Homework03Images - Problem 5.7 A circular loop of line...

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ECE 3030: Electromagnetic Fields and Waves Fall 2009 Homework 3 SOLUTIONS for GRADER: Images Reminder: These details are for the students to work out! Date: Tuesday 9/22 Problem 4.15: a. The location of the image is at z = - d , parallel to the x - y plane, and carries a charge of - λ C/m. b. The x component at point P is zero (by symmetry), i.e., E x = 0. c. The y component at point P is zero (by symmetry), i.e., E y = 0. d. The z component is produced by two circular loops of opposite charge: Φ( x = 0 ,y = 0 ,z ) = + λ 4 πε 0 p c 2 + ( z - d ) 2 2 π Z 0 c d φ + - λ 4 πε 0 p c 2 + ( z + d ) 2 2 π Z 0 c d φ = λc 2 ε 0 p c 2 + ( z - d ) 2 + - λc 2 ε 0 p c 2 + ( z + d ) 2 Note: Do not substitute z = L into the potential before taking its gradient; we need to differentiate with respect to z for the electric field: ~ E P = ~ E( ~ r = L ˆ z) = ~ E( x = 0 ,y = 0 ,z = L ) = - Φ ∂z ˆ z = - 1 2 λc [ - 2( z - d )] 2 ε 0 [ c 2 + ( z + d ) 2 ] 3 / 2 ˆ z z = L - - 1 2 λc [ - 2( z - d )] 2 ε 0 [ c 2 + ( z + d ) 2 ] 3 / 2 ˆ z z = L = λc 2 ε 0 " L - d [ c 2 + ( z - d ) 2 ] 3 / 2 - L + d [ c 2 + ( z + d ) 2 ] 3 / 2 # ˆ z using the cosine trick from last week’s homework.
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Unformatted text preview: Problem 5.7 A circular loop of line charge. From Problem 5.6 (see your notes for Demo 3), a charge q next to a dielectric medium of permittivity produces an image charge of-q - + . From this, you should deduce that the loop with a charge density of + will have an image loop with a charge density of- - + . By symmetry, the electric eld will have only a z component at the point P due to the two loops of charge: E z = 2 a 4 [( d-L/ 2) 2 + a 2 ] d-L/ 2 p ( d-L/ 2) 2 + a 2- - + 2 a 4 [( d + L/ 2) 2 + a 2 ] d + L/ 2 p ( d + L/ 2) 2 + a 2 CORNELL UNIVERSITY c WES SWARTZ (09/8/29) 31...
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