716_Physics ProblemsTechnical Physics

716_Physics ProblemsTechnical Physics - 56 Electric...

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56 Electric Potential P25.14 Arbitrarily take V = 0 at point P . Then (from Equation 25.8) the potential at the original position of the charge is −⋅=− Es EL cos θ . At the final point a , VE L =− . Suppose the table is frictionless: KU if += + af 0 1 2 21 2 2 00 10 300 1 50 1 60 0 00100 0300 2 6 −= = = ×− ° = qEL mv qEL v qEL m cos cos .. c o s . . . ej bg a f C N C m kg ms Section 25.3 Electric Potential and Potential Energy Due to Point Charges P25.15 (a) The potential at 1.00 cm is Vk q r e 1 91 9 2 7 8 99 10 1 60 10 100 10 144 10 == ×⋅ × × . . Nm C C m V 22 e j . (b) The potential at 2.00 cm is q r e 2 9 2 7 8 99 10 1 60 10 200 10 0719 10 × × . . C m V e j . Thus, the difference in potential between the two points is VV V =−= × 8 719 10 . V . (c) The approach is the same as above except the charge is −× 160 10 19 C . This changes the sign of each answer, with its magnitude remaining the same.
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