solution_pdf - phan (hbp253) – Homework #1 –...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: phan (hbp253) – Homework #1 – antoniewicz – (58855) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A particle of mass 85 g and charge 72 μ C is released from rest when it is 39 cm from a second particle of charge − 23 μ C. Determine the magnitude of the initial ac- celeration of the 85 g particle. Correct answer: 1151 . 2 m / s 2 . Explanation: Let : m = 85 g , q = 72 μ C = 7 . 2 × 10 − 5 C , d = 39 cm = 0 . 39 m , Q = − 23 μ C = − 2 . 3 × 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 = ma bardbl vectora bardbl = k e bardbl vectorq bardblbardbl vector Q bardbl md 2 = k e vextendsingle vextendsingle 7 . 2 × 10 − 5 C vextendsingle vextendsingle vextendsingle vextendsingle − 2 . 3 × 10 − 5 C vextendsingle vextendsingle (0 . 085 kg) (0 . 39 m 2 ) = 1151 . 2 m / s 2 . 002 10.0 points Two identical small charged spheres hang in equilibrium with equal masses as shown in the figure. The length of the strings are equal and the angle (shown in the figure) with the vertical is identical. The acceleration of gravity is 9 . 8 m / s 2 and the value of Coulomb’s constant is 8 . 98755 × 10 9 N m 2 / C 2 . . 9 m 4 ◦ . 04 kg . 04 kg Find the magnitude of the charge on each sphere. Correct answer: 2 . 19281 × 10 − 8 C. Explanation: Let : L = 0 . 09 m , m = 0 . 04 kg , and θ = 4 ◦ . L a θ m m q q From the right triangle in the figure above, we see that sin θ = a L . Therefore a = L sin θ = (0 . 09 m) sin(4 ◦ ) = 0 . 00627808 m . The separation of the spheres is r = 2 a = . 0125562 m . The forces acting on one of the spheres are shown in the figure below. θ θ m g F T e T sin θ T cos θ Because the sphere is in equilibrium, the resultant of the forces in the horizontal and vertical directions must separately add up to zero: summationdisplay F x = T sin θ − F e = 0 summationdisplay F y = T cos θ − mg = 0 . phan (hbp253) – Homework #1 – antoniewicz – (58855) 2 From the second equation in the system above, we see that T = mg cos θ , so T can be eliminated from the first equation if we make this substitution. This gives a value F e = mg tan θ = (0 . 04 kg) ( 9 . 8 m / s 2 ) tan(4 ◦ ) = 0 . 0274113 N , for the electric force. From Coulomb’s law, the electric force be- tween the charges has magnitude | F e | = k e | q | 2 r 2 , where | q | is the magnitude of the charge on each sphere. Note: The term | q | 2 arises here because the charge is the same on both spheres. This equation can be solved for | q | to give | q | = radicalBigg | F e | r 2 k e = radicalBigg (0 . 0274113 N) (0 . 0125562 m) 2 (8 . 98755 × 10 9 N m 2 / C 2 ) = 2 . 19281 × 10 − 8 C ....
View Full Document

This note was uploaded on 09/10/2009 for the course PHY 302l taught by Professor Morrison during the Spring '08 term at University of Texas.

Page1 / 11

solution_pdf - phan (hbp253) – Homework #1 –...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online