Electric Fields
i.e. The Electric Charge,
Electric Force, and
Electrostatics
Electric Charge
History
600 BC
1600 AD
Greeks first discover attractive
properties of amber when rubbed.
Electric bodies repel as well as attract.
Chapter 23
Electric Charge
Electric Charge
The Charge
The Transfer of Charge
Two types of charge:
+ protons
- electrons
SILK
Like charges repel each other.
Opposites attract each other.
Glass Rod
Some materials attract electrons
more than others.
Electric Charge
The Transfer of Charge
+ -
SILK
Electric Charge
The Transfer of Charge
+ + -
SILK
Glass Rod
Glass Rod
Here electrons are pulled off the glass onto the silk.
Usually matter is charge neutral because the
number of electrons and protons are equal. But
here the silk has an excess of electrons and the
rod a deficit.
Electric Charge
Electric Charge
The Transfer of Charge
The Transfer of Charge
+ - +
+ -
++
++ - - - +
SILK
Glass Rod
SILK
Glass Rod
Glass and silk are insulators:
charges stuck on them stay put.
Electric Charge
Electric Charge
History
Conductors Vs. Insulators
In conductors, charges are free to move about. The charges
(usually electrons) arrange themselves into a static situation
(Ftot = 0). Conductors with a net charge (positive or negative)
have the excess charge move to the surface, if no other forces
are present. Metals are conductors!
In insulators, charges cannot move freely. The charges stay
where they are put.
Electric Charge
Summary of things we know:
– There is a property of matter called electric
charge. (In the SI system its units are
Coulombs.)
– Charges can be negative (like electrons) or
positive (like protons).
– In matter, the positive charges are stuck in
place in the nuclei. Matter is negatively
charged when extra electrons are added, and
positively charged when electrons are
removed.
– Like charges repel, unlike charges attract.
– Charges travel in conductors, not in insulators
– Force of attraction or repulsion ~ 1 / r2
600 BC
Greeks first discover attractive
properties of amber when rubbed.
Electric bodies repel as well as attract
1600 AD
1735 AD
1750 AD
1770 AD
1890 AD
du Fay: Two distinct types of electricity
Franklin: Positive and Negative Charge
Coulomb: “Inverse Square Law”
J.J. Thompson: Quantization of
electric charge - “Electron”
and ... Charge is Quantized
q = multiple of an elementary charge e:
e = 1.6 x 10-19 Coulombs
electron
proton
neutron
positron
Charge
-e
+e
0
+e
Mass
1
1836
1839
1
Diameter
0
~10-15m
~10-15m
0
(Protons and neutrons are made up of
quarks, whose charge is quantized in
multiples of ± ne/3. Quarks can’t be
isolated under normal conditions.)
!
F12
Gravitational and Electric Forces of the
Hydrogen atom
Coulomb’s Law
r12
q1
q2
M
+e
r"12
!
kq q
F12 = 1 2 2 r"12
r12
Force on 1 due to 2
r12
-e
m
Gravitational force
m = 9.1 10-31 kg
M = 1.7 10-27 kg
r12 = 5.3 10-11 m
Electric Force
πε0)-1 = 9.0 x 109 Nm2/C2
k = (4πε
ε0 = permitivity of free space = 8.86 x 10-12 C2/Nm2
Gravitational and Electric Forces of the
Hydrogen atom
10-31 kg
M
+e
r12
q
m
Gravitational force
m = 9.1
M = 1.7 10-27 kg
r12 = 5.3 10-11 m
Q = -q = 1.6 10-19C
Electric Force
!
Mm
Fg = G 2 r"
r12
!  1  Qq
 2 rˆ
Fe = 
 4πε 0  r12
Fg = 3.6 10-47 N
Superposition of forces from two charges
Blue charges fixed , negative, equal charge (-q)
What is force on positive red charge +q ?
y
--qq
-q
-q
+q
x
+q
Fe = 8.2 10-8N
Superposition of forces from two charges
Blue charges fixed , negative, equal charge (-q)
What is force on positive red charge +q ?
