Transcript 1

1
Today’s agendum:
Electric potential energy.
You must be able to use electric potential energy in work-energy calculations.
Electric potential.
You must be able to calculate the electric potential for a point charge, and use the electric
potential in work-energy calculations.
Electric potential and electric potential energy of a system of
charges.
You must be able to calculate both electric potential and electric potential energy for a
system of charged particles (point charges today, charge distributions next lecture).
The electron volt.
You must be able to use the electron volt as an alternative unit of energy.
2
This lecture introduces electric potential energy and something
called “electric potential.”
Electric potential energy is “just like” gravitational potential
energy.
Except that all matter exerts an attractive gravitational force, but charged particles exert
either attractive or repulsive electrical forces—so we need to be careful with our signs.
Electric potential is the electric potential energy per unit of
charge.
If you understand the symbols in the starting equations, and
avoid sign and direction mistakes, homework and exams are
not difficult.
3
Electric Potential
Electric Potential Energy
Electric Potential Energy
Work done by Coulomb force when q1
moves from a to b:
rb
rb
ra
ra
W   FE  ds   
W  k q1q 2

rb
ra
k q1q 2
r
2
r
dr
q1 (+) ds
dr
1
dr  k q1q 2
2
r
rb
 1
 
 r  ra
 1 1
W  k q1q 2      k q1q 2
 rb ra 
b
FE
rb
a q1 (+)
ra
1 1
   q2 (-)
 rb ra 
You don’t need to worry about the details of the math. They are provided for anybody who wants to study them later.
4
1 1
W  k q1q 2   
 rb ra 
r
dr
I did the calculation for a + charge
moving away from a – charge; you
could do a similar calculation for ++,
-+, and ++.
The important point is that the work
depends only on the initial and final
positions of q1.
a
ra
FE
b
ds
q1 (+)
rb
q2 (-)
In other words, the work done by the electric force is
independent of path taken. The electric force is a
conservative force.
5
Disclaimer: this is a “demonstration” rather than a rigorous proof.
The next two slides are intended to draw a parallel between the
electric and gravitational forces.
Instead of saying “ugh, this is confusing new stuff,” you are
supposed to say, “oh, this is easy, because I already learned the
concepts in Physics 103.”
6
A bit of review:
Consider an object of mass m in a gravitational field. It has
potential energy U(y) = mgy and “feels” a gravitational force
FG = GmM/r2, attractive.
y
If released, it gains kinetic
energy and loses potential
energy, but mechanical energy
is conserved: Ef=Ei. The
change in potential energy is
Uf - Ui = -(Wc)if.
Ui = mgyi
yi
x
Uf = 0
What force does Wc? Force due to gravity.
graphic “borrowed” from http://csep10.phys.utk.edu/astr161/lect/history/newtongrav.html 7
A charged particle in an electric
field has electric potential
energy.
++++++++++++++
It “feels” a force (as given by
Coulomb’s law).
It gains kinetic energy and loses
potential energy if released. The
Coulomb force does positive
work, and mechanical energy is
conserved.
+
E
F
-------------------
8
The next two slides define electrical potential energy.
9
Now that we realize the electric force is conservative, we can
define a potential energy associated with it.
The subscript “E” is to
remind you this is electric
potential energy. After
this slide, I will drop the
subscript “E.”
UE  UEf  UEi    WE if
The change in potential energy when a charge
q0 moves from point a to point b in the electric
field of another charge q is
rb
rb
ra
ra
U E    FE  d   
k qq 0
dr
2
r
The minus sign in this equation comes from the
definition of change in potential energy. The
sign from the dot product is “automatically”
correct if you include the signs of q and q0.
a
ra
q
r
dr
FE?
b
ds
q0
rab
? on FE means the
direction depends on the
signs of the charges.
10
rb
U    FE  d
(from the previous slide)
ra
is equivalent to your starting equation
f
Uf  Ui  q  E  d
i
“i” and “f” refer to the two points for which we are calculating the potential energy difference. You could also
use “a” and “b” like your text does, or “0” and “1” or anything else convenient. I use “i” and “f” because I
always remember that (anything) = (anything)f – (anything)i.
The next two slides use this definition of electrical potential
energy to derive an equation for the electrical potential energy
of two charged particles.
11
starting with an equation from two slides back…
U b  Ua  k qq 0 
rb
ra
r
rb
1
 1
dr  k qq 0   
2
r
 r  ra
1 1
U b  Ua  k qq 0   
 rb ra 
By convention, we choose electric
potential energy to be zero at infinite
separation of the charges.
dr
a
ra
FE?
b
ds
q0
rab
this diagram shows q0 after
it has moved from a to b
q
If there are any math majors in the room, please close your
eyes for a few seconds. We should be talking about limits.
0
 1 1 0
U b  U  k qq 0   
 rb  
12
This provides us with the electric potential energy for a system
of two point charges q and q0, separated by a distance r:
1
1 qq 0
U  r   k qq 0 
.
r
40 r
You can call the charges q and q0, or q1 and q2, or whatever you want.
If you have more than two charged particles, simply add the potential energies for each unique pair of particles.
Example: calculate the electric potential energy for two protons
separated by a typical proton-proton intranuclear distance of
2x10-15 m.
What is the meaning of the + sign in the result?
13
Really Important fact to keep straight.
U  Uf  Ui    Wconservative if
The change in potential energy is the negative of the work done
by the conservative force which is associated with the potential
energy (the electric force).
If an external force* moves an object “against” the conservative
force, then
 Wexternal if    Wconservative if
Always ask yourself which work you are calculating.
*for example, if you push two negatively charged balloons together
14
Today’s agendum:
Electric potential energy.
You must be able to use electric potential energy in work-energy calculations.
Electric potential.
You must be able to calculate the electric potential for a point charge, and use the electric
potential in work-energy calculations.
Electric potential and electric potential energy of a system of
charges.
You must be able to calculate both electric potential and electric potential energy for a
system of charged particles (point charges today, charge distributions next lecture).
The electron volt.
You must be able to use the electron volt as an alternative unit of energy.
15
Electric Potential
Previously, we defined the electric field by the force it exerts on
a test charge q0:
F0
E = lim
q 0 0 q
0
Similarly, it is useful to define the potential of a charge in
terms of the potential energy of a test charge q0:
Ur
V  r  = lim
q 0 0
q0
The electric potential V is independent of the test charge q0.
16
1 qq 0
From U  r  
40 r
we see that the electric potential of a point charge q is
1 q
V r 
.
4 0 r
The electric potential difference between points a and b is
rb
rb F
rb
U  ra FE  d
E
V 

