Fields in Physics
A field is a physical quantity that has a value for
every point in space and time.
The strength of all known fields falls off with distance,
diminishing to the point of being undetectable.
A field creates a “condition in space” or generates
“action-at-a-distance.” An object placed in a field
experiences a force transmitted to it by the field
which causes the object to accelerate.
Conversely, mechanical forces must directly contact
an object to cause acceleration.
Examples of vector fields of importance in
physics are gravitational, electric, and magnetic fields.
Acceleration of Earth’s Gravitational Field
Assuming the Earth is spherical,
objects on its surface experience
force generated as if all the Earth’s
mass was concentrated at its center.
rEarth
mEarth
G  m object  m Earth
According to Newton’s Law
of Universal Gravitation:
Fobject 
According to Newton’s
Second Law:
Fobject  m object  a
Comparing the two
Equations:
2
rEarth
G  m Earth
a
2
rEarth
a is the acceleration caused by the Earth’s gravitational
field at its surface (commonly symbolized as g).
Acceleration of Earth’s Gravitational Field
mEarth = 5.981024 kg
rEarth = 6.37106 m
GmEarth
g 2
rEarth
2


N

m
11
24
 6.67 10

5
.
98

10
kg
2 
kg 
N


 9.83
2
kg
6.37 106 m


 kg  m 
 2 
N
m
s 

g  9.83  9.8
 9.83 2
kg
kg
s


This quantity is
the acceleration
due to the Earth’s
gravitational field.
Electric Fields
A similar procedure can be carried out on Coulomb’s
Law to develop an expression for the electric field.
Coulomb’s Law for the
force on q2 due to the q1
q1
F21  k 2 q 2
r
Factoring out q2, the
remaining terms are
the “electric field” of q1:
The force exerted on a charge
q2 in the electric field of q1
can therefore expressed as:
q1
E1  k 2
r
Units:
Newtons
Coulomb
q1
F21  k 2 q 2  E1  q 2
r
Summary of Gravitational and Electric Fields
Gravitational force on
m2 due to the Earth
mEarth
F  G 2 m2
r
Electrostatic force on
q2 due to the q1
pull out terms
Gravitational field of Earth
mEarth
g G 2
r
Units of g
Newtons
kg
q1
F  k 2 q2
r
Electric field of q1
q1
Ek 2
r
Units of E
Newtons
Coulomb
Electric Field Lines
Electric field lines are defined relative to what force an imaginary
positive charge would experience if it were brought into the influence of
the electric field of an actual charge.
Positive charges have field lines
emanating radially outward.
Negative charges have field lines
emanating radially inward.
Electric Field Strength and Distance
q1
Ek 2
r
Electric field strength is independent of the angle outward from charge.
Electric field strength decreases with the square of distance outward.
Field strength at a
distance 2r is 1/4E
anywhere on this
circle circumference
Suppose he field strength
at a distance r is E. It is
the same anywhere on the
circle circumference
Field strength at a
distance 3r is 1/9E
anywhere on this
circle circumference
Electric Field Lines and Field Magnitude
The field lines qualitatively show the strength of field.
Positive charge field lines outward
Field lines densest closer to charge
indicates electric field greater nearer
the charge.
Negative charge field lines inward.
Field lines denser in picture (b)
indicating it is a stronger electric
field that that in picture (a)
Field Lines Due to Opposite Charges
Field lines emanate outward
from positive charges and
go inward toward negative.
Isolated point charges would
have radially emanating field
lines, but the lines of each are
distorted due to the presence
of a second charge.
Field Lines Due to Opposite Charges
Field lines are twice
as dense in the region
of twice as much charge
Field Lines Due to Like Charges
Repulsive field lines
generate a region of
space where there is
no net electric field.
Using Induction to Create Positive Charge
We have seen that conductors can
be negatively-charged by contacting
the object with negative charge. How can
the object be positively-charged?
If the negatively-charged rod is
held near the conductor, the field
polarizes charge on the conductor.
An attached wire allows negative
charge to migrate through to ground.
If the ground wire is removed,
the conductor now possesses
excess positive charge, which
migrates by repulsion uniformly to
the outer edge of the conductor.
Conductors and Electric Fields
The charge on a conductor repels
and spreads to its outermost
surface, whether the conductor
is hollow or solid.
Conductors and Electric Field Lines
What
about
inside?
Field lines emanate
outward from the
surface of the conductor
A point at the center
of the conductor
experiences no
net force
Careful
analysis shows
that any point
anywhere within the
conductor will
experience no net force
Fields From Irregularly-Shaped Conductors
Electric charges and
field lines are more
densely-packed near
a sharp point of an
irregularly-shaped
conductor
Charge still concentrates at
the outermost extremities of
the irregularly-shaped conductor.
The electric field
within the object
cancels at all points
within.
Lightning Striking a Car - The Faraday Cage
Parallel Plate Capacitors
Parallel plate capacitors have
two conductors of opposite
charge separated by a space
which is small compared to
their length. Electric field lines
between the plates are all
approximately uniform and
parallel between the two plates.
The electric field between the
plates of a parallel plate capacitor
is uniform (or the same value)
anywhere between the plates. The
field is zero outside the plates.
Electric Field Example 1
What is the electric field of a 2.0 nC point charge at a
distance of a.) 50 cm b.) 100 cm and c.) 150 cm?
1 nC = 1 nanoCoulomb = 1  10-9 C
2
9
N

