Electric Field
Electric field
a region of space in which electric
charges will feel a force
The Concept of a Field
A field is defined as a property of space in which a
material object experiences a force.
Above earth, we say there is
a gravitational field at P.
m .
P
F
Because a mass m experiences a
downward force at that point.
No force, no field; No field, no force!
The direction of the field is determined by the force.
The Gravitational Field
A

F
 B
F
If g is known at
every point above
the earth then the
force F on a given
mass can be found.
Consider
A and
above
Note that points
the force
F isBreal,
but
the field
surface
of the
earth—justway
is just
a convenient
points
in space.
of describing
space.
The field at points A or B might
be found from:
F
g
m
The magnitude and direction of
the field g depends on the
weight, which is the force F.
Representing electric field
• Electric field is represented by field lines
(lines of force)
Tells us two things:
1. direction of field
2. strength of field
Representing electric field
1. Radial field
 spreads out in all directions
a. Positive charge
(outward)
b. Negative charge
(inward)
Representing electric field
2. Electric field of two oppositely
charged objects
Representing electric field
3. Electric field of two like charged
objects
Representing electric field
4. Uniform electric field
 has the same strength at all
points and produced
between oppositely
charged parallel plates
 Electric field lines always begin
on a positive charge and end on a
negative charge and do not stop
in midspace.
The Electric Field
1. Now, consider point P a
distance r from +Q.
2. An electric field E exists
at P if a test charge +q
has a force F at that point.
3. The direction of the E is
the same as the direction of
a force on + (pos) charge.
4. The magnitude of E is
given by the formula:
+q
P +.
F
E
r
+ ++
+
++Q++
Electric Field
F
N
E  ; Units
q0
C
Field is Property of Space
F .
+q +
E
r
++
+
+
++Q++
Electric Field
Force on +q is with
field direction.
Force on -q is
against field
direction.
-q -.
F
E
r
++
+
+
++Q++
Electric Field
The field E at a point exists whether there is a
charge at that point or not. The direction of the
field is away from the +Q charge.
Field Near a Negative Charge
E
+q +.
F r
- --- -Q
-Electric Field
Force on +q is with
field direction.
Force on -q is
against field
direction.
F
-q -.
E
r
-- -Q --Electric Field
Note that the field E in the vicinity of a negative
charge –Q is toward the charge—the direction that a
+q test charge would move.
The Magnitude of E-Field
The magnitude of the electric field intensity at a
point in space is defined as the force per unit
charge (N/C) that would be experienced by any
test charge placed at that point.
Electric Field
Intensity E
F
N
E  ; Units  
q0
C
q0 - test charge (has a small magnitude so it does
not affect the other charges)
Example 1. The positive test charge has a
magnitude of 3.0x10-8C and experiences a
force of 6.0x10-8N.
(a)Find the force per coulomb that the test charge
experiences.
(b)Predict the force that a charge of +12x10-8C would
experience if it replaced the test charge.
(a)
F 6.0 10 8 N

 2.0 N C
8
qo 3.0 10 C


(b) F  2.0 N C 12.0 108 C  24 108 N
Example 2. The charges on the two metal
spheres and the ebonite rod create an electric
field at the spot indicated. The field has a
magnitude of 2.0 N/C. Determine the force on
the charges in (a) and (b)




(a)
F  qo E  2.0 N C 18.0 108 C  36 108 N
(b)
F  qo E  2.0 N C 24.0 108 C  48 108 N
The E-Field at a distance r
from a single charge Q
Consider a test charge +q placed
at P a distance r from Q.
The outward force on +q is:
kQq0
F 2
r
FE
+q +.. P
P
r
kQ
++ E  2
+
+
+
r
+Q +
++
The electric field E is therefore:
F kQq0 r 2
E 
q0
q0
kQ
E 2
r
Example 3. What is the electric field
intensity E at point P, a distance of 3 m
from a negative charge of –8 nC?
.
E=?
r
3m
-Q
-8 nC
First, find the magnitude:
P
9 Nm 2
C2
-9
)(8 x 10 C)
kQ (9 x 10
E 2 
r
(3 m) 2
E = 8.00 N/C
The direction is the same as the force on a positive
charge if it were placed at the point P: toward –Q.
E = 8.00 N, toward -Q
Example 4: The Electric field of a Point Charge
The isolated point charge of q=+15μC is in a vacuum.
The test charge is 0.20m to the right and has a charge
qo=+0.80μC. Determine the electric field at point P.
The Electric Field
F k
q qo
r2

8.99 10

9
N  m 2 C 2 0.80 10 6 C 15 10 6 C
0.20m 2
F
2.7 N
6
E


3
.
4

10
NC
-6
qo 0.80 10 C

 2.7 N
Example 5: The Electric Fields from Separate
Charges May Cancel
Two positive point charges, q1=+16μC and q2=+4.0μC are
separated in a vacuum by a distance of 3.0m. Find the
spot on the line between the charges where the net electric
field is zero.
Ek
q
r2
Example 6. A constant E field of 40,000 N/C
is maintained between the two parallel
plates. What are the magnitude and
direction of the force on an electron that
passes horizontally between the plates.c
+ + + + + + + + +
The E-field is downward,
and the force on e- is up.
e- -
F
E  ; F  qE
q
Fe -.
E
e- -
- - - - - - - - -
F  qE  (1.6 x 10 C)(4 x 10
-19
F = 6.40 x 10-15 N, Upward
4 N
C
)
The Resultant Electric Field
The resultant field E in the vicinity of a number
of point charges is equal to the vector sum of the
fields due to each charge taken individually.
Consider E for each charge.
Vector Sum:
E = E1 + E2 + E3
Magnitudes are from:
kQ
E 2
r
q1 ER
E2
A
E1
q3 -
E3
+
q2
Directions are based
on positive test charge.
Example 7. Find the resultant field at point A
due to the –3 nC charge and the +6 nC
charge arranged as shown.
-3 nC
q1 3m
E1
5m
+6 nC

E2 A 4 m
E1 
(9 x 10
E for each q is shown
with direction given.
9 Nm 2
C2
+
q2
-9
)(3 x 10 C)
(3 m)
kq1
kq2
E1  2 ; E2  2
r1
r2
2
E2 
(9 x 10
9 Nm 2
C2
)(6 x 10-9C)
(4 m) 2
Signs of the charges are used only to find direction of E
Example 7. (Cont.)Find the resultant field at
point A. The magnitudes are:
-3 nC
q1 3m
E1
E1 
5m
+6 nC

E2 A 4 m
+
q2
E2 
(9 x 10
9 Nm 2
C2
(3 m) 2
(9 x 10
9 Nm 2
C2
E1 = 3.00 N, North E2 = 3.38 N, West
Next, we find vector resultant ER
E1
ER  E  R ; tan  
E2
2
2
2
1
)(3 x 10-9 C)
)(6 x 10-9C)
(4 m) 2
ER

E2
E1
Example 7. (Cont.)Find the resultant field at
point A using vector mathematics.
ER
E1 = 3.00 N, North

E2
E1
E2 = 3.38 N, West
Find vector resultant ER
E  (3.00 N) 2  (3.38 N) 2  4.52 N;
3.00 N
tan ∅ =
3.38 N
 = 41.60 N of W; or q = 48.40 W of N
Resultant Field: ER = 4.52 N at 138.41°