73-79. Electric Potential and Potential Energy (Application 1)

# 73-79. Electric Potential and Potential Energy (Application 1)

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Electric Potential and Potential Energy: Application (1) Consider a point charge Q = 2 μ C fixed at position x = 0 . A particle with mass m = 2g and charge q = - 0 . 1 μ C is launched at position x 1 = 10 cm with velocity v 1 = 12 m/s. x = 0 q = -0.1 μ C Q = 2 μ C v 1 m = 2g x = 10cm 1 x = 20cm 2 (fixed) Find the velocity v 2 of the particle when it is at position x 2 = 20 cm. 17/1/2008 [tsl73 – 1/7]

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Electric Potential and Potential Energy: Application (2) Electric potential at point P 1 : V = kq 1 0 . 04m + kq 2 0 . 04m = 1125V + 1125V = 2250V . Electric potential at point P 2 : V = kq 1 0 . 06m + kq 2 0 . 10m = 750V + 450V = 1200V . 17/1/2008 [tsl74 – 2/7]
Point charges q 1 = - 5 . 0 μ C and q 2 = +2 . 0 μ C are positioned at two corners of a rectangle as shown. 15cm 5cm A q =+2.0 μ C 2 q =-5.0 μ C B 1 (a) Find the electric potential at the corners A and B . (b) Find the electric field at point

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## This note was uploaded on 04/30/2008 for the course PHYS 204 taught by Professor Andrevantonder during the Spring '07 term at Rhode Island.

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73-79. Electric Potential and Potential Energy (Application 1)

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