Transcript Lecture16

Chapter 16: Electric Energy and
Capacitance
Homework assignment: 9,18,28,34,44,51 (Six problems!)
Potential Difference and Electric Potential
 Work
and electric potential energy
• Consider a small positive charge placed in a uniform electric field E.
A
+ + + ++ + + ++ +
F  qE ; it does not depend the test charge location
d
q

E
- - - - - - - - - B
WAB  Fd  qEd  0
the force is in the same direction as the
net displacement of the test charge
conservative
force
The work done by a conservative force can be reinterpreted as the
negative of the change in a potential energy associated with that force.
PE  WAB  qEd SI unit : joule (J)
Electric Potential Energy
 Work
and electric potential energy (cont’d)
• In analogy to the gravitational force, a potential can be defined as:
PE  qEy (c. f . PEg  mgy)
• When the test charge moves from height yA to height yB , the work done
on the charge by the field is given by:
 PE  WAB  ( PEB  PE A )  (qEyB  qEy A )  qEy
If y A  y B , PE  0 and the potential decreases.
• U increases (decreases) if the test charge moves in the direction
opposite to (the same direction as) the electric force
PE<0

E
+
+
PE>0
A


F  qE
B

E
+
+
PE<0
PE>0
B
 A
F  qE


F  qE

E
-
A
B

E
-
B


F  qE
A
Electric Potential Energy
 Work
and electric potential energy (cont’d)
• In analogy to the gravitational force, a potential can be defined as:
PE  qEy (c. f . PEg  mgy)
• When the test charge moves from height yA to height yB , the work done
on the charge by the field is given by:
 PE  WAB  ( PEB  PE A )  (qEyB  qEy A )  qEy
If y A  y B , PE  0 and the potential decreases.
• Define an electric potential difference as:
PE
V  VB  VA 
SI unit : joule per coulomb or volt (J/C or V)
q
The change in electric potential energy as a charge q moves from A to B
divided by charge q.
Then for a special case of a uniform electric field:
PE
  E x x  V   E x x
q
SI unit : J/C=V=N/C
Electric Potential Energy
 Example
16.1 : Potential energy differences in an electric
field
• A proton is released from rest at x=-2.00 cm in a constant electric
field with magnitude 1.50x103 V/C pointing the positive x-direction.
(a) Calculate the change in the electric potential energy associated
with the proton when it reaches x=5.00 cm.
PE  qEx x  qEx ( x f  xi )  1.68 1017 J : q  e
(b) Find the change in electric potential energy associated with an
electron fired from x=-2.00 cm and reaching x=12.0 m.
PE  qEx x  qEx ( x f  xi )  3.36 1017 J : q  e
(b) Find the change in electric potential energy associated with an
electron fired from x=3.00 cm to x=7.00 cm if the direction of the
electric field is reversed.
PE  qEx x  qEx ( x f  xi )  9.60 1018 J : q  e
Electric Potential Energy
 Example
16.2 : Dynamics of charged particles
• Continuation of Example 16.1.
(a) Find the speed of the proton at x=0.0500 m in part (a) of Example
16.1.
1

KE  PE  0   mv 2  0   PE  0
2

2
2
v   PE  v   PE  1.42 105 m/s
m
m
(b) Find the electron’s initial speed, given that its speed has fallen
by half at x=0.120 m.
1
1

KE  PE  0   mv 2f  mvi2   PE  0
2
2

1 1 2 1 2
3 2
m( vi )  mvi  PE  0   mvi  PE
2 2
2
8
8PE
vi 
 9.92 106 m/s
3m
2
Electric Potential Energy
 Example
16.3 : TV tubes and atom smasher
• Charged particles are accelerated through potential difference.
Suppose a proton is injected at a speed of 1.00x106 m/s between
two plates 5.00 cm apart. The proton subsequently accelerates
across the gap and exits through the opening.
(a) What must the electric potential difference be if the exist speed
is to be 3.00x106 m/s?
KE  PE  KE  qV  0
1 2 1 2
mv  mv
KE 2 f 2 i
V  

 4.18 104 V
q
q
(b) What is the magnitude of the electric field between the plates?
E
V
 8.36  105 N/C
x
Electric Potential and Potential Energy
 Electric
potential by a point charge
• The electric field of a point charge extends throughout space, so does
its electric potential.
• The zero point can be anywhere but it is convenient to choose the
point at infinity. Then it can be shown that the electric potential due
to a point charge q at a distance r is given by :
V
1
q
4 r
• The 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.
Electric Potential and Potential Energy
 Electric
potential energy of a pair of point charges
• If V1 is the electric potential due to charge q1 at a point P, then
the work required to bring charge q2 from infinity to P without
acceleration is q2V1.
This work is, by definition, equal to the potential energy PE of the
two-particle system when the particles are separated by a distance r.
1
q1q2
PE  q2V1 
40 r
 Electric
If q1q2>0, PE>0
If q1q2<0, PE<0
potential energy with several point charges

