From last time
Attention
More on electric potential and connection to E-field
How to calculate E-field from V
Capacitors and Capacitance
Today:
• switch off computers in the room and be
prepared to a very loud noise!!
More on Capacitors and Capacitance
Energy stored in Capacitors
Current
Resistance
Ohm’s Law
Capacitance
How do we charge a capacitor?
!Battery has fixed electric potential difference
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
across its terminals
!Conducting plates connected to battery
terminals produce a charge separation on
plates
!Electrons move from negative battery terminal
to right plate and from left plate to positive
battery terminal
!This charge motion requires work. The work is
supplied by the battery.
!Between plates there is an E-field and a
potential difference is estableshed
!Vplates = !Vbattery
-
Eleft
Eright
d
+
E+
E-
A=surface area
!=Q/A surface charge
density
!
-Q
+Q
Infinite plates: we normally mean to neglect border
effets (uniform field)
E left + E right = " /2#o + " /2#o = " /#o
-
E+
E-
"V =
1
Q
C
Work of the electric force to move q from positive to negative plate =
d r
r
W el = "#U = "q(V" " V+ ) = q $ E • ds = qEd % #V = Ed
# !Qd Q
"V = d =
=
$0
$0 A C
!
E+
C=
E!
-
"o A
d
0
"V = V+ # V#
!
This is a geometrical factor
!
Energy stored in capacitors
Spherical capacitor
!
Charge Q moved from outer to inner sphere
During the charging of a capacitor, when a charge q
is on the plates, the work needed to transfer further
dq from one plate to the other is:
Gauss’ law says E=kQ/r2
until
second
between
the sphere
inner sphere and the shell
Potential difference
Gaussian surface
to find E
Along path shown
b
"V =
#
a
C=
!
Path to find "V
!
!
+ + +
+
+
+
+ + +
-
-
-
The total work required to charge the capacitor is
W =
-Q
-
% 1 1 ($1
Q
= k' $ *
& a b)
"V
-
-
b
% 1 1(
kQ
1
= $kQ = kQ' $ *
& a b)
r2
ra
!
-
Q
0
q
Q2
dq =
C
2C
The energy stored in any capacitor is:
-
"
For a parallel capacitor:
!
U = 1/2 !oAdE2
Quick Quiz
Energy density
-q
pull
+
-
+
-
d
pull
+
-
!
+q
+
A parallel plate capacitor given a charge q. The plates are then pulled a small distance
further apart. Which of the following apply to the situation after the plates have been moved?
The energy stored per unit volume is
U/(Ad) = 1/2 !oE2
!
!
!
!
This is a fundamental relationship for the local energy
stored in an electric field
1)The charge decreases The charge cannot decrease because C is isolated
A)
C = !0A/d " C decreases!
B)2)The capacitance increases
E= Q/(!0A) " E constant
C)3)The electric field increases
4)The
voltage
between
the
D)
V= Ed ! V increases
Not restricted to capacitors: E-field can be from any
source
Interpretation: energy is stored in the field
Capacitors are devices to store electric energy and
charge (used in radio receivers, filters of power supplies,
electronic flashes)
plates increases
E)5)The energy stored in the
U = Q2/(2C)
U increases
capacitor increases
Modeling a cell membrane
Human capacitors: cell membranes
K+ Na+ Na+
Extracellular fluid
100 !m
Lipid bilayers form a capacitor
with charge separation at about
7-8nm
+ + + + + +
7-8
nm
Plasma membrane
- - - - - Cytoplasm
outside cell
Na+ K+
• Charges are +/- ions instead of electrons
(atoms with different numbers of electrons and
protons)
• Charge motion is through cell membrane (ion
channels) rather than through wire
-0.07 VV•Channels are selective for each kind of ion
!V~-0.07
•Across the cell membrane about !V=-70 mV
(resting potential) since more Na+ outside and
gates are closed while K+ ions diffuse out and
in at the same rate
•There is more positive charge outside the cell
Extracellular fluid
100 !m sphere ~ 3x10-4 cm2
surface area
Capacitance:
"o A
Plasma membrane
Cytoplasm
d
inside cell
(8.85 #10
$12
=
F /m) 4 % (50 #10$6 m)
$9
8 #10 m
2
= 3.5 #10$11 F = 35 pF
~0.1!F/cm2
!
Depolarization in a
nerve cell
Cell membrane depolarization
An influx of Na+ ions thru
Na+ channel can cause a
depolarization of the membrane.
-+- +
- -+
- - +- - +
- - -+
The potential difference becomes
!V~-0.07
"V 0.03 V about 0.03 mV.
Plasma membrane
Potential change = about 0.1 V
About 100 channels/µm2 (each
-+ +
- +
- +
- +
- +
Cytoplasm
has radius of 50 µm)
K+
Extracellular fluid
7-8
nm
Na+ K+
How many ions flow per channel?
