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Transcript
Chapter 17
Electric Potential, Electric Energy
and Capacitance
© 2002, B.J. Lieb
Giancoli, PHYSICS,5/E © 1998. 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
PEa
Va 
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 a has more
potential energy than at b.
Ch 17
+
+
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+ b
+
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+ _
a _
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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 KE.
The same is true for electric potential.
•
Wba
Vab  Va  Vb  
q
Wba
Vab  
q
Ch 17
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: The two
parallel plates shown are
connected to a 9 V battery.
If the charge is +4 C, then
the potential
PE = (9V) (4 C)= 36 J
If the charge were
released at point b it
would “fall” to a and gain
36 J of kinetic energy if
there was no friction.
•
Ch 17
+
+
+
+
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+ b
+
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+ _
a _
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5
Electric Potential and Electric
Field
•
•
•
•
•
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
This equation is usually
written
Vba
E
d
Ch 17
+
+
+
+
+
+ b
+
+
+
+
+
+
+
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+
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+ _
a _
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6
Example 1
Ch 17
7
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 green lines are
actually surfaces in 3D
The equipotentials are
always perpendicular to
the electric field lines.
9 volts
7.75 V
Ch 17
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+
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4.5 V
2.25 V
8
Equipotential Lines of Two Point
Charges
Ch 17
9
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
+
+
+
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+
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+
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10
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
Example: An electron in a TV tube is
accelerated through a potential difference of
25,000 V. What is the kinetic energy of the
electron?
Answer: KE = 25,000 eV
KE = (25,000 eV) (1.6x10-19 J/eV)
= 4.0x10-15 J
•
•
•
Ch 17
11
Electric Potential due to Point
Charges
•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:
•
•
Q
V k
r
V
r
Ch 17
12
Example 2
Ch 17
13
Capacitor Equations I
We will now derive several
equations that describe the
capacitor.
Q  CV
•The unit of capacitance is the
farad (F).
•Q is the charge on one plate (and
there is an equal but opposite
charge on the other plate) and V is
the voltage.
•1F=1C/V
•We refer to C as the capacitance.
•Most capacitors are F or smaller
Ch 17
+
+
+
+
+
+
+
+
+
+
+
+
+
+
_
_
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14
Capacitor Equations II
Another not-derived 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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15
Dielectrics
Most capacitors have an dielectic 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 called the dielectric constant.
Dielectric Constants (20 degrees C)
Ch 17
Vacuum
1.0000
Air
1.0006
Vinyl (plastic)
2.8-4.5
Mica
7
Water
80
16
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.
Q
Q
QK
C 

 KC0
V V0
V0
K
+
+
+
+
+
+
+
+
Ch 17
-+ -+
-+
-+
-+
-+
-+
-+
-+
-+
-+
-+
-+
-+
-+
-+ -+ -+
-+ -+ -+
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17
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
Ch 17
18
Example 3
Ch 17
19
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
20