m
2
.
0

10
C
N
9
a.) E  8.99 10
 72
2
2
C
(0.50 m)
C
2
9
N

m
2
.
0

10
C
N
9
Distance 2r gives
b.) E  8.99 10
 18
2
2
C
(1.0 m)
C field 1/4E
2
9
N

m
2
.
0

10
C
N
9
c.) E  8.99 10
 8.0
Distance 3r gives
2
2
C
(1.5 m)
C field 1/9E
Electric Field Example 1
Field strength at
100 cm (2r)
is 18 N/C (1/4E)
Field strength
at 50 cm
is 72 N/C
Field strength at
150 cm (3r)
is 8.0 N/C (1/9E)
Electric Field Example 2
How far to the right of charge q2 is the electric field zero?
1.50 cm
q1=+9.40 C
q1 field
q2 field
q2=-1.60 C
To cancel, the fields must be equal in magnitude: E1 = E2
Unknown distance to the right of charge q2 = x
Unknown distance to the right of charge q1 = 1.5 + x
9.40
1.60
k
k 2
2
(1.50  x)
x
square root of
both sides
3.07
1.26

1.50  x
x
3.07x = 1.89 + 1.26x
1.81x = 1.89
x = 1.04 cm
Electric Field Example 3
q1=+0.25 nC
6.0 cm
E2
12 cm

6.0 cm
q2=+0.50 nC
13.4 cm
6.0  12.0  13.4 cm
2
2
E1
What is the electric field
due to the charges at this
spot 12.0 cm to the right?
 6.0 
o
  tan 

26
.
6

 12.0 
1
2
9
N

m
0
.
25

10
C
N
9
E1  8.99 10
 125
2
2
C
0.134
C
2
9
N

m
0
.
50

10
C
N
9
E 2  8.99 10
 250
2
2
C
0.134
C
Electric Field Example 3
250 cos(26.6) = 224 N/C
250 sin(26.6)
E2 = 250 N/C
= 112 N/C
26.6 o
26.6 o
26.6 o
26.6 o
125 sin(26.6)
E1 = 125 N/C
= 61 N/C
125 cos(26.6) = 112 N/C
E(x) = 112 + 224 = 336 N/C
Enet = 340 N/C
8.63 o
336 N/C
E(y) = 112 -61 = 51 N/C
E net  3362  512  340 N/C
51 N/C
 51 
o
  tan 

