End of
Lecture 12
Uniform Electric Fields:
Motion of a charge particle
The force on a charged particle q in a uniform electric field
E
Fe =q E
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)
=m Fe =q E
a
So:
qE
a =
m
1
Electric Field Lines: Questions
Electric field lines from a point positive charge point outward.
Electric field lines from a point negative charge point inward.
q
E
+
In which direction will the charge move?
2
Uniform Electric Fields:
Motion of a charge particle
qE
i
a =a i=
m
We chose our field along the positive x direction.
We can apply all the kinematics we have learnt.
x
E
q
+
1 2
x f = x iv i t a t
2
and
v f =v ia t
and
v f 2=v i22 a x f x i 3
Uniform Electric Fields:
Motion of a charge particle
Lets assume the particle starts at rest,
And the final position
v i=0
and at
x i=0
x f= x
x
q
E
+
qE
x
v f =2 a x f =2
m
2
and the kinetic energy
1
2
K = m v f =qE x
2
4
Lecture 13
Motion of a Charge in Electric Fields
Lets do another kinematics problem with electrostatic forces.
Consider a uniform electric field
E= E0 j
The electrostatic force on a
charge:
Fe=q E
An electron has charge -e
Fe=e E
-------------------------
E0
+++++++++++++++++++++++++++
L
What is the speed of the electron when it exits the metal plates?
5
Motion of a Charge in Electric Fields
We have initial speed and length of the region with the field.
-------------------------
Initial velocity is along x
vi=v i i
-
Acceleration is Force/Mass
E0
Fe
j
e
E
=
a=
me
me
ax =0
and
++++++++++++++++++++++++++
eE
ay =
me
L
Work the rest out in class
6
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 1
Define a “surface normal” to an
element of area.
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.
3
1
2
5
4
8
Electric Flux
Step 2
If the “surface normal” is parallel to the
direction of the electric field,
then the flux through an area A is
E =E A
A
n
E
E
9
Electric Flux
Step 2
If the “surface normal” is at an angle
the direction of the electric field,
then the flux through an area A is
E =E Acos
A
E
n
E
10
Electric Flux
Step 3
How to deal with curved surfaces ?
Define an infinitesimal area dA
n
dA
dA=
The flux through this area is
E
. n dA
d E= E. dA=
E
dA
11
Electric Flux
Step 4
To get the flux through the entire area
we must integrate over the entire
surface
The total electric flux over the entire
curved surface is:
E =surface E. dA
dA
E. n
=
surface
E
dA
12
Electric Flux Through A
Closed Surface
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.
dA
E
13
Electric Flux Through A
Closed Surface
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. dA=E
dAcos 0
dA
E
14
Electric Flux Through A
Closed Surface
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. dA=E
dAcos =0
dA
E
15
Electric Flux Through A
Closed Surface
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. dA=E
dAcos 0
dA
E
16
Electric Flux Through A
Closed Surface
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.
. dA
E =closed surface E
dA
= E. n
17