Download Unit 14 - HKU Physics

Unit 14
Electric Forces, Fields and Circuits
14.1
Electric charge
14.2
Coulomb’s law
14.3
Shell theorems for electrostatics
14.4
Electric field
14.5
Electric field lines
14.6
Shielding and charging by induction
14.7
Electric Circuits
14.1 Electric charge
There are two kinds of charges, namely, positive (+) charge and negative (−) charge.
•
Like charges repel
+
•
+
−
−
Unlike charges attract
−
+
•
Objects with zero net change are said to be electrically neutral.
•
Electric charges are generated after rubbing between materials.
Example
After rubbing plastic rod (or amber rod) with fur, the plastic rod (or amber
rod) becomes negatively charged and the fur is positively charged. After
rubbing glass rod with silk, the glass rod becomes positively charged and
the silk is negatively charged.
1
A familiar example of an electrically neutral object is the atom. Atoms
have a small, dense nucleus with a positive charge surrounded by a
cloud of negatively charged electrons. All electrons have exactly the
same electric charge. This charge is very small, and is defined to have a
magnitude, e = 1.60 × 10−19 C. S.I. unit of charge is coulomb, C. Clearly,
the charge on an electron, which is negative, is −e. This is one of the
defining, or intrinsic, properties of the electron. Another intrinsic
property of the electron is its mass, m e :
m e = 9.11 × 10−31 kg
In contrast, the charge on a proton – one of the main constituents of nuclei – is exactly +e. As
a result, since atoms have equal numbers of electrons and protons, their net charge is
precisely zero. The mass of the proton is
m p = 1.673 × 10−27 kg.
Note that this is about 2000 times larger than the mass of the electron. The other main
constituent of the nucleus is the neutron, which, as its name implies, has zero charge. Its mass
is slightly larger than that of the proton:
m n = 1.675 × 10−27 kg.
Example
How is it that rubbing a piece of amber with fur gives the amber a charge?
Answer
Rubbing the fur across the amber simply results in a transfer of charge
from the fur to the amber – with the total amount of charge remaining
unchanged. Before charging, the fur and the amber are both neutral.
During the rubbing process some electrons are transferred from the fur to
the amber, giving the amber a net negative charge, and leaving the fur
with a net positive charge. At no time during this process is charge ever
created or destroyed. This, in fact, is an example of one of the
fundamental conservation laws of physics: Conservation of electric
charge. When charge is transferred from one object to another it is
generally due to the movement of electrons. In a typical solid, the nuclei
of the atoms are fixed in position. The outer electrons of these atoms,
however, are often weakly bound and fairly easily separated. The atom
2
that loses an electron is a positive ion, and the atom that receives an extra electron becomes a
negative ions. This is charging by separation.
Remark: A neutral glass rod carries positive charges after rubbing with a piece of silk.
Example
Find the amount of positive electric charge in one mole of helium atoms.
Answer
Note that the nucleus of a helium atom consists of two protons and two neutrons. The total
positive charge in a mole is
N A (2e) =
(6.02 × 1023 )(2)(1.60 × 10−19 C ) =
1.93 × 105 C .
14.1.1 Polarization
We know that charges of opposite sign attract, but it is also possible
for a charged rod to attract small objects that have zero net charge.
The mechanism responsible for this attraction is called polarization.
When a charged rod is far from a neutral object the atoms in the
object are undistorted. As the rod is brought closer, however, the
atoms distort, producing an excess of one type of charge on the
surface of the object (in this case a negative charge). This induced charge is referred to as a
polarization charge. Since the sign of the polarization charge is the opposite of the sign of the
charge on the rod, there is an attractive force between the rod and the object.
14.1.2 Conductor and insulator
Insulators: materials in which charges are not free to move, e.g. nonmetallic
substances, say, amber.
Conductors: materials that allow electric charges to move more or less freely, e.g.
metals.
On a microscopic level, the difference between conductors and insulators is that
the atoms in conductors allow one or more of their outermost electrons to become detached.
These detached electrons, often referred to as “conduction electrons,” can move freely
throughout the conductor. The figures show the charging of a conductor by touching it with a
charged rod. The transferred charge quickly spreads out the entire surface of the sphere.
3
14.2 Coulomb’s law
Electric force – Coulomb’s law
Fe = k
q1 q 2
, where k (an electrostatic constant > 0) is a constant.
r2
The electrostatic constant k =
1
4πε 0
= 8.99 × 10 9 N ⋅ m 2 / C 2 , where ε 0 is called permittivity
constant of free space, and ε 0 =
8.85 × 10−12 C 2 N −1 m −2 =
8.85 × 10−12 Fm −1 , where Farad is a SI
unit and will be discussed in the chapter about capacitor. ( 1 Farad
= 1=
F 1C 2 N −1 m −1 ).
Gravitational force – Newton’s gravitational law
Fg = −G
m1 m2
, where G = 6.67 × 10 −11 N ⋅ m 2 / kg 2 .
2
r
The negative sign is inserted to represent an attractive force.
Remarks:
1.
Fundamental laws cannot be derived!
Coulombs law,
 Newtons law,




