PHYSICS 1B – Fall 2009
Electricity
&
Magnetism
Professor Brian Keating
SERF Building. Room 333
Wednesday October 14, 2009
Wednesday, October 21, 2009
16.1 PART 2 & 16.2
ELECTRIC POTENTIAL
(CONTINUED)
Quiz grades: on the web by last 5 digits of your PID
number
Average was an 7 with a standard deviation of 2.
Wednesday, October 21, 2009
Hydrogen Bond
N
H
O C
N
H O C
The hydrogen bond energy can be
estimated by partial charges
-0.3e +0.3e -0.4e +0.4e
N
H
0.1
O
0.2
C
0.25
nm
DNA
Wednesday, October 21, 2009
Hydrogen Bond
N
H
O C
N
H O C
The hydrogen bond energy can be
estimated by partial charges
-0.3e +0.3e -0.4e +0.4e
N
H
0.1
O
0.2
0.25
bond energy =sum
Wednesday, October 21, 2009
C
nm
(scalar sum)
DNA
Hydrogen Bond
N
H
O C
N
H O C
The hydrogen bond energy can be
estimated by partial charges
-0.3e +0.3e -0.4e +0.4e
N
H
0.1
O
0.2
0.25
bond energy =sum
Wednesday, October 21, 2009
C
nm
(scalar sum)
DNA
Hydrogen Bond
N
H
O C
N
H O C
The hydrogen bond energy can be
estimated by partial charges
-0.3e +0.3e -0.4e +0.4e
N
H
0.1
O
0.2
0.25
bond energy =sum
Wednesday, October 21, 2009
C
nm
(scalar sum)
DNA
Hydrogen Bond
N
H
O C
N
H O C
The hydrogen bond energy can be
estimated by partial charges
-0.3e +0.3e -0.4e +0.4e
N
H
0.1
O
0.2
0.25
bond energy =sum
ΔPE=-0.22 eV
Wednesday, October 21, 2009
C
nm
(scalar sum)
DNA
Hydrogen Bond
N
H
O C
N
H O C
The hydrogen bond energy can be
estimated by partial charges
-0.3e +0.3e -0.4e +0.4e
N
H
0.1
O
0.2
0.25
bond energy =sum
ΔPE=-0.22 eV
Wednesday, October 21, 2009
C
nm
(scalar sum)
DNA
Weaker than a ionic bond but
still significant.
Two charges of +q each are placed at corners of an
equilateral triangle, with sides of 10 cm. If the Electric
field due to each charge is 100 V/m at the A find the
potential at A
A
d=10 cm
B
Wednesday, October 21, 2009
C
Two charges of +q each are placed at corners of an
equilateral triangle, with sides of 10 cm. If the Electric
field due to each charge is 100 V/m at the A find the
potential at A
V at A due to each charge
A
d=10 cm
B
Wednesday, October 21, 2009
C
Two charges of +q each are placed at corners of an
equilateral triangle, with sides of 10 cm. If the Electric
field due to each charge is 100 V/m at the A find the
potential at A
V at A due to each charge
A
d=10 cm
B
Wednesday, October 21, 2009
C
Two charges of +q each are placed at corners of an
equilateral triangle, with sides of 10 cm. If the Electric
field due to each charge is 100 V/m at the A find the
potential at A
V at A due to each charge
A
d=10 cm
B
C
Vtotal=VBA +VCA=2V =20 V
Potential is a scalar
Wednesday, October 21, 2009
Two charges of +q each are placed at corners of an
equilateral triangle, with sides of 10 cm. The Electric
field due to each charge is 100 V/m at A.
What is the potential at A?
A
d=10 cm
A. 10V
B. 100V
C. 1000V
B
Wednesday, October 21, 2009
C
Two charges of +q each are placed at corners of an
equilateral triangle, with sides of 10 cm. The Electric
field due to each charge is 100 V/m at A.
What is the potential at A?
