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Physics 2B:
Electricity and
Magnetism
Eric L. Michelsen
emichels at physics etc.
Part I: Electricity
Slightly tailored for Randall Knight’s
Physics for Scientists and Engineers, 2nd
ed.
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Charges
• Do particles push on each other because theyare charged? Or ...
• Do we say that particles are charged because
they push on each other?
+ - particle 1 particle 2
F21 F12
+ +1 2
F21 F12
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More on charges• Charge is measured in Coulombs (C)
• A coulomb is like a “dozen,” only bigger:6.242e18 electrons-worth of charge
• ~10-5 moles of electrons
• Or, an electron has 1.602e-19 C of charge• Very small
+ -1 2
F21 F12
+2 -
1 2
F12
?
+2 -3
1 2
F12
?
Force is proportional
to both charges:
|F| = k e q1q2 / r 2
fromparticle 2 on particle 1
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Coulomb’s Law
• Force weakens with distance• α 1/r 2 (like gravity)
• Obeys superposition
• New charges don’t interfere with forces fromexisting charges, so forces simply add vectorially
+ -1 2F21 F12
+ -F21 F12
?
1 212 122
1212 12
0
ˆ
ˆ,
1
4
e
e
q qk
r
where r r
k K
F r
rr rr12
r12
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What are the units of ?
A dimensionless
B m
C m2
D m-1
E m-2
1212ˆ
r
rr
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Aside:
Superposition
• Kinematic example• First: Take conditions one
at a time
• Only initial position
• Only initial velocity• Only constant acceleration
• Superposition:
• The effect of all the
conditions together is thesum of each separately
• Often used in E&M
x(t )
t
v(t )
t
v(t )
t
t
xi x f = xi
x f = area = vit
0
t 0
t 0
v f =
a t
x f = area = (1/2)at 2
vi = 0, a = 0
xi = 0, a = 0
xi = 0, vi = 0
x f = xi + vit + (1/2)at 2
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Source charges and test charge
• Source charges usually considered fixed in space• Test charge moves around
• No physical difference between “source” and “test”
• Coulomb’s Law obeys superposition
+
-
-
+
test chargeq0 > 0
12
3
4 F1
F2
F3
F4
Fesource
charges
1-4
40
21
4
0 21
ˆ
ˆ
i
e e iii
ie i
ii
q q
k r
qq k
r
F r
r
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Electric Fields
• Consider a distribution of source charges• The force on a test charge at any point is
proportional to the charge:
0 0
0
At a point: constant vector
Extending to any point in space: ( ) ( )
e e
e
q q where
q
F F E E
F r E r
+
--
+
q0 < 0
12
3
4 F1
F2
F3
F4
Fe
+
--
+
q0 > 0
12
3
4 F1
F2
F3
F4
FeUnits
of E?
Direction
of E?
There’s no
physical
difference
between source
and test charges
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Electric field from source charges
• Source charges create an E-field4
0 021
4
21
ˆRecall: ,
ˆ( )
ie e i e
ii
ie i
ii
qq k and q
r
qk r
F r F E
E r r
+
-
-
+
q0 > 0
12
3
4 F1
F2
F3
F4
Fe
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At P, which way does
the E-field point?
A
B
C
D
E
-
-
P
●
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Lines of Force
• Field of vectors in space: the E-field• Lines of force follow the arrows
E
How does lines-of-
force density relate
to E-field strength?
E-field strength is
proportional to lines-
of-force density.
+ -
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Do objects always move in the
direction of the force applied to them?
• No. Objects bend toward the direction ofapplied force, but they generally don’t move
parallel to it
• Lines of force• Not trajectories of motion
• g-field?
F g
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Simulated particle trajectory
• What sign is the test charge?
line offorce
trajectory
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If I release a negative test charge at rest
on a line of force, how will it move?
A To the right, following the line of force
B To the right, crossing the lines of force
C It won’t moveD To the left, following the line of force
E To the left, crossing the lines of force
+ -
When does the
test charge move
along the line of
force?
