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Spring 11 Phys 121 - HW 04

# Spring 11 Phys 121 - HW 04 - dzifanu(sd26397 – hw 4 –...

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Unformatted text preview: dzifanu (sd26397) – hw 4 – opyrchal – (121102) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 0.0 points WITHDRAWN 002 (part 1 of 3) 10.0 points Consider a solid conducting sphere with a radius 0 . 8 cm and charge − 5 . 7 pC on it. There is a conducting spherical shell concentric to the sphere. The shell has an inner radius 3 . 8 cm (with 3 . 8 cm > . 8 cm) and outer radius 5 . 1 cm and a net charge 26 . 1 pC 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 . . 8 cm 3 . 8 cm , Q ′ 2 inside 5 . 1 cm Q ′′ 2 outside − 5 . 7 pC 26 . 1 pC P Find the charge Q ′′ 2 . Correct answer: 20 . 4 pC. Explanation: Let : Q 1 = − 5 . 7 pC = − 5 . 7 × 10 − 12 C , Q 2 = 26 . 1 pC = 2 . 61 × 10 − 11 C , and k e = 1 4 π ǫ = 8 . 98755 × 10 9 N · m 2 / C . Sketch a concentric Gaussian surface S (dashed line) within the shell. r Since the electrostatic field in a conducting medium is zero, according to Gauss’s Law, Φ S = Q 1 + Q ′ 2 ǫ = 0 Q ′ 2 = − Q 1 . But the net charge on the shell is Q 2 = Q ′ 2 + Q ′′ 2 , so the charge on the outer surface of the shell is given by Q ′′ 2 = Q 2 − Q ′ 2 = Q 2 − ( − Q 1 ) = Q 2 + Q 1 = − 5 . 7 × 10 − 12 C + 2 . 61 × 10 − 11 C = 2 . 04 × 10 − 11 C = 20 . 4 pC . 003 (part 2 of 3) 10.0 points Find the magnitude of the electric field at point P, midway between the outer surface of the solid conducting sphere and the inner surface of the conducting spherical shell. Correct answer: 96 . 8413 N / C. Explanation: Let : r a = 0 . 8 cm = 0 . 008 m and r b = 3 . 8 cm = 0 . 038 m r c = 5 . 1 cm = 0 . 051 m r P = . 008 m + 0 . 038 m 2 = 0 . 023 m . Choose the spherical surface S centered at the same center as the conducting spherical surfaces, which passes through P ( i.e. , with radius r c = r P ). Here, 4 π r 2 P E = Q 1 ǫ E = Q 1 4 π ǫ r 2 P = k e Q 1 r 2 P dzifanu (sd26397) – hw 4 – opyrchal – (121102) 2 = (8 . 98755 × 10 9 N · m 2 / C) × − 5 . 7 × 10 − 12 C (0 . 023 m) 2 = − 96 . 8413 N / C bardbl vector E bardbl = 96 . 8413 N / C . 004 (part 3 of 3) 10.0 points Find the potential V P at point P. Assume the potential at r = ∞ is zero. Correct answer: 2 . 7158 V. Explanation: Using the superposition principle, adding the 3 concentric charges distributions; i.e. , Q 1 at a , − Q 1 at b and Q 1 + Q 2 at c , gives V P = − integraldisplay vector E r · dvectorr , by symmetry , = − integraldisplay E r dr = − integraldisplay c ∞...
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Spring 11 Phys 121 - HW 04 - dzifanu(sd26397 – hw 4 –...

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