Transcript Example 17-2

Chapter 17
Electric Potential, Electric Energy and
Capacitance
© 2006
Giancoli, PHYSICS,6/E © 2005. Electronically reproduced by permission
of Pearson Education, Inc., Upper Saddle River, New Jersey
Ch 17
1
Review of Work and Energy
Electric potential is based on the concept of work and
energy.
•Work = (force) (distance) (cos )
•Units are J (joules) 1 J = N m
•Potential Energy is energy of position such as the energy
stored in a stretched spring or a roller coaster at the top of
the first hill.
•You have to do work to move a positive charge in an
electric field to point a, so that charge has electrical
potential energy (PEa).
•If you release the charge it will “fall” away from the other
charge thus gaining kinetic energy KE = ½ m v2
Ch 17
2
•
Electric Potential
Electric potential Va is potential
energy per unit charge
Va 
PEa
q
•Often “potential” is used instead
of electric potential
•Unit is a volt (V) which is a
joule/coulomb
•1V=1J/C
Point b has a higher electric
potential than a and thus a positive
particle at b has more potential
energy than at a.
Ch 17
+
+
+ +
+ b
+
+
+
+
+
+
+
+
_
_
_
_
+ _
a _
_
_
_
_
_
_
_
3
Potential Difference
•The difference in potential energy between two points a
and b is the work done in moving a charge from one
point to the other Wba ( assuming the charge is moved
slowly so there in no change in kinetic energy.)
It is similar with electric potential.
•
Wba
Vab  Va  Vb  
q
Vab  
Ch 17
Wba
q
4
•
Potential Energy Difference
Since electric potential is potential
energy per unit charge, we can
find the change in potential energy
PE  PEb  PEa  qVba
Example 17-1: The two parallel
plates shown are connected to a 9 V
battery. If a charge of +4 C is
located at point b then the potential
energy difference between b and a
is PE = (9V) (4 C)= 36 J
•If the charge were released at
point b it would accelerate towards
a and gain 36 J of kinetic energy if
there was no friction.
Ch 17
+
+
+
+ +
b
+
+
+
+
+
+
+
_
_
_
_
a _
_
_
_
_
_
_
_
5
The Electron Volt
•Many devices accelerate electrons and protons through a
given potential difference
•Because all charges are e or multiples of e, a special unit of
energy has been created called the electron volt
•Definition: An electron volt (eV) is the energy gained or
lost when a particle of charge e moves through a potential
difference of 1 V.
• 1 eV = q V = (1.60x10
-19
C)( 1.0V)
= 1.60x10-19 J
Ch 17
6
Example 17-2: An electron in a TV tube is accelerated
through a potential difference of 25,000 V. What is the
kinetic energy of the electron in electron volts?
Answer: KE = 25,000 eV
What is the kinetic energy in J? We use the conversion factor
KE = 25,000 eV (1.6x10-19J/eV)
= 4.0x10-15 J
Ch 17
7
The Electron Volt
Example 17-3: A 42He nucleus is accelerated from rest
through a potential difference of 750,000 V. What is the
kinetic energy of the nucleus?
The standard notation is AZX. where A is the number of
nucleons (protons and neutrons) and Z is the number of
protons. Thus this nucleus has a charge = +2 e . We could
also refer to this as a He2+ ion.
Answer: In this case the charge is 2 e and thus the kinetic
energy is twice as great as when a singe proton is accelerated.
KE = 2 (750,000 eV) = 1.5x106 eV = 1.5 MeV
Ch 17
8
Electric Field of Parallel Plates
•
•
•
•
•
Consider the uniform electric field between
parallel plates
Ignoring signs the work done by the field to
move the charge from b to a is W= q Vba
Since the E field is uniform, we can also
calculate W from
W=Fd=qEd
We can combine these two equations to
give Vba = E d
The electric field between parallel
conducting plates is thus
Vba
E
d
+
+
+ +
b
+
+
+
+
+
+
+
+
_
_
_
+ _
a
_
_
_
_
_
_
_
This equation can be used anywhere the electric field is uniform.
Ch 17
9
Example 17-4A. Two parallel conducting plates are separated by a distance of
5.00 mm. If one plate is grounded and the other plate is at a voltage of +200 V,
calculate the electric field between the plates. The region between the plates is
vacuum.
Note: The term ‘ground’ refers to zero electric potential or voltage. Actually
ground is defined as the electric potential of the earth and often ground is
achieved by connecting a wire to a metal stake driven into the earth.
VBA
E
d
200 V  0 V