Consider effect of each charge separately:
Superposition of forces from two charges
Blue charges fixed , negative, equal charge (-q)
What is force on positive red charge +q ?
Take each charge in turn:
y
y
--qq
--qq
+q
+q
-q
-q
-q
-q
x
+q
+q
x
Superposition of forces from two charges
Superposition of forces from two charges
Blue charges fixed , negative, equal charge (-q)
Blue charges fixed , negative, equal charge (-q)
What is force on positive red charge +q ?
What is force on positive red charge +q ?
Find resultant:
Create vector sum:
y
y
--qq
NET
FORCE
--qq
-q
-q
-q
-q
x
+q
+q
x
+q
+q
Superposition Principle
!
F13
q1
q2
F13 y
!
F12
F13 x
F12 y
q3
→
Example: electricity balancing gravity
!
F13
!
F
∧
!
F12
∧
• What forces are acting on the
charged balls ?
• Write the forces in Cartesian
coordinates.
• Equate forces (Newton’s 3rd
law)
• Solve for unknowns!
Example: electricity balancing gravity
!
FE
!
T
!
!
FG = mg
q
m
m
F12 x
F = ( F12 x + F13 x ) i + ( F12 y + F13 y ) j
• Draw vector force
diagram while
identifying the forces.
• Apply Newton’s 3rd
Law for a system in
equilibrium to the
components of the
forces.
• Solve!
q
!
T
Example: electricity balancing gravity
If θ = 20o, m = 3 mg, and L =
0.75 m, what is q and the
forces are acting on the
charged balls?
!
FE
!
!
FG = mg
!
FE
!
q1 q2
FE = k 2
1
!
FE = k
2Li θ
2
2
2
q
i
2
L
L
!
T
θ
q
q
m
!
!
FG = mg
m
Example: electricity balancing gravity
Example: electricity balancing gravity
!
FG = mg = (3 ×10 −6 kg ) 9.8 sm2 = 2.94 × 10 −6 N
(
!
! FG
also, T =
)
!
FE = k
mg
=
cosθ cosθ
L
L
2.94 × 10 −6 N
=
= 3.13 × 10 −6 N
cos(20o )
!
FE
!
T
q=
θ
q
q2
4L sin 2 θ
2
(1.07 ×10
−6
!
FE 4 L2 sin 2 θ
q=
k
N )4(0.75m ) sin ( 20o )
2
9 × 109 Nm C 2
2
m
!
!
FE = T sin θ = 3.13 × 10 −6 N sin(20 o ) = 1.07 ×10 −6 N
!
T
!
FE
q
q ≅ 7.19 × 10 −9 C
L
L
2
q
m
m
!
!
FG = mg
θ
q
!
!
FG = mg
example problem
example problem
Example: Three point charges are arranged as shown in the figure. Find the
total force on 3 due to 1 and 2.
Example: Three point charges are arranged as shown in the figure. Find the
total force on 3 due to 1 and 2.
q1 = -1 x 10-8 C,
q2 = 5 x 10-8 C,
q3 = 2 x 10-8 C,
Assume:
m
position (0, 0.5 m)
position (0.8 m, 0)
position (0, 0)
q1 = -1 x 10-8 C,
q2 = 5 x 10-8 C,
q3 = 2 x 10-8 C,
Assume:
position (0, 0.5 m)
position (0.8 m, 0)
position (0, 0)
!
kq 3 q 1
(9 × 10 9 )(2 × 10 −8 )(− 1 × 10 −8 ) N − ˆj ≅ 7 .2 × 10 −6 N ˆj
rˆ 31 =
F 31 =
2
r31
(0 . 5 )2
( )
!
kq 3 q 2
(9 × 10 9 )(2 × 10 −8 )(5 × 10 −8 ) N − ˆi ≅ −1.4 × 10 −5 N ˆi
rˆ 32 =
F 32 =
2
y
r32
(0 .8 )2
! !
