730_Physics ProblemsTechnical Physics

# 730_Physics ProblemsTechnical Physics - 70 Electric...

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70 Electric Potential (b) From part (a), when the outer cylinder is considered to be at zero potential, the potential at a distance r from the axis is V k r r e a = F H G I K J 2 λ ln . The field at r is given by E V r k r r r r k r e a a e = − = − F H G I K J F H G I K J = 2 2 2 λ λ . But, from part (a), 2 k V r r e a b λ = ln b g . Therefore, E V r r r a b = F H G I K J ln b g 1 . P25.62 (a) From Problem 61, E V r r r a b = ln b g 1 . We require just outside the central wire 5 50 10 50 0 10 0 850 1 6 3 . . ln . × = × F H G I K J V m V m r r b b b g or 110 0 850 1 1 m m F H G I K J = e j r r b b ln . . We solve by homing in on the required value r b m a f 0.0100 0.00100 0.00150 0.00145 0.00143
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