Homework 4 - gonzales (pag757) – Homework 4 – Hoffmann...

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Unformatted text preview: gonzales (pag757) – Homework 4 – Hoffmann – (56745) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Four charges are fixed 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, where the potential energy of the fifth charge due to the four point charges is negligible. 1. bardbl vectorv bardbl = q radicalBigg 2 √ 5 k ma 2. bardbl vectorv bardbl = q radicalBigg 3 √ 2 k ma 3. bardbl vectorv bardbl = q radicalBigg 3 √ 5 k ma 4. bardbl vectorv bardbl = q radicalBigg 6 √ 3 k ma 5. bardbl vectorv bardbl = q radicalBigg 3 √ 3 k ma 6. bardbl vectorv bardbl = q radicalBigg 6 √ 6 k ma 7. bardbl vectorv bardbl = q radicalBigg 6 √ 5 k ma 8. bardbl vectorv bardbl = q radicalBigg 2 √ 2 k ma 9. bardbl vectorv bardbl = q radicalBigg 6 √ 2 k ma 10. bardbl vectorv bardbl = q radicalBigg 3 √ 6 k ma 002 10.0 points Point P is located midway between the surface of the sphere and the inside surface of the shell, or at r = (a+b)/2 Consider a solid conducting sphere with a radius a and charge Q 1 on it. There is a conducting spherical shell concentric to the sphere. The shell has an inner radius b (with b > a ) and outer radius c and a net charge Q 2 on the shell. Denote the charge on the inner surface of the shell by Q ′ 2 and that on the outer surface of the shell by Q ′′ 2 . Q 1 , a b, Q ′ 2 Q ′′ 2 , c Q 1 Q 2 P Assume: The potential at r = ∞ is zero. Find the potential V P at point P . 1. V P = 2 k e Q 1 a + b − 2 k e Q 2 b 2. V P = 2 k e Q 1 a + b + k e Q 1 b − k e ( Q 1 − Q 2 ) c 3. V P = 0 4. V P = 2 k e Q 1 a + b − k e Q 2 c 5. V P = 2 k e Q 1 a 6. V P = 2 k e ( Q 1 − Q 2 ) a + b 7. V P = 2 k e Q 1 a + b 8. V P = k e Q 1 a + b − k e Q 2 b 9. V P = 2 k e Q 1 a + b + k e Q 2 c 10. V P = 2 k e Q 1 a + b − k e Q 1 b + k e ( Q 1 + Q 2 ) c 003 (part 1 of 3) 10.0 points A proton is released from rest in a uniform electric field of magnitude 1 . 4 × 10 5 V / m di- rected along the positive...
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This note was uploaded on 01/27/2012 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.

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Homework 4 - gonzales (pag757) – Homework 4 – Hoffmann...

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