Physics 2112 Unit 7: Conductors and Capacitance Today’s Concept: Conductors Capacitance Electricity & Magnetism Lecture 7, Slide 1
Physics 2112
Unit 7: Conductors and Capacitance
Today’s Concept:
Conductors
Capacitance
Electricity & Magnetism Lecture 7, Slide 1
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.
3. Definition of Potential: =
b
a
baba ldE
q
UV
CONCEPTS DETERMINE THE CALCULATION !
Comments
2. Gauss’ Law: If charge distributions have sufficient symmetry (spherical,
cylindrical, planar), then Gauss’ law can be used to determine the electric field everywhere.
=0
enclosedQAdE
Electricity & Magnetism Lecture 7, Slide 2
Conductors
Charges free to move
E = 0 in a conductor
Surface = Equipotential
E at surface perpendicular to surface
The Main Points
Electricity & Magnetism Lecture 7, Slide 3
CheckPoint: Two Spherical Conductors 1
Electricity & Magnetism Lecture 7, Slide 4
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
CheckPoint: Two Spherical Conductors 2
Electricity & Magnetism Lecture 7, Slide 5
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
CheckPoint: Two Spherical Conductors 3
Electricity & Magnetism Lecture 7, Slide 6
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. QA increases B. QA decreases C. QA does not
change
Electric Circuit Element
Capacitor
Unit 7, Slide 7
Same uses as spring in mechanical system • smooth out rough spots • store energy • cause controlled oscillations
Simplest Example: Parallel plate capacitor
Q
E
Capacitor (II)
Unit 7, Slide 8
QCV
d
+Q
|| Fkx
Define capacitance, C, such that:
Q
V Units of Farad, F = Coulomb/Volt
Key Points
Unit 7, Slide 9
• +Q and –Q always have
same magnitude
• Charges don’t more
directly from one plate to
the other
• Charged from the
outside
First determine E field produced by charged conductors:
Second, integrate E to find the potential difference V
x
y
As promised, V is proportional to Q ! • Method good for
all cases • Formula good for
parallel plate only
Review of Capacitance Example
o
E
=
=
d
ydEV0
What is ?
A = area of plate
A
Q=
+Q
Q
d E
dA
QdyEEdyV
o
d d
===
0 0
)(
d
AC 0
=AQd
Q
V
QC
o/=
Unit 7, Slide 10
Example 7.1 (Capacitor)
Unit 7, Slide 11
A flat plate capacitor has a
capacitance of C = 10pF and an area
of A=1cm2. What is the distance
between the plates?
CheckPoint Results: Charged Parallel Plates 1
Electricity & Magnetism Lecture 7, Slide 12
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
CheckPoint Results: Charged Parallel Plates 1
Electricity & Magnetism Lecture 7, Slide 13
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
Conceptual Idea:
Find V in terms of some general Q and divide Q out.
Example 7.2 (Linear Capacitor)
metal
metal
a1
a2
a3
a4 cross-section 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 ?
Electricity & Magnetism Lecture 7, Slide 14
V
QC
Plan: • 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
Example 7.2 (Linear Capacitor)
metal
metal
a1
a2
a3
a4 cross-section 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 ?
Unit 7, Slide 15
Do Limiting Cases Work? • L gets bigger, C gets bigger
• a2 –> a3, C gets bigger
Energy in Capacitors
Electricity & Magnetism Lecture 7, Slide 16
= VdqU = dqC
q
U is equal to the amount of work took to put all the
charge on the two plates:
Example 7.3 (Energy in Capacitor)
Unit 7, Slide 17
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?