732_Physics ProblemsTechnical Physics

732_Physics ProblemsTechnical Physics - 72 Electric...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
72 Electric Potential P25.66 (a) From Gauss’s law, E A = 0 (no charge within) Ek q rr r qq r Be A Ce AB == × × = F H G I K J = + −× =− F H G I K J 2 9 8 22 9 9 899 10 100 10 89 9 500 10 45 0 . . . . . . ej bg Vm ±Vm (b) Vk r = + F H G I K J 45 0 9 9 . . . V At r 2 , V 45 0 0300 150 . . V Inside r 2 , V r dr B r r + + F H G I K J =− + F H G I K J z 150 89 9 150 89 9 11 450 89 9 2 2 V V . . . . At r 1 , V + =+ 450 89 9 0150 150 . . V so V A 150 V . P25.67 From Example 25.5, the potential at the center of the ring is V kQ R i e = and the potential at an infinite distance from the ring is V f = 0 . Thus, the initial and final potential energies of the point charge-ring system are: UQ
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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 .

Ask a homework question - tutors are online