8
.
63

 336 
1
Electric Field Example 4
q1 = -1.0 nC
6.0 cm
6.0 cm
E2
E1
12.0 cm

13.4 cm
q2 = +1.0 nC
What is the electric field a
distance 12.0 cm away?
6.02  12.02  13.4 cm
 6.0 
o
  tan 
  26.6
 12.0 
1
2
9
N

m
1
.
0

10
C
N
9
E1  E 2  8.99 10
 501
2
2
C
0.134
C
E(x)net = 0
E(y)net = 2501sin(26.6) =449 N/C
E1(x) = -E2(x)
E1(y) E2(y)
501 N
501 N
Enet = 449 N/C
Electric Flux
The electric flux (E) is a measure of the density of
electric field lines which flow through a given
cross-sectional area.
A
N


E
A
 E
N  m2
SI units:
C
E = EAcos
 is the angle that the normal
vector N makes with the
electric field E direction.
The electric flux increases with electric field strength,
and cross-sectional area, and decreases with angle .
Electric Flux Through a Loop
Area perpendicular to E field
(N at 0o to E field): E = EA
Area parallel to E field
(N at 90o to E field): E = 0
Area tilted in E field
(N at angle  to E field):
E = EA cos 
Electric Flux Examples
E
Circle of radius
28.0 cm E = 2.0 N/C
N = 0o relative to E
E = EAcos(0o)
E =Er2
E = (2.0 N/C)(3.14)(0.28 m)2
E = 0.492 (N·m2)/C
N
Triangle of base
12.0 cm and height
30.0 cm in E = 4.8 N/C
N = 58o relative to E
E = EAcos(58o)
E =E(1/2 b×h)(0.530)
E = (4.8 N/C)(0.0180 m2)(0.530)
E = 0.0458 (N·m2)/C
Rectangle of length
180 cm and width
350 cm in E = 3.6 N/C
N = 78o relative to E
E = EAcos(78o)
E =E(l × w)(0.208)
E = (3.6 T)(6.30 m2)(0.208)
E = 4.72 (N·m2)/C
E
N
E
N
The Permittivity of Free Space
Consider a point charge of value q and
a spherical surface a distance r from it.
The electric field on that surface is:
q
Ek 2
r
The electric field passes perpendicular through all points on
the circular surface. The electric flux through
that
circle
is:
Surface area
q
 E  EA  k 2 (4r 2 )  4kq of a sphere
r
The permitttivity of free space 0 is defined:
1
C
12
0 
 8.85 10
4k
N  m2
Using 0, the electric flux generated by point charge q through
q
the surface of a sphere enclosing it is:
E 
0
The Permittivity of Free Space
The Coulomb’s law and the electric field equation are often
written in terms of 0 instead of the proportionality constant k.
q1q2
F12  k 2
r
q1
E1  k 2
r
Coulomb’s law
Permitttivity of free space:
Electric field equation
0 
k
q1q2
F12 
40 r 2
1
4k
1
40
E1 
q1
40 r
2
Electric Flux Through a Surface
The flux through a surface
can be positive or negative.
field lines exiting a surface field lines entering a surface
E 
The flux through a spherical
surface enclosing a point
charge is the same independent of the size of the
sphere enclosing the charge.
q
0
E 
E 
q
0
q
0
independent of r
Gauss’s Law
Using the techniques of calculus, it
can be shown that the relationship
between the flux through any surface
is a function of only the charge enclosed
and the permittivity of free space.
Positive flux for field
lines exiting a surface
E 
q
0
Gauss’s law holds independent of the size or shape of the
surface enclosing a charge.
Johann Carl Friedrich Gauss
(1777 - 1855)
Gauss’s Law Application
What is the electric flux
through the surface due
to the point charge?
Since the surface
does not enclose
the point charge,
there is no net flux
E = 0
Field lines exiting
surface - positive flux
+q
Field lines entering
surface - negative flux
Careful mathematical
consideration shows that
the negative and positive
flux exactly cancel.
If this were a conducting surface, Gauss’s
ΦE
E
0
law says it would contain no electric field
A
Gauss’s Law Application
Consider a point charge of +Q that is
surrounded by a hollow conducting
spherical shell as shown. Use
Gauss’s law to find the electric field
enclosing the charge at a.) r1 b.) r2
and c.) r3.
a.) Φ E1
Q
 E1 4r 
ε0
2
1
Q
Q
E1 
k 2
2
4πε 0 r1
r1
The same result that would be expected if no conducting shell
is surrounding the point charge.
Gauss’s Law Application
Consider a point charge of +Q that is
surrounded by a hollow conducting
spherical shell as shown. Use
Gauss’s law to find the electric field
enclosing the charge at a.) r1 b.) r2
and c.) r3.
b.) The Gaussian surface is inside the conducting sphere:
RA < r2 < RB
The field within a conductor is zero, therefore:
Φ E2  E 2A  0
There is no net charge within the conductor. The central point
charge +Q polarizes the conductor such that its inner surface
acquires charge -Q and its outer surface acquires charge +Q.
Gauss’s Law Application
Consider a point charge of +Q that is
surrounded by a hollow conducting
spherical shell as shown. Use
Gauss’s law to find the electric field
enclosing the charge at a.) r1 b.) r2
and c.) r3.
total charge enclosed
by surface
Q-QQ Q
c.) r3 > RB
Φ E 3  E 3 (4πr3 ) 

