Capacitance
Capacitor- a device for storing charge and energy,
can be discharged rapidly to release energy.
Applications
•Camera flash
•automobile starting system
•capacitors in electronic devices
•computer memories (store information)
•Laser flash lamp
laser fusion
Energy stored in a capacitor 2 x106 J released in 1-2 x10-6 s.
High Power ~ 1012 W
Parallel plate capacitor
-
+
-
-
-
+
+
-
+
+
FIELD LINES IN BLACK (VECTORS)
POTENTIAL CONTOURS IN RED
(NO ARROWS, BECAUSE NOT A VECTOR)
+q
conductors
+ + + + + + + + + + +
+ + + + + + + + + + +
- - - - - - - - - - - - - - - - - - - - - - - - -
-q
potential source
Eg. battery
Parallel Plate Capacitor
Work is done to separate charges
Capacitor Stores Electrical Energy
ΔV
Circuit Diagram
Wire = conductor
Voltage
source
+
ΔV
-
conductor
+q capacitor
C
-q
Measuring voltage
C
V
V
Voltmeter
Ideal voltmeter
Draws no charge
Perfect insulator
Parallel Plate Capacitor
Metal plates with area A, separated by a gap, d
containing an insulating material. (eg. Air)
A
+q
ΔV
d
-q
Capacitance – describes the
ability to store separated charge
q
C=
!V
Units C/V, farad (F)
A 100 microfarad capacitor is charged to 100 V. How
much charge does it store?
q
C=
"V
q = C( "V ) = 100 x10!6 (100) = 10!2 C
This is a lot of charge. Recall that 10-2 C on a sphere of
1m radius generates a potential of
q
kq
r=1m
V = e = 9 x107V
r
How does the capacitor store this charge without
high potentials? How do you get a large capacitance?
Gaussian surface- a cylinder with sides perpendicular
to the plane. E is constant at ends. Flux through
sides is zero.
q
" E = 2 AE =
!o
E
+
+
+
+
+
+
q
E=
2 A! o
+
+
+
+
+
+
+
"
E=
2! o
+
+
E
A
where
q
! =
A
Parallel plate capacitor
two “infinite” planes of charge area A separated
by distance d where d<< A, carry charge +q, -q
A
+q/A=+σ
++++++++++++
----------------The charges are at the inner
surface of the capacitor
d
-q/A=-σ
Field inside the capacitor plates
By superposition of charges due to sheet of charge
Eout =0
E+
!
Ein=
"o
A
q/A=σ
++++++++++++
-----------------
EEout =0
!
!
Ein=E+ +E- = 2
=
2" o " 0
d
What is the capacitance of a parallel plate capacitor
with plates of area A separated by distance d? (in air)
+q
A
ΔV
d
-q
Recall from
Gauss’ law
rearrange
thus
!
q
#V
E=
=
=
" o A" o
d
A! o
q
"V
=
E field
increases with
charge density
d
A! o
C =
d
to increase C
Increase A
Decrease d
How do you stabilize the separated charge
i.e. make ΔV as small as possible for a given q
(high capacitance)
+q
A
+ + + + + + + + + + +
+ + + + + + + + + + +
- - - - - - - - - - - - - - - - - - - - - - - - -
-q
area A
as Large as possible
distance d
as SMALL as possible
d
Thin film capacitors
Metal film separated by thin insulators
metal
insulator
Metal
Making the area large and the insulating gap small
increases C
A parallel plate capacitor with 2 plates each with area 1.0 m2
separated by a distance of 1.0 mm holds +q,-q, q=10-6C
(a) Find the capacitance. (b) Find the E field in the capacitor
(c) Find ΔV across the plates
A=1 m2
q
ΔV
d=1mm
-q
C=
A! o
d
E=
q
A! o
ΔV=
Ed
1(8.9 x10!12 )
=
= 8.9 x10!9 F = 8.9nF
0.001
1x10!6
5
=
=
1.1
x
10
V /m
!12
(1)(8.9 x10 )
= 1.1x105 (1x10 !3 ) = 1.1x102V
A parallel plate capacitor (infinite) with a constant charge q
has its separation increased by a factor of 2.
How does this change E and ΔV?
q
q
Eo
do
-q
C
ΔVo
E= Eo ΔV= 2ΔVo
2do
C
E=
-q
q
is constant
A! o
!V = dE = 2!Vo
A parallel plate capacitor (infinite) with a constant voltage
difference ΔV has its separation increased by a factor of 2.
How does this change E and q?
