# HW7Sol - Platt David Homework 7 Due 4:00 am Inst Ken Shih...

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Platt, David – Homework 7 – Due: Sep 23 2005, 4:00 am – Inst: Ken Shih 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 4) 10 points Consider two “ solid ” conducting spheres with radii r 1 = 4 R and r 2 = 9 R ; i.e. , r 2 r 1 = 9 R 4 R = 9 4 . The two spheres are separated by a large distance so that the field and the potential at the surface of sphere #1 only depends on the charge on #1 and the corresponding quan- tities on #2 only depend on the charge on #2. Place an equal amount of charge on both spheres, q 1 = q 2 = Q . r 1 q 1 #1 r 2 q 2 #2 After the electrostatic equilibrium on each sphere has been established, what is the ratio of the potentials V 2 V 1 at the “ centers ” of the two solid conducting spheres? 1. V 2 V 1 = 4 9 correct 2. V 2 V 1 = 9 4 3. V 2 V 1 = 81 16 4. V 2 V 1 = 9 2 5. V 2 V 1 = 81 32 6. V 2 V 1 = 1 7. V 2 V 1 = 9 8 8. V 2 V 1 = 16 81 9. V 2 V 1 = 81 8 Explanation: For a solid conducting sphere, the charge is uniformly distributed at the surface. From Gauss’ Law, the electric field outside the sphere is given by E ( r ) = k Q r 2 , where Q is the total charge on the sphere and r is the distance from the center of the sphere. By in- tegration with respect to r , the potential can be expressed as V ( r ) = k Q r , so the potential at the surface of the sphere is V ( r ) = k Q r , (1) where R is radius of the sphere and r R . For the electrostatic case, the potential is constant throughout a conducting body, so the potential at the center is the same as anywhere on the conductor. Thus at two centers V 2 V 1 = k q 2 r 2 k q 1 r 1 = r 1 r 2 = 4 R 9 R = 4 9 . 002 (part 2 of 4) 10 points What is the ratio of the electric fields E 2 E 1 at the “ surfaces ” of the two spheres? 1. E 2 E 1 = 81 16 2. E 2 E 1 = 4 9 3. E 2 E 1 = 9 2

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