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# Midterm1 - Version 076 Midterm 01 Yao(59110 This print-out...

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Version 076 – Midterm 01 – Yao – (59110) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An alpha particle (charge = +2 . 0 e ) is sent at high speed toward a copper nucleus (charge = +29 e ). The Coulomb constant is 8 . 99 × 10 9 N · m 2 / C 2 . What is the electric force acting on the alpha particle when the alpha particle is 4 . 0 × 10 14 m from the copper nucleus? 1. Unable to determine 2. 9 . 20576 N, repulsive 3. 8 . 34272 N, attractive 4. 8 . 34272 N, repulsive correct 5. 9 . 20576 N, attractive 6. None of these Explanation: Let : q 1 = 2 q p , q 2 = 29 q p , q p = 1 . 60 × 10 19 C , r = 4 . 0 × 10 14 m , and k C = 8 . 99 × 10 9 N · m 2 / C 2 . F electric = k C q 1 q 2 r 2 = k C (2 q p ) (29 q p ) r 2 = k C 58 ( q p ) 2 r 2 = 8 . 99 × 10 9 N · m 2 / C 2 × 58 ( 1 . 6 × 10 19 C ) 2 (4 × 10 14 m) 2 = 8 . 34272 N . This is a repulsive force. 002 (part 1 of 2) 10.0 points Four point charges are placed at the four cor- ners of a square. Each side of the square has length L . The charges are q 1 = q , q 2 = q 3 = q 4 = q . A B q 1 q 2 q 4 q 3 L L The potential at point A in the absence of q 1 is 1. V A = parenleftbigg 2 + 1 2 parenrightbigg k q 2 L 2. V A = parenleftbigg 2 + 1 2 parenrightbigg k q L correct 3. V A = 2 k q 2 L 4. V A = 2 2 k q L 2 5. V A = 2 k q L 6. V A = parenleftbigg 2 + 1 2 parenrightbigg k q L 2 7. V A = 2 k q L 2 8. V A = 2 2 k q L 9. V A = 2 2 k q 2 L Explanation: Let : Q 1 = q and q 2 = q 3 = q 4 = q . V = summationdisplay i V i , with V i = k q i r i . At point A V A = V 2 + V 3 + V 4

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Version 076 – Midterm 01 – Yao – (59110) 2 = k q L + k q 2 L + k q L = k q L parenleftbigg 2 + 1 2 parenrightbigg . 003 (part 2 of 2) 10.0 points Now leave q 1 , q 3 , and q 4 fixed and bring q 2 from infinity to point B . The work W B required is given by 1. W B = k q 2 2 L 2 2. W B = k q 2 L 3. W B = k q L 2 4. W B = k q 2 L 2 5. W B = k q 2 2 L 6. W B = k q 2 L 2 7. W B = k q 2 L 2 8. W B = k q 2 2 L correct 9. W B = k q 2 2 L 2 Explanation: At point B , in the absence of q 2 , V = V 1 + V 3 + V 4 = k q parenleftbigg 1 L + 1 L + 1 2 L parenrightbigg = k q 2 L . Therefore, the work from infinity to B is W B = q V = k q 2 2 L . 004 10.0 points Find the greatest surface charge density that can exist on a conductor before dielectric breakdown of the air occurs. The dielectric strength of the air is 3 × 10 6 V / m and permittivity of free space is 8 . 85 × 10 12 C 2 / N · m 2 . 1. σ max = 79 . 65 μ C / m 2 2. σ max = 159 . 3 μ C / m 2 3. σ max = 106 . 2 μ C / m 2 4. σ max = 132 . 75 μ C / m 2 5. σ max = 26 . 55 μ C / m 2 correct 6. σ max = 53 . 1 μ C / m 2 Explanation: Let : E b = 3 × 10 6 V / m and ǫ 0 = 8 . 85 × 10 12 C 2 / N · m 2 . The electric field due to a surface charge den- sity σ is E = σ ǫ 0 , so the maximum surface charge density is σ max = ǫ 0 E b = (8 . 85 × 10 12 C 2 / N · m 2 ) × (3 × 10 6 V / m) parenleftbigg 10 6 μ C C parenrightbigg = 26 . 55 μ C / m 2 .
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