etc.
2.
are concluded according to results in experiments and have
survived in every experimental test.
Objects are considered as point particle or point charge, if d1 , d 2 << r .
r
3.
+
−
d1
d2
Newton’s gravitational law is concluded for point particles. Similarly, Coulomb’s law
is also for the point charge.
4.
The magnitude of the force of interaction between two point
changes is directly proportional to the product of the charges
and inversely proportional to the square of the distance

F1
+
q1

F2
+
q2
between them.
F1 = − F2
Obey Newton’s third law
4
5.
Unit: International System of Units or Metric System (SI)
Charge q: measured in Coulomb or C.
Remark:
Electrostatic constant is related to the speed of light c:
1
c=
ε 0 µ0
,
where c = 2.998 × 10 8 m / s , and µ 0 = 4π × 10-7 N/A2, the permeability of free space.
Example
Compare the electric and gravitational forces between a proton and an electron in a hydrogen
atom.
Answer
Taking the distance between the two particles to be the radius of hydrogen, r = 5.29 × 10 −11 m ,
we find that the electric force has a magnitude
Fe =k
qe q p
r2
=(8.99 × 109 N ⋅ m 2 / C 2 )
(1.60 × 10−19 C )(1.60 × 10−19 C )
=8.22 × 10−8 N .
−11
2
(5.29 × 10 m)
Similarly, the magnitude of the gravitational force between the electron and the proton is
Fg =
G
me m p
r2
=
(6.67 × 10−11 N ⋅ m 2 / kg 2 )
(9.11 × 10−31 kg )(1.673 × 10−27 kg )
=
3.63 × 10−47 N .
−11
2
(5.29 × 10 m)
Hence, we obtain the ratio of the two forces
Fe
8.22 × 10−8 N
=
= 2.26 × 1039 = 2, 260, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000
Fg 3.63 × 10−47 N
y
Point charge
14.2.1 Superposition of Coulomb’s force
q1
The force exerted on charge 1 by charge 2:

F12 =
1
q1 q 2
4πε 0 r12 2

r1
rˆ12

where r̂12 : unit vector along r12 .

 
r12 = r1 − r2

r2
O
q2
x
5
sign → repulsion
 same
q1 and q 2 
opposite sign → attraction
Similarly, The force exerted on charge 2 by charge 1:

F21 =
q1 q 2
1
4πε 0 r21 2
rˆ21

where r̂21 : unit vector along r21 .
The total force acting on charge q due to coulomb’s forces F 1 , F 2 and F 3 .
F = F1 + F2 + F3
(Principle of superposition)
The direction of forces shown in the right figure, representing that the
charge q is of opposite charge of q 1 , q 2 and q 3 .
q2

F2
q1
q3

F3

F1
q
Example
Three charges, each equal to +2.90 µC, are placed at three corners of a square 0.500 m on a
side. Find the magnitude and direction of the net force on charge number 3.
Answer
y

F

F31
2
θ
r=0.500 m
3
2r

F32
r=0.500 m
x
1
The magnitude of force exerted on charge 3 by charge 1:
(2.90 × 10−6 C ) 2
F31 = k
= (8.99 × 10 N ⋅ m / C )
= 0.151 N
2(0.500 m) 2
( 2r ) 2
q2
9
2
2
The magnitude of force exerted on charge 3 by charge 2:
−6
2
q2
9
2
2 (2.90 × 10 C )
=
(8.99
×
10
N
⋅
m
/
C
)
=0.302 N
r2
(0.500 m) 2