q0  q1 q2
q0
qi
   ... 
U
From the superposition principle

i
40  r1 r2
ri
 40
ri : the distance between charge qi and charge q 0
Electric Potential and Potential Energy
 Example
16.4 : Finding the electric potential
y(m)
• A 5.00-mC point charge is at
(0,4.00) P
origin and a point charge q2=-2.00mC
is on the x-axis at (3.00,0) m.
r1
(a) Find the electric potential at point
P due to these charges.
q1
2
6
q 
N  m  5.00 10 C 
+

  1.12 104 V
V1  ke 1   8.99 109
2
0
r1 
C  4.00 m 
2
6
q2 
C
9 N  m   2.00 10
4



V2  ke
  8.99 10


0
.
360

10
V
2

r2 
C 
5.00 m

r2
q2
-
x(m)
(3.00,0)
VP  V1  V2  7.60 103 V
(b) Find the work needed to bring the 4.00-mC charge from infinity to P.
W  PE  q3V  q3 (VP  V )  (4.00 10 6 C)(7.60 103  0)  3.04 10 2 J
Potentials and Charged Conductors
 Surface
electric potential of a conductor in electrostatic
equilibrium
• Work done to move a charge from point A to point B
W  PE
PE  q(VB  VA )
W  q(VB  VA )
No net work is needed to move a charge between
two points that are at the same electric potential.
• We will learn that when a charged conductor is in electrostatic
equilibrium:
-A net charge placed on it resides entirely on its surface.
-The electric field just outside its surface is perpendicular to
the surface and that the field inside the conductor is zero.
-All points on its surface are at the same potential.
Potentials and Charged Conductors
 Equipotential
surface
• E = 0 everywhere inside a conductor
- At any point just inside the conductor the component of E tangent to
the surface is zero
- The tangential
component of E is also zero just outside the surface

E
vacuum
E
E//
conductor

E 0
If it were not, a charge could move around a
rectangular path partly inside and partly outside
and return to its starting point with a net amount
of work done on it.
• When all charges are at rest, the electric field just outside a conductor must
be perpendicular to the surface at every point
• When all charges are at rest, the surface of a conductor is always an
equipotential surface
Potentials and Charged Conductors
 Examples
of equipotential surface
The equipotentials are perpendicular to the electric field lines
at every point.
Potentials and Charged Conductors
 Electron
volt
The electron volt is defined as the kinetic energy that an electron
gains when accelerated through a potential difference of 1V.
1 V = 1 J/C
1 eV = 1.60 x 10-19 C.V = 1.60 x 10-19 J
In atomic, nuclear and particle physics, the electron volt is used
commonly to express energies.
- Electrons in normal atoms have energies of tens of eV’s.
- Excited electrons in atoms that emit x-rays have energies
of thousands of eV’s ( keV = 103 eV).
- High energy gamma rays emitted by the nucleus have
energies of millions of eV’s (MeV = 106 eV).
- The world most energetic accelerator near Chicago accelerates
protons/anti-proton up to Tera eV’s (TeV = 1012 eV )
Capacitance
 Capacitor
• Any two conductors separated by an insulator (or a vacuum) form a
capacitor
• In practice each conductor initially has zero net charge and electrons
are transferred from one conductor to the other (charging the conductor)
• Then two conductors have charge with equal
magnitude and opposite sign, although the net
charge is still zero
• When a capacitor has or stores charge Q , the
conductor with the higher potential has charge
+Q and the other -Q if Q>0
Capacitors and Capacitance
 Capacitance
• One way to charge a capacitor is to connect these conductors to opposite
terminals of a battery, which gives a fixed potential difference Vab between
conductors ( a-side for positive charge and b-side for negative charge). Then
once the charge Q and –Q are established, the battery is disconnected.
• If the magnitude of the charge Q is doubled, the electric field becomes
twice stronger and Vab=V is twice larger.
• Then the ratio Q/V is still constant and it is called the capacitance C.
C
Q
V
units
1 F  1 farad  1 C/V  1 coulomb/vo lt
• When a capacitor has or stores charge Q , the conductor
with the higher potential has charge +Q and the other
-Q if Q>0
-Q
Q
b
a
Parallel-Plate Capacitance
 Parallel-plate
• Charge density:
• Electric field:
capacitor in vacuum
q

