Force Fields
Objective:
TSW understand and apply the concept of a force field by
calculating the field, the force and motion of a particle in a
field.
Gravitational Force Field: A property of space around a
mass that causes forces on other masses. The gravitational
force per unit mass exerted on a point mass (g = Fg /m)
F
F
F
F
g = the gravitational field
The gravitational force is always in the same direction as the gravitational field
The gravitational field is a vector. (Magnitude and direction)
The gravitational Force = (The mass) x (The Gravitational Field)
Fg  mg
M em
Fg  G 2
r
M e Fg
g G 2 
r
m
The units for a gravitational field are N/kg
Let’s zoom in to our physics room:
Fg
Constant Field
 N
Fg  mg  m10 
 kg 
Example 1:
Calculate the gravitational field 30000km from a planet with a
mass of 4x1026kg and a radius of 20000km
M
g G 2
r
26
4

10
g  6.67 1011
3 2
(50000 10 )
N
g  10.7
kg
30000km
r
Example 2:
How long will a 4kg object starting from rest take to accelerate 25meters
through a constant gravitational field of 22N/kg?
 F  ma
88  ( 4) a
m
a  22 2
s
4kg
Fg = mg = (4)(22) = 88N
25m
1 2
y  y0  v0t  at
2
1
25  (22)t 2
2
t  1. 5 s
Electric Fields: A property of space around a charge that
causes forces on other charges. The electric force per positive
unit test charge (E = FE / q)
+ FE
Use a positive test charge
FE
_
FE
+
+
FE
FE
+
+
FE
+
+
Use a positive test charge
The
gravitational
Force
= (The
mass)xx(The
(TheElectric
Gravitational
The
Electric Force
= (The
charge)
Field) Field)
FgE  mg
qE
q1eqm2
M
FFg EGk 22
rr
M e FEFg
q
Eg  kG 2  
r r2 q m
The
units
forfor
a gravitational
field
N/kg
The
units
a electric field
areare
N/C
Things to Know
• Electric field lines go out of a positive
charge and into a negative charge.
• Positive charges experience a force with
the field.
• Negative charges experience a force
against the field.
• The Electric Field inside a conductor is
zero.
Example 3
What is the electric field at a point 25cm away from a point
charge of 2µC?
q
Ek 2
r
6
2 10
E  9 10
(.25) 2
9
5 N
E  2.88  10
C
Away from the charge
2µC
25cm
E
Example 4
A 5 gram mass with a charge of -3µC is released from rest in a
constant field of 20000N/C directed in the positive x direction.
Find its position and velocity after 4.0 seconds.
Negative charges experience a force against the field
FE = qE
 F  ma
FE  ma
qE  ma
qE
m
a
 12 2
m
s
5g
-3µC
m
v  v0  at  (12)( 4)  48
s
1 2
x  x0  v0t  at
2
1
x  (12)( 4) 2  96m
2
Example 6
What is the electric field at the origin?
E3
EE21
kq
kq111
EE32 1 222
rr
 666
10
10
(9(910
10999)()(121..5010
) ))
EE32 1
345..500222
NN
N
720
EE32 1844
735
CC
C
To the right (+)
Up (+)
To the right (+)
The electric field due to each charge
must be calculated individually and
then added together as vectors.
Now add the vectors together:
E3=844N/C
E1 = < 735, 0
>
E2 = < 720, 0
>
E3 = <
E
E1=735N/C
θ
1455N/C
E2=720N/C
844N/C
0 , 844 >
E = < 1455, 844 >
E  14552  844 2
N
E  1682
C
1  844 
  tan 

 1455 
  30

The electric field due to more than one point charge.
F
+
F
Two charged parallel plates
The Electric field is constant between the two plates.
The Electric Field inside a
conductor is zero
E = 0N/C
An electron has a velocity of 6x105m/s as it enters a constant electric
field of 10000N/C as shown in the diagram below. How far will the
electron travel in the x direction before striking one of the plates?
2.82 x 10-3m
_
v
4cm
FNow
First
mawe
have
projectile motion
1a simple
2 acceleration
the
force
and
of the1
y  y0 electron
v0t  must
at
x

x

v
t

at
problem!
0
0
be
calculated.
FE  ma
2
2
5 shown
9
 ma
The qE
Electron
path
as
1 will follow15a parabolic
2 x  (6  10 )( 4.7  10 )
.02  (1.8 10 19
)t
qE2 (1.6 10 )(10000)
 315 m
x  2.82
a

 1
.810
10 m 2

31
s
t  4m
.7 10 9 s9.1110
2