The components of F31 and F32 :
F32 =k
o
=
F31, x F=
=
N )(0.707) 0.107 N
(0.151
31 cos 45.0
6
o
=
F31, y F=
(0.151=
N )(0.707) 0.107 N
31 sin 45.0
o
=
=
F32, x F=
N )(1) 0.151 N
(0.302
32 cos 0
o
=
F32, y F=
(0.151
=
N )(0) 0 N
32 sin 0
The components of the resultant force:
Fx = F31, x + F32, x =0.107 N + 0.302 N = 0.409 N
Fy = F31, y + F32, y = 0.107 N + 0 N = 0.107 N
The resultant force acting on charge 3:
F=
Fx 2 + Fy 2 = 0.423 N
The direction of the resultant force on charge 3:
 F3, y 
−1
o
=
θ tan
=

 14.7 .
 F3, x 
14.3 Shell theorems for electrostatics
Theorem 1:
A uniform spherical shell of charge behaves, for external points, as if all its charge were
concentrated at its center.
F=
q1
q1 q
4πε 0 r 2
1
Total charge q
on spherical shell
Theorem 2:
A uniform spherical shell of charge exerts no force on a charged particle placed inside the
shell.
F=0
Remarks: The theorems are similar to the gravitational case.
7
14.4 Electric field
Gravitational field (a vector field)
Test body
 − G M e m rˆ
M
 F
r2
g= =
= −G 2e rˆ
m
m
r
m
Gravitational field: Gravitational force per unit mass

• Electric field E

 Fe
1 q
rˆ
E≡
=
q 0 4πε 0 r 2
Earth
q0
Electric field: Electrostatic force per unit charge
SI unit of electric field: Newton/Coulomb or N/C
Charged
particle q

Fe
A test charge
with positive
charge
Remark:
1.
Why do we need to introduce the concept of Electric field?
Introducing the field as an intermediary between the charges, we can represent the interaction
as:
charge
field
charge
Our problem of determining the interaction between the charges is therefore reduced to two
separate problems: (1) determine, by measuring or calculation, the electric field established
by the first charge at every point in space, and (2) calculate the force that the field exerts on
the second charge placed at a particular point in space.
2.
Principle of superposition in electric field:
The resultant electric field E at a point is given by E = E1 + E2 + E3 , where E 1 , E 2 , and E 3
are the electric fields experienced at that point due to charge 1, 2 and 3 respectively.
14.4.1 Discrete and continuous charge distribution
a)
Discrete case (the total electric field E at a point is contributed by individual charges
q 1 , q 2 , q 3 , …, etc.)
E = E1 + E2 + E3 + ⋅⋅⋅⋅⋅⋅
8
b)
Continuous case
i)
When charge is uniformly distributed along a line.
l
Total
Linear charge density
λ≡
q
l
charge per unit length
(λ =
q
dq
)
dl
Total charge q
ii)
Charge on a surface (uniformly distributed)
Surface charge density
q
σ≡
S
iii)
Surface area S
charge per unit area
dq
(σ =
)
dS
Total charge q
On a volume
Volume charge density
ρ≡
q
V
charge per unit volume
ρ=
dq
dV
Volume V
Example (Challenging)
Find the electric field at a point P, which is at the top of the center of a charged ring. The
total charge on the ring is q.

dE '

