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Lesson_2.2_Homework_and_Solutions

# Lesson_2.2_Homework_and_Solutions - Lesson 2.2 Homework...

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Lesson 2.2 Homework Solutions 1. The charge on a proton = 1.602 × 10 -19 C. If r is the distance of the point from the proton when the field is: r = 1.8 × 10 -4 NC -1 , 4 2 9 19 2 6 4 4 1.8 10 9 10 1.602 10 8.0 10 1.8 10 1.8 10 8 kq r kq r m m μ - - - - - × = × × × = = = × = × × 2. (i) E = 500 NC -1 . F = qE = 1.2 × 600 = 720 N (ii) The force of friction between the floor and the professor’s shoes is: f k = μ k mg = 0.62 × 60 × 9.8 = 365 N The net force on the professor = 720 – 365 = 355 N 2 365 60 5.9 net F a m ms - = = = (iii) Assuming that the velocity of motion of the professor at the entrance is zero, we have v 2 = v o 2 + 2ax = 0 + 2 × 5.9 × 10 = 118 v = 10.9 ms -1 . K = (1/2)mv 2 = (1/2) × 60 × 10.9 2 = 3564 J 3. +3 μ C -2 μ C E 1 E 2 r = 5.0 m P 143 o Field at P due to +3 μ C is: E 1 = 9 6 1 2 9 10 3 10 1687.5 4 N C i - - × × × =

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Field at P due to –2 μ C is: E 2 = 9 6 1 2 9 10 2 10 720 5 N C - - × × × = This is directed towards the –2 μ C and makes an angle 143
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Lesson_2.2_Homework_and_Solutions - Lesson 2.2 Homework...

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