A
d=10 cm
A. 10V
B. 100V
C. 1000V
B
Wednesday, October 21, 2009
C
Two charges of +q each are placed at corners of an
equilateral triangle, with sides of 10 cm. The Electric
field due to each charge is 100 V/m at A.
What is the potential at A?
A
d=10 cm
A. 10V
B. 100V
C. 1000V
B
C
Vtotal=VBA +VCA=2V =20 V
Potential is a scalar
Wednesday, October 21, 2009
3 charges of 1x10-9 C are placed at the corners of a
equilateral triangle Each side of the triangle has a length
of 1.0 cm. Find the work needed to bring the charges
together from a long distance away.
q1
PE due to Coulomb interaction
How many interactions?
PE =
q2
Wednesday, October 21, 2009
q3
3 charges of 1x10-9 C are placed at the corners of a
equilateral triangle Each side of the triangle has a length
of 1.0 cm. Find the work needed to bring the charges
together from a long distance away.
q1
PE due to Coulomb interaction
How many interactions?
PE =
q2
Wednesday, October 21, 2009
q3
3
3 charges of 1x10-9 C are placed at the corners of a
equilateral triangle Each side of the triangle has a length
of 1.0 cm. Find the work needed to bring the charges
together from a long distance away.
q1
PE due to Coulomb interaction
How many interactions?
PE =
q2
Wednesday, October 21, 2009
q3
PE12+PE13+PE23
3
3 charges of 1x10-9 C are placed at the corners of a
equilateral triangle Each side of the triangle has a length
of 1.0 cm. Find the work needed to bring the charges
together from a long distance away.
q1
PE due to Coulomb interaction
How many interactions?
PE =
q2
Wednesday, October 21, 2009
q3
PE12+PE13+PE23
3
3 charges of 1x10-9 C are placed at the corners of a
equilateral triangle Each side of the triangle has a length
of 1.0 cm. Find the work needed to bring the charges
together from a long distance away.
q1
PE due to Coulomb interaction
How many interactions?
PE =
q2
Wednesday, October 21, 2009
q3
PE12+PE13+PE23
3
The following charges are brought together from a large
distance away. What is the change in PE? Is the charge
distribution stable? (i.e. does it have a negative PE)
+q
1
a
a
2
3
-q
+q
a
Wednesday, October 21, 2009
The following charges are brought together from a large
distance away. What is the change in PE? Is the charge
distribution stable? (i.e. does it have a negative PE)
How many interactions?
+q
1
a
a
2
+q
How many positive?
a
Wednesday, October 21, 2009
3
-q
How many negative?
What is the total change in PE?
The following charges are brought together from a large
distance away. What is the change in PE? Is the charge
distribution stable? (i.e. does it have a negative PE)
How many interactions?
+q
1
a
+q
How many positive?
a
2
a
Wednesday, October 21, 2009
3
3
-q
How many negative?
What is the total change in PE?
The following charges are brought together from a large
distance away. What is the change in PE? Is the charge
distribution stable? (i.e. does it have a negative PE)
+q
1
a
a
2
+q
a
Wednesday, October 21, 2009
3
-q
How many interactions?
3
How many positive?
1
How many negative?
What is the total change in PE?
The following charges are brought together from a large
distance away. What is the change in PE? Is the charge
distribution stable? (i.e. does it have a negative PE)
+q
1
a
a
2
+q
a
Wednesday, October 21, 2009
3
-q
How many interactions?
3
How many positive?
1
How many negative?
2
What is the total change in PE?
The following charges are brought together from a large
distance away. What is the change in PE? Is the charge
distribution stable? (i.e. does it have a negative PE)
+q
1
a
a
2
+q
a
Wednesday, October 21, 2009
3
-q
How many interactions?
3
How many positive?
1
How many negative?
2
What is the total change in PE?
The following charges are brought together from a large
distance away. What is the change in PE? Is the charge
distribution stable? (i.e. does it have a negative PE)
+q
1
a
a
2
+q
a
Wednesday, October 21, 2009
3
-q
How many interactions?
