Chapter 25
Electric Charges: Conductors and Isolators
¾ Electrical conductors are materials in which
some of the electrons are free electrons
‰ These electrons can move relatively freely
through the material
‰ Examples of good conductors include copper,
aluminum and silver
Electricity and Magnetism
Electric Fields: Coulomb
Coulomb’ss Law
¾ Electrical insulators are materials in which all
of the electrons are bound to atoms
‰ These electrons can not move relatively freely
through the material
‰ Examples of good insulators include glass, rubber
and wood
Reading: Chapter 25
1
¾ Semiconductors are somewhere between
insulators and conductors
Electric Charges
Electric Charges
Electric Charges
Electric Charges
2
1
Electric Charges
Electric Charges
Electric Charges
Electric Charges
Electric Charges
Electric Charges
2
Electric Charges
Electric Charges
Electric Charges
Electric Charges: Conductors and Isolators
¾ There are two kinds of electric charges
¾ Electrical conductors are materials in which
some of the electrons are free electrons
- Called positive and negative
ƒ Negative charges are the type possessed by
electrons
ƒ Positive charges are the type possessed by protons
‰ These electrons can move relatively freely
through the material
‰ Examples of good conductors include copper,
aluminum and silver
¾ Charges of the same sign repel one another and
charges with opposite signs attract one another
¾Electric charge is always conserved in isolated system
¾ Electrical insulators are materials in which all
of the electrons are bound to atoms
‰ These electrons can not move relatively freely
through the material
‰ Examples of good insulators include glass, rubber
and wood
Neutral – equal number of positive
and negative charges
Positively charged
15
Conservation of Charge
¾ Semiconductors are somewhere between
insulators and conductors
16
Coulomb’s Law
Electric charge is always conserved in isolated system
Two identical sphere
• Mathematically, the force between two electric charges:
q1 = 1μC
r
qq r
F12 = ke 1 2 2 rˆ
r
q2 = −2 μC
They are connected by conducting wire. What is the electric
g of each sphere?
p
charge
The same charge q. Then the
conservation of charge means that :
2q = q1 + q2
For three spheres:
q1 = 1μC q2 = −2 μC
q=
q1 + q2 1 − 2
=
μC = −0.5 μC
2
2
q3 = 3 μC
• The SI unit of charge is the coulomb (C)
• ke is called the Coulomb constant
– ke = 8.9875
8 9875 x 109 N.m2/C2 = 1/(4πe
1/(4 o)
– eo is the permittivity of free space
– eo = 8.8542 x 10-12 C2 / N.m2
Electric charge:
‰
electron
e = -1.6 x 10-19 C
‰
proton
