Transcript Lecture 9
Physics 2102 Jonathan Dowling
Physics 2102 Lecture: 11 FRI 06 FEB
Capacitance I
25.1
–3 QuickTime™ and a decompressor are needed to see this picture.
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
Potential DIFFERENCE between conductors = V
Q = CV
Units of capacitance:
Farad
where
C =
capacitance (F) = Coulomb/Volt
+Q –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)
E
0
Q
0
A
Area of each plate
= A
Separation
= d
charge/area
=
= Q/A
Relate
E V
to potential difference
V
: 0
d
E
d x
d
0
Q
0
A dx
Qd
0
A
What is the capacitance
C
?
C
Q V
0
A d
-Q
Units : C 2 Nm 2 m 2 m C 2 Nm CC J C V
+Q
Capacitance and Your iPhone!
C
Q V
0
A 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 =
0 A/d
= (8.85 x 10 –12 F/m)(0.25 m 2 )/(0.001 m) = 2.21 x 10 –9 F (Very Small!!)
Lesson: difficult to get large values of capacitance without special tricks!
Units : C 2 Nm 2 m 2 m C 2 Nm CC J C V Farad
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 (= 0 A/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.
+Q –Q
Does the Electric Field Inside: (a) Increase?
(b) Remain the Same?
(c) Decrease?
• V is fixed by battery!
• C decreases (=e • E = Q/ 0 0 A/d) • Q=CV; Q decreases A decreases
Spherical Capacitor
V
What is the electric field inside the capacitor? (Gauss’ Law)
b
Relate
E
to potential difference between the plates:
a E
Q
4 0
r
2
E
d r
a b
kQ dr r
2
kQ r
b a
Radius of outer plate
= b
Radius of inner plate
= a
Concentric spherical shells: Charge
+Q –Q
on inner shell, on outer shell
kQ
1
a
1
b
Spherical Capacitor
What is the capacitance?
C = Q/V =
Q Q
4 0 1
a
1
b
4 0 (
b
ab a
) Radius of outer plate
= b
Radius of inner plate
= a
Concentric spherical shells: Charge
+Q –Q
on inner shell, on outer shell Isolated sphere: let
b >> a, C
4 0
a
Cylindrical Capacitor
QuickTime™ and a decompressor are needed to see thi s pi ctur e.
What is the electric field in between the plates? Gauss’ Law!
Radius of outer
V
Relate
E
to potential difference between the plates:
b
a
E
d r
C
Q
/
V a b
E
Q
2 0
rL
2
Q
0
rL dr
2 0
L
ln
b a
Q
ln
r
2 0
L
b a
Length of capacitor
= L +Q
on inner rod,
–Q
on outer shell
Q
2 0
L
ln
b a
plate
= b
Radius of inner plate
= a
cylindrical Gaussian surface of radius
r
Summary
• Any two charged conductors form a capacitor.
•Capacitance :
C= Q/V
•Simple Capacitors:
Parallel plates
:
Spherical
:
Cylindrical
:
C =
0 A/d C = 4
e 0 ab/(b-a)
C = 2 0 L/ln(b/a)]