Download Lect07

Definitions & Examples
a
b
C
A
++++
L
d
a
C3
b
Q
V
C1
C2

-----
a
b
C
Today…
• Calculate E from V
• Definition of Capacitance
• Example Calculations-Parallel Plate Capacitor
• Combinations of Capacitors
• Capacitors in Parallel
• Capacitors in Series
• Lightning!
• Appendices
A. Calculate electric field of dipole from potential
B. Cylindrical capacitor examples
Last time…
• If we know E(x,y,z) we can calculate V(x,y,z):
r
V ( x, y, z )  V ( x0 , y0 , z0 )    dl E ( x, y, z )
r0
• Potential energy of a test charge in external field:
potential x test charge (q)
U  qV
• Conductors in E-fields “become” equipotential
surfaces/volumes
– E-field always normal to surface of conductor
E from V?
• We can obtain the electric field E from the potential V
by inverting our previous relation between E and V:

r

r  xˆ dx
V
V+dV

dV   E  xˆ dx   E x dx
• Expressed as a vector, E is the negative gradient of V
• Cartesian coordinates:
• Spherical coordinates:
• Note: The units of E [= N/C] can be expressed [V/m].
Preflight 7:
This graph shows the electric
potential at various points
along the x-axis.
2) At which point(s) is the electric field zero?
A
B
C
D
E from V: an Example
• Consider the following electric potential:
• What electric field does this describe?
... expressing this as a
vector:
• Something for you to try:
Can you use the dipole potential to obtain the
dipole field? Try it in spherical coordinates ...
you should get (see Appendix A):
1
Lecture 7, ACT 1
1
The electric potential in a region of space is given by
The x-component of the electric field Ex at x = 2 is
(a)
Ex = 0
(b)
Ex > 0
(c)
Ex < 0
Lecture 7, ACT 1
1
The electric potential in a region of space is given by
The x-component of the electric field Ex at x = 2 is
(a)
Ex = 0
(b)
Ex > 0
We know V(x) “everywhere”
To obtain Ex “everywhere”, use
(c)
Ex < 0
The Bottom Line
If we know the electric field E everywhere,

allows us to calculate the potential function V everywhere (keep
in mind, we often define VA = 0 at some convenient place)
If we know the potential function V everywhere,
allows us to calculate the electric field E everywhere
• Units for Potential! 1 Joule/Coul = 1 VOLT
Capacitance
• A capacitor is a device whose purpose is to store electrical
energy which can then be released in a controlled manner
during a short period of time.
• A capacitor consists of 2 spatially separated conductors
which can be charged to +Q and -Q respectively.
• The capacitance is defined as the ratio of the charge on
one conductor of the capacitor to the potential difference
between the conductors.
[The unit of capacitance is
Q
C
the Farad: 1 F = 1C/V]
V
• The capacitance belongs only to the capacitor,
independent of the charge and voltage.
Example:
Parallel Plate Capacitor
• Calculate the capacitance. We
assume +s, -s charge densities
on each plate with potential
difference V:
C
A
++++
d
-----
Q
V
• Need Q:
Q  sA
• Need V:
from def’n:
 
Vb  Va   E  dl
b
– Use Gauss’ Law to find E
a
Recall:
Two Infinite Sheets
(into screen)
-
+
s
• Field outside the sheets is zero
• Gaussian surface encloses
zero net charge
• Field inside sheets is not zero:
• Gaussian surface encloses
non-zero net charge Q  sA
 
 E  dS  AEinside
E
s
0
s
- E=0
-
+
+
E=0
+
+
A +
+
+
+
+
A
+
+
E
Example: Parallel Plate Capacitor
(see Appendix B for other examples)
•
•
Calculate the capacitance:
A
Assume +Q, -Q on plates with
potential difference V.
++++
d
s
Q
E

