Re: [題目] 定體電荷密度 某兩點之電位差

[nightkid (固有結界：無限の劍製)]

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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