3
How many positive?
1
How many negative?
2
What is the total change in PE?
The following charges are brought together from a large
distance away. What is the change in PE? Is the charge
distribution stable? (i.e. does it have a negative PE)
+q
1
a
a
2
+q
a
3
-q
How many interactions?
3
How many positive?
1
How many negative?
2
What is the total change in PE?
STABLE
Wednesday, October 21, 2009
Which of the charge distributions is the most stable?
(has the lowest PE)
+q
-q
+q
-q
+q
-q
-q
+q
A. This one
Wednesday, October 21, 2009
B. This one
Which of the charge distributions is the most stable?
(has the lowest PE)
+q
a
+q
A. This one
Wednesday, October 21, 2009
-q
+q
-q
-q
-q
+q
B. This one
Which of the charge distributions is the most stable?
(has the lowest PE)
+q
a
+q
A. This one
PE0
PE0/
Total PE
Wednesday, October 21, 2009
-q
+q
-q
-q
-q
+q
B. This one
Which of the charge distributions is the most stable?
(has the lowest PE)
+q
a
+q
A. This one
PE0
PE0/
Total PE
Wednesday, October 21, 2009
+2
-2
-q
+q
-q
-q
-q
+q
B. This one
Which of the charge distributions is the most stable?
(has the lowest PE)
+q
a
+q
A. This one
PE0
PE0/
Total PE
Wednesday, October 21, 2009
+2
-2
-2
-q
+q
-q
-q
-q
+q
B. This one
Which of the charge distributions is the most stable?
(has the lowest PE)
+q
a
+q
A. This one
PE0
PE0/
Total PE
Wednesday, October 21, 2009
+2
-2
-2
-q
+q
-q
-q
-q
+q
B. This one
Which of the charge distributions is the most stable?
(has the lowest PE)
+q
a
+q
A. This one
PE0
PE0/
Total PE
Wednesday, October 21, 2009
+2
-2
-2
-q
+q
-q
-q
-q
+q
B. This one
-4
Which of the charge distributions is the most stable?
(has the lowest PE)
+q
a
+q
A. This one
PE0
PE0/
Total PE
Wednesday, October 21, 2009
+2
-q
+q
-q
-q
-q
+q
B. This one
-2
-2
-4
+2
Which of the charge distributions is the most stable?
(has the lowest PE)
+q
a
+q
A. This one
PE0
PE0/
Total PE
Wednesday, October 21, 2009
+2
-q
+q
-q
-q
-q
+q
B. This one
-2
-2
-4
+2
Which of the charge distributions is the most stable?
(has the lowest PE)
+q
a
+q
A. This one
PE0
PE0/
Total PE
Wednesday, October 21, 2009
+2
-q
+q
-q
-q
-q
+q
B. This one
-2
STABLE
-4
-2
+2
Which of the charge distributions is the most stable?
(has the lowest PE)
+q
-q
+q
-q
+q
-q
-q
+q
#+V? #-V?
Wednesday, October 21, 2009
#+V?
#-V?
Which of the charge distributions is the most stable?
(has the lowest PE)
+q
a
+q
#+V? #-V?
Wednesday, October 21, 2009
-q
+q
-q
-q
-q
+q
#+V?
#-V?
Which of the charge distributions is the most stable?
(has the lowest PE)
+q
a
+q
#+V? #-V?
PE0
PE0/
Total PE
Wednesday, October 21, 2009
-q
+q
-q
-q
-q
+q
#+V?
#-V?
Which of the charge distributions is the most stable?
(has the lowest PE)
+q
a
+q
PE0
PE0/
Total PE
Wednesday, October 21, 2009
#+V? #-V?
+2
-2
-q
+q
-q
-q
-q
+q
#+V?
#-V?
Which of the charge distributions is the most stable?
(has the lowest PE)
+q
a
+q
PE0
PE0/
Total PE
Wednesday, October 21, 2009
#+V? #-V?
