j0106 - Jackson Problem 1.6 B J Mattson 1 Part a Outside...

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Unformatted text preview: Jackson Problem 1.6 B. J. Mattson 1 Part a Outside both of the plates, the total E field is zero (a Gaussian pillbox including both plates conatins no net charge). Between the plates, the E filds from each plate add. So, for the magnitude of the field between the plates, I just need to double the contribution for one plate. Use a pillbox on the upper plate with area, a, and squish the box so that the contri— butions from the sides are zero. E fida = — p(:Z"’) d33: (1) 60 Eoutszdea+Em31dea — fl 60 2E = 5 (3) 60 So, the field between the plates is E = —%2. Choosing a path in the —2 direction, the potential difference between the plates is: v = —/dOE-d7 (5) = Adieu (6) = id (7) _Q _Qd But,0—Z,soV—€0—A. And, the capcitance is: __Q__6A :C—V—a 2 Part b The field between a and b is: *ex U11 §> & a || | r—t ‘o A Si 2; w a 47r7“2E Then7 And the eapcitance is: :> C : g : 47reofl 3 Part C Using a cylinder between the conducting cylinders as the Gaussian surface, a 1 E fida = — p(:fi’)d3x 60 27TTLE = Q 60 Q E = 27TL60T So the potential is: And the capacitance: :> C : g : 27TL60 [Zn (2)]71 ...
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j0106 - Jackson Problem 1.6 B J Mattson 1 Part a Outside...

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