Ch14-h1-solutions

# Ch14-h1-solutions - pan (zp695) – Ch14-h1 – chiu –...

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Unformatted text preview: pan (zp695) – Ch14-h1 – chiu – (56565) 1 This print-out should have 29 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An electron in a region where there is an electric field experiences a force of magnitude 5 × 10- 16 N. What is the magnitude of the electric filed at the location of the electron? The charge on an electron is- 1 . 602 × 10- 19 C Correct answer: 3121 . 1 N / C. Explanation: The field is defined to be the force per unit charge experienced by a particle (so long as the particle has a charge small enough that it does not change the background field significantly). Thus, we have: E = F q = 5 × 10- 16 N 1 . 602 × 10- 19 C = 3121 . 1 N / C 002 (part 1 of 8) 5.0 points In the region shown below there is an elec- tric field due to charged objects not shown in the diagram. A tiny glass ball with a charge of 6 × 10- 9 C placed at location A experiences a force of 3 . 5 × 10- 5 N , 3 . 5 × 10- 5 N , 0 N , as shown in the diagram. + F A Which of the below arrows indicates the direction of the electric field at location A ? enter j if the field has zero magnitude. Take the z-component of the field to be zero. a b c d e f g h Correct answer: b . Explanation: The force on a positively charged particle is in the same direction as the electric field, so the correct answer is b . 003 (part 2 of 8) 3.0 points What is the x-component of the electric field at location A ? Correct answer: 5833 . 33 N / C. Explanation: The electric field is the force on a test parti- cle divided by the charge of that test particle, so E = F q E x = 3 . 5 × 10- 5 N 6 × 10- 9 C = 5833 . 33 N / C 004 (part 3 of 8) 3.0 points What is the y-component of the electric field at location A ? Correct answer: 5833 . 33 N / C. Explanation: The electric field is the force on a test parti- cle divided by the charge of that test particle, so pan (zp695) – Ch14-h1 – chiu – (56565) 2 E = F q E y = 3 . 5 × 10- 5 N 6 × 10- 9 C = 5833 . 33 N / C 005 (part 4 of 8) 3.0 points What is the magnitude of this electric field? Correct answer: 8249 . 58 N / C. Explanation: The magnitude of any vector is the square root of the sum of the squares of its compo- nents, as per the Pythagorean Theorem. So, we have: E = | E x , E y , E z | = E 2 x + E 2 y + E 2 z = (5833 . 33 N / C) 2 + (5833 . 33 N / C) 2 + 0 2 = 8249 . 58 N / C 006 (part 5 of 8) 5.0 points Now, the glass ball is moved very var away, a tiny plastic ball with charge- 8 × 10- 9 C is placed at location A . Which arrow best indicates the direction of the electric force on this negatively charged plastic ball? Correct answer: f . Explanation: The force on a negatively charged particle points in the opposite direction as the field, so the correct answer is f ....
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## This note was uploaded on 02/21/2012 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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Ch14-h1-solutions - pan (zp695) – Ch14-h1 – chiu –...

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