 
 d   E  d .
ra q
ra
q0
q0
0
17
Things to remember about electric potential:
 Electric potential and electric potential energy are related, but
are not the same.
Electric potential difference is the work per unit of charge
that must be done to move a charge from one point to
another without changing its kinetic energy.
 The terms “electric potential” and “potential” are used
interchangeably.
Ur
.
 The units of potential are joules/coulomb: V  r  =
q0
1 joule
1 volt =
1 coulomb
18
Things to remember about electric potential:
 Only differences in electric potential and electric potential
energy are meaningful.
It is always necessary to define where U and V are zero.
Here we defined V to be zero at an infinite distance from the
sources of the electric field.
Sometimes it is convenient to define V to be zero at the
earth (ground).
It should be clear from the context where V is defined to be
zero, and I do not foresee you experiencing any confusion
about where V is zero.
19
Two more starting equations:
1 qq 0
1 q
and V  r  
U r 
4 0 r
4 0 r
so
U r
1 qq 0 1
1 q


 V(r)
q0
40 r q 0 40 r
(potential is equal to potential energy per unit of charge)
Potential energy and electric potential are defined relative to
some reference point, so it is “better” to use
U
V  Vf  Vi 
q
20
f
Uf  Ui  q  E  d
i
f
U f Ui

  E  d
i
q
q
f
Vf  Vi   E  d
i
21
Two conceptual examples.
Example: a proton is released in a region in space where there
is an electric potential. Describe the subsequent motion of the
proton.
The proton will move towards the region of lower potential. As it moves, its
potential energy will decrease, and its kinetic energy and speed will increase.
Example: an electron is released in a region in space where
there is an electric potential. Describe the subsequent motion of
the electron.
The electron will move towards the region of higher potential. As it moves,
its potential energy will decrease, and its kinetic energy and speed will
increase.
22
Today’s agendum:
Electric potential energy.
You must be able to use electric potential energy in work-energy calculations.
Electric potential.
You must be able to calculate the electric potential for a point charge, and use the electric
potential in work-energy calculations.
Electric potential and electric potential energy of a
system of charges.
You must be able to calculate both electric potential and electric potential energy for a
system of charged particles (point charges today, charge distributions next lecture).
The electron volt.
You must be able to use the electron volt as an alternative unit of energy.
23
Electric Potential Energy of a System of Charges
To find the electric potential energy for a system of two
charges, we bring a second charge in from an infinite distance
away:
r
q1
q1
U 0
before
q2
q1q 2
U k
r
after
24
To find the electric potential energy for a system of three
charges, we bring a third charge in from an infinite distance
away:
r12
q1
q2
before
q1q 2
U k
r12
q1
r12
q2
r13
r23
q3
after
 q1q 2 q1q3 q 2q3 
U  k



r13
r23 
 r12
25
Electric Potential and Potential Energy of a Charge
Distribution
1
Collection of charges: VP 
4 0
qi
i r .
i
P is the point at which V is to be calculated, and ri is the distance of the ith
charge from P.
Charge distribution:
1
dq
V
.

40 r
dq
P
r
Potential at point P.
26
Example: a 1 C point charge is located at the origin and a -4
C point charge 4 meters along the +x axis. Calculate the
electric potential at a point P, 3 meters along the +y axis.
y
 q1 q 2 
qi
VP = k  = k  + 
i ri
 r1 r2 
-6
-6


1×10
-4×10
9
= 9×10 
+

5 
 3
P
3m
q1
4m
q2
x
= - 4.2×103 V
27
Example: how much work is required to bring a +3 C point
charge from infinity to point P? An external force moves q3
slowly from an infinite distance to the point P.
0
y
Wexternal  E  K  U
q3
Wexternal  U  q3V
P
3m
q1
0
Wexternal  q3  VP  V 
4m
q2
x
Wexternal  3 106  4.2 103 
Wexternal  1.26 10 2 J
The work done by the external force was negative, so the work done by the electric field was
positive. The electric field “pulled” q3 in (keep in mind q2 is 4 times as big as q1).
Positive work would have to be done by an external force to remove q3 from P.
28
Example: find the total potential energy of the system of three
charges.
y
q3
 q1 q 2 q1 q3 q 2 q3 
U = k
+
+

r
r
r
13
23 
 12
P
3m
q2
4m
q1


x
 

 