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Birth, life, and death of
lines of force
• No matter how close we get to a positive charge, lines of force point away from it• Lines of force originate on positive charges
• No matter how close we get to a negative charge, lines of force point toward it
• Lines of force terminate on negative charges
• Can’t cross or split• ... because they follow the E-field, which can only point one way from
every point
• But they can turn quickly at a point of zero field
• •
+ −
X • •E = 0
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Electric conductors
• Conductors (good or poor) have mobile charges• Good conductors: metal
• Poor conductors: carbon, electrolytes (e.g. water)
• If immersed in an electric field, mobile charges move
• While they are moving, we have an electric current
• current is the motion of charge
q
-q I
velocity
Much more
on this later
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Insulators
• Insulators do not have long-range mobile charges
• E.g., glass, plastic
• May have dipoles that can be
induced or rotate a little
• All insulators are dielectric at
some level
• Some insulators have negligible
polarization (gases)
• Don’t worry about dielectrics
until a later chapter Much more
on this later
appliedE
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Charging by induction
• Start withneutralconductors
• Induce• Connect
spheres by wireor touching
• Separate• (E-field
not shown)
+-+
+-+
Connect
with wire
A l t h + h it
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An electroscope has a + charge, so its
leaves are apart. I bring a negative rod
near it. What happens to the leaves?A Leaves get closer
B Leaves spread farther
C One gets higher, the other lower
D Nothing
E Leaves collapse completely
+
- -
-
The electroscope question
(stop to think 26.3 p799)How can we be
confident this is true?
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dQ
E-field of ring of charge
• What is E-field at distance z from center?• Axial symmetry implies E parallel to axis
R
z
3 / 22 2
ˆ,e z
z k Q E
R z
E z
Q
r
d E
dE z
θ θ
2 2ˆ cose z e
dq dqd k dE k
r r E r
2 2 2 2 2
z e e
dq z z dE k k dq
r r R z R z
3 / 20 02 2
z E Q
z e
z dE k dq
R z
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Flux• Loosely: “counting” lines of force through the area
• This only works because E α 1/r 2
• More precisely: Flux is E-field component through the area,times the area:
• Area vector is perpendicular to the area, so the parallel E-field
component gives the E-field perpendicular to the area• So flux is a dot-product
E
A(area vector)
E E area
or d E A E A
E
AWhat kind of quantity
is Φ ?
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Why flux?• Surface charge distributions (sheet
charge)• A uniform density of charge in a large
flat sheet
• What is the E-field near the surface?• Critically important, but ...
• Can integrate over Coulomb’s Law, but it’stedious
• Gauss’ Law (flux) makes this simple
+ + + +
+ + + +
+ + + +
sheet charge:η C/m2
side
view
E
+ + + +
F th |E| d |A| h d th
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For the same |E| and |A|, how does the
flux through the area on the left
compare to that on the right?A Flux is greater on the left
B They are the same
C Flux is greater on right
E
A
E
A
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What is the flux through a sphere?
A
B
C
D
E
q
E E A
2
e E
k q
r
2ˆe
E
k q
r
r
r
24 E er k q
4 E ek q
0 E d Ω?
d A E
surface
cosd d E dA E A
2
2
surface
4 4e e
qdA k r k q
r
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What is the flux through the funny-
shaped surface?
q
A
B
C
D
E
4 E ek q
0 E
4 E ek q
4 E ek q
r
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What is the flux through the bag?
A
B
C
D
Eq
4 E ek q
0 E
4 E ek q
4 E ek q
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Gauss’ Law
• The total outward flux from a closed surfaceequals 4πk e times all the charge enclosed:
• Does superposition apply?
• Superposition of what?
• What is qin here?
0
0 0
14 , will be handy later
4
in E e in e
qk q k
q
2q
Why are arrows
different lengths?
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Gauss’ Law (part deux)
q
out +
in -
out +
out +
S f h di ib i
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Surface charge distributions:
the beauty of Gauss’ Law
• A uniform density of charge in a large flatsheet
• Gazillions of tiny particles (electrons, ions)
• What is the E-field near the surface?• What “shape” is it?
+ + + +
+ + + +
+ + + ++ + + +
sheet charge
density, η C/m2
gaussian
surface (box)
side
view
E
0
00
Gauss' Law: /
2 / ,2
E inq
EA A E
A
A
A
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What is the E-field above a surface-
charged conductor?
A zero
B η/2ε0
C η/ε0
D 2η/ε0
E 4η/ε0
conductor
+ + + + + + + + +
E?
How is this possible?η
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Three ways to make E-fields easy
• Using Gauss’ Law relies on any of 3 methods• E-field is constant (though unknown)
• E-field is zero
• E-field is tangent (to gaussian surface)
+ + + + + + + + +
E? A
AE = 0
conductor
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Volume charge distributions
• Consider a spherically symmetric volumedistribution, in C/m3
• What is E-field at distance r , outside thesphere?