+2
-2
-2
-q
+q
-q
-q
-q
+q
#+V?
#-V?
Which of the charge distributions is the most stable?
(has the lowest PE)
+q
a
+q
PE0
PE0/
Total PE
Wednesday, October 21, 2009
#+V? #-V?
+2
-2
-2
-q
+q
-q
-q
-q
+q
#+V?
#-V?
Which of the charge distributions is the most stable?
(has the lowest PE)
+q
a
+q
PE0
PE0/
Total PE
Wednesday, October 21, 2009
#+V? #-V?
+2
-2
-2
-q
+q
-q
-q
-q
+q
#+V?
#-V?
-4
Which of the charge distributions is the most stable?
(has the lowest PE)
+q
a
+q
PE0
PE0/
Total PE
Wednesday, October 21, 2009
#+V? #-V?
+2
-2
-2
-q
+q
-q
-q
-q
+q
#+V?
+2
#-V?
-4
Which of the charge distributions is the most stable?
(has the lowest PE)
+q
a
+q
PE0
PE0/
Total PE
Wednesday, October 21, 2009
#+V? #-V?
+2
-2
-2
-q
+q
-q
-q
-q
+q
#+V?
+2
#-V?
-4
Which of the charge distributions is the most stable?
(has the lowest PE)
+q
a
+q
PE0
PE0/
Total PE
Wednesday, October 21, 2009
#+V? #-V?
+2
-2
-2
-q
+q
-q
-q
-q
+q
#+V?
#-V?
-4
+2
more stable
16.2 Equipotentials
Equipotential surfaces
Wednesday, October 21, 2009
Equipotential Surface - positions in space at
which the electrical potentials are equal
Example 1- A sphere centered around a point
charge
Every point on the surface of
the sphere of radius r has
the same potential
q
r
The surface of the sphere is
an equipotential surface
Wednesday, October 21, 2009
Equipotential surfaceExample 2: a charged conductor
The surface of a conductor is
an equipotential surface.
E
+
+
E field is perpendicular to the surface.
+
+
+
Component of E =0 parallel to the surface
ΔV=Ed=0
equipotential
Wednesday, October 21, 2009
Thus, No change in potential along the
surface
The interior of the conductor is an
equipotential and at the same potential as
the surface.
+
+
+
V
+
Wednesday, October 21, 2009
+
V
The interior of the conductor is an
equipotential and at the same potential as
the surface.
+
+
+
V
+
Wednesday, October 21, 2009
+
V
E=0 in the conductor
Thus, the potential
doesn’t change from the
surface potential
14
Wednesday, October 21, 2009
PHYSICS 1B – Fall 2009
Electricity
&
Magnetism
Professor Brian Keating
SERF Building. Room 333
Friday October 16, 2009
Wednesday, October 21, 2009
The equipotential surfaces are perpendicular to E field lines.
E field line
Equipotential
Wednesday, October 21, 2009
Equipotential lines: point charge
Wednesday, October 21, 2009
Equipotential lines: dipole
Which line type/style represents the electric field?
A. solid
B. dashed
Wednesday, October 21, 2009
Draw a sketch of the equipotential surfaces for a electric
dipole (+q, –q) in a plane through both charges
+q
Wednesday, October 21, 2009
-q
Draw a sketch of the equipotential surfaces for a electric
dipole (+q, –q) in a plane through both charges
+q
Wednesday, October 21, 2009
-q
Draw a sketch of the equipotential surfaces for a electric
dipole (+q, –q) in a plane through both charges
V>0
+q
Wednesday, October 21, 2009
-q
Draw a sketch of the equipotential surfaces for a electric
dipole (+q, –q) in a plane through both charges
V>0
+q
Wednesday, October 21, 2009
-q
Draw a sketch of the equipotential surfaces for a electric
dipole (+q, –q) in a plane through both charges
V>0
V<0
+q
Wednesday, October 21, 2009
-q
Draw a sketch of the equipotential surfaces for a electric
dipole (+q, –q) in a plane through both charges
V>0
V<0
+q
Wednesday, October 21, 2009
-q
Draw a sketch of the equipotential surfaces for a electric
dipole (+q, –q) in a plane through both charges
V=0
V>0
V<0
+q
Wednesday, October 21, 2009
-q
Suppose the two charges are 10 cm apart and the
equipotential surfaces are as labeled estimate the E field
between the two charges.
