Chapter 23 Electric Charge and Electric Fields
• What is a field?
• Why have them?
• What causes fields?
Field Type Caused By
gravity mass
electric charge
magnetic moving charge
Electric Charge• Types:
– Positive• Glass rubbed with silk • Missing electrons
– Negative• Rubber/Plastic rubbed with fur• Extra electrons
• Arbitrary choice – convention attributed to ?
• Units: amount of charge is measured in [Coulombs]
• Empirical Observations:– Like charges repel– Unlike charges attract
Charge in the Atom
• Protons (+)
• Electrons (-)
• Ions
• Polar Molecules
Charge Properties• Conservation
– Charge is not created or destroyed, only transferred.
– The net amount of electric charge produced in any process is zero.
• Quantization– The smallest unit of charge is that on an electron or
proton. (e = 1.6 x 10-19 C)
• It is impossible to have less charge than this
• It is possible to have integer multiples of this charge
Q Ne
Conductors and Insulators
• Conductor transfers charge on contact• Insulator does not transfer charge on contact• Semiconductor might transfer charge on contact
Charge Transfer Processes
• Conduction
• Polarization
• Induction
The Electroscope
Coulomb’s Law• Empirical Observations
• Formal Statement
1 2F q q
2
1F
r
Direction of the force is along the line joining the two charges
1 212 212
21
kq qˆF r
r
Active Figure 23.7
(SLIDESHOW MODE ONLY)
Coulomb’s Law Example
• What is the magnitude of the electric force of attraction between an iron nucleus (q=+26e) and its innermost electron if the distance between them is 1.5 x 10-12 m
Hydrogen Atom Example
• The electrical force between the electron and proton is found from Coulomb’s law– Fe = keq1q2 / r2 = 8.2 x 108 N
• This can be compared to the gravitational force between the electron and the proton– Fg = Gmemp / r2 = 3.6 x 10-47 N
Subscript Convention
1 212 212
21
kq qˆF r
r
12 1 2F force on charge q due to charge q
21 2 1r distance from charge q to charge q
21 2 1r̂ unit vector oriented from charge q to charge q
+q1+q2
21r̂ 12F
21r
More Coulomb’s Law
+q1+q2
21r̂
Coulomb’s constant:2 2
9 92 2
o
N m N m 1k 8.988x10 9.0x10
C C 4
permittivity of free space:2
12o 2
1 C8.85x10
4 k N m
1 212 212
21
kq qˆF r
r
12F
21r
21 21r r
2121
21
rr̂
r
Charge polarity: Same sign Force is rightOpposite sign Force is Left
Electrostatics --- Charges must be at rest!
Superposition of Forces
0 01 02 03F F F F ....
0 1 0 2 0 30 10 20 302 2 2
10 20 30
kq q kq q kq qˆ ˆ ˆF r r r ....
r r r
N31 2 i
0 0 10 20 30 0 i02 2 2 2i 110 20 30 i0
qq q qˆ ˆ ˆ ˆF kq r r r .... kq r
r r r r
+Q3
+Q2
+Q1
+Q0
01F
03F
02F
10r
20r
30r
Coulomb’s Law Example
• Q = 6.0 mC• L = 0.10 m• What is the magnitude
and direction of the net force on one of the charges?
F1
+
F2
F3
Q
x
y
L
++
+
L
Q
Zero Resultant Force, Example
• Where is the resultant force equal to zero?– The magnitudes of the
individual forces will be equal
– Directions will be opposite
• Will result in a quadratic
• Choose the root that gives the forces in opposite directions
Electrical Force with Other Forces, Example
• The spheres are in equilibrium
• Since they are separated, they exert a repulsive force on each other– Charges are like charges
• Proceed as usual with equilibrium problems, noting one force is an electrical force
Electrical Force with Other Forces, Example cont.
• The free body diagram includes the components of the tension, the electrical force, and the weight
• Solve for |q| • You cannot
determine the sign of q, only that they both have same sign
The Electric Field• Charge particles create forces on each other without ever
coming into contact. » “action at a distance”
• A charge creates in space the ability to exert a force on a second very small charge. This ability exists even if the second charge is not present.
• We call this ability to exert a force at a distance a “field”• In general, a field is defined:
• The Electric Field is then:
test quantity 0
ForceField lim
test quantity
q 0
FE lim
q
N
C
Why in the limit?