 0 A 0
 
Q
Vb  Va   E  dl  Ed 
d
A 0
a
b

C
-----
Q A 0

V
d
• As hoped for, the capacitance of this capacitor
depends only on its geometry (A,d).
• Note that C ~ length; this will always be the case!
Practical Application: Microphone
(“condenser”)
d
Current sensor
Battery
Moveable plate
Fixed plate
Sound waves incident
 pressure oscillations
 oscillating plate separation d
 oscillating capacitance ( C ~ 1 )
 oscillating charge on plate d
 oscillating current in wire ( I  dQ )
dt
 oscillating electrical signal
2
See this in action at http://micro.magnet.fsu.edu/electromag/java/microphone/ !
Lecture 7, ACT 2
• In each case below, a charge of +Q is placed on a solid
spherical conductor and a charge of -Q is placed on a
concentric conducting spherical shell.
– Let V1 be the potential difference between the spheres with (a1, b).
– Let V2 be the potential difference between the spheres with (a2, b).
– What is the relationship between V1 and V2? (Hint – think about
parallel plate capacitors.)
-Q
-Q
+Q
+Q
a1
b
(a) V1 < V2
(b) V1 = V2
a2
b
(c) V1 > V2
Lecture 7, ACT 2
• In each case below, a charge of +Q is placed on a solid
spherical conductor and a charge of -Q is placed on a
concentric conducting spherical shell.
– Let V1 be the potential difference between the spheres with (a1, b).
– Let V2 be the potential difference between the spheres with (a2, b).
– What is the relationship between V1 and V2? (Hint – think about
parallel plate capacitors.)
-Q
-Q
+Q
+Q
a1
b
(a) V1 < V2
(b) V1 = V2
a2
b
(c) V1 > V2
• What we have here are two spherical capacitors.
• Intuition: for parallel plate capacitors: V = (Q/C) = (Qd)/(A0).
Therefore you might expect that V1 > V2 since (b-a1) > (b-a2).
• In fact this is the case as we can show directly from the definition of V!
Capacitors in Parallel
a
a
VC
1
Q1
-Q1
b
Q2
-Q2
C2

Q
V
C
-Q
b
• Find “equivalent” capacitance C in the sense that
no measurement at a, b could distinguish the
above two situations.
• Aha! The voltage across the two is the same….

C2
Q1 Q2
Q2  Q1
Parallel Combination: V 

C1
C1 C2
Q Q1  Q2 Q1 (C1  C2 )
C



Equivalent Capacitor:
V
V
C1V

C  C1  C2
Preflight 7:
Two identical parallel plate
capacitors are shown in an endview in A) of the figure. Each
has a capacitance of C.
4) If the two are joined together as shown in B), forming a
single capacitor, what will be the final capacitance?
a) C/2
b) C
c) 2C
Capacitors in Series
+Q -Q
a
C1
+Q
C2
b

+Q
-Q
a
b
C
-Q
• Find “equivalent” capacitance C in the sense that no
measurement at a, b could distinguish the above two situations.
• The charge on C1 must be the same as the charge on
C2 since applying a potential difference across ab
cannot produce a net charge on the inner plates of
C1 and C2
– assume there is no net charge on node between C1 and C2
Q

C
RHS:
Vab
LHS:
Q Q
Vab  V1  V2  
C1 C2

1
1
1


C C1 C2
Examples:
Combinations of Capacitors
a

C3
b
C1
a
C2
b
C
• How do we start??
• Recognize C3 is in series with the parallel
combination on C1 and C2. i.e.,
1 1
1
 
C C3 C1  C2

C3 (C1  C2 )
C
C1  C2  C3
3
Preflight 7:
C
C
C
C
C
Configuration A
Configuration B
Configuration C
Three configurations are constructed using identical capacitors
6) Which of these configurations has the lowest overall capacitance?
a) Configuration A
b) Configuration B
c) Configuration C
Preflight 7:
A circuit consists of three unequal capacitors C1, C2, and C3 which are
connected to a battery of emf E. The capacitors obtain charges Q1 Q2,
Q3, and have voltages across their plates V1, V2, and V3. Ceq is the
equivalent capacitance of the circuit.
8) Check all of the following
that apply:
a) Q1= Q2
b) Q2= Q3
e) V1 < V2
f) Ceq > C1
c) V2= V3
d) E = V1
Lecture 7, ACT 3
• What is the equivalent capacitance, Ceq, of the
combination shown?
o
Ceq
C
C
C
o
(a) Ceq = (3/2)C
(b) Ceq = (2/3)C
(c) Ceq = 3C
Lecture 7, ACT 3
• What is the equivalent capacitance, Ceq, of the
combination shown?
o
Ceq
C
C
C
o
(b) Ceq = (2/3)C
(a) Ceq = (3/2)C
C
C
1 1 1
 