V=0
V<0
+q
10 V
Wednesday, October 21, 2009
-q
-10 V
Suppose the two charges are 10 cm apart and the
equipotential surfaces are as labeled estimate the E field
between the two charges.
V=0
V<0
+q
10 V
Wednesday, October 21, 2009
-q
-10 V
Rutherford Scattering experiment
Determination of the size of the nucleus
α particle He nucleus
q=+2e
m =6.64x10-27 kg
v=2.0x107 m/s
recoil
gold
foil
gold nuclei
Q=+79e
Wednesday, October 21, 2009
Rutherford Scattering experiment
Determination of the size of the nucleus
α particle He nucleus
q=+2e
m =6.64x10-27 kg
v=2.0x107 m/s
recoil
gold
foil
gold nuclei
Q=+79e
Wednesday, October 21, 2009
Rutherford Scattering experiment
Determination of the size of the nucleus
α particle He nucleus
q=+2e
m =6.64x10-27 kg
v=2.0x107 m/s
recoil
gold
foil
gold nuclei
Q=+79e
Wednesday, October 21, 2009
Rutherford Scattering experiment
Determination of the size of the nucleus
α particle He nucleus
q=+2e
m =6.64x10-27 kg
v=2.0x107 m/s
recoil
gold
foil
gold nuclei
Q=+79e
Wednesday, October 21, 2009
Rutherford Scattering experiment
Determination of the size of the nucleus
α particle He nucleus
q=+2e
m =6.64x10-27 kg
v=2.0x107 m/s
d
closest approach
recoil
gold
foil
gold nuclei
Q=+79e
Wednesday, October 21, 2009
Rutherford Scattering experiment
Determination of the size of the nucleus
α particle He nucleus
q=+2e
m =6.64x10-27 kg
v=2.0x107 m/s
d
closest approach
recoil
ΔKE=-ΔPE
gold
foil
gold nuclei
Q=+79e
Wednesday, October 21, 2009
Rutherford Scattering experiment
Determination of the size of the nucleus
α particle He nucleus
q=+2e
m =6.64x10-27 kg
v=2.0x107 m/s
d
closest approach
recoil
ΔKE=-ΔPE
gold
foil
gold nuclei
Q=+79e
nuclear size < d, much smaller than size of an atom~ 0.3x10-9 m
Wednesday, October 21, 2009
Parallel plate capacitor
-
-
+
-
-
+
+
-
+
+
FIELD LINES IN BLACK (VECTORS)
POTENTIAL CONTOURS IN RED
(NO ARROWS, BECAUSE NOT A VECTOR)
Wednesday, October 21, 2009
Deflection of an electron beam in an electric field
eE
e+
Calculation – velocity, acceleration – Next Slide:
calculate the angle the electron exits at…
Wednesday, October 21, 2009
An electron beam passes through two parallel plates
of a length 10 cm having an electric field of E.
The initial velocity of the electron is 1.0x107 m/s. Find
the angle through which the beam is deflected.
L=10 cm
evx
=1.0x107
m/s
Wednesday, October 21, 2009
E
evy
vx
θ
An electron beam passes through two parallel plates
of a length 10 cm having an electric field of E.
The initial velocity of the electron is 1.0x107 m/s. Find
the angle through which the beam is deflected.
L=10 cm
evx
=1.0x107
m/s
Wednesday, October 21, 2009
E
evy
vx
θ
An electron beam passes through two parallel plates
of a length 10 cm having an electric field of E.
The initial velocity of the electron is 1.0x107 m/s. Find
the angle through which the beam is deflected.
