HW#2-solutions - ali(saa396 – HW#2 – Erskine –(57215...

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Unformatted text preview: ali (saa396) – HW#2 – Erskine – (57215) 1 This print-out should have 32 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Two alpha particles (helium nuclei), each consisting of two protons and two neu- trons, have an electrical potential energy of 6 . 31 × 10 − 19 J . Given: k e = 8 . 98755 × 10 9 N m 2 / C 2 , q p = 1 . 6021 × 10 − 19 C , and g = 9 . 8 m / s 2 . What is the distance between these parti- cles at this time? Correct answer: 1 . 46235 × 10 − 9 m. Explanation: Let: U electric = 6 . 31 × 10 − 19 J , k e = 8 . 98755 × 10 9 N m 2 / C 2 , q p = 1 . 6021 × 10 − 19 C , q α = 2 q p = 3 . 2042 × 10 − 19 C , and q n = 0 C . q 1 = q 2 = 2 q p + 2 q n = 2 (1 . 6021 × 10 − 19 C) + 2 (0 C) = 3 . 2042 × 10 − 19 C U electric = k e q 1 q 2 r r = k e q 1 q 2 U electric = k e q 2 1 U electric = (8 . 99 × 10 9 N · m 2 / C 2 ) × (3 . 2042 × 10 − 19 C) 2 6 . 31 × 10 − 19 J = 1 . 46235 × 10 − 9 m . 002 10.0 points Four charges are at the corners of a square centered at the origin as follows: q at ( − a, + a ); 2 q at (+ a, + a ); − 3 q at (+ a, − a ); and 6 q at ( − a, − a ). A fifth charge + q with mass m is placed at the origin and released from rest. Find the speed when it is a great distance from the origin. 1. bardbl vectorv bardbl = q radicalBigg 6 √ 2 k ma correct 2. bardbl vectorv bardbl = q radicalBigg 3 √ 3 k ma 3. bardbl vectorv bardbl = q radicalBigg 3 √ 2 k ma 4. bardbl vectorv bardbl = q radicalBigg 6 √ 3 k ma 5. bardbl vectorv bardbl = q radicalBigg 6 √ 6 k ma 6. bardbl vectorv bardbl = q radicalBigg 3 √ 6 k ma Explanation: x a +6 q a − 3 q +2 q + q + q, m √ 2 a The initial energy of the charge is E i = K i + U i = U i = q parenleftbigg k q √ 2 a + 2 k q √ 2 a + ( − 3 q ) k √ 2 a + 6 k q √ 2 a parenrightbigg = 6 k q 2 √ 2 a . The final energy is E f = 1 2 mv 2 . From energy conversation, we have E i = E f 6 k q 2 √ 2 a = 1 2 mv 2 ali (saa396) – HW#2 – Erskine – (57215) 2 v = q radicalBigg 6 √ 2 k ma . keywords: 003 (part 1 of 2) 10.0 points An electric field does 8 J of work on a 0 . 0005 C charge. What is the voltage change? Correct answer: 16000 V. Explanation: Let : W = 8 J and q = 0 . 0005 C . Work is W = qV V = W q = 8 J . 0005 C = 16000 V . 004 (part 2 of 2) 10.0 points The same electric field does 16 J of work on a . 001 C charge. What is the voltage change? Correct answer: 16000 V. Explanation: Let : W = 16 J and q = 0 . 001 C . The voltage change is V = W q = 16 J . 001 C = 16000 V . 005 10.0 points Suppose that the Earth (assumed spherical) has a net charge that is not zero....
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This note was uploaded on 05/08/2011 for the course PHY 302l taught by Professor Morrison during the Spring '08 term at University of Texas.

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HW#2-solutions - ali(saa396 – HW#2 – Erskine –(57215...

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