ε0
ε0
Q
Q
E3 
k 2
2
4πε 0 r3
r3
The conducting sphere shields its interior from electric fields
but does not shield the external world from fields within it.
2
Work and Force
Work is the energy due to an applied force acting over
a distance.
Work done
(scalar)
 
W  F d
Force Distance that the
(Vector) force acts (vector)
In magnitude : W = Fd cos
kg  m
kg  m
Units : 2  m 
 Joules
2
s
s
2
Angle of action
between force and
displacement
Conservative Forces
Conservative forces:
1.Springs
2.Gravity
3.Electric Fields
4.Magnetic Fields
Conservative Fields
Conservative forces generate potential energy in an
object due to its position relative to the conservative
force (or conservative force field).
Non-conservative forces
1.Friction
2.Air Resistance
3.Other contact forces (e.g. pushing or pulling)
Potential Energy, Work, and Electric Fields
The change in potential energy is equal in magnitude
but opposite in sign to the work done by the
conservative force.
Change in
Work done
U = Uf -Ui = -W
potential energy
by force field
q
Electric field due to charge q: E  k
r2
q q0
Force charge q exerts on charge q0: F  k
 Eq0
2
r
Work done by a uniform electric
field to move positive test charge q0
a distance d parallel to field lines
Potential energy change in the field
while moving the positive test charge:
W = F·d = Eq0d
U = -W = -Eq0d
The Electric Potential Difference
Earlier it was convenient to define an electric
field E to describe the force F per unit charge q0
that an arbitrary charge experiences
In a similar fashion it is convenient to define a
quantity describing the change in potential
energy per unit charge q0 that is called the
electric potential difference:
F
E
q0
U
W
V 

q0
q0
Units of electric potential:
Joules/Coulomb or Volts
Work done by a uniform electric
field to move positive test charge q0
a distance d parallel to field lines
W = F·d = Eq0d
Electric potential difference as an
arbitrary charge is moved a distance
d in a uniform electric field E.
q0 Ed
V  
  Ed
q0
Electric potential is a scalar quantity (magnitude only).
Electric Potential Due to a Point Charge
The techniques of calculus can be used to find the electric
potential due to a point charge (non-uniform electric field):
kq
V
r
Units of Volts or
Joules/Coulomb
Note that this can also be expressed by replacing k with the
permittivity of free space 0: V  q
40 r
The above expression requires that an arbitrary charge is
initially located at infinity (if ri =  then E = 0 and Vi = 0)
and it is moved from zero electric potential to its value at the
final distance r.
The sign of charge must be explicitly placed in the voltage
expression. Voltages can be positive or negative quantities.
Electric Potential Due To a Point Charge
What is the electric potential a.) 10 cm, b.) 20 cm and
c.) infinitely away from a charge of +2.8 nC?
9
kq (8.99 10 N  m C ) (2.8 10 C)
Va 

