“if there is one- they have to come (or go)!” electrostatic force field

“If there is one-they have to come (or go)!”

Electrostatic Force Field

Your comfort level with your responses to the following two question assessment tool should indicate if the presentation that follows will increase you knowledge base on the topic outlined by the questions in this tool.

Pre-presentation Self Assessment Activity

e-

-

Ex

+

y

0

ux

Va

Vb

+z

- z

x

+z

L

the horizontal length of the electrostatic field.

Problem #1:Using the reference graphic shown below, develop a model that will predict the coordinate position when the electron is exiting the electrostatic field.

(Assume that the electron’s horizontal velocity has a constant value while the electron is in the electrostatic field. Also assume that “d” is the distance between the two charged plates.)

“If there is one- they have to come (or go)!”


Using your answer from review problem #1, develop the model equation that will predict the electric field needed to have the electron collide with the upper charged plate just as the electron is leaving the electric field.


Problem #2:


e-

-

Ex

+

y

0

ux

Va

Vb

+z

- z

x

+z

L


Electric force field

and

the resultant motion

When a force exists where it exists is called a force field

For engineers and technicians it is usually very convenient to describe the force field instead of the force.

“If there is one- electrons have to come (or go)!”

Electric Field between two charged Plates

[ ](q )1

R1,2

2E= (4)o

-1

Newton/Coulomb

[ ]E=

=

Magnitude of Electric field strength

Volts/meters

y

z

x

VaVb

V = Vb

Va

By historic definition, voltage change occurs because positive charge moves from higher energy environment to lower energy environment.

-( )

+-

Electric field strength vector

V d

dv

dv =

E dl

( )

[ ]= E (q)

felectric

= dv/dl (q)felectric

[ ]

- -

= dv (q)f [ ]dl

-

dv =

f dl

q1( )

dv =

f dl q( )

- -

q= e

-


dvwhere is the sum of all of the infinitesimal voltage changes as the positive test charge moves from Va to Vb.

As the charge moves an infinitesimal distance in the electrostatic field the voltage value changes an infinitesimal amount.

Electron in an Electrostatic Field

y

z

x

VaVb

+-

E

e-

electron accelerates along the “x” direction toward positive plate

2

1x = ax t 2+u t + x0 0

Newton’s model as developed for distance traveled by an object with mass in a uniform gravitational force field. (same or opposite direction as field lines)

Distance traveled model for electron moving in the x direction in a uniform electrostatic force field.

x = a2

1x t 2

+u t + x0 0

This acceleration term is for the acceleration of the electron in the electrostatic field.

This acceleration term is for the acceleration of gravity


Change in velocity Change

in time

x du = a dt

x

if an electron is accelerating then its velocity is changing all the time.

Electron in Electrostatic Field

x

z

VaVb

+-

E

e-


Newton discovered that in a gravitational field the force (the objects weight) was proportional to the objects mass.

force m

Gravitational force = a m

xmass of the object

Electrostatic force = a m

x

This acceleration term is constant as long as the electrostatic field strength is constant.

emass of the electron

Electrostatic force = E q

This acceleration term is constant as long as the gravitational field strength is constant.

a mx eE q =

qm

ax E= ( )

charge to mass ratio


Since both gravitational and electrostatic forces follow an inverse square distance relationship, by analogy:

y


y

z

VaVb

+-

E

e-


Electrostatic force = a m

x

This acceleration term is constant as long as the electrostatic field strength is constant.

emass of the electron

electron velocity in the x direction of the electric field.

qm

E t +( )ux = u0

current position of electron in “x” direction when the electron started at negative plate.

x

x = [ ]t1

t2dt

qm

E t +( ) u0

= 0)(When t1

+xx = tqm

E t +( ) u 0

20

12



e-

Typical situation:

-

Ex

+z+

y

An electron in a vacuum environment has a constant velocity, ux , in the x direction and is about to enter an electrostatic field as shown below.

0

ux

Va

Vb

- z

What is the predicted path of the electron as it travels through the electric field if the horizontal velocity, ux, remains constant?



e-

Typical situation:

-

Ex

+z+

y


0

ux

Va

Vb

- z

What is the exact path the electron will travel as it goes through the electric field?

In this situation, the electron will be directed up (in the +Z direction) as it moves through the electric field in the X direction.

+zz = tqm

E t +( ) u 0

20

12

Electron distance traveled in the Z direction as a function of the time the electron is in the electric field.

From the co-ordinate system for this situation:

0z = 0

Setting zo = 0 allows the model to follow the position of the electron as it enters half way between the top and the bottom charged plates.

Note:

z = tqm

E t +( ) u 0

212

e -



e-

Typical situation:

-

Ex

+

y


0

ux

Va

Vb

+z

- z


z = tqm

E t +( ) u 0

212

In this case, while the electron is in the electric field it will move in the up (+z) while it moves to the right.

A) Model for upward motion.

B) Model for motion to the right.

t1

x = t2 dtu0( )

Electron distance traveled in the “x” direction.

x = tu0 When t1 =0

x

+z

+z =


= a constant value =

xu 0u


e-

Typical situation:

-

Ex

+

y


0

ux

Va

Vb

+z

- z


In this case, while the electron is in the electric field it will move in the up (+z) while it moves to the right.

Travel upward.