5.00  10 3 m
V
E  4.00 10
m
4
Ch 17
N
(or
)
C
10
Example 17-4B. If an electron is released near the grounded plate in
Ex. 17-4A, what is its kinetic energy in eV and J when it reaches the
+ 200 V plate.. Also calculate the velocity of the electron at that time.
KE  200 eV

KE  (200eV ) 1.60 1019 J

 3.20 10
eV
17
J
1
KE  me v 2
2
 2 KE 
v 

 m 
1
2
 (2)3.2 10 J  

 
31
 9.1110 kg 
17
1
2
v  8.4 106 m s
Ch 17
11
Equipotential Lines
•
•
•
•
•
•
•
A 9 volt battery is connected to the
parallel plates
Thus the positive plate is at a potential of
9 V and the negative plate is a 0 V.
The green lines are called equipotential
lines because every point on the line is at
the same electrical potential
Notice how the voltage drops as you go
from the 9 V plate, which is also an
equipotential to the negative plate at 0 V
The negative plate has a wire that is
connected to the earth. This is referred to
as grounding. It establishes the negative
plate at V=0
The green lines are actually surfaces in
3D
The equipotentials are always
perpendicular to the electric field lines.
9 volts
+
+
+
+
+
+
+
+
+
+
+
7.75 V
Ch 17
_
_
_
_
_
_
_
_
_
_
_
4.5 V
2.25 V
Ground
symbol
12
Electric Potential due to a Point Charge
•To derive an expression for the electric potential of a
point charge +Q, we calculate the work done per charge to
move a test charge from infinity to a point that is a distance
r from Q.
This derivation requires calculus because the electric
repulsion increases as the test charge moves closer.
The result of this derivation is:
•
•
V k
Ch 17
Q
r
13
Electric Potential due to a Point Charge
V k
Q
r
Note that V is positive if Q is positive.
V is negative if Q is negative.
Ch 17
14
Equipotential Lines of Two Point Charges
Equal in Magnitude
V 0
Ch 17
V 0
V 0
15
Example 17-5 A. Calculate the electric potential of the upper right corner of
a rectangle that is 3.0 cm high and 6.0 cm long if there are charges of +8.0
C in the upper left corner, +2.0 C in the lower left corner and +4.0 C in
the lower right corner. Assume V = 0 at infinity.
V
Q1  8.0 C
is
r2 
Q2  2.0 C
Q3  4.0 C
not
a vector
3.0 cm 
2
 6.0 cm 
2
 6.7 cm
Q3
Q1
Q2
V  k
k
k
r1
r2
r3
2
6
C
2 106 C
4 106 C 
9 Nm  8.0 10
V  9.00 10