L=10 cm
evx
=1.0x107
m/s
Wednesday, October 21, 2009
E
evy
vx
θ
An electron beam passes through two parallel plates
of a length 10 cm having an electric field of E.
The initial velocity of the electron is 1.0x107 m/s. Find
the angle through which the beam is deflected.
L=10 cm
evx
=1.0x107
m/s
Wednesday, October 21, 2009
E
evy
vx
θ
An electron beam passes through two parallel plates
of a length 10 cm having an electric field of E.
The initial velocity of the electron is 1.0x107 m/s. Find
the angle through which the beam is deflected.
L=10 cm
evx
=1.0x107
m/s
Wednesday, October 21, 2009
E
evy
vx
θ
An electron beam passes through two parallel plates
of a length 10 cm having an electric field of E.
The initial velocity of the electron is 1.0x107 m/s. Find
the angle through which the beam is deflected.
L=10 cm
evx
=1.0x107
m/s
Wednesday, October 21, 2009
E
evy
vx
θ
Capacitance
Capacitor- a device for storing charge and energy,
can be discharged rapidly to release energy.
Applications
•Camera flash
•automobile starting system
•capacitors in electronic devices
•computer memories (store information)
•Laser flash lamp
Wednesday, October 21, 2009
laser fusion
Energy stored in a capacitor 2 x106 J released in 1-2 x10-6 s.
High Power ~ 1012 W
Wednesday, October 21, 2009
Parallel plate capacitor
-
-
+
-
-
+
+
-
+
+
FIELD LINES IN BLACK (VECTORS)
POTENTIAL CONTOURS IN RED
(NO ARROWS, BECAUSE NOT A VECTOR)
Wednesday, October 21, 2009
PHYSICS 1B – Fall 2009
Electricity
&
Magnetism
Professor Brian Keating
SERF Building. Room 333
October 19, 2009
Wednesday, October 21, 2009
+q
conductors
+ + + + + + + + + + +
+ + + + + + + + + + +
- - - - - - - - - - - - - - - - - - - - - - - - -
-q
potential source
Eg. battery
Parallel Plate Capacitor
Work is done to separate charges
Capacitor Stores Electrical Energy
Wednesday, October 21, 2009
ΔV
Circuit Diagram
Wire = conductor
Voltage
source
+
ΔV
-
conductor
Wednesday, October 21, 2009
+q capacitor
C
-q
Wednesday, October 21, 2009
Measuring voltage
C
V
Wednesday, October 21, 2009
V
Voltmeter
Ideal voltmeter
Draws no charge
Perfect insulator
Parallel Plate Capacitor
Metal plates with area A, separated by a gap, d
containing an insulating material. (eg. Air)
A
+q
ΔV
d
-q
Capacitance – describes the
ability to store separated charge
Units C/V, farad (F)
Wednesday, October 21, 2009
A 100 microfarad capacitor is charged to 100 V. How
much charge does it store?
This is a lot of charge. Recall that 10-2 C on a sphere of
1m radius generates a potential of
q
r=1m
How does the capacitor store this charge without
high potentials? How do you get a large capacitance?
Wednesday, October 21, 2009
Gaussian surface- a cylinder with sides perpendicular
to the plane. E is constant at ends. Flux through
sides is zero.
E
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
E
A
Wednesday, October 21, 2009
where
Gaussian surface- a cylinder with sides perpendicular
to the plane. E is constant at ends. Flux through
sides is zero.
