Phys121-2sol

Phys121-2sol - Anderson, Matthew Homework 2 Due: Feb 5...

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Anderson, Matthew – Homework 2 – Due: Feb 5 2008, noon – Inst: Vitaly 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. The due time is Central time. 001 (part 1 o± 1) 20 points Forces between two objects which are in- versely proportional to the square o± the dis- tance between the objects include which o± the ±ollowing? I) Gravitational ±orce between two celestial bodies II) Electrostatic ±orce between two electrons III) Nuclear ±orce between two neutrons. 1. I only 2. I, II, and III 3. I and II only correct 4. II and III only 5. III only Explanation: I) Gravitational ±orce 1 r 2 . II) Electrostatic ±orce 1 r 2 . III) The nuclear ±orce is very strong com- pared to the gravitational ±orce and electro- static ±orce over a very short range; outside o± that range, however, it is negligible and cannot be proportional to 1 r 2 keywords: 002 (part 1 o± 1) 20 points Two identical small charged spheres hang in equilibrium with equal masses as shown in the fgure. The length o± the strings are equal and the angle (shown in the fgure) with the vertical is identical. The acceleration o± gravity is 9 . 8 m / s 2 and the value Coulomb’s constant is 8 . 98755 × 10 9 N m 2 / C 2 . 0 . 08 m 3 0 . 01 kg 0 . 01 kg Find the magnitude o± the charge on each sphere. Correct answer: 6 . 3301 × 10 - 9 C. Explanation: Let : L = 0 . 08 m , m = 0 . 01 kg , and θ = 3 . L a θ m m q q From the right triangle in the fgure above, we see that sin θ = a L . There±ore a = L sin θ = (0 . 08 m) sin(3 ) = 0 . 00418688 m . The separation o± the spheres is r = 2 a = 0 . 00837375 m . The ±orces acting on one o± the spheres are shown in the fgure below. θ θ m g F T e T sin θ T cos θ Because the sphere is in equilibrium, the resultant o± the ±orces in the horizontal and
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Anderson, Matthew – Homework 2 – Due: Feb 5 2008, noon – Inst: Vitaly 2 vertical directions must separately add up to zero: X F x = T sin θ - F e = 0 X F y = T cos θ - mg = 0 . From the second equation in the system above, we see that T = mg cos θ , so T can be eliminated from the ±rst equation if we make this substitution. This gives a value F e = mg tan θ = (0 . 01 kg) ( 9 . 8 m / s 2 ) tan(3 ) = 0 . 00513596 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 | = s | F e | r 2 k e = s (0 . 00513596 N) (0 . 00837375 m) 2 (8 . 98755 × 10 9 N m 2 / C 2 ) = 6 . 3301 × 10 - 9 C .
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This note was uploaded on 03/11/2010 for the course PHYS 121 taught by Professor Opyrchal during the Spring '08 term at NJIT.

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Phys121-2sol - Anderson, Matthew Homework 2 Due: Feb 5...

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