-6
-6
-6
-6
 1×10-6 -4×10-6
1×10
3×10
-4×10
3×10
U = 9 109 
+
+

4
3
5

qq
U  2.16 10  2 J also  k 1 2  Wexternal  9 10 3 J  1.26 10  2 J
r12
 
29


Today’s agendum:
Electric potential energy.
You must be able to use electric potential energy in work-energy calculations.
Electric potential.
You must be able to calculate the electric potential for a point charge, and use the electric
potential in work-energy calculations.
Electric potential and electric potential energy of a system of
charges.
You must be able to calculate both electric potential and electric potential energy for a
system of charged particles (point charges today, charge distributions next lecture).
The electron volt.
You must be able to use the electron volt as an alternative unit of energy.
30
The Electron Volt
An electron volt (eV) is the energy acquired by a particle of
charge e when it moves through a potential difference of 1 volt.
U= qV
1 eV= 1.6 10-19 C  1 V 
1 eV= 1.6 10-19 J
This is a very small amount of energy on a macroscopic scale,
but electrons in atoms typically have a few eV (10’s to 1000’s)
of energy.
31
Today’s agendum:
Electric potential of a charge distribution.
You must be able to calculate the electric potential for a charge distribution.
Equipotentials.
You must be able to sketch and interpret equipotential plots.
Potential gradient.
You must be able to calculate the electric field if you are given the electric potential.
Potentials and fields near conductors.
You must be able to use what you have learned about electric fields, Gauss’ “Law”, and
electric potential to understand and apply several useful facts about conductors in
electrostatic equilibrium.
32
Electric Potential of a Charge Distribution
Example: potential and electric field between two parallel
conducting plates.
Assume V0<V1 (so we can determine the direction of the
electric field). Also assume the plates are large compared to
their separation, so the electric field is constant and
perpendicular to the plates.
Also, let the plates be separated
by a distance d.
E
V0
d
V1
33
V  V1  V0   
plate 1
plate 0
Ed
x
V    E dx   E  dx  Ed
E
d
d
0
0
V
d
y
E
z
V0
dl
d
V1
, or V  Ed
I’ll discuss in lecture why the
absolute value signs are needed.
34
Important note: the derivation of
V  Ed
did not require rectangular plates, or any plates at all. It works
as long as E is uniform.
In general, E should be replaced by the component of E along
the displacement vector d .
35
Example: A rod of length L located along the x-axis has a total
charge Q uniformly distributed along the rod. Find the electric
potential at a point P along the y-axis a distance d from the
origin.
y
=Q/L
P
d
dq=dx
r
dq
dx
x
L
x
dq
dx
dV  k
k
r
x2  d2
L
V   dV
0
36
V
L
0
dx
Q L dx
k
k 
2
2
L 0 x2  d2
x d
y
A good set of math tables will
have the integral:
P
d
r
dq
dx
x
L
x

dx
x d
2
2

 ln x  x 2  d 2
kQ  L  L2  d 2
V
ln 
L 
d





37
Example: Find the electric potential due to a uniformly charged
ring of radius R and total charge Q at a point P on the axis of
the ring.
dq
r
R
P
x
Every dq of charge on the
ring is the same distance
from the point P.
x
dq
dq
dV  k
k
r
x2  R2
V
ring
dV  k 
ring
dq
x2  R 2
38
dq
r
R
P
x
x
V
V
k
x R
2
2

ring
dq
kQ
x2  R2
Could you use this expression for V to calculate E? Would you
get the same result as we obtained previously?
39
dq
r
a

P

x
dE
E x ,ring 
x
kxQ
2
R
2

3
2
40
Example: A disc of radius R has a uniform charge per unit area
 and total charge Q. Calculate V at a point P along the central
axis of the disc at a distance x from its center.
dq
r
P
R
x
x
The disc is made of
concentric rings. The
area of a ring at a
radius r is 2rdr, and
the charge on each ring
is (2rdr).
We can use the equation for the potential due to a ring, replace
R by r, and integrate from r=0 to r=R.
dVring 
k2rdr
x2  r2
41
dq
r
P
R
x
x
1
V   dV 
ring
40