 252 V
ra
0.10m
kq (8.99 109 N  m 2 C2 ) (2.8 109 C)
Vb 

 126 V
rb
0.20m
9
2
2
9
kq (8.99 10 N  m C ) (2.8 10 C)
Vc 

0V
rb

What is the change in electric potential (voltage drop)
9
moving 10 cm to 20 cm away?
V =Vb - Va = -126 V
2
2
moving from 10 cm infinitely away?
V =Vc - Va = -252 V
Electric Potential Due To a Point Charge
How much work is done by the +2.8 nC charge when
moving q0 = +4.0 C charge:
a.) from 10 to 20 cm away?
W = -U = -q0V = -(+4.0 C)(-126 J/C) = +504 J
b.) from 10 cm to infinity?
W = -U = -q0V = -(+4.0 C)(-252 J/C) = +1008 J
There is a drop in potential energy (U = Uf - Ui < 0)
as the positive charge is moved away.
If the charge being moved away were negative, there
would be negative work and gain in potential energy.
Gravitational vs. Electric Potential Energy
Potential energy difference
when moving a 1.0 mC
charge in the field created
by another 1.0 mC charge.
10 MJ
Potential energy difference
when moving a 0.5 kg
mass in the gravitational
field created by the Earth.
10 MJ
k q
U  qV  q 
r
U  mgr  m 
U
U
0J
0J
0 mm
r
5 mm
0 km
r
G  mE  r
(rE  r) 2
1 × 105 km
Electric Potential of Point Charge Summary
kq
V
r
For positive charge q, electric potential is positive and decreases
(becomes less positive) as you move away from the charge.
For negative charge q, electric potential is negative and increases
(becomes less negative) as you move away from the charge.
For positive charge q, electric potential difference is
negative as you move away from q (rf > ri) and positive
as you move toward q (rf < ri).
For negative charge q, electric potential difference is
f
i
positive as you move away from q (rf > ri) and negative
as you move toward q (rf < ri).
 1 1
V  kq  
 r r 
When an arbitrary charge qo is subjected to a potential difference:
qo
+
+
-
U  q0 V
V
+
+
-
U
+
+
U  W
qo
+
+
-
V
+
+
-
W
+
+
-
Potential Difference Due to Multiple Charges
25 cm
q1 = -3.0 nC
12 cm
q4 = -2.5 nC
x
Point a
q2 = +2.0 nC
12 cm
q3 = +1.5 nC
25 cm
What is the electric potential a.) at the center of the
four charges? b.) halfway between q2 and q3? c.)What
is the electric potential difference between points b
and a?
1
2
2
r
25  12  13.9 cm
2
Potential Difference Due to Multiple Charges
25 cm
q1 = -3.0 nC
x
12 cm
Point a
q2 = +2.0 nC
12 cm
q3 = +1.5 nC
q4 = -2.5 nC
25 cm
Nm
8.99 10
2
9
9
C
Va 
(3.0 10 C  2.0 10 C
0.139m
 1.5 10 9 C  2.5 10 9 C)  -129 V
2
9
Potential Difference Due to Multiple Charges
25 cm
q1 = -3.0 nC
r1
12 cm
r4
q2 = +2.0 nC
r2
x Point b
r3
q3 = +1.5 nC
q4 = -2.5 nC
25 cm
What is the electric potential b.) halfway between q2
and q3?
r = r = 6 cm
2
3
r1  r4  252  62  25.7 cm
Potential Difference Due to Multiple Charges
25 cm
q2 = +2.0 nC
q1 = -3.0 nC
x
12 cm
q4 = -2.5 nC
Vb 
q3 = +1.5 nC
25 cm
2
9
9
9
9

N

m
(2.0

10
C

1.5

10
C)
(2.5

10
C

3.0

10
C) 
9
8.99  10



2
C 
0.06 m
0.257 m

 331 V
c.) V = Vb - Va = 331 - (-129) = +460 V
+460 V is the electrical potential difference independent
of the path chosen to get from point a to point b.
Electric Potential in a Parallel Plate Capacitor
A parallel plate capacitor supplies a uniform electric field
between its plates.
If the capacitor plates are separated by d = 0.75 cm what is the
electric field between the plates connected to a 12.0 V battery?
V= -E·d
Moving from the positive plate to the negative plate decreases
electric potential by 12 volts: V = - 12.0 V
ΔV
 12.0 V
Volts
E

