hw6 - Homework 6 Solutions Spring 2010 Required Problems...

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Homework 6 Solutions Spring 2010 Required Problems Chapter 23 3. The kinetic energy gained by the electron is the work done by the electric force. Use Eq. 23- 2b to calculate the potential difference. ( 29 16 ba ba 19 5.25 10 J 3280V 1.60 10 C W V q - - × = - = - = - × The electron moves from low potential to high potential, so plate B is at the higher potential. 22. Because of the spherical symmetry of the problem, the electric field in each region is the same as that of a point charge equal to the net enclosed charge. ( a )For 3 2 encl 2 2 2 2 0 0 0 1 1 3 : 4 4 8 Q Q Q r r E r r r πε = = = For 2 1 0 , r r r E < = < because the electric field is 0 inside of conducting material. For 1 2 encl 1 2 2 2 0 0 0 1 1 1 0 4 4 8 Q Q Q r r E r r r < < = = = ( b ) For 2 r r , the potential is that of a point charge at the center of the sphere. 3 2 2 0 0 0 1 1 3 4 4 8 , Q Q Q V r r r r r = = = ( c ) For 2 1 r r r < < , the potential is constant and equal to its value on the outer shell, because there is no electric field inside the conducting material. ( 29 2 1 2 0 2 3 8 , Q V V r r r r r r = = = < < ( d 1 0 r r < < , we use Eq. 23-4a. The field is radial, so we integrate along a radial line so that . d Edr = E r r g l
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1 1 1 1 1 2 0 0 1 1 0 1 0 1 0 2 1 1 1 8 8 1 1 1 1 1 1 0 8 8 2 8 , r r r r r r r r r r Q Q V V d E dr dr r r r Q Q Q V V r r r r r r r r πε - = - = - = - = - = + - = + + < < = E r r g l ( e ) To plot, we first calculate ( 29 0 2 0 2 3 8 Q V V r r r = = = and ( 29 0 2 2 0 2 3 8 . Q E E r r r = = = Then we plot 0 V V and 0 E E as functions of 2 . r r For 1 0 r r < < : ( 29 ( 29 2 2 1 2 0 2 2 0 1 1 1 2 2 3 3 3 2 0 0 2 0 2 0 2 1 1 1 8 8 1 ; 3 3 8 8 Q Q r r V E r r r r r r Q Q V E r r r - - + = = + = = = For 2 1 r r r < < : 0 2 0 0 2 0 2 0 2 3 8 1 ; 0 3 3 8 8 0 Q V E r Q Q V E r r = = = = For 2 r r : ( 29 ( 29 2 2 1 2 2 2 0 0 2 2 2 0 0 2 0 2 0 2 3 3 8 8 ; 3 3 8 8 Q Q V r E r r r r r r r Q Q V r E r r r - - = = = = = = The spreadsheet used for this problem can be found on the Media Manager, with filename “PSE4_ISM_CH23.XLS,” on tab “Problem 23.22e.” 0.0 1.0 2.0 3.0 4.0 5.0 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 r / r 2 V / 0
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0.0 1.0 2.0 3.0 4.0 5.0 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 r / r 2 E / 0 31. By energy conservation, all of the initial potential energy will change to kinetic energy of the electron when the electron is far away.
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This note was uploaded on 04/23/2010 for the course PHYSICS ?? taught by Professor Zettle during the Spring '08 term at University of California, Berkeley.

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hw6 - Homework 6 Solutions Spring 2010 Required Problems...

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