V
x2  r2
20
R
0


20

2rdr

ring


x 2  r 2 20

Q
x R x 
20 R 2
2
2


Q

R 2
R
0
rdr
x2  r2
x2  R2  x
42

dq
r
P
R
x
x
Q
V
20 R 2

x2  R 2  x

Could you use this expression for V to calculate E?
43
See your text for other examples of potentials calculated from
charge distributions, as well as an alternate discussion of the
electric field between charged parallel plates.
44
PRACTICAL APPLICATION
For some reason you think practical applications are important.
Well, I found one!
45
Today’s agendum:
Electric potential of a charge distribution.
You must be able to calculate the electric potential for a charge distribution.
Equipotentials.
You must be able to sketch and interpret equipotential plots.
Potential gradient.
You must be able to calculate the electric field if you are given the electric potential.
Potentials and fields near conductors.
You must be able to use what you have learned about electric fields, Gauss’ “Law”, and
electric potential to understand and apply several useful facts about conductors in
electrostatic equilibrium.
46
Equipotentials
Equipotentials are contour maps of the electric potential.
http://www.omnimap.com/catalog/digital/topo.htm
47
Equipotential lines are another visualization tool. They
illustrate where the potential is constant. Equipotential lines
are actually projections on a 2-dimensional page of a 3dimensional equipotential surface. (“Just like” the contour
map.)
The electric field must be perpendicular to equipotential lines.
Why?
Otherwise work would be required to move a charge along an
equipotential surface, and it would not be equipotential.
In the static case (charges not moving) the surface of a
conductor is an equipotential surface. Why?
Otherwise charge would flow and it wouldn’t be a static case.
48
Here are some electric field and equipotential lines I generated
using an electromagnetic field program.
Equipotential lines are shown in red.
49
Today’s agendum:
Electric potential of a charge distribution.
You must be able to calculate the electric potential for a charge distribution.
Equipotentials.
You must be able to sketch and interpret equipotential plots.
Potential gradient.
You must be able to calculate the electric field if you are given the electric potential.
Potentials and fields near conductors.
You must be able to use what you have learned about electric fields, Gauss’ “Law”, and
electric potential to understand and apply several useful facts about conductors in
electrostatic equilibrium.
50
Potential Gradient
(Determining Electric Field from Potential)
The electric field vector points from higher to lower potentials.
More specifically, E points along shortest distance from a higher
equipotential surface to a lower equipotential surface.
You can use E to calculate V:
b
Vb  Va   E  d .
a
You can use the differential version of this equation to calculate
E from a known V:
dV  E  d  E d
dV
 E 
d
51
For spherically symmetric charge distribution:
dV
Er  
dr
In one dimension:
dV
Ex  
dx
In three dimensions:
V
V
V
Ex  
, Ey  
, Ez  
.
x
y
z
V ˆ V ˆ V ˆ
or E  
i 
j 
k   V
x
y
z
52
dV
E 
d
dV
Er  
dr
V
V
V
Ex  
, Ey  
, Ez  
.
x
y
z
Calculate -dV/d(whatever) including all signs. If the result is
+, E vector points along the +(whatever) direction. If the result
is -, E vector points along the –(whatever) direction.
53
Example: In a region of space, the electric potential is
V(x,y,z) = Axy2 + Bx2 + Cx, where A = 50 V/m3, B = 100 V/m2,
and C = -400 V/m are constants. Find the electric field at the
origin.
V
E x (0, 0, 0)  
   Ay 2  2Bx  C 
 C
(0,0,0)
x (0,0,0)
V
E y (0,0,0)  
 (2Axy) (0,0,0)  0
y (0,0,0)
V
E z (0, 0, 0)  
0
z (0,0,0)
V

E(0,0,0)   400  ˆi
m

54
Today’s agendum:
Electric potential of a charge distribution.
You must be able to calculate the electric potential for a charge distribution.
Equipotentials.
You must be able to sketch and interpret equipotential plots.
Potential gradient.
You must be able to calculate the electric field if you are given the electric potential.
Potentials and fields near conductors.
You must be able to use what you have learned about electric fields, Gauss’ “Law”, and
electric potential to understand and apply several useful facts about conductors in
electrostatic equilibrium.
55
Potentials and Fields Near Conductors
When there is a net flow of charge inside a conductor, the
physics is generally complex.
When there is no net flow of charge, or no flow at all (the
electrostatic case), then a number of conclusions can be
reached using Gauss’ “Law” and the concepts of electric fields
and potentials…
56
Summary of key points (electrostatic case):
The electric field inside a conductor is zero.
Any net charge on the conductor lies on the outer surface.
The potential on the surface of a conductor, and everywhere
inside, is the same.
The electric field just outside a conductor must be
perpendicular to the surface.
Equipotential surfaces just outside the conductor must be
parallel to the conductor’s surface.
57
Another key point: the charge density on a conductor surface
will vary if the surface is irregular, and surface charge collects at
“sharp points.”
Therefore the electric field is large (and can be huge) near
“sharp points.”
Another Practical Application
To best shock somebody, don’t touch them with your hand;
touch them with your fingertip.
Better yet, hold a small piece of bare wire
in your hand and gently touch them with
that.
58
Today’s agendum:
Capacitance.
You must be able to apply the equation C=Q/V.
Capacitors: parallel plate, cylindrical, spherical.
You must be able to calculate the capacitance of capacitors having these geometries, and
you must be able to use the equation C=Q/V to calculate parameters of capacitors.
Circuits containing capacitors in series and parallel.
You must be understand the differences between, and be able to calculate the “equivalent
capacitance” of, capacitors connected in series and parallel.
59
Capacitors and Dielectrics
Capacitance
A capacitor is basically two parallel
conducting plates with air or insulating
material in between.
E
V0
A capacitor doesn’t
have to look like
metal plates.
L
V1
Capacitor for use in
high-performance
audio systems.
60
The symbol representing a capacitor in an
electric circuit looks like parallel plates.
Here’s the symbol for a battery, or an external
potential.
+-
When a capacitor is connected to an external potential,
charges flow onto the plates and create a potential difference
between the plates.
Capacitor plates
build up charge.
-
V
-
The battery in this circuit has some voltage V. We haven’t discussed what
that means yet.
61
In a battery, chemical reactions release energy. The energy is expended
by transporting charged particles from one terminal of the battery
to the other. A “charge separation” is maintained.
62
If the external potential is
disconnected, charges remain on the
plates, so capacitors are good for
storing charge (and energy).
+conducting wires
+V
Capacitors are also very good at releasing
their stored charge all at once. The capacitors
in your tube-type TV are so good at storing
energy that touching the two terminals at the
same time can be fatal, even though the TV
may not have been used for months.
High-voltage TV capacitors are supposed to have “bleeder
resistors” that drain the charge away after the circuit is
turned off. I wouldn’t bet my life on it.
63
Graphic from http://www.feebleminds-gifs.com/.
assortment of
capacitors
64
+Q -Q
Here’s this V again. It
is the potential
difference provided
by the “external
potential”. For
example, the voltage
of a battery.
C
+ V
The magnitude of charge acquired by each plate of a capacitor
is Q=CV where C is the capacitance of the capacitor.
Q
C
V
C is always positive.
The unit of C is the farad but most capacitors have values
of C ranging from picofarads to microfarads (pF to F).
micro 10-6, nano 10-9, pico 10-12
(Know for exam!)
65
Today’s agendum:
Capacitance.
You must be able to apply the equation C=Q/V.
Capacitors: parallel plate, cylindrical, spherical.
You must be able to calculate the capacitance of capacitors having these geometries, and
you must be able to use the equation C=Q/V to calculate parameters of capacitors.
Circuits containing capacitors in series and parallel.
You must be understand the differences between, and be able to calculate the “equivalent
capacitance” of, capacitors connected in series and parallel.
66
Parallel Plate Capacitance
-Q
We previously calculated the electric field
between two parallel charged plates:
+Q
E

