{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

solution2 - fullwood(tf5349 Homework 1 Electrostatics...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
fullwood (tf5349) – Homework 1 Electrostatics – Catala – (12054) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the magnitude of the electric field at a point midway between two charges +17 . 2 × 10 9 C and +62 . 9 × 10 9 C separated by a distance of 77.2 cm. The value of the Coulomb constant is 8 . 99 × 10 9 N · m 2 / C 2 . Correct answer: 2757 . 41 N / C. Explanation: Let : q 1 = 17 . 2 × 10 9 C , q 2 = 62 . 9 × 10 9 C , x = 77 . 2 cm , and k C = 8 . 99 × 10 9 N · m 2 / C 2 . r 1 = r 2 = x 2 = 77 . 2 cm 2 · parenleftbigg 1 m 100 cm parenrightbigg = 0 . 386 m The electric field is E = k C q r 2 . The magnitudes of the electric fields are E 1 = 8 . 99 × 10 9 N · m 2 / C 2 × 1 . 72 × 10 8 C (0 . 386 m) 2 = 1037 . 8 N / C , and E 2 = 8 . 99 × 10 9 N · m 2 / C 2 × 6 . 29 × 10 8 C (0 . 386 m) 2 = 3795 . 21 N / C . If the 17 . 2 × 10 9 C charge is the left one and the positive direction is to the right, E net = E 1 E 2 = 1037 . 8 N / C 3795 . 21 N / C = 2757 . 41 N / C bardbl vector E net bardbl = 2757 . 41 N / C . The electric field at midpoint between the charges +17 . 2 × 10 9 C and +62 . 9 × 10 9 C is 2757 . 41 N / C toward the +17 . 2 × 10 9 C charge. 002 10.0 points A charge of +2 . 20 × 10 9 C is placed at the origin, and another charge of +6 . 50 × 10 9 C is placed at x = 1 . 7 m. Find the point (coordinate) between these two charges where a charge of +3 . 80 × 10 9 C should be placed so that the net electric force on it is zero. The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Correct answer: 0 . 625258 m. Explanation: Let : q 1 = 2 . 20 × 10 9 C at the origin , q 2 = 6 . 50 × 10 9 C , q 3 = 3 . 80 × 10 9 C , and x 2 = 1 . 7 m . q 3 is between q 1 and q 2 , so if r 1 , 3 = P , then r 2 , 3 = x 2 P . F electric = k C q 1 q 2 r 2 The net electric force is zero, so F 1 , 3 = F 2 , 3 k C q 1 q 3 r 2 1 , 3 = k C q 2 q 3 r 2 2 , 3 q 1 r 2 1 , 3 = q 2 r 2 2 , 3 q 1 P 2 = q 2 ( x 2 P ) 2 ( x 2 P ) 2 P 2 = q 2 q 1 x 2 P P = radicalbigg q 2 q 1 x 2 P = P radicalbigg q 2 q 1 P parenleftbigg 1 + radicalbigg q 2 q 1 parenrightbigg = x 2
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
fullwood (tf5349) – Homework 1 Electrostatics – Catala – (12054) 2 P = x 2 1 + radicalbigg q 2 q 1 = 1 . 7 m 1 + radicalbigg 6 . 5 × 10 9 C 2 . 2 × 10 9 C = 0 . 625258 m . 003 (part 1 of 2) 10.0 points One gram of platinum has 3 . 09 × 10 21 atoms, and each platinum atom has 78 elec- trons. How many electrons are contained in 6.00 g of platinum? Correct answer: 1 . 44612 × 10 24 electrons. Explanation: Given : m = 6 . 00 g , N a = 3 . 09 × 10 21 atoms / g , and N e = 78 electrons / atom . N el = m N a N e = (6 g)(3 . 09 × 10 21 atoms / g) · (78 electrons / atom) = 1 . 44612 × 10 24 electrons . 004 (part 2 of 2) 10.0 points What is the total charge of these electrons? Correct answer: 2 . 31379 × 10 5 C. Explanation: Given : q e = 1 . 60 × 10 19 C / electron Q tot = N el q e = (1 . 44612 × 10 24 electrons) · ( 1 . 6 × 10 19 C / electron) = 2 . 31379 × 10 5 C . 005 (part 1 of 3) 10.0 points Three point charges,+5.1 μ C, +2.7 μ C, and -3.3 μ C, lie along the x -axis at 0 cm, 2.6 cm, and 5.5 cm, respectively.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern