Download 1 slide per page() - Wayne State University Physics and Astronomy

General Physics (PHY 2140)
Lecture 2
¾ Electrostatics
9 Electric flux and Gauss’s law
9 Electrical energy
9 potential difference and
electric potential
9 potential energy of charged
conductors
http://www.physics.wayne.edu/~alan/
Chapters 15-16
5/9/2007
1
Lightning Review
Last lecture:
F = ke
1. Coulomb’s law
9
q1 q2
r2
the superposition principle
2. The electric field
E=
JG
F
q0
Review Problem: A “free” electron and “free” proton are placed in an
identical electric field. Compare the electric force on each particle.
Compare their accelerations.
5/9/2007
2
Review Solution:
Recall
F = qE
For a proton or an electron the size of the force is the same!
„
„
„
Charges are the same in magnitude.
Opposite in sign.
The direction of the electric forces are opposite.
However the accelerations are different!
„
„
a=F m
Mass of electron is 9.11x10-31 Kg
Mass of a proton is 1.67x10-27 Kg
So, the acceleration of the proton is smaller by me/mp = 5.5x10-4
5/9/2007
3
15.5 Electric Field Lines
A convenient way to visualize field patterns is to draw
lines in the direction of the electric field.
Such lines are called field lines.
Remarks:
1.
2.
Electric field vector, E, is tangent to the electric field lines at
each point in space.
The number of lines per unit area through a surface
perpendicular to the lines is proportional to the strength of the
electric field in a given region.
E is large when the field lines are close together and small
when far apart.
5/9/2007
4
15.5 Electric Field Lines (2)
Electric field lines of single positive (a) and (b) negative
charges.
a)
b)
+ q
5/9/2007
- q
5
15.5 Electric Field Lines (3)
Rules for drawing electric field lines for any charge
distribution.
1.
2.
3.
5/9/2007
Lines must begin on positive charges (or at infinity) and must
terminate on negative charges or in the case of excess charge
at infinity.
The number of lines drawn leaving a positive charge or
approaching a negative charge is proportional to the magnitude
of the charge.
No two field lines can cross each other.
6
15.5 Electric Field Lines (4)
Electric field lines of a dipole.
+
5/9/2007
-
7
15.6 Conductors in Electrostatic Equilibrium
Good conductors (e.g. copper, gold) contain charges
(electron) that are not bound to a particular atom, and
are free to move within the material.
When no net motion of these electrons occur the
conductor is said to be in electro-static equilibrium.
5/9/2007
8
15.6 Conductors in Electrostatic Equilibrium
Properties of an isolated conductor (insulated from the
ground).
1.
2.
3.
4.
5/9/2007
Electric field is zero everywhere within the conductor.
Any excess charge on an isolated conductor resides entirely on
its surface.
The electric field just outside a charged conductor is
perpendicular to the conductor’s surface.
On an irregular shaped conductor, the charge tends to
accumulate at locations where the radius of curvature of the
surface is smallest – at sharp points.
9
Faraday’s ice-pail experiment
+++++
+
+
+
+
+
-
-
-
+
+++++
-
+
+
+
+
+
+
+
+
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Demonstrates that the charge resides on the surface of a conductor.
5/9/2007
10
Mini-quiz
Question:
Suppose a point charge +Q is in empty space. Wearing rubber gloves,
we sneak up and surround the charge with a spherical conducting shell.
What effect does this have on the field lines of the charge?
?
+ q
5/9/2007
+
11
Question:
Suppose a point charge +Q is in empty space. Wearing rubber gloves, we sneak up and surround the
charge with a spherical conducting shell. What effect does this have on the field lines of the
charge?
Answer:
Negative charge will build up on the inside of the shell.
Positive charge will build up on the outside of the shell.
There will be no field lines inside the conductor but the field lines will remain outside the shell.
+
+
+
+
+
-
-
+ q
-
+
-
-
-
+
-
-
+
+
5/9/2007
+
-
-
+
+
12
15.9 The oscilloscope
Changing E field applied on the deflection plate (electrodes)
moves the electron beam.
V2
θ
d
L
5/9/2007
V1
13
Oscilloscope: deflection angle (additional)
V2
θ
d
L
V1
vx =
electron gun
between plates
vy = a yt
vy =
eV 2
eE
=
me
me d
L = vxt
eV 2 L
me d v x
tan θ =
5/9/2007
ay =
2 eV1
me
vy
vx
=
eV 2 L
eV 2 Lm e
V2 L
=
=
m e d v x 2 m e d 2 eV1 d 2 eV
14
15.9 Electric Flux and Gauss’s Law
A convenient technique was introduced by Karl F. Gauss
(1777-1855) to calculate electric fields.
Requires symmetric charge distributions.
Technique based on the notion of electrical flux.
5/9/2007
15
15.9 Electric Flux
To introduce the notion of flux, consider
a situation where the electric field is
uniform in magnitude and direction.
Consider also that the field lines cross a
surface of area A which is
perpendicular to the field.
The number of field lines per unit of
area is constant.
The flux, Φ, is defined as the product of
the field magnitude by the area crossed
by the field lines.
Φ = EA
5/9/2007
Area=A
E
16
15.9 Electric Flux
Units: Nm2/C in SI units.
Find the electric flux through the area A = 2 m2, which is
perpendicular to an electric field E=22 N/C
Φ = EA
Answer: Φ = 44 Nm2/C.
5/9/2007
17
15.9 Electric Flux
If the surface is not perpendicular to the field, the
expression of the field becomes:
Φ = EA cos θ
Where θ is the angle between the field and a normal to
the surface.
N
θ
θ
5/9/2007
18
15.9 Electric Flux
Remark:
When an area is constructed such that a closed surface
is formed, we shall adopt the convention that the flux
lines passing into the interior of the volume are negative
and those passing out of the interior of the volume are
positive.
5/9/2007
19
Example:
Question:
Calculate the flux of a constant E field (along x) through a
cube of side “L”.
y
1
2
E
x
z
5/9/2007
20
Question:
Calculate the flux of a constant E field (along x) through a cube of side “L”.
Reasoning:
z Dealing with a composite, closed surface.
z Sum of the fluxes through all surfaces.
z Flux of field going in is negative
z Flux of field going out is positive.
z E is parallel to all surfaces except surfaces labeled 1 and 2.
z So only those surfaces (1 & 2) contribute to the flux.
y
1
2
E
x
z
5/9/2007
21
Question:
Calculate the flux of a constant E field (along x) through a cube of side “L”.
Reasoning:
z Dealing with a composite, closed surface.
z Sum of the fluxes through all surfaces.
z Flux of field going in is negative
z Flux of field going out is positive.
z E is parallel to all surfaces except surfaces labeled 1 and 2.
z So only those surface contribute to the flux.
Solution:
Φ1 = − EA1 cos θ1 = − EL2
Φ 2 = EA2 cos θ 2 = EL2
Φ net = − EL2 + EL2 = 0
y
1
2
E
x
z
5/9/2007
22
15.9 Gauss’s Law
The net flux passing through a closed surface
surrounding a charge Q is proportional to the magnitude
of Q:
Φ net = ∑ EA cos θ ∝ Q
In free space, the constant of proportionality is 1/εo
where εo is called the permittivity of of free space.
1
1
−12
2
2
8.85
10
εo =
=
=
×
C
N
⋅
m
4π ke 4π 8.99 × 109 Nm 2 / C 2
(
5/9/2007
)
23
15.9 Gauss’s Law
The net flux passing through any closed surface is equal
to the net charge inside the surface divided by εo.
Φ net = ∑ EA cos θ =
Q
εo
Can be used to compute electric fields. Example: point
charge
Q
Q
2
Φ net = ∑ EA cos θ = 4π r E → E =
= ke 2
2
r
4πε 0 r
5/9/2007
24
16.0 Introduction
The Coulomb force is a conservative force
A potential energy function can be defined for any
conservative force, including Coulomb force
The notions of potential and potential energy are
important for practical problem solving
5/9/2007
25
16.1 Potential difference and electric potential
The electrostatic force is
conservative
As in mechanics, work is
JG
E
B
A
d
W = Fd cos ϑ
Work done on the positive charge
by moving it from A to B
W = Fd cos ϑ = qEd
5/9/2007
26
Potential energy of electrostatic field
The work done by a conservative force equals the
negative of the change in potential energy, ΔPE
ΔPE = −W = −qEd
This equation
„
„
5/9/2007
is valid only for the case of a uniform electric field
allows us to introduce the concept of the electric potential
27
Electric potential
The potential difference between points A and B, VB-VA,
is defined as the change in potential energy (final minus
initial value) of a charge, q, moved from A to B, divided
by the charge
ΔPE
ΔV = VB − VA =
q
Electric potential is a scalar quantity
Electric potential difference is a measure of electric
energy per unit charge
Potential is often referred to as “voltage”
5/9/2007
28
Electric potential - units
Electric potential difference is the work done to move a
charge from a point A to a point B divided by the
magnitude of the charge. Thus the SI units of electric
potential
1V = 1 J C
In other words, 1 J of work is required to move a 1 C of
charge between two points that are at potential
difference of 1 V
5/9/2007
29
Electric potential - notes
Units of electric field (N/C) can be expressed in terms of the
units of potential (as volts per meter)
1 N C = 1V m
Because the positive tends to move in the direction of the
electric field, work must be done on the charge to move it in
the direction, opposite the field. Thus,
„
„
A positive charge gains electric potential energy when it is moved in
a direction opposite the electric field
A negative charge looses electrical potential energy when it moves
in the direction opposite the electric field
5/9/2007
30
Analogy between electric and gravitational fields
The same kinetic-potential energy theorem works here
A
JG
E
A
d
q
B
JG
g
d
m
B
If a positive charge is released from A, it accelerates in
the direction of electric field, i.e. gains kinetic energy
If a negative charge is released from A, it accelerates in
the direction opposite the electric field
KEi + PEi = KE f + PE f
5/9/2007
31
Example: motion of an electron
What is the speed of an electron accelerated from rest across a
potential difference of 100V? What is the speed of a proton
accelerated under the same conditions?
Given:
ΔV=100 V
me = 9.11×10-31 kg
mp = 1.67×10-27 kg
|e| = 1.60×10-19 C
Vab
Observations:
1. given potential energy
difference, one can find the
kinetic energy difference
2. kinetic energy is related to
speed
KEi + PEi = KE f + PE f
KE f − KEi = KE f = ΔPE = qΔV
Find:
ve=?
vp=?
5/9/2007
1 2
2 qΔV
mv f = qΔV → v f =
2
m
ve = 5.9 ×106 m , v p = 1.3 ×105 m
s
s
32
16.2 Electric potential and potential energy
due to point charges
Electric circuits: point of zero potential is defined by
grounding some point in the circuit
Electric potential due to a point charge at a point in
space: point of zero potential is taken at an infinite
distance from the charge
With this choice, a potential can be found as
q
V = ke
r
Note: the potential depends only on charge of an object,
q, and a distance from this object to a point in space, r.
5/9/2007
33
Superposition principle for potentials
If more than one point charge is present, their electric
potential can be found by applying superposition
principle
The total electric potential at some point P due to several
point charges is the algebraic sum of the electric
potentials due to the individual charges.
Remember that potentials are scalar quantities!
5/9/2007
34
Potential energy of a system of point
charges
Consider a system of two particles
If V1 is the electric potential due to charge q1 at a point P,
then work required to bring the charge q2 from infinity to P
without acceleration is q2V1. If a distance between P and
q1 is r, then by definition
q2
P
q1
r
A
q1q2
PE = q2V1 = ke
r
Potential energy is positive if charges are of the same
sign and vice versa.
5/9/2007
35
Mini-quiz: potential energy of an ion
Three ions, Na+, Na+, and Cl-, located such, that they
form corners of an equilateral triangle of side 2 nm in
water. What is the electric potential energy of one of the
Na+ ions?
Cl?
qNa qCl
qNa qNa
qNa
PE = ke
+ ke
= ke
[ qCl + qNa ]
r
r
r
but : qCl = −qNa !
Na+
5/9/2007
Na+
qNa
PE = ke
[ −qNa + qNa ] = 0
r
36
Recall from last chapter:
Electric field lines of a dipole.
+
5/9/2007
-
37
16.3 Potentials and charged conductors
Recall that work is opposite of the change in potential
energy,
W = − PE = −q [VB − VA ]
No work is required to move a charge between two points
that are at the same potential. That is, W=0 if VB=VA
Recall:
1. all charge of the charged conductor is located on its surface
2. electric field, E, is always perpendicular to its surface, i.e. no work is
done if charges are moved along the surface
Thus: potential is constant everywhere on the surface of a
charged conductor in equilibrium
… but that’s not all!
5/9/2007
38
Because the electric field in zero inside the conductor,
no work is required to move charges between any two
points, i.e.
W = −q [VB − VA ] = 0
If work is zero, any two points inside the conductor have
the same potential, i.e. potential is constant everywhere
inside a conductor
Finally, since one of the points can be arbitrarily close to
the surface of the conductor, the electric potential is
constant everywhere inside a conductor and equal to its
value at the surface!
Note that the potential inside a conductor is not necessarily zero,
even though the interior electric field is always zero!
5/9/2007
39
The electron volt
A unit of energy commonly used in atomic, nuclear and
particle physics is electron volt (eV)
The electron volt is defined as the energy that electron
(or proton) gains when accelerating through a potential
difference of 1 V
Relation to SI:
Vab=1 V
1 eV = 1.60×10-19 C·V = 1.60×10-19 J
5/9/2007
40
Problem-solving strategy
Remember that potential is a scalar quantity
Superposition principle is an algebraic sum of potentials due
to a system of charges
Signs are important
Just in mechanics, only changes in electric potential are
significant, hence, the point you choose for zero electric
potential is arbitrary.
5/9/2007
41
Example : ionization energy of the electron
in a hydrogen atom
In the Bohr model of a hydrogen atom, the electron, if it is in the
ground state, orbits the proton at a distance of r = 5.29×10-11 m. Find
the ionization energy of the atom, i.e. the energy required to remove
the electron from the atom.
Note that the Bohr model, the idea of electrons as tiny balls orbiting the nucleus, is not
a very good model of the atom. A better picture is one in which the electron is spread
out around the nucleus in a cloud of varying density; however, the Bohr model does
give the right answer for the ionization energy
5/9/2007
42
In the Bohr model of a hydrogen atom, the electron, if it is in the ground state,
orbits the proton at a distance of r = 5.29 x 10-11 m. Find the ionization energy,
i.e. the energy required to remove the electron from the atom.
The ionization energy equals to the total energy of the
electron-proton system,
Given:
r = 5.292 x 10-11 m
me = 9.11×10-31 kg
mp = 1.67×10-27 kg
|e| = 1.60×10-19 C
Find:
E=?
E = PE + KE
e2
v2
, KE = m
with PE = − ke
2
r
The velocity of e can be found by analyzing the force
on the electron. This force is the Coulomb force;
because the electron travels in a circular orbit, the
acceleration will be the centripetal acceleration:
mac = Fc
or
v2
e2
m = ke 2 ,
r
r
or
e2
v = ke
,
mr
2
Thus, total energy is
e 2 m ⎛ ke e 2 ⎞
e2
−18
E = − ke + ⎜
k
2.18
10
J ≈ -13.6 eV
=
−
=
−
×
⎟
e
r 2 ⎝ mr ⎠
2r
5/9/2007
43
16.4 Equipotential surfaces
They are defined as a surface in space on which the potential
is the same for every point (surfaces of constant voltage)
The electric field at every point of an equipotential surface is
perpendicular to the surface
convenient to represent by drawing
equipotential lines
5/9/2007
44
5/9/2007
45