Q
E 
.
0 0 A
This is valid when the separation is small
compared with the plate dimensions.
V0
d
V1
We also showed that E and V are related:
d
d
0
0
V   E  d  E  dx  Ed .
0 A
Q
Q
Q
This lets us calculate C for C 



V Ed  Q 
d
a parallel plate capacitor.

d
67
 0 A 
A
Reminders:
Q
C
V
Q is the magnitude of the charge on either plate.
V is actually the magnitude of the potential difference
between the plates. V is really |V|. Your book calls it
Vab.
C is always positive.
68
Parallel plate capacitance depends “only”
on geometry.
-Q
+Q
0 A
C
d
This expression is approximate, and must
be modified if the plates are small, or
separated by a medium other than a
vacuum.
E
V0
d
V1
A
 0 A
C
d
Greek letter Kappa. For
today’s lecture use
Kappa=1.
69
Isolated Sphere Capacitance
An isolated sphere can be thought of as concentric spheres
with the outer sphere at an infinite distance and zero potential.
We already know the potential outside a conducting sphere:
Q
V
.
4 0 r
The potential at the surface of a charged sphere of radius R is
Q
V
4 0 R
so the capacitance at the surface of an isolated sphere is
Q
C
 40 R.
V
70
Capacitance of Concentric Spheres
Let’s calculate the capacitance of a concentric spherical
capacitor of charge Q.
In between the spheres
Q
E
4 0 r 2
Q
V 
40
b
a

b
a
dr
Q

2
r
40
4 0
Q
C

V
1 1 
 a  b 
1 1 
 a  b 
+Q
-Q
You need to do this derivation if you have a
problem on spherical capacitors!
71
alternative calculation of capacitance of isolated sphere
4 0
Q
C

V
1 1 
 a  b 
b
a
+Q
-Q
Let aR and b to get the capacitance of an isolated
sphere.
72
Coaxial Cylinder Capacitance
We can also calculate the capacitance of a
cylindrical capacitor (made of coaxial
cylinders).
The next slide shows a cross-section view of
the cylinders.

L
73
b
b
a
a
ΔV = Vb - Va = -  E  d = -  E r dr
Gaussian
surface
b
dr
b
ΔV = - 2k λ  = - 2k λ ln  
r
a
a
b
2kλ
E=
r
r
a
Q
Q
λL
C=
=
=
ΔV
ΔV
λL
b
2k λ ln  
a
2πε 0 L
L
C=
=
b
b
2k ln  
ln  
a
a
E
dl
-Q
This derivation is sometimes needed
for homework problems!
Lowercase c is capacitance per unit length: c =
2πε 0
b
ln  
a
74
Example: calculate the capacitance of a capacitor whose plates
are 20 cm x 3 cm and are separated by a 1.0 mm air gap.
0 A
C
d
C
12
8.85

