lecture 12 motion of a charge particleanand/teaching/p1051ay/... · motion of a charge particle 1...
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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
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�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