• What does Gauss say?
• How does it compare toa point charge at distance r ?
q
r
2
2
4 4 4 E e in r earea
r e
k q E r k q
q E k
r
H d h E fi ld f h i ll
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How does the E-field of a spherically
symmetric volume distribution
compare to that of a point charge?A It’s bigger, because some of the charge is
closer
B It’s the same
C It’s smaller, because more of the charge is
farther away
qr r
F l d t k t ti l
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Forces lead to work, potential energy,
and voltage (aka potential)
• 1D:• Independent of the trajectory: time, speed
• Work can be + or -
• Potential energy is proportional to test charge
• Voltage is defined as electric potential energy perunit charge (aka potential difference)• In volts (V), equivalent to joules/coulomb
0( ) ( ) ( ) ( )
B B
e A AW F x dx U B U A W q E x dx
q0 E > 0
F
A B
0
( ) ( )( ) ( ) ( 0)
B
A
U B U AV B V A E x dx
q
x
q0 E > 0
F
A B
+
-
“El t i P t ti l” i diff t th
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“Electric Potential” is different than
“Potential Energy”
• Electric Potential (aka “Potential”) = Voltage• Potential is defined as electric potential energy per
unit charge
• Potential is independent of test charge
• “Potential energy” of the test charge depends
on its charge:
0
( ) ( )( ) ( ) e eU B U A
V B V Aq
0( ) ( ) ( ) ( )e eU B U A q V B V A
H d th k d (f A t B)
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How does the work done (from A to B)
by the field on a fast-moving charge
compare to the work done on a slow-moving charge?
A It’s greater
B It’s the same
C It’s less
D Depends on the charge
E > 0 A B
q0
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Potential of a point charge
• In other words, what is the work needed to bring +1 C ofcharge from far away to a distance r from the source charge?• Gets steeper closer in
• Which way does a positive charge move?
+ r
V (r )
-
V (r )
r E(r )
PE of +
test charge
PE of -
test charge V (r ) < 0
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Potential of a point charge (2)
• In other words, what is the work needed to bring +1 C ofcharge from far away to a distance r from the source charge?
• We can use a 1D analysis
• Voltage referenced to ∞ is defined as absolute potential
• Still a function of distance, r :
• If source charge is negative, potential is negative
qF A at ∞
Bq0displacement
direction from A to B
2
1( ) ( ) ( )
B B
r B r
e e e A B
q qV B V A E r dr k dr k q k
r r r
( ) B B r e
B
qV r V k
r
source
charge
V (r )
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What is the potential midway
between two point charges?
A positive
B zero
C negativeD depends on the test charge
q -qq0
V ?
Wouldn’t I have to do work to bring a
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Wouldn t I have to do work to bring a
charge from infinity to the midpoint?
• At first, from some directions, yes• But once past the + charge, the E-field does as muchwork (returns as much energy) to finish the job as itcost to bring the charge in from infinity• Net energy cost = 0
q -q
maximum potential
equipotentialsurfaces
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How does the work done on q0
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How does the work done on q0
from A to B along path 1
compare to that of path 2?A It’s more
B It’s the same
C It’s less
E
A •
• B
q0
•
p a t h 1
p a t h
2What if it were
different?
H d i h
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How do negative test charges move
in an electric potential?
A Down the potential hill
B They don’t move
C Up the potential hill
How do negative
charges move in a
potential energy field?
V ( x)
?
U (r)?
Different setup
than above
x
When acted on only by the potential field.
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E and V
• E is to F as V is to U e• Electric field E is force per unit charge
• V is potential energy per unit charge
0 0
( ),( ) )( () e eU q q
V rF rE r r electric potential energy per unit charge
electric force per unit charge
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Hard work: grading your potential
x
y
z
dx
dy
dz
,e x xđW F dx
,e z z đW F dz
,e y yđW F dy
any path
( ) ( ) ( ) B
A
dV d
V B V A d
E s
E r s
ˆ ˆ ˆd dx dy dz s x y z ( ) ( ) ( )ˆ ˆ ˆ( )
( )
U U U
x y z
U
r r rF r x y z
r
The gradient produces a vector
field from a scalar field
0 0
( ) ( )
( ) ( )
U
q q
F r r
E r V r
d s
E
A •
• B
d s
Equivalence of statements
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Equivalence of statements
about conservative fields• “The E-field is conservative”
• ≡ “The electric potential is a function of position only: V (r )”
• ≡ “The electric work done on a charge between two points isindependent of path”
• We expect this from conservation of energy
• Different work on different paths + reversing a path reverses the work= infinite energy for free!
