2._Motion_in_Electric_Fields.ppt

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

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1 1 1000 1000 1 . 0 100 NC or Vm m V d V E

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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
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:

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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
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

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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
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

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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
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

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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)
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

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22 Motion Perpendicular to the Electric Field Vertical Component

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