Re: [問題] 大學普物電場求法

看板Physics作者Frobenius (▽．(▽×▽φ)=0)時間13年前 (2011/06/14 02:11)推噓4(4推 0噓 5→)

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作者 Frobenius (▽．(▽×▽φ)=0) 看板 trans_math 標題 Re: [張爸] 算物理遇到的時間 Mon Aug 24 17:39:58 2009 ─────────────────────────────────────── 綜合版詳解 ( Honor1984、t86xu3 and me ) a/2 a/2 ∫ ∫ (x^2 + y^2 + d^2)^(-3/2)dxdy -a/2 -a/2 π/4 (a/2)secθ = 8∫ ∫ (r^2 + d^2)^(-3/2)rdrdθ 0 0 π/4 (a/2)secθ = 8∫ ∫ (1/2)(r^2 + d^2)^(-3/2) d(r^2+d^2) dθ 0 0 π/4 1 │r = (a/2)secθ = 8∫ [ - --------------│ ]dθ 0 √(r^2 + d^2) │r = 0 π/4 1 1 = 8∫ [ - - --------------------------- ]dθ 0 d √[(a/2)^2(secθ)^2 + d^2)] 2π 8 π/4 dθ = --- - - ∫ -------------------------- d d 0 √[(a/2d)^2(secθ)^2 + 1)] 2π 8 π/4 dθ = --- - - ∫ ----------------------------- d d 0 √[(a/2d)^2 (secθ)^2 + 1)] 2π 8 π/4 cosθ dθ = --- - - ∫ --------------------------- d d 0 √[(a/2d)^2 + cos^2(θ)] 2π 8 π/4 dsinθ = --- - - ∫ ------------------------------- d d 0 √[(a/2d)^2 + 1 - sin^2(θ)] 2π 8 π/4 dsinθ/√((a/2d)^2 + 1) = --- - - ∫ ----------------------------------- d d 0 √[1 - sin^2(θ)/((a/2d)^2 + 1)] 2π 8 -1 1 = --- - - Sin [ ------------------- ] d d √[2 (a/2d)^2 + 2)] 2π 8 -1 1 = --- - - Tan [ ------------------ ] d d √[2 (a/2d)^2 + 1] 2π 8 -1 1 = --- - - Tan [ ------------------ ] d d √[1 + 2 (a/2d)^2] 4 π -1 1 = - { --- - 2 Tan [ ------------------ ] } d 2 √[1 + 2 (a/2d)^2] ------------------------------------------------------------------------- 2Tanφ Tan2φ = ------------- 1 - (Tanφ)^2 -1 2Tanφ 2φ = Tan [ ------------- ] 1 - (Tanφ)^2 -1 1 1 φ = Tan [ ------------------ ] => Tanφ = ------------------ √[1 + 2 (a/2d)^2] √[1 + 2 (a/2d)^2] 1 2 (a/2d)^2 => (Tanφ)^2 = ---------------- => 1 - (Tanφ)^2 = ---------------- 1 + 2 (a/2d)^2 1 + 2 (a/2d)^2 -1 1 -1 2 / √[1 + 2 (a/2d)^2] 2 Tan [ ------------------ ] = Tan [ ------------------------------ ] √[1 + 2 (a/2d)^2] 2 (a/2d)^2 /[1 + 2 (a/2d)^2] -1 √[1 + 2 (a/2d)^2] -1 = Tan [ ------------------- ] = Tan [ (2d/a) √[2 + (2d/a)^2] ] (a/2d)^2 ------------------------------------------------------------------------- 4 π -1 = - { --- - Tan [ (2d/a) √[2 + (2d/a)^2] ] } d 2 4 -1 = - Cot [ (2d/a) √[2 + (2d/a)^2] ] d 4 -1 2d √(2a^2+4d^2) = - Cot [ ---------------- ] d a^2 4 -1 a^2 = - Sin [ ---------- ] d a^2 + 4d^2 有限長正方形帶電平板(面電荷密度σ)正上方距離d的點P的電場是 → ^ σ a/2 a/2 d dx dy E(a,d) = z -------- ∫ ∫ ------------------------- 4πε0 -a/2 -a/2 (x^2 + y^2 + d^2)^(3/2) ^ σ 4 -1 a^2 = z -------- d - Sin [ ---------- ] 4πε0 d a^2 + 4d^2 ^ σ -1 a^2 = z -------- 4 Sin [ ---------- ] 4πε0 a^2 + 4d^2 ^ σ -1 a^2 = z -------- Sin [ ---------- ] πε0 a^2 + 4d^2 無限長帶電平板(面電荷密度σ)正上方距離d的點P的電場是 → ^ σ π ^ σ lim E(a,d) = z -------- --- = z ------ a→∞ πε0 2 2ε0 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 114.43.241.200 ※ 編輯: Frobenius 來自: 114.43.241.200 (08/24 17:57)

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Honor1984

08/24 21:36,

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ntust661

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midarmyman

08/25 15:59,

08/25 15:59