• Can prove it explicitly from Coulomb’s Law & vector calculus• Specifically: the curl of the electrostatic E-field = 0
0 E
0
0
d
B s
B
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Good conductors
• Good conductors (typically metals) have huge(essentially infinite) numbers of mobile electrons(~104 C/cm3)
• Electrons free to move within the conductor
• But they can’t leave the surface
• E-field drives negative electrons• leave behind positive ions
• Allows for a redistribution of charge
• Charge always moves to reduce E-field
• Molarity:
• 1 C = 10-5 mol of electrons
• Copper mobile electron density= 10-1 mol/cm3 = 100 mol/L
applied E
+
+
+
-
-
-
I d t hi h d th
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In a conductor, which way do the
positive charges move?
A left
B they don’t move
C rightD depends on the conductor
applied E
+
+
+
-
-
-
In a static E-field, after equilibrium,
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In a static E field, after equilibrium,what is the E-field inside a good
conductor?A Depends on the source charges around it
B Enhances any applied E-field
C Reduces any applied E-fieldD zero
applied E+
+
+
-
-
-E?
conductor
I ilib i h d h i l
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In equilibrium, how does the potential
at B, V (B), compare to that at A, V (A)?
A It’s less
B It’s the same
C It’s greater D Depends on the conductor
applied E+
+
+
-
-
-A •
• B
conductor
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Fluid conductors
• Have smaller number of mobile charges• Mobile charges may be electrons or ions
• Conductor may be liquid or gas
• Plasma (ion/electron gas, not blood)• Electrolyte
Whi h d th iti h
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Which way do the positive charges
move?
A left
B they don’t move
C right
applied E
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Alternate units for E
• [E]=V/m• Equivalent to N/C
• Can view E two ways:
• For simplicity, consider
a uniform E-field:
0 0e V U q q FE d d
force
voltage
, or fromV V d E d E x
E
d
+ ΔV − A B
potential energy
per unit charge
Sign convention for ΔV :
positive end of ΔV has
higher PE
for positive test charge
?
sign convention
Relationship between E-field and V-field
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Relationship between E-field and V-field
• E points between equipotentials
• E is perpendicular to equipotentials• Tighter equipotential surfaces = stronger E (more volts/meter)
• Wider equipotential surfaces = weaker E (less volts/meter)
q -q
equipotentialsurfaces
E
)( () V E r r
1 V
2 V 3 V
)( () U F r r
0 V
−3 V
decreasing
potential: E points “downhill”
E
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Potential energy of a bunch of charges
• Imagine bringing charges, q j, in one at a time• The 2nd and subsequent charges ( j)interact with all previous charges (i)
• Also written (as in Knight):
• But it’s really a double sum
• And even:
1
2 1
,
jni j
e e ij ijij j i
q qU k r
r
r
i je e
iji j
q qU k
r
q1 q2
q3 q4
1
2
i je e
iji j
q qU k
r
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Capacitor
• Two parallel thin plates• Small gap =>
neglect edges
• Charges symmetrically• By induction!
• (not to be confused withinductance)
• Stores charge
• Q (charge on one plate)is proportional to ΔV
• Defines capacitance C :• Q = C ΔV
• Measured in C/V ≡ F,Farads
E+
+
+
+
−
−
−
−
wire
large plate
smallgap, d
+ ΔV −
What is the E field outside the
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What is the E-field outside the
plates of a capacitor?
A left
B zero
C right
D depends on which plate has more charge
E not enough information +
+
+
+
-
-
-
-
wire
+ ΔV −
i i
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Energy in a capacitor
• Given a capacitor charged to voltage ΔV , howmuch work does it take to increase the charge by a small amount, dq?
• How much work tocharge the wholething from 0 to ΔV ?