10

  0.2  0.03
0.001
C  53 1012 F
d = 0.001
area = 0.2 x 0.03
C  53 pF
75
If you keep everything in SI (mks) units, the result is “automatically” in SI units.
Example: what is the charge on each plate if the capacitor is
connected to a 12 volt* battery?
Q  CV
0V
Q   53 1012  12 
V= 12V
Q  6.4 1010 C
+12 V
*Remember, it’s the potential difference that matters.
76
If you keep everything in SI (mks) units, the result is “automatically” in SI units.
Example: what is the electric field between the plates?
V
E
d
0V
12V
E
0.001 m
E
V= 12V
d = 0.001
V
E  12000
,"up."
m
+12 V
77
If you keep everything in SI (mks) units, the result is “automatically” in SI units.
Today’s agendum:
Capacitance.
You must be able to apply the equation C=Q/V.
Capacitors: parallel plate, cylindrical, spherical.
You must be able to calculate the capacitance of capacitors having these geometries, and
you must be able to use the equation C=Q/V to calculate parameters of capacitors.
Circuits containing capacitors in series and parallel.
You must be understand the differences between, and be able to calculate the “equivalent
capacitance” of, capacitors connected in series and parallel.
78
Capacitors in Circuits
Recall: this is the symbol representing a
capacitor in an electric circuit.
And this is the symbol for a battery…
+-
…or this…
…or this.
79
Circuits Containing Capacitors in Parallel
Vab
Capacitors connected in parallel:
C1
a
C2
b
C3
+ V
The potential difference (voltage drop) from a to b must equal V.
Vab = V = voltage drop across each individual capacitor.
Note how I have introduced the idea that when circuit components are connected in parallel, then the voltage
80
drops across the components are all the same. You may use this fact in homework solutions.
C1
Q=CV
 Q1 = C1 V
Q1
a
C2 -
+
& Q2 = C2 V
& Q3 = C3 V
Q2
C3
Q3
+ V
Now imagine replacing the parallel
combination of capacitors by a single
equivalent capacitor.
a
By “equivalent,” we mean “stores the same
total charge if the voltage is the same.”
Q1 + Q2 + Q3 = Ceq V = Q
Ceq
Q
+ V
Important!
81
Summarizing the equations on the last slide:
Q1 = C1 V
Q2 = C2 V
Q3 = C3 V
C1
C2
a
Q1 + Q2 + Q3 = Ceq V
C3
+ -
Using Q1 = C1V, etc., gives
V
C1V + C2V + C3V = Ceq V
C1 + C2 + C3 = Ceq
b
(after dividing both sides by V)
Generalizing:
Ceq = Ci
(capacitors in parallel)
82
Circuits Containing Capacitors in Series
Capacitors connected in series:
C1
C2
C3
+ -
+Q V -Q
An amount of charge +Q flows from the battery to the left plate
of C1. (Of course, the charge doesn’t all flow at once).
An amount of charge -Q flows from the battery to the right
plate of C3. Note that +Q and –Q must be the same in
magnitude but of opposite sign.
83
The charges +Q and –Q attract equal and opposite charges to
the other plates of their respective capacitors:
C1
+Q -Q
A
C2
+Q -Q
B
C3
+Q -Q
+ V
These equal and opposite charges came from the originally
neutral circuit regions A and B.
Because region A must be neutral, there must be a charge +Q
on the left plate of C2.
Because region B must be neutral, there must be a charge --Q
on the right plate of C2.
84
Vab
a
C1
+Q -Q
V1
C2
A
+Q -Q
V2
C3
B
b
+Q -Q
V3
+ V
The charges on C1, C2, and C3 are the same, and are
Q = C1 V1
Q = C2 V2
Q = C3 V3
But we don’t know V1, V2, and V3 yet.
We do know that Vab = V and also Vab = V1 + V2 + V3.
Note how I have introduced the idea that when circuit components are connected in series, then the voltage
drop across all the components is the sum of the voltage drops across the individual components. This is 85
actually a consequence of the conservation of energy.
Let’s replace the three capacitors by a single equivalent
capacitor.
Ceq
+Q -Q
V
+ V
By “equivalent” we mean V is the same as the total voltage
drop across the three capacitors, and the amount of charge Q
that flowed out of the battery is the same as when there were
three capacitors.
Q = Ceq V
86
Collecting equations:
Q = C1 V1
Q = C2 V2
Q = C3 V3
Important!
Vab = V = V1 + V2 + V3.
Q = Ceq V
Substituting for V1, V2, and V3:
Q
Q
Q
V=
+
+
C1 C 2 C 3
Substituting for V:
Q
Q
Q
Q
=
+
+
Ceq C1 C2 C3
Dividing both sides by Q:
1
1
1
1
=
+
+
Ceq C1 C2 C3
87
Generalizing:
1
1
=
Ceq
Ci
i
(capacitors in series)
88
Example: determine the
capacitance of a single capacitor
that will have the same effect as
the combination shown. Use
C1 = C2 = C3 = C.
C2
C1
C3
I don’t see a series combination of capacitors, but I do see a
parallel combination.
C23 = C2 + C3 = C + C = 2C
89
Now I see a series combination.
1
1
1
=
+
Ceq C1 C23
C23 = 2C
C 1= C
1
1
1
2
1
3
= +
=
+
=
Ceq C 2C 2C 2C 2C
C eq
2
= C
3
90
Example: for the capacitor circuit shown, C1 = 3F, C2 = 6F,
C3 = 2F, and C4 =4F. (a) Find the equivalent capacitance. (b)
if V=12 V, find the potential difference across C4.
C1
C2
C4
I’ll work this at the blackboard.
C3
V
91
Overview
Electric charge and electric force
Coulomb’s Law
Electric field
calculating electric field
motion of a charged particle in an electric field
Gauss’ “Law”
electric flux
calculating electric field using Gaussian surfaces
properties of conductors
92
Overview
Electric potential and electric potential energy
calculating potentials and potential energy
calculating fields from potentials
equipotentials
potentials and fields near conductors
Capacitors
capacitance of parallel plates, concentric cylinders,
(concentric spheres not for this exam)
equivalent capacitance of capacitor network
Don’t forget the concepts from Physics 103 that were frequently
used!
93
Three charges +Q, +Q, and –Q, are located at the corners of
an equilateral triangle with sides of length a. What is the force
on the charge located at point P (see diagram)?
y
F1
P
+Q
+Q
  F1 sin   F2 sin   ˆj


