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309-2008-Solutions3

# 309-2008-Solutions3 - ECE 309 Electromagnetic Fields...

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ECE 309 — Electromagnetic Fields University of Virginia Fall 2008 Homework # 3 Solutions 1. Electric Potential of Charge Distributions (a) (b) (c) (a) Consider configuration (a) above with two point charges separated by d . Using superpo- sition, Φ = 1 4 π 0 n X i =1 q i r i = q 4 π 0 n 1 r + 1 r o where Pythagoras tells us that r 2 = z 2 + ( d/ 2) 2 . Thus, Φ( z ) = q 2 π 0 1 p z 2 + ( d/ 2) 2 The corresponding electric field is given by, ~ E = - ~ Φ = - ˆ z d dz Φ( z ) = qz 2 π 0 ( z 2 + d 2 4 ) 3 2 ˆ z as we found in the previous homework! (b) for configuration (b), we integrate over the line of charge, Φ = 1 4 π 0 Z L - L λd‘ r = λ 4 π 0 Z L - L dx x 2 + z 2

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This integral is a little tricky (unless you just look it up!). Using the substitution, y - x p x 2 + z 2 , dy dx - 1 = x x 2 + z 2 dx = x 2 + z 2 x + x 2 + z 2 dy The integral (in terms of the new variable y ) becomes, λ 4 π 0 Z x 2 + z 2 x + x 2 + z 2 dy x 2 + z 2 = λ 4 π 0 Z dy y = λ 4 π 0 log ( y ) Thus Φ( z ) = λ 4 π 0 log x + p x 2 + z 2 L - L Φ( z ) = λ 4 π 0 log L 2 + z 2 + L L 2 + z 2 - L Taking the gradient of Φ , we have E z = - d Φ dz = - λ 4 π 0 L 2 + z 2 + L L 2 + z 2 - L L 2 + z 2 - L - ( L 2 + z 2 + L ) ( L 2 + z 2 - L ) 2 z L 2 + z 2 = λ 4 π 0 2 Lz L 2 + z 2 - L 2 1 L 2 + z 2 E z = 1 2 π 0 λL z L 2 + z 2 (c) For configuration (c), we have Φ = 1 4 π 0 Z S σda r = σ 4 π 0 Z R 0 Z 2 π 0 rdrdφ r 2 + z 2 = σ 0 Z R 0 rdrdφ r 2 + z 2
Φ( z ) = σ 2 0 p r 2 + z 2 R 0 = σ 2 0 p R 2 + z 2 - z The corresponding field is, E z = - d Φ( z ) dz = - σ 2 0 z R 2 + z 2 - 1 = σ 2 0 1 - z z 2 + R 2 If the right-hand charge in part (a) is changed to - q , then Φ( z ) = q 4 π 0 n 1 r - 1 r o = 0 which means the potential is zero everywhere along the z axis. Since, ~ E = - ~ Φ = - ˆ z d Φ( z ) dz = 0 The z -component of ~ E must be zero on the z -axis (meaning the field is perpendicular to that axis).

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309-2008-Solutions3 - ECE 309 Electromagnetic Fields...

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