P26_038 - 38(a We use Eq 26-14 C = 20 L(4.7(0.15 m = 0.73...

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38. (a) We use Eq. 26-14: C =2 πε 0 κ L ln( b/a ) = (4 . 7)(0 . 15 m) 2 ( 8 . 99 × 10 9 N · m 2 C 2 ) ln(3 . 8cm / 3 . 6cm) =0 . 73 nF . (b) The breakdown potential is (14 kV
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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