A cube of side 6.50 cm is placed in a uniform field E = 7.50 × 10^3 N/C with edges parallel to the field lines (field enters the right face, passes through, then exits the left face).(a) What is the net flux through the cube?(b) What is the flux through the right face?(c) What is the flux through the left face?

The expression of the flow we can find the electric  flow through the faces are:

   A) Net flow fi = 0

   B) Left face flow is: Ф = 31.69 Nm ^ 2 / C

   C) Right face flow is: Ф = -31.69

Given parameters

To find

    a) The net flow

    b) The flow of the right face

    c) The flow of the left face

The flux is defined by the scalar product of the normal to the area and the electric field.

           Ф = E . A

Where the bold letters indicate vectors, E is the electrioc field and A the area.

The normal of the area is a vector perpendicular to the area.

a) The net flow.

In this case, the normal of one face points to the left and the normal of the other surface points in the opposite direction, bone to the right.

        Ф  = E A - E A

Since the magnitude of the field is constant, this expression gives zero.

b) The flow on the left face.

The area of the face that is square is:

           A = l²

          A = 0.065²

           A = 4,225 10⁻³ m²

            Ф = E A

            Ф = - 7.50 10³ 4.225 10⁻³

            Ф = - 31.69  N / m² C

c) The flow on the right face

This face points away from the left face

           Ф = + 7.5 10³ 4.225 10⁻³

           Ф = 31.69 Nm² / C

In conclusion using the expression of the Electric flow we can find the flows through the faces are:

   A) net flow  Ф = 0

   B) left face flow Ф = 31.69 Nm² / C

   C) right face flow is: Фf= -31.69 Nm² / C

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a) , b) , c)

Explanation:

a) The net flux through the cube is:

b) The flux through the right face is:

c) The flux through the left face is:

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