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762_Physics ProblemsTechnical Physics

762_Physics ProblemsTechnical Physics - 102 P26.76...

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102 Capacitance and Dielectrics P26.76 The electric field due to the charge on the positive wire is perpendicular to the wire, radial, and of magnitude E r + = λ π 2 0 . The potential difference between wires due to the presence of this charge is V d dr r D d d D d d 1 0 0 2 2 = − = − = F H G I K J + z z E r wire wire λ π λ π ln . The presence of the linear charge density λ on the negative wire makes an identical contribution to the potential difference between the wires. Therefore, the total potential difference is V V D d d = = F H G I K J 2 1 0 b g λ π ln and the capacitance of this system of two wires, each of length A , is C Q V V D d d D d d = = = = λ λ λ π π A A A 0 0 b g a f a f ln ln . The capacitance per unit length is: C D d d A = π 0 ln a f . *P26.77 The condition that we are testing is that the capacitance increases by less than 10%, or, < C C 1 10 . . Substituting the expressions for C and C from Example 26.2, we have, = = < C C k k e b a e b a b a b a A A 2 2 1 10
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