e = 1.6 x 10-19 C
3q = q1 + q2 + q3
q=
q1 + q2 + q3 1 − 2 + 3
=
μC = 1μC
3
3
17
18
3
Coulomb’s Law
F12 = F21 = ke
Coulomb’s Law: Superposition Principle
| q1 || q2 |
r2
• The force exerted by q1 on q3 is F13
1
Direction depends on the sign of
the product
r
F21
q1q2
r
r
F21 = −F12
+
r
F21
r
F12
q1q2 > 0
2
1
opposite
pp
directions,,
the same magnitude
• The force exerted by q2 on q3 is F23
2
+
−
−
q1q2 > 0
1
+
r
F21
r
F12
• The resultant force exerted on q3 is the vector
sum of F13 and F23
r
F12
2
−
q1q2 < 0
The force is attractive if the charges are of opposite sign
The force is repulsive if the charges are of like sign
Magnitude:
F12 = F21 = ke
| q1 || q2 |
r2
19
Coulomb’s Law
r
qq r
F21 = ke 1 2 2 rˆ21
r
+
20
Coulomb’s Law
5m
6m
−
2
1
+
r
F23 3
Resultant force:
r
r
r
F3 = F21 + F31
r
F13
q1 = 1μC q2 = −2 μC
Magnitude:
−6
1
r −
F12 2
r
F32
6m
Resultant force:
r
r
r
F2 = F12 + F32
r
r
F2
F12
r
F32
+
3
q3 = 3 μC
Magnitude:
−6
| q || q |
2 ⋅ 10 ⋅ 3 ⋅ 10
F23 = ke 3 2 2 = 8.9875 ⋅ 109
N = 15 ⋅ 10−4 N
r
62
F13 = ke
5m
+
q1 = 1μC q2 = −2 μC
r
r
F13 F3
r
F23
q3 = 3 μC
r
qq r
F21 = ke 1 2 2 rˆ21
r
| q3 || q1 |
10 −6 ⋅ 3 ⋅ 10 −6
= 8.9875 ⋅ 109
N = 2.2 ⋅ 10−4 N
2
r
112
F3 = F23 − F13 = (15 ⋅ 10−4 − 2.2 ⋅ 10 −4 )N = 12.8 ⋅ 10 −4 N
F32 = ke
| q3 || q2 |
2 ⋅ 10 −6 ⋅ 3 ⋅ 10 −6
= 8.9875 ⋅ 109
N = 15 ⋅ 10−4 N
r2
62
F12 = ke
| q1 || q2 |
10 −6 ⋅ 2 ⋅ 10−6
= 8.9875 ⋅ 109
N = 7.2 ⋅ 10 −4 N
2
52
r
F2 = F32 − F12 = (15 ⋅ 10−4 − 7.2 ⋅ 10 −4 )N = 7.8 ⋅ 10 −4 N
21
Coulomb’s Law
r
qq r
F21 = ke 1 2 2 rˆ21
r
r +
F31 1
q1 = 1μC q2 = −2 μC
q3 = 3 μC
Magnitude:
Coulomb’s Law
5m
r
F21
−
2
6m
+
3
Resultant force:
r
r
r
F1 = F21 + F31
r
r
F1
F31
r
F21
| q || q |
10 −6 ⋅ 3 ⋅ 10−6
F31 = ke 3 2 1 = 8.9875 ⋅ 109
N = 2.2 ⋅ 10−4 N
r
112
F21 = ke
22
| q1 || q2 |
10−6 ⋅ 2 ⋅ 10−6
= 8.9875 ⋅ 109
N = 7.2 ⋅ 10 −4 N
52
r2
r
qq r
F21 = ke 1 2 2 rˆ21
r
q1 = 1μC q2 = −2 μC
Resultant force:
r
r
r
F1 = F21 + F31
r
r
F31
F21
r
F1
6m
−
+
q3 = 3 μC
7m
3
2
Magnitude:
F21 = ke
| q1 || q2 |
10 −6 ⋅ 2 ⋅ 10 −6
= 8.9875
8 9875 ⋅ 109
N = 5 ⋅ 10 −4 N
r2
62
F31 = ke
| q3 || q1 |
10 −6 ⋅ 3 ⋅ 10−6
N = 1.1⋅ 10 −3 N
= 8.9875 ⋅ 109
r2
52
r
F1 = F21 − F31 = (7.2 ⋅ 10 −4 − 2.2 ⋅ 10−4 )N = 5 ⋅ 10−4 N
+
5m
r
F13
23
r
F31
+
1 r
F21
5m
+
3
r
F23
1
F13
r
F23
6m
−
7m
2
r
F3
+
5m
1
+
3
7m
6m
r
F32
r
F12
r
F32
r
F1
r
F12
−
2
24
4
Chapter 25
Electric Field
¾ An electric field is said to exist in the region of space around a
charged object
¾ This charged object is the source charge
¾ When another charged object, the test charge, enters this electric
field, an electric force acts on it.
