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

This preview shows pages 1–4. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## 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.

### Page1 / 7

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online