作者nightkid (固有結界：無限の劍製)
看板Physics
標題Re: [題目] 定體電荷密度 某兩點之電位差
時間Wed Mar 30 23:10:06 2011
使用cylindrical coordinate
Gauss's Law (Integrate form) : ∮E‧dA = Qenc/ε0 ; Qenc: 高斯面內所包之電荷量
在圓柱外之電場: (選定圓柱中心為z軸) 高斯面選取為z軸往外s、高度為L的圓柱
由於此問題之電場於z方向上必定為零
並且本題目具有azimuthal symmetry 所以在電場在s處必定為定值
且方向必定平行於所選定之高斯面之面積法向量
所以: ∮E‧dA = E ×2πsL = ρ×π(R^2)L/ε
=> E = ρ(R^2)/2εs ; 其方向為e_s
dV = E‧dl => △V = ∫E‧dl ; in this case: dl = (ds)e_s
=> △V = ∫Eds = ρ(R^2)/2ε ∫ds/s = ρ(R^2)/2ε ln(s2/s1)
接下來就自己代入吧
※ 引述《ComeonLuLuLu (盧彥勳加油)》之銘言：
: 電學
: 給定體電荷密度 求某兩點之電位差
: Charge of uniform density 90 nC/m3 is distributed throughout the inside of a
: long nonconducting cylindrical rod (radius = 2.0 cm). Determine the magnitude
: of the potential difference of point A (2.0 cm from the axis of the rod) and
: point B (4.0 cm from the axis).
: 一個長圓柱 圓柱面上的某點 與 該點外兩公分之一點 求電位差
: 線電位差 就已經很難積了
: 體電位差 要怎算啊 還有角度的....
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◆ From: 140.115.220.100
1F：推 ntust661:推@@ 03/30 23:10
※ 編輯: nightkid 來自: 140.115.220.100 (03/30 23:11)
※ 編輯: nightkid 來自: 140.115.220.100 (03/30 23:14)
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