Electric Field near a Point Charge
oq 0o
FE lim
q
o2
kq qˆF r
r
o
o2
2q 0o
kqqr̂ kqr ˆE lim r
q r
+Q -Q
Electric Field Vectors
Electric Field Lines
Active Figure 23.13
(SLIDESHOW MODE ONLY)
Rules for Drawing Field Lines• The electric field, , is tangent to the field lines.• The number of lines leaving/entering a charge is
proportional to the charge.• The number of lines passing through a unit area normal
to the lines is proportional to the strength of the field in that region.
• Field lines must begin on positive charges (or from infinity) and end on negative charges (or at infinity). The test charge is positive by convention.
• No two field lines can cross.
E
# of electric field linesE
Area
Electric Field Lines, General
• The density of lines through surface A is greater than through surface B
• The magnitude of the electric field is greater on surface A than B
• The lines at different locations point in different directions– This indicates the field is
non-uniform
Example Field Lines
+ + + + +
Dipole Line Charge
Q dq
dx
Linear Charge Density:
For a continuous linear charge distribution,
Active Figure 23.24
(SLIDESHOW MODE ONLY)
More Field Lines
Q dq
A dA Surface Charge Density:
Q dq
V dV Volume Charge Density:
Superposition of Fields
0 01 02 03E E E E ....
31 20 10 20 302 2 2
10 20 30
kqkq kqˆ ˆ ˆE r r r ....
r r r
N31 2 i
0 10 20 30 i02 2 2 2i 110 20 30 i0
qq q qˆ ˆ ˆ ˆE k r r r .... k r
r r r r
+q3
+q2
+q1
01E
03E
02E
10r
20r
30r0
Superposition Example
• Find the electric field due to q1, E1
• Find the electric field due to q2, E2
• E = E1 + E2
– Remember, the fields add as vectors
– The direction of the individual fields is the direction of the force on a positive test charge
Electric Field of a Dipole (ex. 23.6)
p
2a-q +q
E
y
E
E E E
22
kqE E
y a
x xE E E 2E cos
32 2 2 22 2 2
kq a kpE 2
y a y a y a
y 2a3
kpE
yp 2aQ
P23.19
Three point charges are arranged as shown in Figure P23.19.
(a) Find the vector electric field that the 6.00-nC and –3.00-nC charges together create at the origin.
(b) (b) Find the vector force on the 5.00-nC charge.
Figure P23.19
P23.52
Three point charges are aligned along the x axis as shown in Figure P23.52. Find the electric field at
(a) the position (2.00, 0) and
(b) the position (0, 2.00).
Figure P23.52
9 2 2 9
1 2 2
8.99 10 N m C 4.00 10 Cˆˆ
2.50 m
ˆ5.75 N C
ek q
r
E r i
i
9 2 2 9
2 2 2
8.99 10 N m C 5.00 10 Cˆ ˆˆ 11.2 N C
2.00 mek q
r
E r i i
9 2 2 9
3 2
8.99 10 N m C 3.00 10 Cˆ ˆ18.7 N C
1.20 m
E i i
1 2 3 24.2 N CR E E E E
FIG. P23.52(a)
in +x direction.
