Physics 2112
Unit 7: Conductors and Capacitance
Today’s Concept:


Conductors
Capacitance
Electricity & Magnetism Lecture 7, Slide 1
Comments
THERE ARE ONLY THREE THINGS YOU NEED TO KNOW TO DO ALL OF
HOMEWORK
1. E = 0 within the material of a conductor:
Charges move inside a conductor in order to cancel out the fields that would be there in the
absence of the conductor. This principle determines the induced charge densities on
the surfaces of conductors.
2. Gauss’ Law:
If charge distributions have sufficient symmetry (spherical,
cylindrical, planar), then Gauss’ law can be used to determine
the electric field everywhere.
  Qenclosed
 E  d A = 0
 
U ab

=   E  dl
q
a
b
3. Definition of Potential:
Vab
CONCEPTS DETERMINE THE CALCULATION !
Electricity & Magnetism Lecture 7, Slide 2
Conductors
The Main Points




Charges free to move
E = 0 in a conductor
Surface = Equipotential
E at surface perpendicular to
surface
Electricity & Magnetism Lecture 7, Slide 3
CheckPoint: Two Spherical Conductors 1
Two spherical conductors are separated by a large distance.
They each carry the same positive charge Q. Conductor A has
a larger radius than conductor B.
Compare the potential at the surface of conductor A with the potential
at the surface of conductor B.
A. VA > VB
B. VA = VB
C. VA < VB
Electricity & Magnetism Lecture 7, Slide 4
CheckPoint: Two Spherical Conductors 2
Two spherical conductors are separated by a large distance.
They each carry the same positive charge Q. Conductor A has
a larger radius than conductor B.
The two conductors are now connected by a wire. How do the
potentials at the conductor surfaces compare now?
A. VA > VB
B. VA = VB
C. VA < VB
Electricity & Magnetism Lecture 7, Slide 5
CheckPoint: Two Spherical Conductors 3
Two spherical conductors are separated by a large distance. They each carry the same
positive charge Q. Conductor A has a larger radius than conductor B.
What happens to the charge on conductor A after it is connected to conductor B by
the wire?
A.
B.
C.
QA increases
QA decreases
QA does not
change
Electricity & Magnetism Lecture 7, Slide 6
Capacitor
Electric Circuit Element
Same uses as spring in mechanical
system
•
smooth out rough spots
•
store energy
•
cause controlled oscillations
Unit 7, Slide 7
Capacitor (II)
Simplest Example: Parallel plate
capacitor
d
Define capacitance, C, such that:
Q
CV  Q
+Q
kx | F |
E
Q
V
Units of Farad, F = Coulomb/Volt
Unit 7, Slide 8
Key Points
• +Q and –Q always have
same magnitude
• Charges don’t more
directly from one plate to
the other
• Charged from the
outside
Unit 7, Slide 9
Review of Capacitance Example
y
First determine E field produced by charged conductors:
+Q
What is  ?
d
E
E=
Q
x

o
Q
=
A
A = area of plate
Second, integrate E to find the potential difference V
 
V =   E  dy
d
0
d
d
Q
V =   ( Edy ) = E  dy =
d
o A
0
0
As promised, V is proportional to Q !
Q
Q
C =
V Qd /  o A
C=
0 A
d
• Method good for
all cases
• Formula good for
parallel plate only
Unit 7, Slide 10
Example 7.1 (Capacitor)
A flat plate capacitor has a
capacitance of C = 10pF and an area
of A=1cm2. What is the distance
between the plates?
Unit 7, Slide 11
CheckPoint Results: Charged Parallel Plates 1
Two parallel plates of equal area carry equal and opposite charge Q0.
The potential difference between the two plates is measured to be V0.
An uncharged conducting plate (the green thing in the picture below)
is slipped into the space between the plates without touching either
one. The charge on the plates is adjusted to a new value Q1 such that
the potential difference between the plates remains the same.
Compare Q1 and Q0.
A. Q1 < Q0
B. Q1 = Q0
C. Q1 > Q0
Electricity & Magnetism Lecture 7, Slide 12
CheckPoint Results: Charged Parallel Plates 1
An uncharged conducting plate (the green thing in the picture below) is slipped into
the space between the plates without touching either one. The charge on the plates
is adjusted to a new value Q1 such that the potential difference between the plates
remains the same.
Compare the capacitance of the two configurations in the
above problem.
A. C1 > C0
B. C1 = C0
C. C1 < C0
Electricity & Magnetism Lecture 7, Slide 13
Example 7.2 (Linear Capacitor)
cross-section
a
a34
a2
a1
metal
A capacitor is constructed from two
conducting cylindrical shells of radii a1, a2,
a3, and a4 and length L (L >> ai).
What is the capacitance C of this capacitor ?
metal
 Conceptual Idea:
 Plan:
•
•
•
•
Q
C
V
Find V in terms of some general Q and divide Q out.
Put +Q on outer shell and Q on inner shell
Cylindrical symmetry: Use Gauss’ Law to calculate E everywhere
Integrate E to get V
Take ratio Q/V (should get expression only using geometric parameters (ai, L))
 Limiting Case:
•
•
L gets bigger, C gets bigger
a2 –> a3, C gets bigger
Electricity & Magnetism Lecture 7, Slide 14
Example 7.2 (Linear Capacitor)
cross-section
a
a34
a2
a1
metal
A capacitor is constructed from two
conducting cylindrical shells of radii a1, a2,
a3, and a4 and length L (L >> ai).
What is the capacitance C of this capacitor ?
metal
 Do Limiting Cases Work?
•
•
L gets bigger, C gets bigger
a2 –> a3, C gets bigger
Unit 7, Slide 15
Energy in Capacitors
U is equal to the amount of work took to put all the
charge on the two plates:
q
U =  Vdq = 
C
dq
Electricity & Magnetism Lecture 7, Slide 16
Example 7.3 (Energy in Capacitor)
A 8uF parallel plate capacitor is has a
potential different of 120V between its
two sides. The distance between the
plates is d=1mm.
What is the potential stored in the
capacitor?
What is the energy density of the
capacitor?
Unit 7, Slide 17