Transcript The Electric Field

Lecture 01: Electric Fields &
Forces

Origin of Charge

Conductors and Insulators

Coulomb’s Law

The Electric Field
Charge

Charge is an intrinsic property of matter.
Units: Coulombs (C)

+
Two types:
-
Positive Charge: Protons (Q = +1.6 x
C)
Negative Charge: Electrons (Q = -1.6 x 10-19 C)
Opposites Attract - Likes Repel
10-19
+

In general, atoms are neutral:
Negatively charged electrons “orbit”…
» Atomic Radius approximately
10-10
m
…Positively charged central nucleus.
» Nucleus Radius approximately 10-15 m
-
Conductors & Insulators

Perfect Conductors:
Electrons free to move throughout the conductor.

Perfect Insulators:
Electrons fixed to atoms throughout the insulator.

Most materials are somewhere in-between.
Coulomb’s Law

Gives the magnitude of the force between two charges:
Q1Q2
F k 2
R



Coulomb’s constant k = 9 x 109 N-m2/C2
R is the distance between the two charges.
Remember:
 Opposite charges ATTRACT
 Like charges REPEL
Summary of Concepts
 Charge:
a property of matter
 Conductors
 Coulomb’s
vs. Insulators
Law: a formula to calculate
the electric force between two charges
Example

Given: two positive charges and a negative charge.
Q1 = +1.5C
-
+
d1 = 2.0 m

Q2 = -2.5C
Q3 = +3.5C
+
d2 = 1.5 m
What is the magnitude and direction of the force on the
middle charge due to the other two?
Example

First, the direction of the force:
+



-
+
Both positive charges attract the middle negative charge.
However, since the charge on the right has a greater magnitude
and is closer, the force toward the right will be greater.
The NET force is to the RIGHT.
Example

Second, the magnitude of the force:
Q1 = +1.5C
-
+
d1 = 2.0 m

Q2 = -2.5C
Q3 = +3.5C
+
d2 = 1.5 m
We need to use Coulomb’s Law for each charge:
6
6
Q1Q2
(
1
.
5

10
)(
2
.
5

10
)
9
3
F  k 2  9 10

8
.
44

10
N
2
R
2.0
6
6
Q3Q2
(
3
.
5

10
)(
2
.
5

10
)
9
3
F  k 2  9 10

35
.
0

10
N
2
R
1.5
Example

Finally, subtract the two forces since they are in opposite
directions:
Q2 = -2.5C
Q1 = +1.5C
-
+
d1 = 2.0 m

Q3 = +3.5C
+
d2 = 1.5 m
The NET force is:
26.6 x 10-3 N to the right
Example

First, the direction of the force:
+



-
+
Both positive charges attract the middle negative charge.
However, since the charge on the right has a greater magnitude
and is closer, the force toward the right will be greater.
The NET force is to the RIGHT.
The Electric Field

A charged particle creates an Electric Field.

Electric Field Lines are used to represent fields.
Field lines point in the direction a positive charge
would accelerate.
Density of field lines is proportional to the strength of
the field.

The Electric Field around a single point charge is:
Q
Ek 2
R
The Electric Field


Fields lines point away form positive charges.
Fields lines point toward negative charges
+

-
The force on a charge in an Electric Field is:


F  E q
q
E
Conductors

Electrons are free to move in a Conductor.

The Electric Field in a conductor is Zero.

Electric Field Lines are always Perpendicular
to the surface of a conductor.
Summary of Concepts
 Coulomb’s
Law: a formula to calculate
the electric force between charges.
 The
Electric Field: created by charges
 Conductors
Electrons are free to move
E = 0 inside
Field Lines 
Example

Given: two positive charges and a negative charge on three
corners of a square with side length 3.8 m.
+
Q1 = +2 C
Q2 = -2 C
-

+
Q3 = +2 C
What is the magnitude and direction of the electric field at
the 4th corner of the square?
Example

First the direction of the Electric Field:
+
Q1 = +2 C
Q2 = -2 C
-


+
Q3 = +2 C
Since electric fields point away from positive charges and
toward negative charges, the three vectors shown give the
field from each individual point charge at the 4th corner.
Since the negative charge is farthest away, its electric field
is the weakest, and the NET electric field will therefore be
UP and to the RIGHT.
Example

Second, the magnitude of the field. We will use:
Q
Ek 2
R

We need to calculate the field from each of the three
charges:
6
Q1
(
2
.
0

10
)
9
E1  k 2  9 10
 1247 N/C
2
R
Q2
E2  k 2  9 109
R
3.8
(2.0 10 6 )
3.8 2 
2
 623 N/C
6
Q3
(
2
.
0

10
)
9
E3  k 2  9 10
 1247 N/C
2
R
3.8
Example

Finally, add the fields as VECTORS by adding the
corresponding components:
+
Q1 = +2 C
Q2 = -2 C
-

+
Q3 = +2 C
x-component:
1247 - 623cos45 = +806 N/C
y-component:
1247 - 623sin45 = +806 N/C
The NET force is:
8062  8062  1140 N/C up and to the right