ΔVo d
qo
Eo
o
qo
q=
2
ΔVo
-qo
C
Eo
E= 2
2do
C
ΔV=dE
q
E=
A! o
ΔV= ΔVo
Capacitor combinations
Capacitors connected in series
and parallel
Electrical circuit elements
+
Voltage source
Circuit diagram
resistor
R
capacitor
C
V
Conductor
one capacitor
ΔV
+
C
-
q
C=
!V
+q
-q
Vc=ΔV
Two Capacitors in Parallel
q
q
ΔV
C1
q1
q2
C2
=
ΔV
Ceq
Ceq
q
q
=
!V
equivalent
capacitance
What single capacitor has the same properties as two
capacitors in parallel?
Two Parallel Capacitors
equivalent
capacitance
q
C1
ΔV
q1
q2
C2
!V = !V1 = !V2
q = q1 + q2 = C1!V + C2 !V
Ceq
C1!V + C2 !V
q
=
=
!V
!V
Ceq = C1 + C2
Ceq
q
=
!V
Parallel Capacitors
For N capacitors in parallel
Ceq = C1 + C2 + ........CN
Ceq is the sum of capacitances
Like a larger capacitor, larger area
Find the equivalent capacitance
ΔV
5µF
3µF
10µF
A. 15 uF
Ceq =C1 +B.
C217+ uF
C3
Ceq = 5 + C.
3 +1810uF= 18 µ F
D. 20 uF
Find the equivalent capacitance
ΔV
5µF
3µF
10µF
Ceq =C1 + C2 + C3
Ceq = 5 + 3 + 10 = 18 µ F
Two Capacitors in Series
What is the equivalent
capacitance?
q
C1
ΔV
C2
q
q
+q ΔV
1
-q
+q ΔV
2
-q
Ceq
q
=
!V
the charge on both
capacitors in series
is q
Two Capacitors in Series
q
C1
ΔV
C2
q
q
q
=
!V
+q
-q
Ceq
+q
-q
q=q1=q2
q
q
!V = !V1 + !V2 =
+
C1 C2
1
!V 1 q
q
=
= ( + )
Ceq
q
q C1 C2
1
1
1
=
+
Ceq C1 C2
For N capacitors in series
1
1
1
1
=
+
+ .........
Ceq C1 C2
CN
Capacitors in series
Ceq is smaller than the smallest capacitance.
You store less charge on series capacitors
than you would on either one of them alone
with the same voltage!
Physical Argument
Take a parallel plate capacitor and place a thin metal plate
with the same area in the middle of the gap.
+++++++++++++
++++++++++++
ΔV/2
ΔV
-----------
ΔV/2
----------+++++++++++++
-----------
C
the component capacitances
are larger than the total
C1=C2= 2C
Ceq = C
The equivalent capacitance is less than the
component capacitances
Ceq <C1 or C2
C1
C2
Find the equivalent capacitance
3µF
5µF
ΔV
10µF
1
1
1
1 1 1 1
=
+
+
= + +
= 0.633
Ceq C1 C2 C3 5 3 10
Ceq = 1.58 µ F
34. Find the equivalent capacitance.
X
Cseries=6µF
X
1
Cseries
Ceq= 4.00+2.00+6.00=12.00 µF
1 1 4
1
=
+ =
=
24 8 24 6
34. Find the charge on each capacitor.
C1
C2
X
X
C3
C4
q = C !V
q1=C1ΔV=4x10-6(36)=1.44x10-4 C
q2 =C1ΔV=2x10-6(36)=0.72x10-4 C
q3=q4=CseriesΔV=6x10-6(36)=2.16x10-4C
Cseries =6µF
34.Find the voltage drop across each
capacitor.
ΔV3
ΔV1
ΔV2
ΔV4
q
!V =
C
ΔV1=ΔV2=36V
q
2.16 x10!4
"V3 =
=
= 9.0V
!6
C3
24 x10
q
2.16 x10!4
"V4 =
=
= 27V
!6
C4
8 x10
series capacitors
The larger C has
the smaller
voltage drop
42. Find the equivalent capacitance between a and b.
X
1
1 1
1
= + =
Cser 7 5 35 /12
X
Cser = 2.9 µ F
4.0 µF
a
2.9 µF
6.0 µF
b
Ceq=4.0+2.9+6.0
=12.9 µF
Two identical capacitors. charge one capacitor at
10 V , disconnect, connect the charged capacitor
to the uncharged capacitor. What is the voltage
drop across the each capacitor?
One way to do this problem
q is constant but divided between the two capacitors
ΔV0 =10V
q/2
+q
ΔV
-q
C
q/2
-q/2
-q/2
C
the charge on each capacitor is reduced by 2 fold
thus the voltage across each capacitor is reduced
by 2fold
!V =
!Vo
2
charge one capacitor at 10 V , disconnect, connect
the charged capacitor to the uncharged capacitor.
What is the voltage drop across the each
capacitor?
Another way to do this problem
q is constant but is placed on an equivalent capacitor
ΔVo =10V
Ceq =2C
+q
-q
C
C
The voltage is reduced by 2 fold
!V =
!Vo
q
q
=
=
Ceq 2C
2
ΔV