F2
a
F   F1 cos   F2 cos   ˆi
 kQ 2
kQ 2
F   2 cos 60  2 cos 60
a
 a
-Q
ˆ
i

 kQ 2
kQ 2
  2 sin 60  2 sin 60
a
 a
x
94
ˆ
j

y
kQ2
F  2 2 cos 60 ˆi
a
F1
P
+Q


kQ 2 ˆ
F 2 i
a
F2
a
+Q
-Q
x
95
What is the electric field at P due to the two charges at the
base of the triangle?
y
F1
P
+Q


You can repeat the above calculation,
replacing F by E.
F
Or… E 
q
F2
kQ2
2 2 cos 60 ˆi
E a
Q
a
+Q
-Q
x
kQ
E  2 2 cos 60 ˆi
a
96
An insulating spherical shell has an inner radius a and outer
radius b. The shell has a total charge Q and a uniform charge
density. Find the magnitude of the electric field for r<a.
  qenclosed
 E  dA  o  0
97
An insulating spherical shell has an inner radius a and outer
radius b. The shell has a total charge Q and a uniform charge
density. Find the magnitude of the electric field for a<r<b.
4 3 4 3
Q    b  a 
3
3

  qenclosed
 E  dA  o
4 3 4 3
  r  a 
3
3


2
E  4r  
o
98
An electron has a speed v. Calculate the magnitude and
direction of an electric field that will stop this electron in a
distance D.
Ef  Ei  E   Wother if .
V  Vf  Vi  Ed
U  qV
Do you understand
that these E’s have
different meanings?
Know where this
comes from!
Use magnitudes if you
have determined direction
of E by other means.
99
Two equal positive charges Q are located at the base of an
equilateral triangle with sides of length a. What is the potential
at point P (see diagram)?
P
a
Q
Q
100
Three equal positive charges Q are located at the corners of an
equilateral triangle with sides of length a. What is the potential
energy of the charge located at point P (see diagram)?
P
Q
a
Q
Q
101
For the capacitor system shown, C1=6.0 F, C2=2.0 F, and
C3=10.0 F. (a) Find the equivalent capacitance.
C1
V0
C2
C3
102
For the capacitor system shown, C1=6.0 F, C2=2.0 F, and
C3=10.0 F. The charge on capacitor C3 is found to be 30.0 C.
Find V0.
C1
V0
C2
C3
103
Today’s agendum:
Energy Storage in Capacitors.
You must be able to calculate the energy stored in a capacitor, and apply the energy
storage equations to situations where capacitor configurations are altered.
Dielectrics.
You must understand why dielectrics are used, and be able include dielectric constants in
capacitor calculations.
104
Energy Storage in Capacitors
Let’s calculate how much work it takes to charge a capacitor.
The work required for an external force to move a charge dq
through a potential difference V is dW = dq V.
From Q=CV ( V = q/C):
+
We start with zero charge on the capacitor,
and end up with Q, so
Q
Q
0
0
W   dW  
-
dq
+
q is the amount of charge on
the capacitor during the time
the charge dq is being moved.
q
dW  dq
C
V
+q
-q
2 Q
q
q
Q2
dq 

.
C
2C 0 2C
105
The work required to charge the capacitor is the amount of
energy you get back when you discharge the capacitor
(because the electric force is conservative).
Thus, the work required to charge the capacitor is equal to the
potential energy stored in the capacitor.
Q2
U
.
2C
Because C, Q, and V are related through Q=CV, there are three
equivalent ways to write the potential energy.
Q2 CV 2 QV
U


.
2C
2
2
106
Q2 CV 2 QV
U


.
2C
2
2
All three equations are valid; use the one most convenient for
the problem at hand.
It is no accident that we use the symbol U for the energy
stored in a capacitor. It is just another “version” of electrical
potential energy. You can use it in your energy conservation
equations just like any other form of potential energy!
107
Example: a camera flash unit stores energy in a 150 F
capacitor at 200 V. How much electric energy can be stored?
CV 2
U
2
U
6
2
150

