q02-5887sol

# Q02-5887sol - (2 × 10-6(4 × 10-6 06(9 × 10 9(8 × 10-6(4 × 10-6 √ 6 2 3 2 × 01 = 10 29 J(c 1 2 mv 2 =(9 × 10 9(8 × 10-6(4 × 10-6 √ 6 2

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PHY2054 Section 5887 Quiz 1: Ch.16.1 - 16.5 September 14, 2011 NAME UFID 1. (a) Find the electric potential, taking zero at inﬁnity, at the upper right corner of the rectangle. (1.5 / 5) (b) Find the required work to bring those three charges to the corners. ( This is the same as ﬁnding the electric potential energy of the system ) (1.5 / 5) Note: k e = 9 . 0 × 10 9 Nm 2 / C 2 . (c) If 4 μ C charge is released while others are held ﬁxed in place, ﬁnd the speed of the 4 μ C charge as it goes to inﬁnity. Assume the mass of the charge be 10 - 6 kg. ( 1 / 5 ) (d) How much work is it required to bring the 4 μ C charge from the lower right corner to the upper right corner? ( 1 / 5) solution: (a) V A = 9 . 0 × 10 9 × 8 × 10 - 6 0 . 06 + 9 . 0 × 10 9 × 4 × 10 - 6 0 . 03 + 9 . 0 × 10 9 × 2 × 10 - 6 6 2 + 3 2 × 0 . 01 = 2 . 67 × 10 6 ( V ) (b) PE = (9 . 0 × 10 9 )(8 × 10 - 6 )(2 × 10 - 6 ) 0 . 03 + (9 . 0 × 10 9
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Unformatted text preview: )(2 × 10-6 )(4 × 10-6 ) . 06 + (9 . × 10 9 )(8 × 10-6 )(4 × 10-6 ) √ 6 2 + 3 2 × . 01 = 10 . 29( J ) (c) 1 2 mv 2 = (9 . × 10 9 )(8 × 10-6 )(4 × 10-6 ) √ 6 2 + 3 2 × . 01 + (9 . × 10 9 )(2 × 10-6 )(4 × 10-6 ) . 06 ≈ 5 . 49325 , where m = 10-6 kg. Therefore, v = p 5 . 49325 × 2 / 10-6 = 3 . 31 × 10 3 ( m / s ) . (d) W required =Δ PE = (9 . × 10 9 )(8 × 10-6 )(4 × 10-6 ) . 06 + (9 . × 10 9 )(2 × 10-6 )(4 × 10-6 ) √ 6 2 + 3 2 × . 01 !-(9 . × 10 9 )(2 × 10-6 )(4 × 10-6 ) . 06 + (9 . × 10 9 )(8 × 10-6 )(4 × 10-6 ) √ 6 2 + 3 2 × . 01 ! = . 38( J )...
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## This note was uploaded on 12/15/2011 for the course PHY 2054 taught by Professor Avery during the Spring '08 term at University of Florida.

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