6.1 Electric flux  (Page 4/8)

Flux of a uniform electric field

A constant electric field of magnitude $E_0$ points in the direction of the positive z -axis ( [link] ). What is the electric flux through a rectangle with sides a and b in the (a) xy -plane and in the (b) xz -plane?

Strategy

Apply the definition of flux: $\text{Φ}=\overrightarrow{\text{E}}·\overrightarrow{\text{A}}(\text{uniform}\overrightarrow{\text{E}})$ , where the definition of dot product is crucial.

Solution

1. In this case, $\text{Φ}={\overrightarrow{\text{E}}}_0·\overrightarrow{\text{A}}=E_{0}A=E_{0}ab.$
2. Here, the direction of the area vector is either along the positive y -axis or toward the negative y -axis. Therefore, the scalar product of the electric field with the area vector is zero, giving zero flux.

Significance

The relative directions of the electric field and area can cause the flux through the area to be zero.

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Flux of a uniform electric field through a closed surface

A constant electric field of magnitude $E_0$ points in the direction of the positive z -axis ( [link] ). What is the net electric flux through a cube?

Strategy

Apply the definition of flux: $\text{Φ}=\overrightarrow{\text{E}}·\overrightarrow{\text{A}}(\text{uniform}\overrightarrow{\text{E}})$ , noting that a closed surface eliminates the ambiguity in the direction of the area vector.

Solution

Through the top face of the cube, $\text{Φ}={\overrightarrow{\text{E}}}_0·\overrightarrow{\text{A}}=E_{0}A.$

Through the bottom face of the cube, $\text{Φ}={\overrightarrow{\text{E}}}_0·\overrightarrow{\text{A}}=\text{−}{E}_{0}A,$ because the area vector here points downward.

Along the other four sides, the direction of the area vector is perpendicular to the direction of the electric field. Therefore, the scalar product of the electric field with the area vector is zero, giving zero flux.

The net flux is ${\text{Φ}}_{\text{net}}=E_{0}A-E_{0}A+0+0+0+0=0$ .

Significance

The net flux of a uniform electric field through a closed surface is zero.

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Electric flux through a plane, integral method

A uniform electric field $\overrightarrow{\text{E}}$ of magnitude 10 N/C is directed parallel to the yz -plane at $30\text{°}$ above the xy -plane, as shown in [link] . What is the electric flux through the plane surface of area $6.0{\text{m}}^2$ located in the xz -plane? Assume that $\widehat{\text{n}}$ points in the positive y -direction.

Strategy

Apply $\text{Φ}={\displaystyle {\int }_S\overrightarrow{\text{E}}·\widehat{\text{n}}dA}$ , where the direction and magnitude of the electric field are constant.

Solution

The angle between the uniform electric field $\overrightarrow{\text{E}}$ and the unit normal $\widehat{\text{n}}$ to the planar surface is $30\text{°}$ . Since both the direction and magnitude are constant, E comes outside the integral. All that is left is a surface integral over dA , which is A . Therefore, using the open-surface equation, we find that the electric flux through the surface is

Significance

Again, the relative directions of the field and the area matter, and the general equation with the integral will simplify to the simple dot product of area and electric field.

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