10
200

 
2
U3 J
108
If you keep everything in SI (mks) units, the result is “automatically” in SI units.
Example: compare the amount of energy stored in a capacitor
with the amount of energy stored in a battery.
A 12 V car battery rated at 100 ampere-hours stores 3.6x105 C
of charge and can deliver at least 4.3x106 joules of energy
(we’ll learn how to calculate that later in the course).
A 100 F capacitor that stores 3.6x105 C at 12 V stores an
amount of energy U=CV2/2=7.2x10-3 joules.
If you want your capacitor
to store lots of energy, store
it at a high voltage.
If a battery stores so much more energy, why use capacitors?
106 joules of energy are stored at high voltage in capacitor
banks, and released during a period of a few milliseconds. The
enormous current produces incredibly high magnetic fields. 109
Energy Stored in Electric Fields
Energy is stored in the
capacitor:
1
U  C V 2
2
V
+
E
1  0 A 
2
U 
  Ed 
2 d 
1
U    0 Ad  E 2
2
-
+Q
d
-Q
area A
The “volume of the capacitor” is Volume=Ad
110
Energy stored per unit volume (u):
1
 0 Ad  E 2 1 2
u 2
 0 E
Ad
2
The energy is “stored” in the electric
field!
V
+
E
We’ve gone from the concrete (electric
charges experience forces)…
…to the abstract (electric charges
create electric fields)…
…to an application of the abstraction
(electric field contains energy).
-
+Q
d
-Q
area A
111
“The energy in electromagnetic phenomena is the same as mechanical
energy. The only question is, ‘Where does it reside?’ In the old theories, it
resides in electrified bodies. In our theory, it resides in the electromagnetic
field, in the space surrounding the electrified bodies.”—James Maxwell
1
u  0 E 2
2
V
+
E
This is not a new “kind” of energy. It’s the
electric potential energy resulting from the
coulomb force between charged particles.
+Q
Or you can think of it as the electric energy
due to the field created by the charges.
Same thing.
-
f
-Q
area A
112
Today’s agendum:
Energy Storage in Capacitors.
You must be able to calculate the energy stored in a capacitor, and apply the energy
storage equations to situations where capacitor configurations are altered.
Dielectrics.
You must understand why dielectrics are used, and be able include dielectric constants in
capacitor calculations.
113
Dielectrics
If an insulating sheet (“dielectric”) is
placed between the plates of a
capacitor, the capacitance increases by
a factor , which depends on the
material in the sheet.  is the
dielectric constant of the material.
dielectric
In general, C = 0A / d.  is 1
for a vacuum, and  1 for air.
(You can also define  = 0
and write C =  A / d).
 A
C=
.
d
114
The dielectric is the thin insulating sheet
in between the plates of a capacitor.
dielectric
Any reasons to use a dielectric in a capacitor?
Makes your life as a physics student more complicated.
Lets you apply higher voltages (so more charge).
Lets you place the plates closer together (make d
smaller).
Increases the value of C because >1.
Q = CV
  A
C=
d
115
Example: a parallel plate capacitor has an area of 10 cm2 and
plate separation 5 mm. 300 V is applied between its plates. If
neoprene is inserted between its plates, how much charge
does the capacitor hold.
A=10 cm2
  A
C=
d
C=
 6.7   8.85×10-12 10×10-4 
=6.7
5×10-3
C =1.19 10-11 F
V=300 V
d=5 mm
Q = CV


Q = 1.19 10-11  300   3.56 10-9 C = 3.56 nC
116
Example: how much charge would the capacitor on the
previous slide hold if the dielectric were air?
The calculation is the
same, except replace 6.7
by 1.
A=10 cm2
Or just divide the charge on the
previous page by 6.7 to get.
Q = 0.53 nC
=1
V=300 V
d=5 mm
117
Conceptual Example
V=0
A capacitor connected as shown
acquires a charge Q.
While the capacitor is still connected
to the battery, a dielectric material is
inserted.
V
V
Will Q increase, decrease, or stay the same?
Why?
118
Example: find the energy stored in the capacitor in slide 13.
C =1.19 10-11 F
A=10 cm2
1
2
U = C  V 
2


1
2
-11
U = 1.19 10  300 
2
-7
U = 5.36 10 J
=6.7
V=300 V
d=5 mm
119
Example: the battery is now disconnected. What are the
charge, capacitance, and energy stored in the capacitor?
The charge and capacitance are
unchanged, so the voltage drop
and energy stored are unchanged.
Q = 3.56 nC
A=10 cm2
=6.7
C =1.19 10-11 F
U = 5.36 10-7 J
V=300 V
d=5 mm
120
Example: the dielectric is removed without changing the plate
separation. What are the capacitance, charge, potential
difference, and energy stored in the capacitor?
A=10 cm2
C=
 A
d
8.85×10 10×10 

C=
-12
-4
=6.7
5×10-3
C =1.78 10-12 F
V=?
V=300 V
d=5 mm
121
Example: the dielectric is removed without changing the plate
separation. What are the capacitance, charge, potential
difference, and energy stored in the capacitor?
The charge remains unchanged,
because there is nowhere for it
to go.
A=10 cm2
Q = 3.56 nC
V=?
d=5 mm
122
Example: the dielectric is removed without changing the plate
separation. What are the capacitance, charge, potential
difference, and energy stored in the capacitor?
Knowing C and Q we can
calculate the new potential
difference.


3.56 10-9
Q
V = =
C
1.78 10-12
V = 2020 V
A=10 cm2


V=?
d=5 mm
123
Example: the dielectric is removed without changing the plate
separation. What are the capacitance, charge, potential
difference, and energy stored in the capacitor?
A=10 cm2
1
2
U = C  V 
2


1
2
-12
U = 1.78 10  2020 
2
U = 3.63 10-6 J
V=2020 V
d=5 mm
124
Ubefore = 5.36 10-7 J
Uafter = 3.63 10-6 J
Uafter
= 6.7
Ubefore
Huh?? The energy stored increases by a factor of ??
Sure. It took work to remove the dielectric. The stored energy
increased by the amount of work done.
U = Wexternal
125