\) Since \(\theta = 0\) and \(cos(0) = 1\), the formula becomes: \(\Phi = EA\) So the flux through each of the two perpendicular faces is: \(\Phi_1 = Ea\)
04

Flux through faces parallel to the electric field (90 degrees)

: For the faces parallel to the electric field, we can use the electric flux formula: \(\Phi = EAcos(\theta)\) Since \(\theta = 90\) and \(cos(90) = 0\), the formula becomes: \(\Phi = 0\) So the flux through each of the two parallel faces is: \(\Phi_2 = 0\)
05

Flux through opposite faces perpendicular to the electric field (180 degrees)

: For the faces opposite to and perpendicular to the electric field, we can use the electric flux formula: \(\Phi = EAcos(\theta)\) Since \(\theta = 180\) and \(cos(180) = -1\), the formula becomes: \(\Phi = -EA\) So the flux through each of the two opposite perpendicular faces is: \(\Phi_3 = -Ea\)
06

Find the total electric flux through the cube

: We can now find the total electric flux through the cube by adding the flux through each face: Total Flux \(\Phi_{total} = 2\times\Phi_1 + 2\times\Phi_2 + 2\times\Phi_3\) \(\Phi_{total} = 2(Ea) + 2(0) + 2(-Ea) = 0\) The total electric flux through the cube is 0.

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Most popular questions from this chapter

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