¾The electric field is defined as the electric force on the test charge
per unit charge
r
r F
E=
q0
Electric Field
¾ If you know the electric field you can find the force
r
r
F = qE
25
Electric Field
¾ q is positive, F is directed away
from q
‰ The direction of E is also away
from the positive source charge
¾ q is negative, F is directed toward q
‰ E is also toward the negative
source charge
r
r F
E=
q0
Then
26
Electric Field
¾ The direction of E is that of the force on a positive test charge
¾ The SI units of E are N/C
Coulomb’s Law:
If q is positive, F and E are in the same direction
If q is negative, F and E are in opposite directions
r
qq r
F = ke 20 rˆ
r
r
r F
q r
E=
= ke 2 rˆ
q0
r
r
r F
q r
E=
= ke 2 rˆ
q0
r
r
qq r
F = ke 20 rˆ
r
27
Electric Field: Superposition Principle
28
Electric Field
r
q r
E = ke 2 rˆ
r
• At any point P, the total electric field due to a group of
source charges equals the vector sum of electric
+
1
q1 = 1μC q2 = −2 μC
fields of all the charges
Magnitude:
r
q r
E = ke ∑ 2i rˆi
i ri
E2 = ke
E1 = ke
5m
−
2
6m
r
E2
r
E1
Electric Field:
r r
r
E = E1 + E2
r
r
E1 E
r
E2
| q2 |
2 ⋅ 10−6
= 8.9875 ⋅ 109
N / C = 5 ⋅ 102 N / C
r2
62
| q1 |
10−6
= 8.9875 ⋅ 109 2 N / C = 0.7 ⋅ 102 N / C
11
r2
E = E2 − E1 = (5 ⋅ 102 − 0.7 ⋅ 102 )N / C = 4.3 ⋅ 102 N / C
29
30
5
Electric Field
Electric Field
r
q r
E = ke 2 rˆ
r
r
E2
5m
q1 = 1μC q2 = −2 μC
6m
−
+
7m
1
r
q r
E = ke 2 rˆ
r
Electric field
r r
r
E = E1 + E2
r
r
E1
E2
r
E1
z
q1 = 10 μC q2 = 10 μC
r
E
r
E
2
Direction of electric field?
Magnitude:
E2 = k e
| q2 |
2 ⋅ 10 −6
= 8.9875 ⋅ 109
N / C = 5 ⋅ 102 N / C
62
r2
5m
ϕ
5m
|q |
10 −6
E1 = ke 21 = 8.9875 ⋅ 109 2 N / C = 0.37 ⋅ 102 N / C
r
5
+
+
1
2
6m
31
Electric Field
Example
r
q r
E = ke 2 rˆ
r
r
E
q1 = 10 μC q2 = 10 μC
ϕ
r
E2
5m
r
E2 Electric field
r r
r
E = E1 + E2
r
E1
| q2 |
10 ⋅ 10 −6
N / C = 3.6 ⋅ 103 N / C
= 8.9875 ⋅ 109
2
r
52
cos φ =
52 − 3 2 4
=
5
5
E=
r
F31
+
1 r
F21
r
mg
g
r
g
T cos ϕ 2 = mg
tan ϕ 2 =
8
E1
5
33
T sin ϕ 2 = FE
tan ϕ1 =
FE
qq
= ke 21 2
r mg
mg
ϕ 2 = ϕ1
FE
qq
= ke 21 2
r mg
mg
34
3
r
r
F = qE
Resultant force:
r
r
r
F1 = F21 + F31
r
r
F31
F21
r
F1
6m
−
+
r
q r
E = ke ∑ 2i rˆi
i ri
Electric field
r
r
qq r
F12 = ke 1 2 2 rˆ
r
5m
r
q r
E = ke 2 rˆ
r
r
FE
r
qq r
FE = ke 1 2 2 rˆ
r
Magnitude:
Coulomb’s Law:
2
+
r
r r
T + mg + FE = 0
2
6m
l = 1m
r
T
+
ϕ
+
E = 2E1 cos φ
ϕ1 ϕ 2
1
m1 = m2 = m
+
1
l = 1m
q1 = 10 μC q2 = 10 μC
5m
E2 = E1 = ke
32
7m
2
r r
r
E = E1 + E2
r
r
E1
E
2
r
E
r
E2
5m
r
E1
6m
−
+
1
7m
2
35
6