309-2008-Solutions3

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

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Unformatted text preview: 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 n X i =1 q i r i = q 4 π n 1 r + 1 r o where Pythagoras tells us that r 2 = z 2 + ( d/ 2) 2 . Thus, Φ( z ) = q 2 π 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 ( 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 Z L- L λd‘ r = λ 4 π Z L- L dx √ x 2 + z 2 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 Z √ x 2 + z 2 x + √ x 2 + z 2 dy √ x 2 + z 2 = λ 4 π Z dy y = λ 4 π log( y ) Thus Φ( z ) = λ 4 π log x + p x 2 + z 2 L- L Φ( z ) = λ 4 π 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 √ 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 2 Lz L 2 + z 2- L 2 1 √ L 2 + z 2 E z = 1 2 π λL z √ L 2 + z 2 (c) For configuration (c), we have Φ = 1 4 π Z S σda r = σ 4 π Z R Z 2 π rdrdφ √ r 2 + z 2 = σ Z R rdrdφ √ r 2 + z 2 Φ( z ) = σ 2 p r 2 + z 2 R = σ 2 p R 2 + z 2- z The corresponding field is, E z =- d Φ( z ) dz =- σ 2 z √ R 2 + z 2- 1 = σ 2 1- z √ z 2 + R 2 If the right-hand charge in part (a) is changed to- q , then Φ( z ) = q 4 π n 1 r- 1 r o = 0 which means the potential is zero everywhere along the...
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309-2008-Solutions3 - ECE 309 — Electromagnetic Fields...

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