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 Vk 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 ae =− F H G I K J F H G I K J = 2 2 2 . But, from part (a), 2 k V rr e ab = ln bg . Therefore, E V r = F H G I K J ln 1 . P25.62 (a) From Problem 61, E V r = ln 1 . We require just outside the central wire 550 10 50 0 10 0850 1 6 3 . . ln . ×= × F H G I K J ±Vm V m r r b b or 110 1 1 m m F H G I K J = ej r r b b ln . . We solve by homing in on the required value r b m af 0.0100 0.00100 0.00150 0.00145 0.00143 0.00142 110
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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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