2._Motion_in_Electric_Fields.ppt

1 1 1000 1000 1 100 nc or vm m v d v e 12 motion in

  • No School
  • AA 1
  • 43

This preview shows page 11 - 23 out of 43 pages.

1 1 1000 1000 1 . 0 100 NC or Vm m V d V E
Image of page 11

Subscribe to view the full document.

12 Motion in an Electric Field Charged particles that move through electric fields behave the same way as mass does in a gravitational field. The corolation is as follows, Mass  Charge Gravitational Field  Electric Field
Image of page 12
13 Motion in an Electric Field Consider a positive charge placed in a uniform electric field, as shown in the diagram below. ( Note the direction of the Electric field is the direction that a positive charge would move in that field ) Electric Field + + + + + + + + + + + + - - - - - - - - - - - - F E + 0V 1000V 0.1 m q=10μC M=0.1g Find the velocity of the charge after it has travelled a distance of 5 cm. Use the following information:
Image of page 13

Subscribe to view the full document.

14 Motion in an Electric Field N qE F NC Vm d V E 1 4 6 1 4 1 4 10 10 1 10 10 10 1 10 1 1 . 0 1000 Electric Field + + + + + + + + + + + + - - - - - - - - - - - - F E + 0V 1000V 0.1 m q=10μC M=0.1g 2 3 4 1 10 10 10 ms m F a
Image of page 14
15 Motion in an Electric Field Can use the equations of motion to determine the speed of particle after travelling for 5cm. plate ve -' the towards 10 05 . 0 10 2 2 ) 0 (v 2 2 v ? v 10 a 05 . 0 s 0 1 2 3 2 2 1 1 2 2 2 1 2 2 2 2 3 1 1 ms v v as v ms as v as v ms m ms v
Image of page 15

Subscribe to view the full document.

16 Motion in an Electric Field Can also determine the velocity by using the change in kinetic energy of the particle. Electric Field + + + + + + + + + + + + - - - - - - - - - - - - 1000 V 0 V 0.05 m 0.1 m A B 1 10000 Vm d V E
Image of page 16
17 Motion in an Electric Field Electric Field + + + + + + + + + + + + - - - - - - - - - - - - 1000 V 0 V 0.05 m 0.1 m A B To find the potential difference between A and B, rearrange the equation, V V m Vm V s E V s V E 500 05 . 0 10000 1
Image of page 17

Subscribe to view the full document.

18 Motion in an Electric Field Now calculate the kinetic energy at point B. If the charge is released at rest K at B (Gain in K ) = electric Ep lost plate ve -' the towards 10 10 500 10 10 2 2 1 4 6 2 2 1 ms m V q v V q mv
Image of page 18
19 Motion in an Electric Field The work done by the field on the charge can be calculated easily because it is equal to the gain in kinetic energy by the charge. mJ V q W E K 5 10 5 500 10 3 5
Image of page 19

Subscribe to view the full document.

20 Motion Perpendicular to the Electric Field Assumptions - ignore fringe effects (ie. assume that the field is completely uniform) - ignore gravity (it is quite easy to show that the acceleration due to gravity is insignificant compared with the acceleration caused by the electric field). -Before entering electric field, the charge follows a straight line path (no net force) -As soon as it enters the field, the charge begins to follow a parabolic path (constant force always in the same direction) - As soon as it leaves the field, the charge follows a straight line path (no net force)
Image of page 20
21 Motion Perpendicular to the Electric Field + q Straight line parabola Straight line Horizontal Component of the velocity (H component) 0 as 0 so constant is velocity horizontal 2 1 v a v L t t L v v v v h h h h h
Image of page 21

Subscribe to view the full document.

22 Motion Perpendicular to the Electric Field Vertical Component
Image of page 22
Image of page 23

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern