Download Lecture 9

Physics 2102
Jonathan Dowling
Physics 2102
Lecture: 11 FRI 06 FEB
Capacitance I
25.1–3
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Tutoring Lab Now Open
Tuesdays
• Lab Location: 102 Nicholson (across the
hall from class)
• Lab Hours:
MTWT: 12:00N–5:00PM
F: 12:N–3:00PM
Capacitors and Capacitance
Capacitor: any two conductors,
one with charge +Q, other with
charge –Q
–Q
Potential DIFFERENCE between
conductors = V
Q = CV where C = capacitance
Units of capacitance:
Farad (F) = Coulomb/Volt
+Q
Uses: storing and releasing
electric charge/energy.
Most electronic capacitors:
micro-Farads (F),
pico-Farads (pF) — 10–12 F
New technology:
compact 1 F capacitors
Capacitance
• Capacitance depends only
on GEOMETRICAL factors
and on the MATERIAL that
separates the two
conductors
• e.g. Area of conductors,
separation, whether the
space in between is filled
with air, plastic, etc.
+Q
–Q
(We first focus on capacitors
where gap is filled by AIR!)
Electrolytic (1940-70)
Electrolytic (new)
Paper (1940-70)
Capacitors
Variable
air, mica
Tantalum (1980 on)
Ceramic (1930 on)
Mica (1930-50)
Parallel Plate Capacitor
We want capacitance: C = Q/V
E field between the plates: (Gauss’ Law)

Q
E

0 0 A
Relate E to potential difference V:
Area of each
plate = A
Separation = d
charge/area = 
= Q/A
-Q
  d Q
Qd
dx 
V   E  dx  
 A
0 A
0 0
0
d
What is the capacitance C ?
Q 0 A
C 
V
d
 C2 m2  C2  CC C 
Units :  2
       
Nm m  Nm  J  V 
+Q
Capacitance and Your iPhone!
Q 0 A
C 
V
d
Parallel Plate Capacitor —
Example
• A huge parallel plate capacitor
consists of two square metal
plates of side 50 cm, separated by
an air gap of 1 mm
• What is the capacitance?
C = 0A/d
= (8.85 x 10–12 F/m)(0.25 m2)/(0.001 m)
= 2.21 x 10–9 F
(Very Small!!)
Lesson: difficult to get large values
of capacitance without special
tricks!
 C2 m2  C2  CC C 
Units :  2
       F  Farad
Nm m  Nm  J  V 
Isolated Parallel Plate Capacitor
• A parallel plate capacitor of capacitance C
is charged using a battery.
• Charge = Q, potential difference = V.
• Battery is then disconnected.
• If the plate separation is INCREASED,
does Potential Difference V:
(a) Increase?
(b) Remain the same?
(c) Decrease?
• Q is fixed!
• C decreases (=0A/d)
• V=Q/C; V increases.
+Q
–Q
Parallel Plate Capacitor & Battery
• A parallel plate capacitor of capacitance
C is charged using a battery.
• Charge = Q, potential difference = V.
• Plate separation is INCREASED while
battery remains connected.
Does the Electric Field Inside:
(a) Increase?
(b) Remain the Same?
• V is fixed by battery!
• C decreases (=e0A/d)
(c) Decrease?
• Q=CV; Q decreases
• E = Q/0A decreases
+Q
–Q
Spherical Capacitor
What is the electric field inside
the capacitor? (Gauss’ Law)
E
Radius of outer
plate = b
Radius of inner
plate = a
Q
40 r
2
Relate E to potential difference
between the plates:
Concentric spherical shells:
Charge +Q on inner shell,
–Q on outer shell
b
  b kQ
 kQ 
V   E  dr   2 dr   
r
r

a
a
a
b
1 1
 kQ   
a b
Spherical Capacitor
What is the capacitance?
C = Q/V =
Q

Q 1 1
 

40  a b 
40 ab

(b  a)
Radius of outer
plate = b
Radius of inner
plate = a
Concentric spherical shells:
Charge +Q on inner shell,
–Q on outer shell
Isolated sphere: let b >> a,
C  40 a
Cylindrical Capacitor
QuickTi me™ and a
decompressor
are needed to see thi s pi ctur e.
What is the electric field in between
the plates? Gauss’ Law!
E
Q
2 0 rL
Relate E to potential difference
between the plates:
b
C  Q/V
 
V   E  dr
a
b
Radius of outer
plate = b
Radius of inner
plate = a
Length of capacitor = L
+Q on inner rod, –Q on outer shell
20 L

b 
ln  
a 
b
b 
 Q ln r 
Q

ln  

dr  


2

0 rL
 20 L  a 20 L a 
a
Q
cylindrical
Gaussian
surface of
radius r
Summary
• Any two charged conductors form a capacitor.
•Capacitance : C= Q/V
•Simple Capacitors:
Parallel plates:
C = 0 A/d
Spherical:
C = 4e0 ab/(b-a)
Cylindrical:
C = 20 L/ln(b/a)]