70 Electric Potential(b)From part (a), when the outer cylinder is considered to be at zero potential, the potential at adistance rfrom the axis isVkrrea=FHGIKJ2λln.The field at ris given byEVrkrrrrkreaae=−∂∂FHGIKJ−FHGIKJ=222.But, from part (a), 2kVrreab=∆lnbg.Therefore, EVr=FHGIKJ∆ln1.P25.62(a)From Problem 61,EVr=∆ln1.We require just outside the central wire550 1050 0 100850163..ln .×=×FHGIKJ±VmVmrrbbor11011mm−FHGIKJ=ejrrbbln..We solve by homing in on the required valuerbmaf0.01000.001000.001500.001450.001430.00142110
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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 .