+
+
+
+
-
-
-
-
wire
outside
qdU dW V dq dq
C
E
hypothetical
dq > 0
NEVER-
EDDY
2
0 00
22
1
2
1 1
2 2
QW Q
outside
q qU dW dq
C C
QC V
C
+ ΔV −
final charge
+−
−dW e
NB: dq is not a
geometric quantity
Si i f i
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Sign convention for a capacitor
• Choose reference plate: defines polarities• Eliminates minus sign where possible
+
+
+
+
-
-
-
-
wire
0 0
0
0
Q E
A
d V E d Q
A
AQ V C V
d
E
d
I choose this as myreference plate
charge onreference plate
+ ΔV −
capacitance α area,
inversely α separation
What is the total net charge in a
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What is the total net charge in a
capacitor charged to voltage ΔV ?
A Qnet = C ΔV
B Qnet = (1/2)C ΔV
C Qnet = 0D Qnet = C (ΔV )2
E Qnet = (1/2)C (ΔV )2
If I charge a capacitor so that
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If I charge a capacitor so that
Q < 0, what describes its energy?
A U > 0
B U = 0
C U < 0D depends on ΔV
H l bl E l 1
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How to solve any problem: Example 1
• A bound NaOH molecule is modeled as Na+ bound to O(-2) bound to H+, in a straight line in that order. The Na+ ion is1.0e-10 m from the O(-2), which is 1.0e-10 m from the H+.How much energy does it take to remove the Na+ ion from themolecule (i.e., take it to far away), in J?• What does it look like?
• What does it want?
• What do we know about what it wants?
• What is given?
• What can we find from what is given?
• Build a bridge: what principles/formulas relate what is given or whatwe can find to what it wants?
+ -2
Na O H+
1e-10 m 1e-10 m
U q V
( ) e
qV k
r r
1 2
0 0 10 10
( ) ( ) ( )
20
1 10 2 10
total
p ptotal e
V V V
q qU q V q k
r r r
V total (∞)
ifneeded
Energy
q1, q2, q3, r 1, r 2
H l bl E l 2
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How to solve any problem: Example 2
• An electron in a picture tube starts at rest. It is accelerated by
25,000 V across a distance of 0.50 m to the screen. How longdoes it take to travel to the screen, in s?• What does it look like?
• What does it want?
• What do we know about what it wants?
• What is given?• What can we find from what is given?
• Build a bridge: what principles/formulas relate what is given or whatwe can find to what it wants?
2, ,
F qE x x d a t
m m a
E
screen
Δ x = 0.5 m
21,
2avg x at x v t
e e
V W F d qEd q V E
d
19
15
31
8
15
25,000 V /0.5 m 50,000 V/m
1.6 10 C 50,000 V/m8.8 10 m/s
9.1 10 kg
2(0.5 m)
1.1 10 s8.8 10
E
a
t
time
d , ΔV
ifneeded
Fe
When mobile charges move what
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When mobile charges move, what
do they do to the applied E-field?
A they reduce it
B nothing
C they increase it
+ - + -
Di l t i (1)
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Dielectrics (1)
• Have rotating dipoles(on springs)
• Dipoles rotate to opposeapplied E-field
• For same charge on plates,moving charges reduce E-field
• Dielectrics are not conductors• Charges move a little, but are
bound to a small region
0 0net net
E V E V
applied
E0
Enet
+
+
+
+
−
−
−
−
wire
+ ΔV −
dielectric
little E-fields oppose E0
it’s linear!
vacuum valuevacuum value
When I insert a dielectric between the
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plates of a capacitor, what happens to
the capacitance?A It decreases
B nothing
C it increasesD depends on the dielectric E
+
+
+
+
−
−
−
−
wire
large
plate
smallgap, d
+ ΔV −
dielectric
Di l t i (2)
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Dielectrics (2)
• For same charge on plates,rotating dipoles reduce
E-field
• For same charge on plates, ΔV
decreases compared to vacuum
• Coulombs per volt increases• => capacitance increases
E+
+
+
+
−
−
−
−
wire
large
plate
smallgap, d
+ ΔV −dielectric0Q
C V d
0 0net net
V E V
dielectricconstant
Cellular
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technology
“... local dielectric properties can playcrucial roles in membrane functions.”
-- Local Dielectric Properties Around
Polar Region of Lipid Bilayer
Membranes, J. Membrane Biol.