C1 C C
C

C
C1 
2
C
C 3
Ceq  C   C
2 2
(c) Ceq = 3C
C1
Lightning (a.k.a. the atmosphere is a BIG capacitor!!)
+
_
+
_
+
_
Collisions produce
charged particles.
The heavier
particles (-) sit near
the bottom of the
cloud; the lighter
particles (+) near
the top.
Stepped
Leader
Negatively
charged
electrons
begin
zigzagging
downward.
Attraction
As the stepped
leader nears
the ground, it
draws a
streamer of
positive charge
upward.
Flowing
Charge
As the leader
and the
streamer come
together,
powerful
electric current
begins flowing
Contact!
Intense wave of
positive charge,
a “return stroke,”
travels upward
at 108 m/s
t ~ 30ms
Factoids: V ~ 200 M volts
I ~ 40,000 amp
P ~ 1012 W
Summary
•
•
A Capacitor is an object with two spatially separated conducting
surfaces.
The definition of the capacitance of such an object is:
C
Q
V
• The capacitance depends on the geometry :
-Q
A
++++
d
----Parallel Plates
A
C
d
r
+Q
-Q
+Q
a
a
b
b
L
Cylindrical
Spherical
L
C
ln(b / a )
C
ab
ba
Appendix A: Electric Dipole
• In Lecture 5 (appendix), we found the
potential of an electric dipole to be:
z
+q
aq
a
-q
• Now use
(in spherical coordinates)
to find the electric field everywhere.
the dipole
moment

r1
r
r2
Appendix A: Dipole Field
y=
z
+q
a q
a
Etot
r
E
q
0
Er
-q
0
/



/
x=
Appendix A: Sample Problem
• Consider the dipole shown at the
right.
– Fix r = r0 >> a
z
+q
– Define qmax such that the polar
component of the electric field has its
maximum value (for r = r0).
a q
a
What is qmax?
-q
(a) qmax = 0
(b) qmax = 45
r1
r
(c) qmax = 90
• The expression for the electric
field of a dipole (r >> a) is:
• The polar component of
E is maximum when sinq is maximum.
• Therefore, Eq has its maximum value when q = 90.
r2
Appendix B: Cylindrical Capacitor Examples
• Calculate the capacitance:
• Assume +Q, -Q on surface of
r
cylinders with potential difference V.
a
b
• Gaussian surface is cylinder of
radius r (a < r < b) and length L
 
Q
• Apply Gauss' Law:  E  dS  2rLE 
0

L
E
Q
20 Lr
If we assume that inner cylinder has +Q, then the potential V is
positive if we take the zero of potential to be defined at r = b:
a
b
 
V   E  dl    Edr  
a
b
b
b
dr 
ln  
20 rL
20 L  a 
a
Q
Q

C 
2 0 L
Q

V
b
ln  
a
Appendix B: Another example
-Q
+Q
• Suppose we have 4
concentric cylinders of radii
a,b,c,d and charges +Q, -Q,
+Q, -Q
-Q
b
+Q
a
• Question: What is the
capacitance between a and d?
c
• Note: E-field between b and c is
zero! WHY??
•
A cylinder of radius r1: b < r1< c
encloses zero charge!
b
d
Q
Q
Vad  
dr  0  
dr
20 rL
20 rL
a
c
C
d
Q

Vad
b

a
20 L
d
dr
dr

r
r
c

C
20 L
b
d
ln    ln  
a
c
Note: This is just the result for 2
cylindrical capacitors in series!