!
F = F31 + F32 = 7 .2 × 10 − 6 ˆj - 1.4 × 10 − 5 î N
q1
= - 1.4 î + 0 . 72 ˆj × 10 − 5 N
( )
y
(
(
q1
)
)
F31
q3
x
q2
F32
Example: Two Charged Spheres
a
L
What is the charge on each sphere?
θ
Mass of each sphere is 30 g. L=15 cm.
θ = 50 .
Mass of each sphere is 30 g. L=15 cm.
θ = 50 .
L
The total force on each sphere is zero:
From geometry we see that
a = L sin θ = (0.15 m) sin 50 = 0.013 m
Therefore the separation of the spheres is 2a = 0.026 m
x
q2
Example: Two Charged Spheres
What is the charge on each sphere?
θ
q3
Fy = 0 = T cos θ - mg so T =mg/cosθ
Fx = 0 = T sin θ - Fe
so Fe= Tsinθ = mg tanθ
a
Left-hand
sphere:
Fe
θ T
mg
But also Fe = kq2/r2 (Coulomb’s law),
Equating Fe= Fe gives
kq2/r2 = mg tanθ gives
q2 =(1/k) r2 mg tan θ
q = 4.4 10-8 C
Example: Point Charge
The Electric Field
!
• A set of fixed charges exerts a force F given by !
Coulomb’s law on a test charge qtest at position .r
qtest
!
r
!
F
!
r
Q
!
• The electric field is represented by the symbol E ,
and is given in terms of this force by:
!#
!# !#
F
E( r) =
qtest
• Find the electric field of a point charge Q located
at the origin.
This is a vector
function of position.
!#
1 Qqtest
ˆr
F=
2
r
4
πε
0
##
• Dividing out qtest gives the electric field at
!# !#
1 Q
ˆ
E( r) =
2 r
4πε 0 r
#
!
r:
Radially outward,
falling off as 1/r2
The Electric Field
Example: Point Charge
!# !#
E( r) =
qtest
1 Q
ˆr
4πε 0 r 2
• The concept of electric fields was invented by Michael
Faraday to describe his model of how charges interact.
• Charges interact by exerting forces on each other. Faraday
thought of the fields as “lines of force” (sort of like strings)
with a density in space proportional to the strength of the
force.
!
!
E
+
Vector
! Field
(Show E at each
!
r)
-
Field Lines
(Lines of force)
Positive charge (Source)
Force due to an electric field
!
Just !turn
! the definition of E around.
E
r
If
( ) is known, !the force on a
charge q at point r is:
!#
!# !#
F = q E (r )
##
E
Negative charge (Sink)
Electric fields due to various
charge distributions
q
+
The electric field is a vector which
obeys the superposition principle.
!
The electric field at r
points in the direction
that a positive charge
!
placed at r would be
pushed.
Electric field lines
are bunched closer
where the field is
stronger!
Begin with a simple example with
discrete charges: the dipole
Field Due to an Electric Dipole
Electric fields from continuous distributions
at a point x straight out from its midpoint
Find
+q
r
!
!
!
E ( x ) = E+ + E−
Electric dipole moment
p = qd
θ
Up to now we have only considered the electric field of point
charges.
Now let’s look at continuous distributions of charge --- lines or
surfaces or volumes of charge --- and determine the resulting
electric fields.
d
x
-q
!
E−
!
E+
!
E
Sphere
Electric fields from continuous distributions
For discrete point charges, we can use the superposition
principle and sum the fields due to each point charge:
!!
!
E(r) = ∑ Ei
q2
q3
q1
q4
i
Ring
Sheet
Electric fields from continuous distributions
For discrete point charges, we can use the superposition
principle and sum the fields due to each point charge:
!!
!
E(r) = ∑ Ei
q2
q3
q1
i
q4
What if we now have a continuous charge distribution?
q
Electric fields from continuous distributions
E(r)
Electric fields from continuous distributions
- We divide the distribution up into small pieces, then we
sum the contribution to the field from each piece:
Vol.