 1600
d
0.0075 m
meter
Alternative units:
Newtons
Coulomb
Work Done by a Uniform Electric Field
W = Fd cos
W = Eq0d cos = -U
(Constant E)
For a positive E field and positive test charge:
= 0o  cos  = 1  Displacement in direction of
force  maximum work done by the field 
maximum drop in potential
= 180o  cos  = -1  Displacement opposed to
direction of force  maximum work done on the
field  maximum rise in potential
= 90o cos  = 0  Displacement perpendicular to
force  no work or change in potential
Comparing Gravitational and Electric Fields
+
Gravitational
Field Lines
Earth’s Surface
max. height max. gravitational potential
Electrostatic
Field Lines
+
max distance between plates max. electrical potential
(max. voltage for positive charge)
d
At surface dropped all potential
Earth’s Surface
-
At negative plate, all potential
is dropped for a positive charge
Field Strength: E = -V/d
Comparing Gravitational and Electric Fields
m
m
A mass moving
perpendicular and
towards the Earth’s
surface has a drop
in gravitational potential
Earth’s Surface
++
+
-
A positive moving
perpendicular and
towards the negative
plate has a drop
in electric potential
(voltage drop)
+
m
m
A mass moving parallel to
the Earth’s surface has no
drop in gravitational potential
Earth’s Surface
-
+
+
A positive moving parallel to
the charged plates has no
drop in electric potential
(no voltage drop)
Comparing Gravitational and Electric Fields
+
m
m
d
20o
d
20o
dsin20
Gain in gravitational potential:
U = +mg(dsin20)
Earth’s Surface
+
-
Total potential
m drop conserved:
Utot = -mgd
d2
= -mg(d1 + d2)
d1
d
m
Earth’s Surface
+
d1
+
d
d2
-
dsin20
Gain in electrical potential:
V = +E(dsin20)
+
m
+
+
Total potential
drop conserved:
Vtot = -Ed
= -E(d1 + d2)
Moving Charge in a Parallel Plate Capacitor
a.)The plate of the capacitor
are separated by 0.200 cm
+
and are charged by a voltage
of 10.0 V. What is the electric
field between the plates?
ΔV
 10.0 V
Volts
E

 5000
d
0.0020 m
meter
b.)What is the work, potential energy change, and potential
drop if a positive charge of 2.00 C is moved in the direction of
the field a distance of 0.100 cm?
W = Eq0d cos = (+5000 V/m)(+2.00 C)(0.00100 m)(cos(0))
= 10.0 J
U - 10.0 J
U = -W = -10.0 J
ΔV 

 5.00 V
q0
 2.00 C
Moving Charge in a Parallel Plate Capacitor
+
What is the work, potential energy change, and potential
drop if a positive charge of 2.00 C is moved in the same
electric field a distance of 0.100 cm perpendicular to the field?
W = Eq0d cos
 = 90o
cos = 0
W=0J
U = 0 J
V = 0 V
Moving Charge in a Parallel Plate Capacitor
0.10 cm
+
32o
What is the work, potential energy change, and potential
drop if a positive charge of 2.00 C is moved as shown in the
same electric field a distance of 0.100 cm?
The angle between the field lines E and displacement direction
d is: 90o + 32o = 122o
W = Eq0d cos = (+5000 V/m)(+2.00 C)(0.0010 m)(cos(122))
= -5.3 J
U = -W = +5.3 J
U
 5.3 J
ΔV 

 2.65 V
q0
 2.00 C
Parallel Plate Capacitor Summary
1. Moving a positive charge away from the positive and towards the negative:
V < 0
U < 0
W>0
2. Moving a positive charge away from the negative and towards the positive:
V > 0
U > 0
W<0
3. Moving a negative charge away from the positive and towards the negative:
V < 0
U > 0
W<0
4. Moving a negative charge away from the negative and towards the positive:
V > 0
U < 0
W>0