Physics I
Class 20
The Electric Field
Rev. 05-Nov-03 GB
20-1
What Is a Field?
A field is nothing more than a function of location, either f(X,Y)
or f(X,Y,Z). There is a uniquely defined value of a certain
quantity at each point. Example – a temperature map:
20-2
Temperature Map
(Temperature Field)
20-3
Coulomb’s Law of
Electrostatic Force (Review)

F
1 q1 q 2
( r̂ ) (Prof. B’s version.)
2
4  0 r
The meaning of each term:

F:
Electrostatic force on charge 1 from charge 2.
1
2
2
: Electrostatic force constant = 8.98755 × 10+9 N m /C
4  0
q1 : Value of charge 1, positive or negative.
q 2 : Value of charge 2, positive or negative.
r 2 : Center distance from point charge 1 to point charge 2.
r̂ :
Unit vector from charge 1 to charge 2.
20-4
Superposition of
Electrostatic Forces (Review)
+5.0 C
+1.0 C
Y
1
X
resultant
-3.0 C
N

1 q1 q i
Fon 1  
( r̂i ) (find and add X and Y components)
2
i  2 4   0 ri
20-5
The Idea of Electric Field
(Step 1)
When finding the force on charge #1, we notice that q1 appears
in each term in the sum, so we can factor it out of the sum:
N

 N 1 qi

1 q1 q i
Fon 1  
( r̂i )  q1 
( r̂i )
2
2
i  2 4   0 ri
 i2 4   0 ri

Note that the quantity in { } above is a vector.
20-6
The Idea of Electric Field
(Step 2)
After doing the first problem, suppose we were asked to solve a
second problem with a different charge at position #1, call it q1´.
If we were smart enough to have saved the vector part from the
first problem, we could find the new force just by multiplying by
q1´ instead of q1:
N

 N 1 qi

1 q1 q i
 1
Fon
( r̂i )  q1 
( r̂i )
2
2
i  2 4   0 ri
 i2 4   0 ri

20-7
The Idea of Electric Field
(Step 3)
The vector quantity we have calculated
 depends on the location of the point (#1).
 depends on the unit vectors to the other charges.
 depends on the distances to the other charges.
 depends on the values of the other charges.
It does not depend on the value of the charge at the point.
In fact, it can be calculated even when there is no charge there!

Fon1
N

1 qi
E(point #1)  
(r̂i ) 
Electric Field
2
q1
i  2 4   0 ri


Fon 1  q1 E ( point #1)
Force / Field Relationship
20-8
The Electric Field of a Point
Charge (as a Source)
The electric field is a vector field, meaning at each point in
space the electric field has a magnitude and a direction. We
show that by drawing arrows at representative points in the
correct directions with lengths proportional to the magnitudes.
Away from positive
Toward negative
+
-
Just because we don’t draw an electric field vector at
a point doesn’t mean there is no electric field there.
20-9
Example Problem
A sphere with mass m and charge
+q is suspended in a horizontal
electric field, E, by a string.
What is the angle that the string
makes with the vertical direction?
20-10
Example Problem
The Six-Step Method
1. Identify Forces
Tension from string.
Gravity.
Electrostatic.
2. Coordinate System:
X right, Y up
3. Free-Body Diagram
4. Resolve Off-Axis Forces
5. Newton’s 2nd Law
6. Solve
20-11
Example Problem
Solution
X: q E  T sin( )  m a x  0
Y:
T cos( )  m g  m a y  0
T cos( )  m g
T sin( )  q E
T sin( )
qE
 tan( ) 
T cos( )
mg
20-12
Class #20
Take-Away Concepts
1.
Electric field from point charge sources:
(Total field is the superposition of point source fields.)

1 qi
E
( r̂i )
2
4   0 ri
2.
Force on a charge in an electric field:
3.
4.
Electric field points away from + source charges.
Electric field points toward – source charges.


F  qE
20-13
Class #20
Problems of the Day
___1. An electron is placed in a region of space where the
magnitude of the electric field is 100 N/C and the direction of
the electric field is north. The direction of the electric force on
the electron is:
A.
B.
C.
D.
E.
North.
South.
East.
West.
Undefined unless we know the locations and values of the
charges that create the electric field.
20-14
Answer to Problem 1 for Class #20
The answer is B.
Since the electron has a negative charge, the direction of the vector
force on the electron is opposite the direction of the electric field
at the location of the electron. The whole point of specifying the
electric field is to prevent our having to specify all of the charges
that create the electric field, so that we can find the force on a
charge without resorting directly to Coulomb’s Law.
20-15
Class #20
Problems of the Day
2. An electron begins the problem traveling north at 5 x 106 m/s in
a region of space where the electric field is 100 N/C in the north
direction. How far will the electron travel before it comes to a stop
momentarily? Or will the electron not come to a momentary stop?
Some useful constants:
e = 1.6 x 10-19 C
me = 9.1 x 10-31 kg
20-16
Answer to Problem 2 for Class #20
The force is in the south direction, opposing the motion, so the
electron does in fact stop. Take north as the + direction.
F  ( 1.6  10 19 )(100)  1.6  10 17 N
The acceleration is
a  F / m  ( 1.6  10 17 ) /(9.1 10 31 )  1.76  1013 m / s 2
Using equation #5 from the formula sheet, we have
d   v 0 /( 2a )  (5.0  10 6 ) 2 /( 2  1.76  1013 )  0.71 m
2
20-17
Activity #20
Electric Fields
(Computer Simulation Activity)
Objective of the Activity:
1.
2.
3.
Think about electric fields.
Use the simulation program to draw electric field
vectors and field lines.
Use what you learned to find “hidden” charges
from the electric field they make.
20-18