Transcript Lec08

Lecture 8-1
High Electric Field at Sharp Tips
Two conducting spheres are
connected by a long
conducting wire. The total
charge on them is Q = Q1+Q2.
Potential is the same:
kQ1 kQ2

R1
R2

Q1 R1

Q2 R2
With same potential, sphere with smaller radius carry
smaller amount of charge
kQ1
E1  2
R1
kQ2
E2  2
R2
E1 R2


E2 R1
The smaller the
radius of curvature,
the larger the
electric field.
Lecture 8-2
Potential and Conductors
Equipotential
surfaces
• Entire conductor including its surface(s) has uniform V.
•E  an equipotential surface everywhere
•Equipotential surfaces are drawn at constant intervals of V
•Potential difference between nearby equipotentials is
approximately equal to E times the separation distance.
Lecture 8-3
Again, Electrostatic Potential Energy
The electrostatic potential energy of a system (relative to ∞)
is the (external) work needed to bring the charges from an
infinite separation to their final positions.
q1
r13
r12
q2
kq1q2 kq1q3 kq2 q3
U W 


r12
r13
r23
r23
q3
 kq3 kq1 
 kq1 kq2  
1   kq2 kq3 
  q1 



  q2 
  q3 

2   r12
r13 
r
r
r
r
12 
23  
 23
 13
1
  q1V1  q2V2  q3V3 
2
Generally for n charges:
1 n
U   qV
i i
2 i 1
Lecture 8-4
Electrostatic Potential Energy of Conductors
bring dq in from ∞
q,v
R
q+dq,v+dv
kq
U  W  v dq  dq
R
Spherical
conductor
Total work to build charge from 0 up to Q:
U   U  
Q
0
2
kq
kQ
1
dq 
 QV
R
2R 2
Lecture 8-5
Capacitors
Q
Q
+Q
-Q
Q
• A capacitor is a device that is capable of
storing electric charges and thus electric
potential energy.
=> charging
• Its purpose is to release them later in a
controlled way.
=> discharging
• Capacitors are used in vast majority of
electrical and electronic devices.
Typically made of two
conductors and, when
charged, each holds equal
and opposite charges.
Lecture 8-6
Capacitance
The two conductors
hold charge +Q and
–Q, respectively.
• Capacitor plates hold charge Q
• The capacitance C of a
capacitor is a measure of how
much charge Q it can store for a
given potential difference V
between the plates.
Expect
Q  V
Q
 Let C 
V
C  
Q   Coulomb  F
farad
V  Volt
(Often we use V to mean V.)
Lecture 8-7
Steps to calculate capacitance C
1.
2.
3.
4.
Put charges Q and -Q on the two conductors, respectively.
Calculate the electric field E between the plates due to the
charges Q and -Q, e.g., by using Gauss’s law.
Calculate the potential difference V between
the plates
b
due to the electric field E by Vba    E dl
a
Calculate the capacitance of the capacitor by dividing the
charge by the potential difference, i.e., C = Q/V.
Lecture 8-8
Example: (Ideal) Parallel-Plate Capacitor
1. Put charges +q and –q on plates of
area A and separation d.
2. Calculate E by Gauss’s Law
EA
q
0
b
3. Calculate V by
b
V  Vb  Va    E  dl  Ed
a
a
4. Divide q by V
q
q
q
0 A
C 


V Ed  q 
d

d
 0 A 
• q is indeed prop. to V
• C is prop. to A
• C is inversely prop. to d
Lecture 8-9
Ideal vs Real Parallel-Plate Capacitors
Ideal
• Uniform E between plates
• No fields outside
• Often described as “thin”
Real
• Non-uniform E, particularly
at the edges
• Fields leak outside
Lecture 8-10
Physics 241 –warm-up quiz
Two identical ideal parallel plate capacitor. Both initially carry
charge Q. One is always connected with an ideal battery while
the other is not. After increase the distance between the plates
by a factor of two, which of the following statements is true.
A
B
a) E field is not changed in both A and B .
b) E field in A is half of the original value; E field in B
is not changed.
c) E field in A is not changed; E field in B is half of the
original value.
d) E field is half of the original value in both A and B .
e) One need to know the voltage of the battery to give
the answer
Lecture 8-11
Numerical magnitudes
Let’s say:
Area A= 1 cm² ,
Then
C
separation d=1 mm
0 A
d
8.85  1012 (C 2 / Nm 2 )  10 4 ( m 2 )