2 
2
2
C  6.0 10 m 6.7 10 m 3.0 102 m 
V
Ch 17
  2.7 106 V
16
Example 17-5B. For the problem above, how much work would
be done in bringing a proton from infinity to the upper right
corner?
PE  PEB  PEA  q (VB  VA )
VA  0
( at infinity )
W  PE  q VB
W  (1.6 1019 C ) (2.7 106 V )
W  4.310
Ch 17
13
J
17
The Capacitor
•
•
•
•
•
•
•
What would happen to the charged parallel
plates if the battery were disconnected?
Answer: Nothing: the + and – charges are
attracted to each other, so they would remain.
Notice that this is a charge-storage device.
It also stores energy.
The potential difference or voltage V will
remain constant
This device is called a capacitor and it is often
used in electrical circuits.
Often the plates are made of a thin foil with a
thin insulator between the plates. This can
then be rolled up to form a very compact
device
Ch 17
9 volts
+
+
+
+
+
+
+
+
+
+
+
_
_
_
_
_
_
_
_
_
_
_
18
Capacitor Equations
There are several equations that describe
the capacitor which we will not derive.
• The charge on the plate is proportional
to the voltage:
Q  CV
•Q is the charge on one plate (and there is an
equal but opposite charge on the other plate)
and V is the voltage difference.
•We refer to C as the capacitance.
•The unit of capacitance is the farad ( F ).
•1F=1C/V
•Most capacitors are F or smaller
Ch 17
+
+
+
+
+
+
+
+
+
+
+
_
_
_
_
_
_
_
_
_
_
_
_
19
Capacitor Equations
Another equation allows us to calculate the
capacitance of a particular parallel plate.
C = 0 (A/d)
where A is the area of the plates, d is the
distance between them and 0 is called the
permittivity of free space.
0 = 8.85 x 10 –12 C2/Nm2
The symbol of the capacitor used in circuit
drawings is:
Ch 17
+
+
+
+
+
+
+
+
+
+
+
_
_
_
_
_
_
_
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_
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20
Dielectrics
Most capacitors have a dielectric material between the plates that
•Increases capacitance
•Allows higher voltages
•Maintains distance d
•The formula for a capacitor with a dielectric is (this is the formula to
remember).
C = K 0 ( A / d )
where K is the dielectric constant.
Ch 17
21
Dielectrics
Dielectric Constants of some materials
at 20 oC. The dielectric constant is unitless
Ch 17
Vacuum
1.0000
Air
1.0006
Vinyl (plastic)
2.8  4.5
Mica
7
Water
80
22
How Dielectrics Work
•The molecules in dielectrics become polar which means that
the electrons tend to be located closer to the positive plate as
shown below.
•A test charge inside the dielectric feels a E field reduced by
1 / K and thus a smaller V.
C
Ch 17
+
+
+
+
+
+
+
Q
Q
QK


 KC0
V V0
V0
K
-+
-+
-+
-+
-+
-+
-+
-+
-+
-+
-+
-+
-+
-+
-+
-+
-+
-+
-+
-+
-+
_
_
_
_
_
_
_
23
Energy Storage in a Capacitor
A battery must do work to move electrons from one plate to the other. The
work done to move a small charge q across a voltage V is
W = V
q. As the charge increases, V increases so the work to bring q
increases. Using calculus we find that the energy ( U ) stored on a
capacitor is given by:
1
U  QV
2
Since
Q  CV
1
1
1
U  QV  (CV )V  CV 2
2
2
2
1
1
U  QV  C V 2
2
2
Ch 17
24
Example 17-6A. A parallel-plate capacitor has an area of 5.00
cm2, and the plates are separated by 1.00 mm with air (K =
1.0006) between them. Calculate the capacitance.
A
C  K 0
d
C


2
2
2






C
5
.
00
cm
1
m
100
cm
12


 1.0006  8.85 10
2 
3

N m 
1.00 10 m


C  4.4310 12 F
17-6B. If the above capacitor is storing a charge of 400 pC, what
is the voltage across the plates?
Q  CV
Q
V 
C
Ch 17
V
400 10 12 C

4.43 10 12 F
 90.4 V
25
17-6C What is the energy stored by the capacitor?
1
U  CV 2
2
U
U
1
   4.43 10 12 F 90.4 V 
2
 1.80 108 J
17-6D If mica (K = 7.0) is placed between the plates of the
above capacitor, calculate its capacitance.
C  K C0

C  7.0 4.431012 F
C  3110
Ch 17
12

F
26
Cathode Ray Tube (CRT)
•Electrons are ejected from heated metal called cathode
•Then accelerated through potential difference of as much as 25,000 V
•Electron beam is then bent by horizontal and vertical electric fields created
by the horizontal and vertical deflection plates.
•Electron beam then strikes fluorescent screen
•Points on screen fluoresce in different colors
•Electronics attached to deflection plates cause beam to paint a picture on
screen
•Picture made with 525 horizontal lines
•Picture redrawn at 30 Hz
Ch 17
27