Uniform Electric Fields: Motion of a charge particle

1

The force on a charged particle q in a uniform electric field

But Newton's Law tells us how a particle with mass m moves under the influence of an external force (whatever the force is, so it applies to electric forces too)

So:

�E�Fe=q �E

�Fe=q �E=m�a

�a=q �E

m

End ofLecture 12

Electric Field Lines: Questions

2

Electric field lines from a point positive charge point outward.

Electric field lines from a point negative charge point inward.

In which direction will the charge move?

+

q �E

Uniform Electric Fields: Motion of a charge particle

3

We chose our field along the positive x direction.We can apply all the kinematics we have learnt.

+

q

�a=a �i=q E

m�i

�E

x f=x i�vi t�1

2at

2 vf=v i�a tand

x

vf2=vi

2�2a �x f�xi�and

Uniform Electric Fields: Motion of a charge particle

4

Lets assume the particle starts at rest, and at

And the final position

+

q �E

x

v i=0 xi=0

v f2=2 a x f=2

qE

mx K=

1

2mv f

2=qE x

x f=x

and the kinetic energy

Motion of a Charge in Electric Fields

Lets do another kinematics problem with electrostatic forces.

5

�Fe=q�E

�E=E0�j

Consider a uniform electric field

The electrostatic force on a charge:

An electron has charge -e

�Fe=�e�E

- - - - - - - - - - - - - - - - - - - - - - - - -

+++++++++++++++++++++++++++

-

E0

L

What is the speed of the electron when it exits the metal plates?

Lecture 13

Motion of a Charge in Electric Fields

We have initial speed and length of the region with the field.

6

�v i=v i�i

Initial velocity is along x

Acceleration is Force/Mass

�a=�Fe

me

=�eE �j

me

ax=0 a

y=�eE

me

and

Work the rest out in class

- - - - - - - - - - - - - - - - - - - - - - - - -

++++++++++++++++++++++++++

-

L

E0

Electric Flux

Electric flux is a quantity that is proportional to the number of field lines passing though a given area.

This is a loose definition. We will make this definition more concrete. The concept of electric flux is very important. It will lead

us to the most important result in electrostatics:by looking at the electric flux through a closed surface, you can tell

how much charge is enclosed within it

7

Electric Flux

Step 1Define a �surface normal� to an

element of area.

8

1

3

5

2

4

Consider this cube.It has 6 surfaces:

1 and 2 (left and right)3 and 4 (top and bottom)5 and 6 (front and back)

The �surface normal� is a unit vector that is perpendicular to the surface.

If the surface is closed (like this cube)the unit vector always points outward.

Electric Flux

Step 2If the �surface normal� is parallel to the

direction of the electric field,then the flux through an area A is

9

�E=E A

�E

A

�E�n

Electric Flux

Step 2If the �surface normal� is at an angle

the direction of the electric field,then the flux through an area A is

10

�E=EAcos�

�E

A

�

��E

�n

Electric Flux

Step 3How to deal with curved surfaces ?

11

�dA

Define an infinitesimal area �dA

d�E=�E. �dA=�E. �ndA

�dA= �ndA

The flux through this area is

�E

Electric Flux

12

�dA

Step 4To get the flux through the entire area

we must integrate over the entire surface

The total electric flux over the entirecurved surface is:

�E=surface�E. �dA

=surface

�E. �ndA�E

Electric Flux Through A Closed Surface

13

�dA

Lets break this down into three cases:

1. surface whose normals are at angles between -�/2 and � /2 with

respect to the electric field direction.

2. surface whose normals are at an angles of � /2 with respect to the

electric field direction.

3. surface whose normals are at angles between � /2 and 3 2� with respect to the electric field direction. �E

Electric Flux Through A Closed Surface

14

Lets break this down into three cases:

1. surface whose normals are at angles between -� /2 and � /2 with

respect to the electric field direction.

�E

�E. �dA=EdAcos��0�dA

Electric Flux Through A Closed Surface

15

Lets break this down into three cases:

2. surface whose normals are at an angles of � /2 with respect to the

electric field direction.

�E

�E. �dA=EdAcos�=0

�dA

Electric Flux Through A Closed Surface

16

Lets break this down into three cases:

3. surface whose normals are at angles between � /2 and 3� � with respect to the electric field direction.

�E

�E. �dA=EdAcos��0�dA

Electric Flux Through A Closed Surface

17

The total electric flux through a closed surface will therefore be a sum over positive, negative and zero contributions.

To get the total electric flux, we must do the entire integral.

�E=closed surface�E. �dA

= �E. �ndA