Charge xfer required "Q=C"V=(35 pF)(0.1V)
3.5 x
10-12
C/
(1.6x10-19
C/ion) = 2.2 x
107
=(35x10-12 C/V)(0.1V)
= 3.5 x 10-12 Coulombs
ions flow
(100 channels/!m2)x4!(50 !m)2=3.14x106 ion channels
Ion flow / channel =(2.2x107 ions) / 3.14x106 channels ~ 7 ions/channel
Charge motion: current
Current in a conductor wire
•Cell membrane capacitor:
Found ~7 ions flow through ion each ion channel to depolarize
membrane
•Occurs in ~ 1 ms = 0.001 sec.
•This is an electric current I: amount of charge per unit time flowing
through a plane perpendicular to charge motion
•Units: 1 Coulomb / sec = 1 Ampere
Battery produces
E-field in wire
!
Average current:
!
!
!
!
Instantaneous value:
current in direction of + charge particles though electrons in reality flow in a
wire
Current density
! J = I/A (A/m2)
!J = nqv (direction of + charge carriers)
d
!
!
!
Ohm’s Law:
" = resistivity
r
r # nq 2 t & r
qE
t " j = nqv d = %
(E
m
$ m '
E=
J = # E or E = " J
"V
I
=#
L
A
Define: R = "L/A
Resistance in ohms (")
! Ohm’s
!
law becomes: !V = R I
!
Ohm’s Law: for many materials, the current density is proportional to the
electric field producing the current
J=#E
!
!
!
vd =10-4 m/s
Ohm’s Law
When an electric field is applied free electrons experience an acceleration
(opposite to E): a = F/m = qE/m
Calling t the mean time between 2 collisions electron-atoms electrons achieve a
drift velocity:
v d = at =
-
!
Ohm’s law
!
E
+
n = number of electrons/volume
electrons travel distance L = vd !t in time !t
# of electrons: n x volume = nAL =nA vd !t
Iav = !Q/ !t = nqAL vd /L
Conventionally we assume:
!
Average current:
Current flows in
response to E-field
# = conductivity of the conductor
!
E
ohmic device: relationship current and voltage is linear
Quick Quiz
Dependence on Temperature
L
Two cylindrical conductors are made from the same
material. They are of equal length but one has twice
the diameter of the other.
!
A. R1 < R2
!
B. R1 = R2
C. R1 > R2
$ = $0 +%(&'&0)
L
R="
A
$ of a conductor varies approximately
linearly with T
!
!
1
% x ($-$0)
$0
T-T0
$o is the resistivity at To = 20° C
% = temperature coefficient in SI units of
T0
T
oC-1
2
!
!
The higher T the greater atomic vibrations that
increases collision probability
Similarly: R = Ro[1 + %(T - To)]
17
But other materials…
Resistivity
Resistivity:
!
Semiconductors:
!
!
% is negative ) $ decreases for
$=1/#
SI units of " . m
! The resistance depends
resistivity and
geometry:
increasing T
!
!
!
" = " 0 (1+ #$T )
Mercury
Superconductors
!
Below a certain temperature, TC = critical
temperature
$ is zero
!
!
Once a current is set up in a
superconductor, it persists without any
applied voltage since R = 0
Resistors control
the current level in
circuits
!Resistors can be
composite or wirewound
!
Resistors in Series
I = I2 = I
! 1
Resistors in parallel
2 resistors in series:
R!L
Like summing lengths
R
!
!V = !V1 + !V2
!
Req = R1+R2
on
conductor
R
2R
=
R=#
L
A
!
!V = !V1 = !V2
!
I=I1+I2
A A
I1
I2
!V1
"
Neglect
resistance of
wires in circuits
respect to R
resistance
!V2
I = I1 + I2 =
R
R
"V1 "V2 # 1
1&
"V
+
= % + ("V =
R1
R2 $ R1 R2 '
Req
!
2A
Add areas
"V = "V1 + "V2 = R1I1 + R2 I2 = (R1 + R2 )I = Req I
R="
!
L
A
!
Current Conservation: 1st Kirchoff’s law
I2
•As a charge !Q moves from a to b, the
electric potential energy of the
system increases by !Q!V and the
chemical energy in the battery
decreases by this same amount
I1
Iin
I3
I1=I2+I3
•As the charge moves through the
I1
resistor (c to d), the system loses this
electric potential energy during
collisions of the electrons with the
Power = rate of energy loss
atoms of the resistor
I3
Iout
I2
Iout = Iin
I1+I2=I3
!
Junction Rule:
!
Electrical power
•This energy is transformed into internal
energy in the resistor (vibrational
motion of the atoms in the resistor)
•Power in units of Watts = J/s
( Iin = ( Iout
A statement of Conservation of Charge
23
"U "Q
=
"V
"t
"t
"V 2
P = I"V = RI 2 =
R
P=
24
!
Quick Quiz
How does brightness of bulb B compare to that of A?
A. B brighter than A
B. B dimmer than A
C. Both the same
25