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Phy 122 – Assignment 1 A. 1. Opposites attract, so the electron is pulled toward the origin: The field’s direction is that of the force a positive charge would feel: (It’s a property of the point in space, not of what is located there. It’s similar with gravity: Earth’s gravitational field in a certain place is the same whether an elephant or a bread crumb is located there.) 2. Coulomb’s law gives the force between point-like charges, or charged spheres. With a different charge distribution, like these plates, E would be given by a different formula. 3. Convert to metric: (1.0 x109 Electric field: E r 2 k q E k lb C ) 1N N 4.444x109 .225lb C q r2 9 (8.988x10 ) (26)(1.602x10 19 9 4.444x10 ) = 8.424x10 -18 -9 Take square root: r = 2.90 x 10 m ans. B. 1. Field lines point away from + charges, and toward -. So, since E points toward it, this is a negative charge. E k q r 2 q Er 2 Answer: -2.78x10 ( 40)(.252 ) 9 k 8.988x10 -10 2.78x10 10 C C 2. Field from A: 40.0 N/C, south (given) Field from B: E k q (8.988x109 ) r2 Total: 40 + 10 = 50.0 N/C ans. 2.78x10 10 (.50m) 2 10.0 N / C , south C. a. Left. (Opposites attract, so the electron is pulled toward the ball.) F b. E 1.98 10 11 N 1.24 108 N / C , Right. ( E ’s direction is defined with a q 1.60 10 19 C positive charge in mind. A positive charge at the electron’s location would be repelled by the +2.7 μC.) c. Same answer as (b). E is a property of the point in space, not of what is located there. D. F1 is the force that q feels from q1. F2 is the force that q feels from q2. The directions come from the fact that opposite charges attract: q is pulled toward q1, and q is pulled toward q2. The magnitude of each force comes from Coulomb’s law: F1 k F2 k q1 q r2 q2 q r2 8.988 x109 8.988 x109 (5.00 x10 6 )(5.00 x10 6 ) (1.00m) 2 (3.00 x10 6 )(5.00 x10 6 ) (1.50m) 2 0.2247 N 0.0599 N Add them, remembering to call right positive and left negative: (-.2247) + (.0599) = -.1649 Answer: .165 N to the left E.