1 2ˆ ˆˆ 8.46 N C 0.243 0.970ek q
r E r i j
2 2
3 2
1 3 1 2 3
ˆˆ 11.2 N C
ˆ ˆˆ 5.81 N C 0.371 +0.928
ˆ ˆ4.21 N C 8.43 N C
e
e
x x x y y y y
k q
rk q
r
E E E E E E E
E r j
E r i j
i j
9.42 N CRE 63.4 above axisx
P23.19
9 91 3
1 2 21
8.99 10 3.00 10ˆ ˆ ˆ2.70 10 N C
0.100
ek q
r
E j j j
9 92 2
2 2 22
2 32 1
8.99 10 6.00 10ˆ ˆ ˆ5.99 10 N C
0.300
ˆ ˆ5.99 10 N C 2.70 10 N C
ek q
r
E i i i
E E E i j
9 ˆ ˆ5.00 10 C 599 2700 N Cq F E i j
6 6ˆ ˆ ˆ ˆ3.00 10 13.5 10 N 3.00 13.5 N F i j i j
(a)
(b)
Continuous Charge Distributions
Ni
0 i02i 1 i0
qˆE k r
r
0 2all charge
dqˆE k r
r
Discrete charges Continuous charge distribution
0 2
kqˆE r
r
Single charge
0 2
kdqˆdE r
r
Single piece of a charge distribution
+Q3
+Q2
+Q1
01E
03E
02E
10r
20r
30r 0 0
++
++
0dE
dq
Finding dq
Q dq
dx
Line charge dq dx dq Rd
Cartesian Polar
Surface chargeQ dq
A dA dq dxdy dq rdrd
Volume chargeQ dq
V dV dq dxdydz dq rdrd dz
2dq r sin drd d
2
kdqˆdE r
r
Example – Infinitely Long Line of Charge
+
+
+
+
dE
+
+
+
dy
x
dq dy
y
2 2 2r x y
2
kdqˆdE r
r
xdE
ydE
+
y-components cancel by symmetry
x 2
kdqdE cos
r
2 2 2 2
k dy xdE
x y x y
3 2
2 2 2
dy 2 2kE k x k x
x xx y
Example – Charged Ring (ex 23.8)
dE
x
dq ds ad
2 2 2r x a
xdE
ydE
a
d
+
+
+
+ +
+
+
2
kdqˆdE r
r
y-components cancel by symmetry
x 2
kdqdE cos
r
2 2 2 2
k ad xdE
x a x a
2
3 3 32 2 2 2 2 202 2 2
k xa k xa kQxE d 2
x a x a x a
Check a Limiting Case
3
2 2 2
kQxE
x R
When: x R
The charged ring must look like a point source.
3 3 3 2
2 2 22 22
2
kQx kQx kQx kQE
x xx R R
x 1x
0
Uniformly Charged Disk (ex. 23.9)
dE
x
3
2 2 2
kQxE
x r
r
3
2 2 2
kxdqdE
x r
dq dA rdrd 2 rdr
3
2 2 2
kx 2 rdrdE
x r
R
2 2
2
R R x R
3 3 32 2 2 20 02 2 2x
kx 2 rdr 2rdr duE kx kx
x r x r u
2 2
2 2
2
2
x R1
x R 3 22
2 2 2 2 2x
x
u 1 1 xkx u du kx 2kx k 2 1
1 x R x x R2
Binomial Expansion Theorem
n 2n n 11 x 1 nx x ...
2!
x 1 Quadratic terms and higher are small
n1 x 1 nx
Two Important Limiting Cases
2 2o o
x 1E k 2 1 k 2 2
4 2x R
Large Charged Plate: R x
dE
x
rR
Very Far From the Charged Plate: x R
12 2
22 2 2
2
x x RE k 2 1 k 2 1 k 2 1 1
xx R Rx 1
x
2 2 2
2 2 2 2
1 R 1 R k R kQk 2 1 1 k 2
2 x 2 x x x
Parallel Plate Capacitor
-Q
+Q
o
E2
o
E
E 0
E 0
Motion of Charged Particles in a Uniform Electric Field
-e
-Q +Q
q 0
FE lim
q
F qE ma
qEa
m
2 20 0v v 2a x x
x x
e Ev 2a x 2 x
m
x
Example
• A proton accelerates from rest in a uniform electric field of 500 N/C. At some time later, its speed is 2.50 x 106 m/s.– Find the acceleration of the
proton.– How long does it take for the
proton to reach this speed?– How far has it moved in this time?– What is the kinetic energy?
e
-Q +Q
x
Motion of Charged Particles in a Uniform Electric Field
-e
+Q
-Q
q 0
FE lim
q
F qE ma
y
e Ea
m
x x0 x x0v v a t v
y y0 y
e Ev v a t t
m
x0v
2
2 2 2x y x0
e E tv v v v
m
v
1
x0
e E ttan
mv
Active Figure 23.26
(SLIDESHOW MODE ONLY)
Motion of Charged Particles in a Uniform Electric Field
-e-e
-Q +Q
+Q
-Qx
Phosphor Screen
This device is known as a cathode ray tube (CRT)
Summary
Coulomb’s Law1 2
12 21221
kq qˆF r
r
oq 0o
FE lim
q
Electric Field
Point Charges:
These can be used to find the fields in the vicinity of continuous charge distributions:
2kE
x
Line of Charge:
o
E2
Charged Plate:
32 2 2
kpE
r a
Dipole:p
+
++++
dE
x
rR