HW02-solutions

HW02-solutions - kazi(mik88 – HW02 – tsoi –(57210 1...

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Unformatted text preview: kazi (mik88) – HW02 – tsoi – (57210) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Through what potential difference would an electron need to be accelerated for it to achieve a speed of 5 . 8 % of the speed of light (2 . 998 × 10 8 m / s), starting from rest? The mass of an electron is 9 . 109 × 10 − 31 kg and the charge on an electron is 1 . 602 × 10 − 19 C. Ignore special relativity for this problem. Correct answer: 859 . 602 V. Explanation: Let : s = 5 . 8% = 0 . 058 , c = 2 . 998 × 10 8 m / s , m e = 9 . 109 × 10 − 31 kg , and q e = 1 . 602 × 10 − 19 C . The speed of the electron is v = 0 . 058 c = 0 . 058 ( 2 . 998 × 10 8 m / s ) = 1 . 73884 × 10 7 m / s , By conservation of energy 1 2 m e v 2 =- (- q e ) Δ V Δ V = m e v 2 2 q e = ( 9 . 109 × 10 − 31 kg ) × ( 1 . 73884 × 10 7 m / s ) 2 2 (1 . 602 × 10 − 19 C) = 859 . 602 V . 002 (part 1 of 2) 10.0 points Consider two conducting spheres separated by a large distance, each with a net charge Q . The spheres have radii r 1 and r 2 , where r 2 > r 1 . Assume that the potential at infinity is zero. Also assume that the two spheres are sufficiently far apart so that the potential at the surface of sphere #1 depends only on the charge Q 1 and that at the surface of sphere #2 only on Q 2 . #1 #2 r 1 2 r Find the ratio E 2 E 1 of the electric field at the surface of sphere #2 to that of the electric field at the surface of sphere #1. 1. E 1 E 2 = parenleftbigg r 2 r 1 parenrightbigg 2 2. E 1 E 2 = r 1 r 2 3. E 1 E 2 = parenleftbigg r 1 r 2 parenrightbigg 2 correct 4. E 1 E 2 = r 2 r 1 5. E 1 E 2 = 1 Explanation: E 2 E 1 = k e Q r 2 2 k e Q r 2 1 = parenleftbigg r 1 r 2 parenrightbigg 2 . 003 (part 2 of 2) 10.0 points Connect the spheres by a long conducting wire. After a redistribution of charge, static equilibrium is again established. Denote the new charge on sphere #1 by Q ′ 1 , and the new charge on #2 by Q ′ 2 . Find the ratio Q ′ 2 Q ′ 1 of the charges. 1. Q ′ 2 Q ′ 1 = parenleftbigg r 2 r 1 parenrightbigg 2 2. Q ′ 2 Q ′ 1 = parenleftbigg r 1 r 2 parenrightbigg 2 kazi (mik88) – HW02 – tsoi – (57210) 2 3. Q ′ 2 Q ′ 1 = r 2 r 1 correct 4. Q ′ 2 Q ′ 1 = r 1 r 2 5. Q ′ 2 Q ′ 1 = 1 Explanation: The potentials of the two spheres must be equal when connected. This can be seen if we note that when connected, they form one continuous conductor, and no field can exist within a conductor, so the potential must be the same everywhere. We know that the potential at the surface of a sphere is the same as for a point charge in its center, so V ′ 1 = V ′ 2 k e Q ′ 1 r 1 = k e Q ′ 2 r 2 Q ′ 2 Q ′ 1 = r 2 r 1 ....
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This note was uploaded on 02/21/2011 for the course PHY 302l taught by Professor Morrison during the Spring '08 term at University of Texas.

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HW02-solutions - kazi(mik88 – HW02 – tsoi –(57210 1...

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