P25_039 - 39. (a) Let = 0.15 m be the length of the...

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39. (a) Let ` =0 . 15 m be the length of the rectangle and w . 050 m be its width. Charge q 1 is a distance ` from point A and charge q 2 is a distance w , so the electric potential at A is V A = 1 4 πε 0 h q 1 ` + q 2 w i = ( 8 . 99 × 10 9 N · m 2 / C 2 ) · 5 . 0 × 10 6 C 0 . 15 m + 2 . 0 × 10 6 C 0 . 050 m ¸ =6 . 0 × 10 4 V . (b) Charge q 1 is a distance w from point b and charge q 2 is a distance ` , so the electric potential at B is V B = 1 4 0 h q 1 w + q 2 ` i = ( 8 . 99 × 10 9 N · m 2 / C 2 ) · 5 . 0 × 10 6 C 0 . 050 m + 2 . 0 × 10 6 C 0 . 15 m ¸ = 7 . 8 × 10 5 V . (c) Since the kinetic energy is zero at the beginning and end of the trip, the work done by an external agent equals the change in the potential energy of the system. The potential energy is the product of the charge q 3 and the electric potential. If
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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