85,225-231 (1985)
“An important physical effect of cholesterol is to increase
the membrane's internal electrical dipole potential, which is
one of the major mechanisms by which it modulates ion
permeability. The dipole potential ... arises because of the
alignment of dipolar residues ... in the .. interior of the
membrane. ... its magnitude can vary from ∼100 to >400
mV.... Recent investigations have suggested that it affects
numerous different biological membrane processes.”
-- Cholesterol Effect on the Dipole Potential of Lipid
Membranes, Biophys J. 2006 June 1; 90(11).
I’m a cell, and ions are expensive. I need to
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, p
create a transmembrane potential difference.
What should I choose to perfuse my membrane?
A Cholesterol, κ = 2
B Water, κ = 80
C Air, κ = 1D It doesn’t matter
Electricity with all my heart
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Electricity, with all my heart• Electrical model allows localizing
performance of individual muscles or
small groups from multiplemeasurements on the skin
Jose Bohorquez, An Integrated-Circuit Switched-Capacitor Model and Implementation of
the Heart , Proceedings of the First International Symposium on Applied Sciences on
Biomedical and Communication Technologies, 25-28 October 2008. DOI10.1109/ISABEL.2008.4712624
+
right atrium
body arteries and veins
tricuspid valve
+
right ventricle
pulmonic valve
+
left atrium
mitral valve
+
left ventricle
aortic valve
pulmonary arteries and veins
Charging a capacitor:
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g g p
some things just take time
• To what voltage (ΔV ) does the capacitorcharge?
• Why does it stop charging?
+
+
+
+
-
-
-
-
wire
Econventionalcurrent
NEVER-
EDDY
+ ΔV −
+−
1.5 V
electroncurrent
1.5 V
+
I
Wh t i lt ?
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What is a voltage source?• An ideal voltage source maintains a constant voltage
across its terminals under all conditions
• A battery is a voltage source. So is ...• ... a generator
• Nothing is really ideal, but it’s a useful model
• When charges move to reduce the voltageacross the terminals ...• The voltage source pumps out more charge to keep the voltage up
• Pulls in charge at low potential energy, pumps out a charge at high PE• This may or may not result in a continuous cycle of charge (current)
• An ideal voltage source can ...• Supply or absorb an arbitrarily large charge
• Supply or absorb an arbitrary large current(arbitrarily large charge in arbitrarily short time)
• Supply or absorb arbitrarily lar ge energy
+
I
charge in(low PE)
charge out(high PE)
+
+
PE
Ideal voltage source (2)
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Ideal voltage source (2)
• Usually, voltage sources supply energy to the circuit
• They have a source of energy, the emf ,
which overcomes the electric force
• The voltage source does positive work onthe charges
• The E-field does negative work on the
charges
• But sources can just as well absorb
energy from the circuit
• You can charge a battery
+
I
e- out
(high PE)
e- in
(low PE)
-
-
PE
Electron flow view
Electric current
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Electric current
• Conductors (good or poor) have mobile charges• If immersed in an electric field, mobile charges move
• While they are moving, we have an electric current
• current is the motion of charge
• Conventional current is the flow of hypothetical positive
charges
q
-q I
velocity
current
What describes the electric current
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What describes the electric current
below?
A It is to the left
B It is about zero
C It is to the rightD Depends on the conductor
q
-q
-q
Continuous current
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Continuous current
• If mobile charges are continuously removed
from one side, and resupplied on the other, we
have a continuous current
• Conventional current is the flow of
hypothetical positive charges
• In C/s ≡ amperes (A)
-q
-q
wire
hypothetical
I = dQ/dt
NEVER-
EDDY
+−
0( ) lim
avg
t
Q I
t Q dQ
I t t dt
Most references use
‘i’ and ‘q’ for time-
dependent values
+ ΔV −-q
electron
I
How does the PE of the e- going into
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the poor conductor compare to the PE
of the e- coming out?
A It’s less
B It’s equal
C It’s moreD not enough information
-q
-q
wire
NEVER-
EDDY
+−
+ ΔV −-q
electron
+
-q
poorconductor
How does the PE of the hypothetical (+)
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charge going into the poor conductor
compare to the PE of it coming out?
wire
hypotheticaldQ > 0
NEVER-
EDDY
+−
+ ΔV −
q
q
q
A It’s less
B It’s equal
C It’s moreD not enough information
From now on, we always
use conventional current
when discussing circuits.
+
I
poorconductor
A wire is a good conductor. How
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much work does it take to move a
charge through it?
