Explanation Given: The electric field due to a point charge located at the origin in R 3 is E = k e r r 2 where r = x 2 + y 2 + z 2 constant, radius of the disk through we need to find the flux is 2 and center is (0,0,3) Calculation: Step 1: Let r = x 2 + y 2 + z 2 and r ^ = x 2 + y 2 We parametrize the disc by: Φ ( r ^ , θ ) = ( r ^ cos θ , r ^ sin θ , 3 ) T r ^ = ∂ Φ ∂ r ^ = 〈 cos θ , sin θ , 0 〉 T θ = ∂ Φ ∂ θ = 〈 − r ^ sin θ , r ^ cos θ , 0 〉 n = T r ^ × T θ = ( i j k cos θ sin θ 0 − r ^ sin θ r ^ cos θ 0 ) = 〈 0 , 0 , r ^ 〉 Step 2: Now, E ⋅ n = k e r r 2 ⋅ 〈 0 , 0 , r ^ 〉 = k r ^ r 3 〈 x , y , z 〉 ⋅ 〈

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ISBN: 9781319434526

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 17.5, Problem 21E

To determine

To calculate: The flux of E through the disk D of radius 2 parallel to the xy-plane with center (0, 0, 3).

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Chapter 17.5, Problem 20E

Chapter 17.5, Problem 22E

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Chapter 17 Solutions

CALCULUS 4E (LL) W/ ACHIEVE ACCESS

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The flux of E through the disk D of radius 2 parallel to the xy -plane with center ( 0 , 0 , 3 ) .

Solution Summary: The author calculates the flux of E through the disk D of radius 2 parallel to the xy -plane with center.

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To calculate: The flux of E through the disk D of radius 2 parallel to the xy-plane with center (0, 0, 3).

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Chapter 17 Solutions