Задачи по векторному анализу. Михайлов В.К - 104 стр.

      ρ     ρ ρ
4.53. a = θ er + eθ ;
      ρ               ρ           ρ
4.54. a = −2r cos θ er + r sin θ eθ ;
      ρ                 ρ                  ρ          ρ
4.55. a = sin θ ⋅ cos ϕ er + cos θ ⋅ cos ϕ eθ − sin ϕ eϕ ;
       ρ 2 cos θ ρ sin θ ρ
4.56. a =        er + 3 eθ .
            r3        r
4.57. Âíóòðè áåñêîíå÷íîãî êðóãëîãî öèëèíäðà ρ ≤ R ðàâ-
      íîìåðíî ðàñïðåäåëåí ýëåêòðè÷åñêèé çàðÿä. Èçâåñò-
      íî, ÷òî òàêîé öèëèíäð ñîçäàåò â ïðîñòðàíñòâå ýëåêò-
      ðè÷åñêîå ïîëå
       ρ         ρ                 kρ     при ρ ≤ R ,
       E = E( ρ )eρ , ãäå E( ρ ) =  2
                                   kR / ρ при ρ ≥ R
      (çäåñü k — íåêîòîðûé êîýôôèöèåíò).
      Âû÷èñëèòü ïîòåíöèàë u (ρ) ýòîãî ïîëÿ, ïîëàãàÿ, ÷òî
      u (0) = 0.
4.58. Ïî îáúåìó øàðà ρ ≤ R ðàâíîìåðíî ðàñïðåäåëåí ýëåêò-
      ðè÷åñêèé çàðÿä. Èçâåñòíî, ÷òî òàêîé øàð ñîçäàåò â
                                              ρ        ρ
      ïðîñòðàíñòâå ýëåêòðè÷åñêîå ïîëå E = E(r )er , ãäå

                         kr / R3 при r ≤ R ,
                  E(r) = 
                         k / r при r ≥ R
                               2



      (çäåñü k — íåêîòîðûé êîýôôèöèåíò).
      Âû÷èñëèòü ïîòåíöèàë u (r) ýòîãî ïîëÿ îòíîñèòåëüíî
      áåñêîíå÷íîñòè. ×åìó ðàâåí ïîòåíöèàë u0 â öåíòðå
      øàðà?
4.59. Ïðè êàêîé ôóíêöèè f (ρ, z) îñåñèììåòðè÷íîå (îòíî-
                              ρ     ρ              ρ
      ñèòåëüíî îñè z) ïîëå a = ρ 2 zeρ + f ( ρ, z )ez áóäåò ïî-
      òåíöèàëüíûì?
4.60. Ïðè êàêîé ôóíêöèè f (ρ, z) îñåñèììåòðè÷íîå (îòíî-
                            ρ    ρ              ρ
      ñèòåëüíî îñè z) ïîëå a = ρzeρ + f ( ρ, z )ez áóäåò ïîòåí-
      öèàëüíûì?




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