A exactly zero
B essentially zero
C a lotD it’s impossible to move a charge through it
Kirchoff’s current law (junction rule)
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Kirchoff s current law (junction rule)
• Conservation of charge• Sum of currents into a node
= sum of currents out of a node
• “Node” is aka “junction”
• Sum of currents into a node = 0
• This is the book’s version
• Sum of currents out of a node = 0
• Not used much
I 1 I 2
I 3
I 1 I 2
I 3
I 1 I 2
I 3
1 2 3
in out I I
I I I
1 2 3
0
0
in I
I I I
1 2 3
0
0
out I
I I I
Kirchoff’s voltage law (loop rule)
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Kirchoff s voltage law (loop rule)
• Conservative nature of electric field
• PE of charge depends on its location,
but not how it got there
• Consider PE change of +1 C of test
charge, at each node around a loop
• If possible, choose loop
direction out from battery +
• I find it easiest to start from battery -
• Final PE must equal starting PE
• PE is proportional to charge (and to voltage)
• Final voltage must equal starting voltage
+
loopdirection
1 2 3
1 2 30 0
rises drops V V V
V V V V
+ V 1 -
+ V 2 - + V 3 -
Capacitor combinations
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Capacitor combinations
• Capacitors in parallel
• When charged, how does V 1 compare to
V 2, and to V ?
• How about charges Q1 and Q2 ?
• Capacitors in series
• When charged, how does V 1 compare to
V 2, and to V ?
• How about charges Q1 and Q2 ?
V
+
C 1
C 2
V
+
C 1C 2
1 21 2 1 2
tot tot tot Q Q QQ Q Q C C C
V V V
1 21 2 1 2
1
/ / 1/ 1/
tot tot tot
tot
Q QV V V C
V Q C Q C C C
Recall:Q
C V
Q1
Q2
+ V 1 - + V 2 -
Ohm’s “Law”
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Ohm s Law
• For a large class of conductors (at constant
temperature), the current through it is proportional to the voltage across it:
• The constant of proportionality is resistance
• Requires consistency in reference directions (sign conventions)• measured in ohms (Ω)
• Some things don’t follow Ohm’s “Law”• Light bulbs ?
wire
NEVER-
EDDY
+−
+
I
conductor
V IR
R
+ ΔV −
resistors
I
+ ΔV −
Discharging a capacitor;+ΔV
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Rearranging capacitors
dq R
+ ΔV −
+ΔV −
e
qdW V dq dq
C
+ΔV −
+ΔV −
Q
0 0
0( )
e
i i
W
eq q
qdW V q dq dq
C
21
2
ie
qW
C
• There’s no R in the final result
• There was never an R to start with
• Even a wire dissipates the same energy
• When rearranging capacitors, the
wires dissipate the missing energy
• What is the energy delivered whendischarging a capacitor?
Q/2
Q/2
The nature of resistance
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The nature of resistance
What is the effect of temperature
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p
on a given conducting element?
A Higher temp usually means higher R
B Higher temp usually means lower R
C Temp has no effect on R D There’s no general trend, it varies by material
Without using any formulas, what
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g y ,
is the total resistance?
A R/2
B R
C 2 RD Depends on R
E 10 ohms
R
R
Resistor combinations
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Resistor combinations
• Resistors in parallel
• How does V 1 compare to V 2, and to V ?
• How about currents I 1 and I 2 ?
• Resistors in series
• How does V 1 compare to V 2, and to V ?• How about currents I 1 and I 2 ?
V +
R1
V
+
1 2
1 2 1 2
1/ / 1/ 1/
tot
tot tot
I I I
V V R I V R V R R R
1 2
1 2 1 21 2
tot
tot tot
V V V
V V V IR IR R R
I I I
Recall: V IR
I 1
+ V 1 - + V 2 -
R2 I 2
I tot
R2
I 2
R1
I 1
Ohmic?
What we need is “More power”
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What we need is More power
• Two terminal devices (R, C, diodes, batteries,
etc.) supply or absorb electrical energy• At what rate?
• Resistors convert electrical energy to heat
• Sign convention for sources isopposite that of all other devices
• Capacitors sometimes absorb & sometimes supply
+
I
R
+ ΔV −
J/s J/C C/s
V I
I + ΔV −
Resistors: Positive powerabsorbs energy from circuit
Sources: Positive power
supplies energy to circuit
What is the power dissipated in a
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p p
15 Ω resistor at 120 V?