E
P
θ
r
ds
R


dE ' θ θ dE '
z
Total charge q
P
Answer
Due to the symmetry of the ring, the electric field along ẑ direction can be calculated as
follows.
E = ∫ dE' cos θ , where cos θ =
z
λ ds
1 dq
1
and dE ' =
.
=
2
2
r
4πε 0 r
4πε 0 z + R 2
The linear charge density of the ring
=
λ
q
dq
, where R is the radius of ring.
=
2π R ds
9
E=
=
λz
1
4πε 0 ( z + R 2 )3 / 2
2
∫ ds , where ∫ ds = 2π R .
z (2π Rλ )
4πε 0 ( z 2 + R 2 )3 / 2
1
Plug in the expression, 2π Rλ = q , hence, the electric field at any point P, a perpendicular
distance z from the plane and center of ring is E =
q
z
.
4πε 0 ( z + R 2 )3 / 2
2
Remark:
When z >> R, that is the distance is much larger than the dimension of the ring,
1
1
1
≈ 2 3/ 2 =
2 3/ 2
(z + R )
(z )
z3
2
E≈
q
4πε 0 z 2
This is as if the case where R → 0 , the ring seems to be a point charge!
14.5 Electric field lines
Rules for drawing electric field lines
Electric field lines:
(i)
Point in the direction of electric field vector E at every point;
(ii)
Start at positive (+) charges or at infinity;
(iii)
End at negative (–) charges or at infinity;
(iv)
Are denser where E has a greater magnitude. In particular, the number of lines
entering or leaving a charge is proportional to the magnitude of the charge.
10
Example
Which of the following statements is correct: Electric field lines (a) can or (b) cannot
intersect?
Answer
By definition, electric field lines are always tangent to the electric field. Since the electric
force, and hence the electric field, can point in only one direction at any given location, it
follows that field lines cannot intersect. If they did, the field at the intersection point would
have two conflicting directions.
14.6 Shielding and charging by induction
In a perfect conductor there are enormous numbers f electrons
completely free to move about. This simple fact has some rather
interesting consequences. Consider, for example, a solid metal sphere
attached to an insulating base as shown in figure. Suppose a positive
charge Q is placed on the sphere. The question is: How does this charge
distribute itself on the sphere when it is in equilibrium? In particular, does the charge spread
itself uniformly throughout the volume of the sphere, or does it concentrate on the surface?
The answer is that the charge concentrates on the surface. Why should this be the case? First,
assume the opposite – that the charge is spread uniformly throughout the sphere’s volume. If
this were the case, a charge at location A would experience an outward force due to the
spherical distribution of charge between it and the center of the sphere. Since charges are free
to move, the charge at A would respond to this force by moving toward the surface. Clearly,
then, a uniform distribution of charge within the sphere’s volume is not in equilibrium. In fact,
the argument that a charge at point A will move toward the surface can be applied to any
charge within the sphere. The preceding result holds no matter what the shape of the
conductor. In general, excess charge placed on a conductor, whether positive or negative,
moves to the exterior surface of the conductor.
11
14.6.1 Shielding
When electric charges are in equilibrium, the electric field within a conductor is zero; E = 0.
A straightforward extension of this idea explains the phenomenon of shielding, in which a
conductor “shields” its interior from external electric fields.
We also noted that the electric field lines contact conductor
surfaces at right angles. If an electric field contacted a conducting
surface at an angle other than 90o, the result would be a
component of force parallel to the surface. This would result in a
movement of electrons and hence would not correspond to
equilibrium.
14.6.2 Charging by induction
One way to charge an object is to touch it with a charged rod; but since
electric forces can act at a distance, it is also possible to charge an object
without making direct physical contact. This type of charging is referred to as
charging by induction.
12
14.7 Electric Circuits
14.7.1 DC circuits
Here is a review to the basic knowledge about DC circuit.
•
When a current I passes through a resistor R, then the potential difference across it is
V and it is related by the equation: V = IR and =
P VI
= I 2=
R V2 /R,
where P is the power consumed by the resistor.
•
When n resistors are in series, the equivalent resistor R = R1 + R2 +  Rn .
•
When n resistors are in parallel, the equivalent resistor
1
1
1
1
= +
+ .
R R1 R2
Rn
14.7.2 Kirchhoff’s laws
Kirchhoff’s voltage rule (loop rule): The algebraic sum of the changes in potential
encountered in a complete traversal of any loop of a circuit must be zero.
Kirchhoff’s current rule (junction rule): The sum of the currents entering any junction must
be equal to the sum of the currents leaving that junction.
Example
The current rule for junction d, we can write i1 + i3 =.
i2
Consider the left-hand loop in a counterclockwise
direction from point b:
E1
E1 − i1 R1 + i3 R3 =
0.
Again, the right-hand loop in a counterclockwise
direction from point b gives: −i3 R3 − i2 R2 − E2 =
0.
The three equations are solvable to give i 1 , i 2 and i 3 .
a
+
i1
R1
E2
b
−
i3
−
R3
c
+
R2
d
13
i2