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
--
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◆ From: 114.43.241.200
※ 編輯: Frobenius 來自: 114.43.241.200 (08/24 17:57)
1F：推 Honor1984:集大成 呵呵122.124.107.112 08/24 21:36
2F：推 ntust661:推122.122.217.172 08/24 23:04
3F：→ midarmyman:推囉! 114.41.234.239 08/25 15:59
4F：推 yang770610:我覺得可以m了..... 122.124.84.29 09/13 23:51