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Φ = S E · n ^ d A = S E · d A ( closed surface )

where the circle through the integral symbol simply means that the surface is closed, and we are integrating over the entire thing. If you only integrate over a portion of a closed surface, that means you are treating a subset of it as an open surface.

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?

A rectangular patch is shown in the xy plane. Its side along the x axis is of length a and its side along the y axis is of length b. Arrows labeled E subscript 0 originate from the plane and point in the positive z direction.
Calculating the flux of E 0 through a rectangular surface.

Strategy

Apply the definition of flux: Φ = E · A ( uniform E ) , where the definition of dot product is crucial.

Solution

  1. In this case, Φ = E 0 · A = E 0 A = E 0 a b .
  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?

A cube ABCDKFGH is shown in the center. A diagonal plane is shown within it from KF to BC. The top surface of the cube, FGHK has a plane labeled minus q slightly above it and parallel to it. Similarly, another plane is labeled plus q is shown slightly below the bottom surface of the cube, parallel to it. Small red arrows are shown pointing upwards from the bottom plane, pointing up to the bottom surface of the cube, pointing up from the top surface of the cube and pointing up to the top plane. These are labeled vector E.
Calculating the flux of E 0 through a closed cubic surface.

Strategy

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

Solution

Through the top face of the cube, Φ = E 0 · A = E 0 A .

Through the bottom face of the cube, Φ = E 0 · A = 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 Φ 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 E of magnitude 10 N/C is directed parallel to the yz -plane at 30 ° above the xy -plane, as shown in [link] . What is the electric flux through the plane surface of area 6.0 m 2 located in the xz -plane? Assume that n ^ points in the positive y -direction.

A rectangular surface S is shown in the xz plane. Three arrows labeled n hat originate from three points on the surface and point in the positive y direction. Three longer arrows labeled vector E also originate from the same points. They make an angle of 30 degrees with n hat.
The electric field produces a net electric flux through the surface S .

Strategy

Apply Φ = S E · n ^ d A , where the direction and magnitude of the electric field are constant.

Solution

The angle between the uniform electric field E and the unit normal n ^ to the planar surface is 30 ° . 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

Φ = S E · n ^ d A = E A cos θ = ( 10 N/C ) ( 6.0 m 2 ) ( cos 30 ° ) = 52 N · m 2 /C .

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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Check Your Understanding What angle should there be between the electric field and the surface shown in [link] in the previous example so that no electric flux passes through the surface?

Place it so that its unit normal is perpendicular to E .

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Practice Key Terms 3

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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