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# F95(3) - Physics 4B FINAL EXAM Chapters 23 34 Fall 1995...

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Physics 4B FINAL EXAM Chapters 23 - 34 Fall 1995 Name:_________________________________________ Posting Code _____ _____ _____ Solve the following problems in the space provided. Use the back of the page if needed. Each problem is worth 10 points. You must show your work in a logical fashion starting with the correctly applied physical principles which are on the last page. Your score will be maximized if your work is easy to follow because partial credit will be awarded. 1. A small plastic sphere of mass 5.00x10 -16 kg is held motionless between two charged parallel plates 4.00mm apart. Assume that the plastic sphere has two excess electrons on it. Find (a)the electric force on the sphere, (b)the electric field between the plates and (c)the potential difference between the plates. (d)In the diagram at the right, show the polarity of the plates, sketch the electric field and show all the forces that act on the sphere. (a)According to Newton's Second Law, Σ F = ma F e - F g = 0 F e = F g . Putting in the numbers, F e = mg = (5.00x10 - 16 )(9.80) = 4.90x10 - 15 N . (b)Using the definition of electric field, E F q = 4.90x10 - 15 2(1.60x10 - 19 ) = 1.53x10 4 V m . (c)The electric potential is V = - r E d r s . Since the field between the plates is constant, V = Ed = (1.53x10 4 )(4.00x10 - 3 ) = 61.3V . 2. Two long concentric cylinders of radii a and b have a potential difference V o established between them. Find the electric field between the cylinders as a function of the distance from the center. (Hint: You may want to start by assuming a linear charge density λ exists on the cylinders, but for full credit you must express λ in terms of the potential difference). Applying Gauss's Law to an imaginary cylinder with a radius r between a and b. Since the cylinders are long, symmetry requires the field to be constant over this Gaussian cylinder as it must point radially outward. Therefor, r E d r A = q ε o E2 π r l = λ l ε o E = λ 2 πε o r The electric potential is V = - r E d r s . Integrating from the inner to the outer cylinder, 0 - V o = - λ 2 πε o r a b dr = λ 2 πε o ln b a ⇒ λ = 2 πε o V o ln b a ( 29 . Finally, E

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F95(3) - Physics 4B FINAL EXAM Chapters 23 34 Fall 1995...

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