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Unformatted text preview: zhou (ejz224) – Homework 1 – shih – (57480) 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A particle of mass 25 g and charge 68 μ C is released from rest when it is 73 cm from a second particle of charge − 14 μ C. Determine the magnitude of the initial ac celeration of the 25 g particle. Correct answer: 642 . 229 m / s 2 . Explanation: Let : m = 25 g , q = 68 μ C = 6 . 8 × 10 − 5 C , d = 73 cm = 0 . 73 m , Q = − 14 μ C = − 1 . 4 × 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 bardblbardbl vector Q bardbl m d 2 = k e vextendsingle vextendsingle 6 . 8 × 10 − 5 C vextendsingle vextendsingle vextendsingle vextendsingle − 1 . 4 × 10 − 5 C vextendsingle vextendsingle (0 . 025 kg) (0 . 73 m 2 ) = 642 . 229 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. . 1 1 m 7 ◦ . 03 kg . 03 kg Find the magnitude of the charge on each sphere. 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 . Correct answer: 5 . 37331 × 10 − 8 C. Explanation: Let : L = 0 . 11 m , m = 0 . 03 kg , and θ = 7 ◦ . 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 . 11 m) sin(7 ◦ ) = 0 . 0134056 m . The separation of the spheres is r = 2 a = . 0268113 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 − m g = 0 . zhou (ejz224) – Homework 1 – shih – (57480) 2 F sin θ F cos θ = F e m g F e = m g tan θ = (0 . 03 kg) ( 9 . 8 m / s 2 ) tan(7 ◦ ) = 0 . 0360987 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  q  = radicalBigg  F e  r 2 k e = radicalBigg (0 . 0360987 N) (0 . 0268113 m) 2 (8 . 98755 × 10 9 N m 2 / C 2 ) = 5 . 37331 × 10 − 8 C . 003 10.0 points Consider two charged plates with the same net charge on each. Imagine a proton at rest a certain distance from a negatively charged plate; after being released it collides with the plate. Then imagine an electron at rest the same distance from a positively charged plate....
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This note was uploaded on 04/09/2011 for the course PHY 302l taught by Professor Morrison during the Spring '08 term at University of Texas.
 Spring '08
 morrison
 Physics, Charge, Mass, Work

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