PHYSICS 1B – Fall 2009
Electricity
&
Magnetism
Professor Brian Keating
SERF Building. Room 333
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?
PHYSICS 1B – Fall 2009
Electricity
&
Magnetism
Professor Brian Keating
SERF Building. Room 333
Capacitors
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
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
16.9 Dielectrics, Energy
Dielectric constant-effect on capacitance
Energy stored in a capacitor
Energy density (depends on E2)
Biological Membranes
Dielectric material – insulators such as paper, glass
plastic, ceramic. Used in the gap in capacitors.
“Dielectric Strength” - is the electric field
at which conduction occurs through the material
dielectric material
+q
-q
Electric Fields in Dielectric Filled
Capacitors
Effects of a dielectric material inserted into a capacitor,
with charge q
Vacuum
+q
ΔVo
-q
dielectric material
+q
ΔV
-q
κ= dielectric constant (dimensionless)
"Vo
"V =
!
Potential due to
charge q
decreases by κ
Dielectric Properties of Selected Materials
Material
Vacuum
Air
Polystyrene
Paper
Pyrex
Water
dielectric constant, κ
1.000000
1.00059
2.3
3.4
5.6
80
Dielectric Strength
(Volt/m)
----2x106
24x106
16x106
14x106
-------
How does the capacitance change?
Originally
+q
ΔVo
Co
-q
Add dielectric
+q
ΔV
C=
q
!q
=
"V "Vo
-q
C = !Co
Capacitance increases
How does the E field change?
+q
ΔVo
-q
+q
ΔV
-q
E=
"V "Vo
=
d
!d
Eo
E=
!
Electric field decreases (when not connected to a battery)
+q
ΔVo
-q
q
q
E=
=
!" o A " A
+q
ΔV
-q
! = "! o
Permittivity is increased
Compared to vacuum
Example: A parallel plate capacitor consists of metal sheets
(A= 1.0m2) separated by a Teflon sheet (κ=2.1) with a thickness
of 0.005 mm. (a) find the capacitance. (b) Find the maximum
voltage. The maximum electric field across Teflon is 60x106
V/m. – this is its dielectric strength.
(a)
Κ=2.1
d=0.005m
A=0.25m2
!" o A 2.1(8.8 x10#12 )(1.0)
C=
=
#3
d
0.005 x10
C = 3.7 x10#6 F
A parallel plate capacitor consists of metal sheets(A= 0.25m2)
separated by a Teflon sheet (κ=2.1) with a thickness of 0.005
mm. (a) find the capacitance. (b) Find the maximum voltage.
The maximum electric field across Teflon is 60x106
V/m.(dielectric strength)
(b)
"Vmax = Edsd = 60 x106 (0.005 x10 !3 ) = 300V
Molecular basis for dielectric constant
+
Dipolar molecules.
e.g. water
are oriented in
an E field
E
-
Oriented molecules decrease the net charge near
the plates
The E field in the
Capacitor is reduced
Eo
E=
!
Polarization of Dielectric
Dielectric Screening
High dielectric constant of water allows ions to dissociate
Na+
Cl-
-
+ + -
Vo Vo
V=
=
! 80
+ -
+ -
+ -
+-
+
+
+
-
+ -
+
-
Find the potential energy in electron volts for the
interaction of Na+ and Cl- separated by 0.5 nm in water.
qNa qcl
$(1.6 x10$19 )2
PE =
=
4!"# o r 4! (80)(8.8 x10$12 )(0.5 x10$9 )
PE = $5.8 x10
$21
1eV
%
&
Jx '
(
$19
1.6
x
10
J
)
*
PE= -0.036 electron Volts
comparable to thermal energies (kinetic energy of
the ion) about 0.025 eV at room temperature
Energy stored in a capacitor.
Work
done to
charge a
capacitor
q=0
q
+q
ΔV=0
ΔV
q=0
-q
2
1q
1
W =
= C !V 2
2C 2
So work depends on the square of q or ΔV
A parallel plate capacitor consists of metal sheets (A= 1m2)
separated by a teflon sheet (κ=2.1) with a thickness of 0.005
mm. Find the maximum energy that can be stored.
C=3.7 x10-6F
ΔVmax=300V
C
1
1
2
Energy = C "V = (3.7 x10!6 )(300)2
2
2
Energy = 1.7 x10!1J
(quick quiz 16.6)
Insert the dielectric material with dielectric constant κ into the
capacitor keeping the voltage source connected.
Find C,q,E, PE
ΔVo
+qo
Co
-qo
C=
ΔVo
q
-q
! Co
q= CV = ! CoVo = ! q
!V !Vo
=
= Eo
E=
d
d
1
1
2
C
"
V
=
! C "Vo 2 = ! PEo
PE=
2
2
Energy Density in a Capacitor
Suppose you wanted to store a large amount of energy
in a capacitor with a given volume of 1m3 using Teflon
As the dielectric (dielectric strength of 60x106 V/m).
What is the maximum energy that could be stored?
1
Energy = CV 2
2
A!" o
C=
d
1 A!" o 2 1 Ad!" o V 2 1
2
Energy =
V =
=
!"
E
(volume )
o
2
2 d
2
d
2
Energy 1
= !" o E 2
The energy density depends
volume 2
only on the E field squared.
For the maximum electric field = dielectric strength
1
Energy = !" o E 2 (volume)
2
1
Energy = (2.1)(8.8 x10#12 )(60 x106 )2 (1)
2
For a 1 m3 capacitor at
Energy = 3.4 x104 J
the maximum voltage.
For comparison the energy content of burning
1 gallon of gasoline is 1.3x108 J
Chemical energy has a higher energy density.
Capacitance of Biological Membranes
+
membrane
+
- +
+
+
+
+
Axon - Nerve cells
+
-
Potential difference across the membrane
Nerve transmission – involves a discharge
of membrane potential
Biological membranes –Capacitance
The low dielectric portion of a biological membrane
has a thickness of 2.0 nm. Assume that it has a
dielectric constant of 2.5 (silicone oil) find the
capacitance of 1m2 of membrane.
low
dielectric
high dielectric
" A !" o A
C=
=
d
d
2.5(8.8 x10!12 )(1)
C=
2.0 x10!9
C = 1.1x10!2 F
High Capacitance
Compare to 3.7x10-6 F for
1m2 the Teflon capacitor.
A nerve cell has a potential across it of 60 mV. Find the
density of charges on the membrane that can give rise to
this potential
ΔV=60 mV
+
+
+
+
+
+
-
-
-
-
-
-
"V 60 x10!3
7
E=
=
=
3
x
10
V /m
!9
d
2 x10
q
!
E=
=
A"# o "# o
This is close to the
dielectric strength
! = "# o E = 2.5(8.8 x10$12 )(3 x107 )
! = 6.6 x10$4 C / m 2
This corresponds to an ion density of 4.1x10-3 ions /nm2
or a distance between ions of about 16 nm. A small number
of excess charges.