{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW 1 - segovia(cs39966 Homework 1 Spurlock(14441 This...

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

View Full Document Right Arrow Icon
segovia (cs39966) – Homework 1 – Spurlock – (14441) 1 This print-out should have 34 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Calculate the net charge on a substance con- sisting of a combination of 8 . 6 × 10 13 protons and 6 . 8 × 10 13 electrons. Correct answer: 2 . 88 × 10 6 C. Explanation: Let : N p = 8 . 6 × 10 13 charges , q p = 1 . 60 × 10 19 C , N e = 6 . 8 × 10 13 charges , and q e = 1 . 60 × 10 19 C . Q net = N p q p + N e q e Q net = (8 . 6 × 10 13 )(1 . 6 × 10 19 C) + (6 . 8 × 10 13 )( 1 . 6 × 10 19 C) = 2 . 88 × 10 6 C 002 10.0points A particle of mass 41 g and charge 64 μ C is released from rest when it is 78 cm from a second particle of charge 17 μ C. Determine the magnitude of the initial ac- celeration of the 41 g particle. Correct answer: 392 . 008 m / s 2 . Explanation: Let : m = 41 g , q = 64 μ C = 6 . 4 × 10 5 C , d = 78 cm = 0 . 78 m , Q = 17 μ C = 1 . 7 × 10 5 C , and k e = 8 . 9875 × 10 9 . The force exerted on the particle is F = k e | q 1 | | q 2 | r 2 = m a bardbl vectora bardbl = k e bardbl vectorq bardbl bardbl vector Q bardbl m d 2 = k e vextendsingle vextendsingle 6 . 4 × 10 5 C vextendsingle vextendsingle vextendsingle vextendsingle 1 . 7 × 10 5 C vextendsingle vextendsingle (0 . 041 kg) (0 . 78 m 2 ) = 392 . 008 m / s 2 . 003 10.0points Three charges are arranged in the ( x, y ) plane (as shown in the figure below, where the scale is in meters). 5 nC 4 nC 9 nC y ( m ) 0 1 2 3 4 5 6 7 8 9 10 x ( m ) 0 1 2 3 4 5 6 7 8 9 10 What is the magnitude of the resulting force on the 5 nC charge at the origin? Correct answer: 12 . 8898 nN. Explanation: Let : q o = 5 × 10 9 C , ( x o , y o ) = (0 m , 0 m) , q a = 4 × 10 9 C , ( x a , y a ) = (4 m , 0 m) , q b = 9 × 10 9 C , and ( x b , y b ) = (0 m , 8 m) . Coulomb’s Law for q o and q a is bardbl vector F oa bardbl = k e q o q a radicalbig ( x a x o ) 2 + ( y a y o ) 2 = k e q o q a radicalbig x 2 oa + y 2 oa = 8 . 98755 × 10 9 N C 2 / m 2
Background image of page 1

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

View Full Document Right Arrow Icon
segovia (cs39966) – Homework 1 – Spurlock – (14441) 2 × ( 5 × 10 9 C) ( 4 × 10 9 C) radicalbig (4 m) 2 + (0 m) 2 = 1 . 12344 × 10 8 N . F oa x = k e q o q a r 2 oa cos θ oa = k e q o q a r 2 oa x oa r oa = (8 . 98755 × 10 9 N C 2 / m 2 ) × ( 5 × 10 9 C) ( 4 × 10 9 C) (4 m) 2 × 4 m 4 m = 1 . 12344 × 10 8 N . F oa y = k e q o q a r 2 oa sin θ oa = k e q o q oa r 2 oa y oa r oa = (8 . 98755 × 10 9 N C 2 / m 2 ) × ( 5 × 10 9 C) ( 4 × 10 9 C) (4 m) 2 × 0 m 4 m = 0 N . Coulomb’s Law for q o and q b is bardbl vector F ob bardbl = k e q o q b radicalbig ( x b x o ) 2 + ( y b y o ) 2 = k e q o q b radicalBig x 2 ob + y 2 ob = 8 . 98755 × 10 9 N C 2 / m 2 × ( 5 × 10 9 C) (9 × 10 9 C) radicalbig (0 m) 2 + (8 m) 2 = 6 . 31937 × 10 9 N . F ob x = k e q o q a r 2 oa cos θ ob = k e q o q b r 2 ob x ob r ob = (8 . 98755 × 10 9 N C 2 / m 2 ) × ( 5 × 10 9 C) (9 × 10 9 C) (8 m) 2 × 0 m 8 m = 0 N . F ob y = k e q o q a r 2 oa sin θ ob = k e q o q b r 2 ob y ob r ob = (8 . 98755 × 10 9 N C 2 / m 2 ) × ( 5 × 10 9 C) (9 × 10 9 C) (8 m) 2 × 8 m 8 m = 6 . 31937 × 10 9 N .
Background 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 ]}