103 ( m)
 8.85  1013 (C / N  m)  8.85  10 13 ( F )
This is on the order of 1 pF (pico farad).
Generally the values of typical capacitors are
more conveniently measured in µF or pF.
Lecture 8-12
Long Cylindrical Capacitor
1. Put charges +q on inner cylinder of
radius a, -q on outer cylindrical shell
of inner radius b.
2.
Calculate E by Gauss’ Law
b
a
+q
L
-q
q
q
E  2 rL   E 
0
2 0 rL
3. Calculate V
a
b
q
q
V    E  dl  
dr 
ln  b / a 
2 0rL
2 0 L
b
a
4. Divide q by V
q
C

V
•
2 0 L •

ln  b / a  ln  b / a  • C dep. log. on a, b
q
q
2 0 L
q V
CL
Lecture 8-13
Spherical Capacitor
1. Put charges +q on inner sphere of radius
a, -q on outer shell of inner radius b.
2.
Calculate E by Gauss’s Law
q
E 
4 0 r 2
3. Calculate V from E
a
dr
q 1 1
V  Va  Vb  

  
2

4 0 b r
4 0  a b 
• q is proportional to V
4. Divide q by V
q
q
ab
C   4 0
V
ba
• C only depends on a,b
C  4 0a as b  
(isolated sphere)
Lecture 8-14
Energy of a charged capacitor
How much energy is stored in a charged capacitor?
Calculate the work required
(usually provided by a battery)
to charge a capacitor to Q
Calculate incremental work dW needed
to move charge dq from negative plate
to the positive plate at voltage V.
dW  V (q)  dq   q / C   dq
Total work is then
Q
1
1 Q2
U   dW   qdq 
C0
2 C
2
1
QV
Q
U  CV 2 

2
2
2C
Lecture 8-15
Where is the energy stored?
• Energy is stored in the electric field itself. Think of the
energy needed to charge the capacitor as being the energy
needed to create the electric field.
• To calculate the energy density in the field, first consider the
constant field generated by a parallel plate capacitor, where
-Q
-------- ------
++++++++ +++++++
+Q
1 Q2 1 Q2
U

2 C 2 ( A 0 / d )
• The electric field is given by:

Q
E 
0 0 A

1 2
U  E  0 Ad
2
This is the energy
density, u, of the
electric field….
• The energy density u in the field is given by:
U
U 1 2
u

 0E
volume Ad 2
Units:
J / m3
Lecture 8-16
Physics 241 –Quiz 6a
If we are to make an ideal parallel plate capacitor of
capacitance C = 1 F by using square plates with a
spacing of 1 mm, what would the edge length of the
plates be? Pick the closest value.
a)
b)
c)
d)
e)
10 m
1 cm
100 m
1 mm
1m
 0  8.85 1012 (C 2 / N  m2 )
Lecture 8-17
Physics 241 –Quiz 6b
We have an ideal parallel plate capacitor of
capacitance C=1 nF that is made of two square
plates with a separation of 2 m. A battery keeps a
potential difference of 15 V between the plates. What is
the electric field strength E between the plates?
a)
b)
c)
d)
e)
30 N/C
5 x 109 V/m
15,000 V/m
7.5 x 106 N/C
Not enough information to tell
 0  8.85 1012 (C 2 / N  m2 )
Lecture 8-18
Physics 241 –Quiz 6c
We have am ideal parallel plate capacitor of
capacitance C=1 F that is made of two square
plates. We could double the capacitance by
a)
b)
c)
d)
e)
doubling the separation d
doubling charges on the plates Q
halving potential difference between the plates V
increasing the edge length a of the plates by a factor of 1.4
a
halving the area of the plates A
a
 0  8.85 1012 (C 2 / N  m2 )
Q
Q
A
d