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Inhomogeneous electric field

What is the total flux of the electric field E = c y 2 k ^ through the rectangular surface shown in [link] ?

A rectangle labeled S is shown in the xy plane. Its side along y axis is of length a and that along the x axis measures b. A strip is marked on the rectangle, with its length parallel to x axis. Its length is b and its breadth is dy. Its area is labeled dA equal to b dy. Two arrows are shown perpendicular to S, n hat equal to k hat and vector E equal to cy squared k hat. These point in the positive z direction.
Since the electric field is not constant over the surface, an integration is necessary to determine the flux.

Strategy

Apply Φ = S E · n ^ d A . We assume that the unit normal n ^ to the given surface points in the positive z -direction, so n ^ = k ^ . Since the electric field is not uniform over the surface, it is necessary to divide the surface into infinitesimal strips along which E is essentially constant. As shown in [link] , these strips are parallel to the x -axis, and each strip has an area d A = b d y .

Solution

From the open surface integral, we find that the net flux through the rectangular surface is

Φ = S E · n ^ d A = 0 a ( c y 2 k ^ ) · k ^ ( b d y ) = c b 0 a y 2 d y = 1 3 a 3 b c .

Significance

For a non-constant electric field, the integral method is required.

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Check Your Understanding If the electric field in [link] is E = m x k ^ , what is the flux through the rectangular area?

m a b 2 / 2

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Summary

  • The electric flux through a surface is proportional to the number of field lines crossing that surface. Note that this means the magnitude is proportional to the portion of the field perpendicular to the area.
  • The electric flux is obtained by evaluating the surface integral
    Φ = S E · n ^ d A = S E · d A ,

    where the notation used here is for a closed surface S .

Conceptual questions

Discuss how would orient a planar surface of area A in a uniform electric field of magnitude E 0 to obtain (a) the maximum flux and (b) the minimum flux through the area.

a. If the planar surface is perpendicular to the electric field vector, the maximum flux would be obtained. b. If the planar surface were parallel to the electric field vector, the minimum flux would be obtained.

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What are the maximum and minimum values of the flux in the preceding question?

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The net electric flux crossing a closed surface is always zero. True or false?

true

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The net electric flux crossing an open surface is never zero. True or false?

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Problems

A uniform electric field of magnitude 1.1 × 10 4 N/C is perpendicular to a square sheet with sides 2.0 m long. What is the electric flux through the sheet?

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Calculate the flux through the sheet of the previous problem if the plane of the sheet is at an angle of 60 ° to the field. Find the flux for both directions of the unit normal to the sheet.

Φ = E · A E A cos θ = 2.2 × 10 4 N · m 2 / C electric field in direction of unit normal; Φ = E · A E A cos θ = −2.2 × 10 4 N · m 2 / C electric field opposite to unit normal

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Find the electric flux through a rectangular area 3 cm × 2 cm between two parallel plates where there is a constant electric field of 30 N/C for the following orientations of the area: (a) parallel to the plates, (b) perpendicular to the plates, and (c) the normal to the area making a 30 ° angle with the direction of the electric field. Note that this angle can also be given as 180 ° + 30 ° .

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The electric flux through a square-shaped area of side 5 cm near a large charged sheet is found to be 3 × 10 −5 N · m 2 / C when the area is parallel to the plate. Find the charge density on the sheet.

3 × 10 −5 N · m 2 / C ( 0.05 m ) 2 = E σ = 2.12 × 10 −13 C / m 2

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Two large rectangular aluminum plates of area 150 cm 2 face each other with a separation of 3 mm between them. The plates are charged with equal amount of opposite charges, ± 20 μ C . The charges on the plates face each other. Find the flux through a circle of radius 3 cm between the plates when the normal to the circle makes an angle of 5 ° with a line perpendicular to the plates. Note that this angle can also be given as 180 ° + 5 ° .

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A square surface of area 2 cm 2 is in a space of uniform electric field of magnitude 10 3 N/C . The amount of flux through it depends on how the square is oriented relative to the direction of the electric field. Find the electric flux through the square, when the normal to it makes the following angles with electric field: (a) 30 ° , (b) 90 ° , and (c) 0 ° . Note that these angles can also be given as 180 ° + θ .

a. Φ = 0.17 N · m 2 /C;
b. Φ = 0 ; c. Φ = E A cos 0 ° = 1.0 × 10 3 N / C ( 2.0 × 10 −4 m ) 2 cos 0 ° = 0.20 N · m 2 / C

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A vector field is pointed along the z -axis, v = α x 2 + y 2 z ^ . (a) Find the flux of the vector field through a rectangle in the xy -plane between a < x < b and c < y < d . (b) Do the same through a rectangle in the yz -plane between a < z < b and c < y < d . (Leave your answer as an integral.)

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Consider the uniform electric field E = ( 4.0 j ^ + 3.0 k ^ ) × 10 3 N/C . What is its electric flux through a circular area of radius 2.0 m that lies in the xy -plane?

Φ = 3.8 × 10 4 N · m 2 / C

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Repeat the previous problem, given that the circular area is (a) in the yz -plane and (b) 45 ° above the xy- plane.

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An infinite charged wire with charge per unit length λ lies along the central axis of a cylindrical surface of radius r and length l . What is the flux through the surface due to the electric field of the charged wire?

E ( z ) = 1 4 π ε 0 2 λ z k ^ , E · n ^ d A = λ ε 0 l

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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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