! !
∆q
∆Ei ( r ) = k 2 i rˆ
r
! !
! !
∆q
E (r ) ≈ ∑ ∆Ei (r ) = ∑ k 2 i rˆ
r
i
i
∆qi
In the limit of very small pieces, the sum is an integral
Vol.
! !
E ( r ) = lim
∆ qi → 0
! !
E (r ) =
→dq
∆q→
Hence,
∑k
i , vol
∫
Vol .
k
∆ qi
rˆ
r2
dq
rˆ
r2
! !
! !
dq ∧
E(r ) = ∫ dE, dE = k 2 r
r
Electric fields from continuous distributions
y
P
!!
!
Hence, E(r) = ∫dE
! !
E (r ) =
dq
dq
∫kr
2
What is field at P?
−∞
! !
dq
dE ( r ) = k 2 r"
r
Vol.
Example: An infinite thin line of charge.
+∞
Charge per unit length
is λ
rˆ
Vol .
Example: An infinite thin line of charge.
Example: An infinite thin line of charge.
!
dE+
y
What is field at P?
P
P
r=
y
−∞
+∞
dq
x
Charge per unit length
is λ
Consider small element
of charge, dq, at position x.
!
dE
dq
−∞
!
dq
dE + = k 2 r"
r
• Consider small element
of charge, dq, at position x.
• dq is distance r from P.
• dq produces dE at P.
!
dE
!
dq
dE + = k 2 r"
r
dE y = 2k
θ
dq
cosθ
r2
λ dx, cosθ
θ=y/r
dq=λ
dq
-x
!
"
E = yE
!
dE−
!
dE+
θ
dq
−∞
+∞
x
!
"
E = yE
!
dE−
!
dE+
x2 + y2
+∞
x
dq
−∞
• Consider small element
of charge, dq, at position x.
!
dq
dE + = k 2 r"
r
• dq is distance r from P.
• dq produces dE at P.
dE y = 2k
• For each dq at +x, there
is a dq at -x.
dq
cosθ
r2
λ dx, cosθ
θ=y/r
dq=λ
dq
-x
+∞
x
dE y =
2kλ dx y
⋅
(x2 + y 2 ) r
x=∞
Ey =
∫ (x
2
x=0
2kλy
2kλ
2 3/2 dx =
+y )
y
Example of continuous distribution:
ring of charge
Example of continuous distribution:
ring of charge
Find the electric field at a point along the axis.
Hint: be sure to use the symmetry of the problem!
Find the electric field at a point along the axis.
Hint: be sure to use the symmetry of the problem!
!
dE
!
dE
!
r
z
Thin ring with total
charge q: charge per
length is
λ =
Thin ring with total
charge q: charge per
length is
dq
R
q
2π R
λ =
Chap 23: Example
A uniform electric field exists in a region between two oppositely
charged plates. An electron is released from rest at the surface of the
negatively charged plate and strikes the surface of the opposite plate,
2.0 cm away, in a time 1.5 x 10-6 sec. (a) What is the speed of the
electron as it strikes the second plate? (b) What is the magnitude of
the electric field E?
1
x = x0 + v0 t + at 2
2
+ + + + + + + +
1
2
2 cm = 0 + 0 + a (1 .5 × 10 − 6 s )
2
ea = 1 .78 × 10 12 cms
-2 - 2 - - - - - -
(a) v − v 0 = 2a(x − x 0 )
v 2 = 2a(x) → v = 2a(x)
v = 2.6 × 10 6
cm
s
!
r
z
2
ma
q
9.11× 10-31 kg(1.78 ×1010 sm )
E=
≅ 0.1 NC
1.6 ×10-19 C
(b) F = ma = qE, E =
2
q
2π R
dq
R
Break the charge up
into little bits and
find the field due to
each bit at the
observation point.
Then integrate. This
is an important
example problem in
the Text!