A

q
E

0 0 A
• Potential diff.:
V  Ed 
• Capacitance:
C
1 qd
0 A
q
A
 0
V
d
• The capacitance depends only on the
geometry of the capacitor.
• It is proportional to the area A.
• It is inversely proportional to the separation d
• When matter is present between the plates, its
properties affect the capacitance.
Parallel-Plate Capacitance
 Units
1 F = 1 C2/N m (Note []C2/N m2)
0 = 8.85 x 10-12 F/m
 Example
1 mF = 10-6 F, 1 pF = 10-12 F
: Size of a 1-F capacitor
d  1 mm , C  1.0 F
(1.0 F)(1.0 10 3 m)
8
2
A


1
.
1

10
m
0
8.85 10 12 F/m
Cd
Parallel-Plate Capacitance
 Example
: Properties of a parallel capacitor
A parallel - palte capacitor in vacuum
d  5.00 mm , A  2.00 m 2 , V  10,000 V  10.0 kV
A (8.85 10 12 F/m)(2.00 m 2 )
C  0 
d
5.00 10 3 m
 3.54 10 5 F  0.00354 mF
Q  CVab  (3.54 109 C/V)(1.00 104 V)
 3.54 105 C  35.4 mC

Q
3.54 105 C
E 

 0  0 A (8.85 1012 C2 / N  m 2 )( 2.00 m 2 )
 2.00 106 N/C
Combinations of Capacitors

Symbols for circuit elements and circuits
Combinations of Capacitors

Capacitors in parallel
Q1  C1V
a
Vab  V Q1
b
C1 Q2
C2
Q2  C2 V
Q  Q1  Q2  (C1  C2 )V
Q
 C1  C2
V
The parallel combination is equivalent to a single capacitor with the
same total charge Q=Q1+Q2 and potential difference.
Ceq  C1  C2  Ceq  i Ci
Combinations of Capacitors

Capacitors in series
a
Q
Vab  V
C1 Vac  V1
Q
Q
c
C2
Q
Vcb  V2
Q
Q
Vac  V1 
Vcb  V2 
C1
C2
1
1 

Vab  V  V1  V2  Q  
 C1 C2 
V
1
1


Q C1 C2
b
The equivalent capacitance Ceq of the series combination is defined as
the capacitance of a single capacitor for which the charge Q is the same
as for the combination, when the potential difference V is the same.
Ceq 
Q
V
1
V
1
1
1


 
Ceq
Q
Ceq C1 C2
1
1
 i
Ceq
Ci
Combinations of Capacitors

Capacitor networks
Combinations of Capacitors

Capacitor networks (cont’d)
Combinations of Capacitors

Capacitor networks 2
C
C
C
C
C
A
A
C
C
C
C
C
1
C
3
B
B
C
C
C
C
C
C
C
A
A
15
C
41
4
C
3
C
B
B
C
C
Energy Stored in a Charged Capacitor

Work done to charge a capacitor
• Consider a process to charge a capacitor up to Q with the final potential
difference V.
Q
V 
C
• Let qi and (v)i be the charge and potential difference at an intermediate stage
during the charging process.
qi
(v) i 
C
• At this stage the work (W)i required to transfer an additional element of
q q
charge q is:
(W ) i  (v) i q  i
V
C
• The total work needed to increase the capacitor
charge q from zero to Q is:
N
q q 1
W  lim  i
 QV
qi
N 
C
2
i 1
(v)i
• The energy stored:
1
1
Q2
Q
2
Energy stored  QV  C (V  
q
2
2
2C
Capacitor with Dielectrics

Dielectric materials
• Experimentally it is found that when a non-conducting material
(dielectrics) between the conducting plates of a capacitor, the
capacitance increases for the same stored charge Q.
• Define the dielectric constant k as:
C
C0 : capacitance w/o dielectric
k
C0
C : capacitance w/ dielectric
• When the charge is constant, Q  C0 V0  CV  C / C0  V0 / V
V 
V0
k
E
Material
vacuum
air(1 atm)
Teflon
Polyethelene
E0
k
( E 
k
1
1.00059
2.1
2.25
V
)
d
Material
Mica
Mylar
Plexiglas
Water
C0   0
k
3-6
3.1
3.40
80.4
A
C
,k
d
C0
C  k 0
A
A

d
d
  k 0
Dielectrics

Induced charge and polarization
• Consider a two oppositely charged parallel plates with vacuum
between the plates.
• Now insert a dielectric material of dielectric constant k.
E  E0 / k when Q is constant
• Source of change in the electric field is redistribution of positive
and negative charge within the dielectric material (net charge 0).
This redistribution is called a polarization and it produces induced
charge and field that partially cancels the original electric field.

E0 
0

   ind
E
0
E
E0
k
1
 ind   1   and define the permittivi ty   k 0
 k

E

C  kC0  k 0
A
A
1
1

u  k 0 E 2  E 2
d
d
2
2
Dielectrics

Molecular model of induced charge
Dielectrics

Molecular model of induced charge (cont’d)