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3 - Version 059 Midterm1 li(59050 This print-out should...

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Version 059 – Midterm1 – li – (59050) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Four positive charges of magnitude q are ar- ranged at the corners of a square, as shown. At the center C of the square, the potential due to one charge alone is V 0 , and the electric field due to one charge alone has magnitude E 0 . C + q + q + q + q Which of the following correctly gives the electric potential and the magnitude of the electric field at the center of the square due to all four charges? Electric Electric Potential Field 1. 0 2 E 0 2. 4 V 0 0 correct 3. 0 0 4. 2 V 0 4 E 0 5. 4 V 0 2 E 0 Explanation: The electric potential is scalar, so the elec- tric potential at the center is 4 V 0 . The electric field is a vector with both magnitude and di- rection in the space. All of the electric field components due to the four charges have the same magnitudes but different directions. In the configuration, the sum of the four compo- nent vectors is zero, so the total field at C is zero. 002 10.0 points A hollow metal sphere of radius R is positively charged. Of the following distances r from the center of the sphere, which location will have the greater electric field strength? 1. r = 0 (center of the sphere) 2. r = 5 R 4 correct 3. None of these because the field is of con- stant strength. 4. r = 2 R 5. r = 3 R 4 Explanation: vector E = 0 inside the sphere parenleftbigg r = 0 and r = 3 4 R parenrightbigg and E 1 R 2 outside the sphere parenleftbigg r = 5 4 R and r = 2 R parenrightbigg . 003 10.0 points A charge of 4 pC is uniformly distributed throughout the volume between concentric spherical surfaces having radii of 1 . 4 cm and 3 . 4 cm. Let: k e = 8 . 98755 × 10 9 N · m 2 / C 2 . What is the magnitude of the electric field 2 cm from the center of the surfaces? 1. 8.5761 2. 29.6835 3. 15.4984 4. 36.0868 5. 15.7851 6. 29.2636 7. 45.5982 8. 33.8505 9. 14.0578 10. 12.9208 Correct answer: 12 . 9208 N / C. Explanation:
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Version 059 – Midterm1 – li – (59050) 2 Let : q tot = 4 pC = 4 × 10 12 C , r 1 = 1 . 4 cm , r 2 = 3 . 4 cm , and r = 2 cm = 0 . 02 m . By Gauss’ law, Φ c = contintegraldisplay vector E · d vector A = q in ǫ 0 The tricky part of this question is to deter- mine the charge enclosed by our Gaussian surface, which by symmetry considerations is chosen to be a concentric sphere with radius r . Since the charge q is distributed uniformly within the solid, we have the relation q in q tot = V in V tot where q in and V in are the charge and volume enclosed by the Gaussian surface. Therefore q in = q tot parenleftbigg V in V tot parenrightbigg = q tot bracketleftbigg r 3 r 3 1 r 3 2 r 3 1 bracketrightbigg = (4 pC) × bracketleftbigg (2 cm) 3 (1 . 4 cm) 3 (3 . 4 cm) 3 (1 . 4 cm) 3 bracketrightbigg = 0 . 575055 pC = 5 . 75055 × 10 13 C . And by Gauss’s Law, E = k e q in r 2 = 8 . 98755 × 10 9 N · m 2 / C 2 × 5 . 75055 × 10 13 C (0 . 02 m) 2 = 12 . 9208 N / C .
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