E
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
E
A
Wednesday, October 21, 2009
where
Parallel plate capacitor
two “infinite” planes of charge area A separated
by distance d where d<< A, carry charge +q, -q
A
+q/A=+σ
++++++++++++
----------------The charges are at the inner
surface of the capacitor
Wednesday, October 21, 2009
d
-q/A=-σ
Field inside the capacitor plates
By superposition of charges due to sheet of charge
A
q/A=σ
++++++++++++
-----------------
Wednesday, October 21, 2009
d
Field inside the capacitor plates
By superposition of charges due to sheet of charge
E+
A
++++++++++++
-----------------
Wednesday, October 21, 2009
q/A=σ
d
Field inside the capacitor plates
By superposition of charges due to sheet of charge
E+
A
++++++++++++
-----------------
E-
Wednesday, October 21, 2009
q/A=σ
d
Field inside the capacitor plates
By superposition of charges due to sheet of charge
E+
Ein=
A
++++++++++++
-----------------
E-
Ein=E+ +E-
Wednesday, October 21, 2009
q/A=σ
d
Field inside the capacitor plates
By superposition of charges due to sheet of charge
Eout =0
E+
Ein=
q/A=σ
++++++++++++
-----------------
EEout =0
Ein=E+ +E-
Wednesday, October 21, 2009
A
d
What is the capacitance of a parallel plate capacitor
with plates of area A separated by distance d? (in air)
+q
A
ΔV
d
-q
Wednesday, October 21, 2009
What is the capacitance of a parallel plate capacitor
with plates of area A separated by distance d? (in air)
+q
A
ΔV
d
-q
Recall from
Gauss’ law
Wednesday, October 21, 2009
E field
increases with
charge density
What is the capacitance of a parallel plate capacitor
with plates of area A separated by distance d? (in air)
+q
A
ΔV
d
-q
Recall from
Gauss’ law
rearrange
Wednesday, October 21, 2009
E field
increases with
charge density
What is the capacitance of a parallel plate capacitor
with plates of area A separated by distance d? (in air)
+q
A
ΔV
d
-q
Recall from
Gauss’ law
E field
increases with
charge density
rearrange
thus
Wednesday, October 21, 2009
to increase C
Increase A
Decrease d
How do you stabilize the separated charge
i.e. make ΔV as small as possible for a given q
(high capacitance)
+q
A
+ + + + + + + + + + +
+ + + + + + + + + + +
- - - - - - - - - - - - - - - - - - - - - - - - -
-q
area A
distance d
Wednesday, October 21, 2009
d
How do you stabilize the separated charge
i.e. make ΔV as small as possible for a given q
(high capacitance)
+q
A
+ + + + + + + + + + +
+ + + + + + + + + + +
- - - - - - - - - - - - - - - - - - - - - - - - -
-q
area A
distance d
Wednesday, October 21, 2009
as Large as possible
d
How do you stabilize the separated charge
i.e. make ΔV as small as possible for a given q
(high capacitance)
+q
A
+ + + + + + + + + + +
+ + + + + + + + + + +
- - - - - - - - - - - - - - - - - - - - - - - - -
-q
area A
as Large as possible
distance d
as SMALL as possible
Wednesday, October 21, 2009
d
Thin film capacitors
Metal film separated by thin insulators
metal
insulator
Metal
Making the area large and the insulating gap small
increases C
Wednesday, October 21, 2009
A parallel plate capacitor with 2 plates each with area 1.0 m2
separated by a distance of 1.0 mm holds +q,-q, q=10-6C
(a) Find the capacitance. (b) Find the E field in the capacitor
(c) Find ΔV across the plates
q
-q
C=
E=
ΔV=
Wednesday, October 21, 2009
A=1 m2
ΔV
d=1mm
A parallel plate capacitor with 2 plates each with area 1.0 m2
separated by a distance of 1.0 mm holds +q,-q, q=10-6C
