{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Spring 11 Phys 121 - HW 02

Spring 11 Phys 121 - HW 02 - dzifanu(sd26397 hw 2...

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

dzifanu (sd26397) – hw 2 – opyrchal – (121102) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of2)10.0points Consider three charges arranged as shown. + + 4 . 4 cm 4 . 8 cm 5 μ C 8 . 6 μ C 2 . 9 μ C What is the magnitude of the electric field strength at a point 2 . 1 cm to the left of the middle charge? The value of the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Correct answer: 4 . 96862 × 10 7 N / C. Explanation: Let : q 1 = 8 . 6 μ C = 8 . 6 × 10 6 C , q 2 = 5 μ C = 5 × 10 6 C , q 3 = 2 . 9 μ C = 2 . 9 × 10 6 C , r 1 , 2 = 4 . 4 cm = 0 . 044 m , r 2 , 3 = 4 . 8 cm = 0 . 048 m , x = 2 . 1 cm = 0 . 021 m , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . r 1 = r 1 , 2 x = 0 . 044 m 0 . 021 m = 0 . 023 m r 2 = x = 0 . 021 m r 3 = r 2 , 3 + x = 0 . 048 m + 0 . 021 m = 0 . 069 m vector E net = vector E 1 + vector E 2 + vector E 3 E = k e q r 2 Considering the magnitudes of the electric fields at a point 2 . 1 cm to the left of the middle charge, E 1 = k e q 1 r 2 1 is directed away from the charge since q 1 is positive, E 2 = k e q 2 r 2 2 is directed away from the charge since q 2 is positive, and E 3 = k e | q 3 | r 2 3 is directed toward the charge since q 3 is nega- tive. Thus E net = E 1 E 2 + E 3 = k e parenleftbigg q 1 r 2 1 q 2 r 2 2 + | q 3 | r 2 3 parenrightbigg = ( 8 . 98755 × 10 9 N · m 2 / C 2 ) × bracketleftbigg (8 . 6 × 10 6 C) (0 . 023 m) 2 (5 × 10 6 C) (0 . 021 m) 2 + (2 . 9 × 10 6 C) (0 . 069 m) 2 bracketrightbigg = (8 . 98755 × 10 9 N · m 2 / C 2 ) × [(0 . 0162571 N / C) (0 . 0113379 N / C) +(0 . 000609116 N / C)] = 4 . 96862 × 10 7 N / C , directed along the positive x -axis. 002(part2of2)10.0points What is the magnitude of the force on a 2 . 9 μ C charge placed at this point? Correct answer: 144 . 09 N. Explanation: Let : q = 2 . 9 μ C = 2 . 9 × 10 6 C . The electric force is F electric = q E net = 2 . 9 × 10 6 C × (4 . 96862 × 10 7 N / C) = 144 . 09 N , bardbl vector F electric bardbl = 144 . 09 N .

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

View Full Document
dzifanu (sd26397) – hw 2 – opyrchal – (121102) 2 F electric has a magnitude of 144 . 09 N and is directed along the negative x axis. 003(part1of2)10.0points Three point charges are placed at the vertices of an equilateral triangle. 7 . 5 m 60 4 . 8 C 4 . 8 C 4 . 8 C P ˆ ı ˆ Find the magnitude of the electric field vec- tor bardbl vector E bardbl at P . The value of the Coulomb constant is 8 . 9875 × 10 9 N · m 2 / C 2 . Correct answer: 1 . 02258 × 10 9 N / C. Explanation: Let : a = 7 . 5 m , q = 4 . 8 C , and k = 8 . 9875 × 10 9 N · m 2 / C 2 . a q q q P ˆ ı ˆ BasicConcepts: Electric field. Electric field vectors due to bottom two charges cancel out each other. The magnitude of the field vector due to charge at to top of the triangle, which gives bardbl vector E bardbl = k q parenleftBigg 3 2 a parenrightBigg 2 = 4 3 k q a 2 = 4 3 (8 . 9875 × 10 9 N · m 2 / C 2 ) ( 4 . 8 C) (7 . 5 m) 2 = 1 . 02258 × 10 9 N / C , where h = a cos(30 ) = 3 2 a is the height of the triangle. 004(part2of2)10.0points Find the direction of the field vector vector E at P .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 8

Spring 11 Phys 121 - HW 02 - dzifanu(sd26397 hw 2...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online