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hw-2 - -19 − 1 6 × 10-19(5 × 10-11 2 ˆ r = − 9 216...

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Homework # 2 Solutions Ex.1 We have 4 . 8 × 10 - 6 C 1 . 6 × 10 - 19 C = 3 × 10 13 electrons Ex.4 Let F k q 1 q 2 r 2 = 8 N Then, under r ( r/ 2), a straightforward calculation reveals F = k q 1 q 2 r 2 k q 1 q 2 ( r/ 2) 2 = 4 F = 32 N Ex.6(a) We have F = k q 1 q 2 r 2 = (9 . 0 × 10 9 N · m 2 · C - 2 ) · (4 × 10 - 6 C) · (4 × 10 - 6 C) (0 . 1 m) 2 = 144 × 10 9 - 6 - 6+2 N = 14 . 4 N Ex.6(b) ← ⊕ ⊕ → . Ex.7(a) The magnitude of the force is given by F = k q 1 q 2 r 2 = (9 . 0 × 10 9 ) · (2 × 10 - 6 ) · (4 × 10 - 6 ) (2 . 0 × 10 - 1 ) 2 (in SI units) = 18 × 10 9 - 6 - 6+2 = 1 . 8 N Ex.7(b) ⊕ → ← ⊖ . Ex.8 Let ˆ r be a unit vector pointing from the electron away from the di- rection of the proton. Then, if we perform our calculation in SI units (forces 1
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in Newtons, distances in Metres, electric charges in Coulombs, etc.), and end up with a positive expression (in ˆ r ), we’ll know we have a repulsive force, but if we end up with a negative expression (in ˆ r ), we’ll know we have an attractive force. Also, in standardized units, negative charges carry negative charge, and positive charges carry positive charge, and so F = k q 1 q 2 r 2 ˆ r =
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Unformatted text preview: -19 ) · ( − 1 . 6 × 10-19 ) (5 × 10-11 ) 2 ˆ r = − 9 . 216 × 10-8 N ˆ r Thus, the electron is attracted to the proton by a force of magnitude given by the above expression. SP1(a). F = k q 1 q 2 r 2 = (9 . × 10 9 ) · (1 × 10-1 ) · (2 × 10-2 ) 2 2 = 4 . 5 × 10 6 N SP1(b). F = k q 1 q 2 r 2 = (9 . × 10 9 ) · (4 × 10-2 ) · (2 × 10-2 ) (1) = 7 . 2 × 10 6 N SP1(c). F net = (7 . 2 − 4 . 5) × 10 6 = 2 . 7 × 10 6 (towards left) SP1(d). E net = (9 . × 10 9 ) · ± ± ± ± (1 × 10-1 ) 4 − (4 × 10-2 ) 1 ± ± ± ± = 13 . 5 × 10 7 N 2 The magnitude is given above. The direction points towards the left of point B. SP1(e). Define the +ve direction to be the one pointing towards the right. Then, F = q E = ( − . 06) · ( − 13 . 5 × 10 7 ) = 8 . 1 × 10 6 N 3...
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