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# HW3 - oliver(jmo673 HW03 Tsoi(59090 This print-out should...

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oliver (jmo673) – HW03 – Tsoi – (59090) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 4) 10.0 points For the configuration shown in the figure, sup- pose that a = 7 cm, b = 21 cm, and c = 28 cm. Furthermore, suppose that the electric field at a point 10 cm from the center is measured to be 4400 N / C radially inward while the elec- tric field at a point 40 cm from the center is 260 N / C radially outward. a b c Insulator Conductor Find the charge on the insulating sphere. Correct answer: - 4 . 89566 × 10 9 C. Explanation: Let : a = 7 cm , b = 21 cm , c = 28 cm , r 1 = 10 cm = 0 . 1 m , and E 1 = 4400 N / C , Basic Concept: E = 0 inside conductors, Gauss’ Law: contintegraldisplay vector E · d vector A = Q ǫ 0 Solution: By Gauss’ Law, E 2 ( 4 π r 2 1 ) = q ǫ 0 , where q is the charge contained in the insulat- ing sphere, so q = E 2 ( 4 π r 2 1 ) ǫ 0 = (4400 N / C) bracketleftbig 4 π (0 . 1 m) 2 bracketrightbig × (8 . 85419 × 10 12 C 2 / N · m 2 ) = - 4 . 89566 × 10 9 C , where E is negative because it points radially inward . 002 (part 2 of 4) 10.0 points Find the net charge on the conducting sphere. Correct answer: 9 . 52428 × 10 9 C. Explanation: Let : r 2 = 40 cm = 0 . 4 m and E 2 = 260 N / C . Similarly, by Gauss’ Law, the electric field at r = r 2 , where r 2 > c , will be related to the to- tal charge enclosed by a concentric Gaussian sphere of radius r 2 . If we call the total charge on the spherical shell Q , Gauss’ Law is E 2 (4 π r 2 2 ) = Q + q ǫ 0 , from which we find Q = E 2 ( 4 π r 2 2 ) ǫ 0 - q = (260 N / C) bracketleftbig 4 π (0 . 4 m) 2 bracketrightbig × (8 . 85419 × 10 12 C 2 / N · m 2 ) - ( - 4 . 89566 × 10 9 C) = 9 . 52428 × 10 9 C . 003 (part 3 of 4) 10.0 points Find the total charge on the inner surface of the hollow conducting sphere. Correct answer: 4 . 89566 × 10 9 C. Explanation: Since E = 0 inside a conductor, the to- tal charge contained in a concentric Gaussian sphere of radius r in , where b < r in < c , must be zero. From this we know that the total charge on the inner surface of the spherical conducting shell must be Q in = - q = 4 . 89566 × 10 9 C .

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oliver (jmo673) – HW03 – Tsoi – (59090) 2 004 (part 4 of 4) 10.0 points Find the total charge on the outer surface of the hollow conducting sphere. Correct answer: 4 . 62862 × 10 9 C. Explanation: To find the total charge on the outer surface of the conductor, Q out , we note that Q out + Q in = Q . Then Q out is Q out = Q - Q in = 9 . 52428 × 10 9 C - 4 . 89566 × 10 9 C = 4 . 62862 × 10 9 C . 005 (part 1 of 2) 10.0 points An electron moving parallel to the x axis has an initial speed of 1 × 10 6 m / s at the origin. Its speed is reduced to 6 × 10 5 m / s at the point x P , 2 cm away from the origin. The mass of the electron is 9 . 10939 × 10 31 kg and the charge of the electron is - 1 . 60218 × 10 19 C. Calculate the absolute value of the poten- tial difference between this point and the ori- gin.
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HW3 - oliver(jmo673 HW03 Tsoi(59090 This print-out should...

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