(a) Find the capacitance. (b) Find the E field in the capacitor
(c) Find ΔV across the plates
q
-q
C=
E=
ΔV=
Wednesday, October 21, 2009
A=1 m2
ΔV
d=1mm
A parallel plate capacitor with 2 plates each with area 1.0 m2
separated by a distance of 1.0 mm holds +q,-q, q=10-6C
(a) Find the capacitance. (b) Find the E field in the capacitor
(c) Find ΔV across the plates
q
-q
C=
E=
ΔV=
Wednesday, October 21, 2009
A=1 m2
ΔV
d=1mm
A parallel plate capacitor with 2 plates each with area 1.0 m2
separated by a distance of 1.0 mm holds +q,-q, q=10-6C
(a) Find the capacitance. (b) Find the E field in the capacitor
(c) Find ΔV across the plates
q
-q
C=
E=
ΔV=
Wednesday, October 21, 2009
A=1 m2
ΔV
d=1mm
A parallel plate capacitor with 2 plates each with area 1.0 m2
separated by a distance of 1.0 mm holds +q,-q, q=10-6C
(a) Find the capacitance. (b) Find the E field in the capacitor
(c) Find ΔV across the plates
q
-q
C=
E=
ΔV=
Wednesday, October 21, 2009
A=1 m2
ΔV
d=1mm
A parallel plate capacitor with 2 plates each with area 1.0 m2
separated by a distance of 1.0 mm holds +q,-q, q=10-6C
(a) Find the capacitance. (b) Find the E field in the capacitor
(c) Find ΔV across the plates
q
-q
C=
E=
ΔV=
Wednesday, October 21, 2009
A=1 m2
ΔV
d=1mm
A parallel plate capacitor with 2 plates each with area 1.0 m2
separated by a distance of 1.0 mm holds +q,-q, q=10-6C
(a) Find the capacitance. (b) Find the E field in the capacitor
(c) Find ΔV across the plates
q
-q
C=
E=
ΔV=
Wednesday, October 21, 2009
A=1 m2
ΔV
d=1mm
A parallel plate capacitor (infinite) with a constant charge q
has its separation increased by a factor of 2.
How does this change E and ΔV?
q
q
Eo
do
-q
C
Wednesday, October 21, 2009
ΔVo
E=
2do
C
ΔV=
-q
A parallel plate capacitor (infinite) with a constant charge q
has its separation increased by a factor of 2.
How does this change E and ΔV?
q
q
Eo
do
-q
C
ΔVo
E= Eo ΔV=
2do
C
-q
is constant
Wednesday, October 21, 2009
A parallel plate capacitor (infinite) with a constant charge q
has its separation increased by a factor of 2.
How does this change E and ΔV?
q
q
Eo
do
-q
C
ΔVo
E= Eo ΔV= 2ΔVo
2do
C
-q
is constant
Wednesday, October 21, 2009
A parallel plate capacitor (infinite) with a constant voltage
difference ΔV has its separation increased by a factor of 2.
How does this change E and q?
q=
ΔVo d
qo
Eo
o
-qo
C
Wednesday, October 21, 2009
ΔVo
E=
2do
C
ΔV= ΔVo
A parallel plate capacitor (infinite) with a constant voltage
difference ΔV has its separation increased by a factor of 2.
How does this change E and q?
q=
ΔVo d
qo
Eo
o
ΔVo
-qo
C
C
ΔV=dE
Wednesday, October 21, 2009
E=
2do
ΔV= ΔVo
A parallel plate capacitor (infinite) with a constant voltage
difference ΔV has its separation increased by a factor of 2.
How does this change E and q?
q=
ΔVo d
qo
Eo
o
ΔVo
-qo
C
C
ΔV=dE
Wednesday, October 21, 2009
E=
2do
ΔV= ΔVo
A parallel plate capacitor (infinite) with a constant voltage
difference ΔV has its separation increased by a factor of 2.
How does this change E and q?
q=
ΔVo d
qo
Eo
o
ΔVo
-qo
C
C
ΔV=dE
Wednesday, October 21, 2009
E=
2do
ΔV= ΔVo
A parallel plate capacitor (infinite) with a constant voltage
difference ΔV has its separation increased by a factor of 2.
How does this change E and q?
q=
ΔVo d
qo
Eo
o
ΔVo
-qo
C
C
ΔV=dE
Wednesday, October 21, 2009
E=
2do
ΔV= ΔVo
Capacitor combinations
Capacitors connected in series
and parallel
Wednesday, October 21, 2009