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 Vd dr r Dd d d 1 00 22 =− = F H G I K J + zz Er wire wire ln . The presence of the linear charge density on the negative wire makes an identical contribution to the potential difference between the wires. Therefore, the total potential difference is ∆∆ VV d == F H G I K J 2 1 0 bg ln and the capacitance of this system of two wires, each of length A , is C Q Ddd === ∈− = λλ λπ AA A 0 0 af ln ln . The capacitance per unit length is: C A = 0 ln . *P26.77 The condition that we are testing is that the capacitance increases by less than 10%, or, < C C 110 . . 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 1.10 ln ln ln ln . ch . This becomes,
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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