Travel to the right.x = tu0

x

+zz = t

qm

E t +( ) u 0

212

+z =

C) Model for combined motion.

or t =

Actual (x,z) position of electron as a function of time

When t0 = 0

t 0

xu0

u0

qm

E( )2

12

z (t)=

-

e

( )x



e-

Typical situation:

-

Ex

+

y


0

ux

Va

Vb

+z

- z

x

+z

(remember that for constant speed in the x direction, x = ? )

Note:K = u0

qmE( )2

u0

qm

E( )2

12

z (t)=

-

e

( )x

u t0

2qmE[ ]

2z(x)=

xu0

time “x” position

“z” position

x = u t

t insec

0

0x =

t =0u0 0

t =22

t =0 0

0x2 = u

(2)

2z(x)= K(

x0

)= 0u0

z(x)=K( 2)=4 K2

u0u0

2

t =1 1

x1 0= u (1)

z(x)=K( 1)=1 K2

u0u0

2

t =3 3

x3 0= u (3)

z(x)=K( 3)=9 K2

u0u0

2

Time (seconds)

1 2 3 4



e-

Typical situation:

-

+

y

0

Va

Vb

- z

(remember that for constant speed in the x direction, x = ? )

Note:K = u0

qmE( )2

u0

qm

E( )2

12

z (t)=

-

e

( )x

u t0

Time (seconds)

1 2 3 4

0ux

+zKu0

2 t insec

1 0

2 3

0ux z(x)

Ku02

1 0

2 3

1 0

4 9

4 4 16 5 5 25 6 6 36

t t2

t



e-

Typical situation:

-

+

y

0

Va

Vb

- z

t insec

u0

qm

E( )2

12

z (t)=

-

e

( )x

1 0

2 3

Time (seconds)

1 2 3 4

0ux

0ux z(x)

Ku02

1 0

2 3

1 0

4 9

4 4 16 5 5 25 6 6 36

t t2

t

+zKu0

2

The data indicates a parabolic path

Note:K = u0

qmE( )2



t insec

1 0

2 3

0ux z(x)

Ku02

1 0

2 3

1 0

4 9

4 4 16 5 5 25 6 6 36

t t2

t

The data indicates a parabolic path

Note:

e-

-

+

y

0

Va

Vb

- z

Time (seconds)

1 2 3 4

0ux

+zKu0

2

u0

qm( )

212

z(x)=

( )x( ) Vd

d( )( )1

VdVE = ( )VdV = V -

Vb a

d is the distance between the two charged plates.

d


e-

Problem #1:

-

Ex

+

y

Using the reference graphic shown below, develop a model that will predict the coordinate position when the electron is exiting the electrostatic field. Assume that the electron’s horizontal velocity has a constant value while the electron is in the electrostatic field. Also assume that “d” is the distance between the two charged plates.

0

ux

Va

Vb

+z

- z

x

+z

L


Write your answer down before you proceed.



e-

-

Ex

+

y

0

ux

Va

Vb

+z

- z

x

+z

L


Using the reference graphic shown below, develop a model that will predict the coordinate position when the electron is exiting the electrostatic field. Assume that the electron’s horizontal velocity has a constant value while the electron is in the electrostatic field. Also assume that “d” is the distance between the two charged plates.


u0

qm( )

212

z(x)=

( )L( ) Vd

d( )( )1

Problem #1:

Post-Presentation Self Assessment Activity

e-

-

Ex

+

y

0

ux

Va

Vb

+z

- z

x

+z

L



u0

qm( )

212

z(x)=

( )L( ) Vd

d( )( )1


Problem #2:


Write your answer down before you proceed.


e-

-

Ex

+

y

0

ux

Va

Vb

+z

- z

x

+z

L


u0

qm( )

212

z(x)=

( )L( ) Vd

d( )( )1

Problem #2:Using your answer from review problem #1, develop the model equation that will predict the electric field needed to have the electron collide with the upper charged plate just as the electron is leaving the electric field.



Since the distance “d” between the plates is fixed the only variable available is the electric field strength.

e-

-

Ex

+

y

0

ux

Va

Vb

+z

- z

x

+z

L


u0

qm( )

212

z(x)=

( )L( ) Vd

d( )( )1

Problem #2:

= z(x)

Vd

d( ) u0

qm( )

212 ( )L( ) ( )1[ ]

-1



e-

-

Ex

+

y

0

ux

Va

Vb

+z

- z

x

+z

L


Note:

=Vd

d( ) u0

qm( )

212 ( )L( ) ( )1[ ]

-1d2( )

The charge, “q”, on a single electron =1.6 x10

Coulomb

-19

The mass, “m”, of a single electron =9.1 x10

kg

-31

=Vd

d

u0 x 10

9.1 2

L[ d( )

1.6

-12

]


Problem #2:Using your answer from review problem #1, develop the model equation that will predict the electric field needed to have the electron collide with the upper charged plate just as the electron is leaving the electric field.


End of Presentation


Things to really remember:

1) The definition of E including its units. 2) The name for E. 3) The direction of E.

4) The charge (in Coulomb) on a single electron. 5) The unit of force in an electrostatic field.

Run this presentation over (and over) until you at least remember these 5 things.

End of Presentation

“if there is one- they have to come (or go)!” electrostatic force field

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