A 1800 W
B 80 W
C 960 WD 8 W
E Depends on the current
V +
I
What is the power dissipated in the
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p p
diode, in W?
A 0.10
B 0.06
C 0.04D 0.02
E not enough information
3V
100 + 2V -
+
What is the power in the diode?
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p
Solution
3V
100
+
+ V R
−
+ 2V -
From the loop rule: 3 2 0 1 volt
From Ohm's law: /10
Note the
0 0.01
polarities on
amperes
Power law on diode: 2 0.01 0.02 wa
the loop rule
tts
R R
R
V V
I V
P
I
loopdirection
Defibrillator
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EKG
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EKG
• Voltages measured between pairs of leads• Sample of one voltage below right
• Different pairs measure different regions of the
heart
Home electrical
i
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service
• Drip loops keep water out
• Grounding prevents arcing
• Find your grounding rod at home
drip loops
prevent this
Old-fashioned 120 V service
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15 A
earth
ground
circuitbreakers
neutral
safetyground
hot 120 V
white
black
neutral/
ground bond
15 A
:
hotneutral
etc.
Are you beingserved?
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served?• Phase-to-neutral = 120 V
• Phase-to-phase = 240 V
15 A
phase B
phase A
earth
ground
circuit
breakers
neutral
safetyground
phases A and B areopposite polarity
neutral
phase B phase A
white
black
neutral/
ground bond
etc.
Why two phases?
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Why two phases?
120 V +
+
0.25
0.25
4 A 4 A
+
0.25
0.25 4 A
0.25
4 A
120 V120 V
240 V
receptacle- ΔVwire +
Power in a resistor
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Power in a resistor
• For all (2-terminal) devices, P = (ΔV ) I
• For resistors:
• Losses in a wire vary as the square of the current inthe wire
• Less current = less loss
• It’s easy to know the current in a wire• But hard to know the voltage across it
• High V → low I → low losses
2
2/ /
V IR P IR I I R
V I V R P V V R
What is the power lost in the wires?
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What is the power lost in the wires?
A 2 W
B 4 W
C 8 WD 16 W
E 32 W
120 V +
0.25
0.25
4 A 4 A
wires
What percentage is that ofthe supplied voltage?
What is the voltage
drop in the wires?
Tip: pick your reference
directions first:
What do the currentarrows look like?
What is I ?
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What is I 2?
A 0.0 A
B 0.50 A
C 1.0 AD 2.0 A
E not enough information
+
+
0.25
0.25 4 A
0.25
4 A120 V
120 V
I 2
What is the power lost in the wires?
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What is the power lost in the wires?
+
+
0.25
0.25 4 A
0.25
4 A120 V
120 V
I 2 What percentage isthat of the load?
What is the voltage
drop?
A 2 W
B 4 W
C 8 WD 16 W
E 32 W
What is earth?
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• What does the light bulb do?
• No steady-state current: bulb is dark • There can be transient current
• Charges move to decrease potentialdifferences, then current stops
• Essentially, earth is a big blob ofresistive material• Two stakes in the ground have a
resistance ~100’s to 1000’s Ω• Try it!
• This makes earth a convenientreference point for voltages
• The entire power grid is tied to earthto avoid arcing (sparks)
V +
steady state I = 0
Which path does the current take?
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Which path does the current take?
A Current takes the path of least resistance, RB Current takes the path of most resistance, 2 R
C Current divides equally between both paths
D Current divides in proportion to the resistancesE Current divides in inverse proportion to the
resistances
V +
I
R
2 R
Electrical measurements V +
R
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• Measuring voltage is easy
• Just stick the probe between the points
• Measuring current is harder
• Must insert meter into the current path
• Measuring resistance requires one end of resistor beopen (so it’s similar difficulty to measuring current)
V +
R1
R3 R2V
+ I+
R1
R3 R2
V +
R1
R3 R2
Ω open
I disagree with the book’s quantitativei i d l f d ti
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microscopic model of conduction• The (average) final speed of an accelerated electron
is:
• Which means the average speed is ½ that• Book is off by a factor of 2
• For their equations to work, we must define τ as ½ themean time between collisions
• All the books do the same thing
f
F qE v a
m m
time since collision
v(t )
vaverage
τ
v f
Current-voltage relation for a resistor
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g
• What is the shape?• What is the slope?
• Knight 2nd ed. p969b is wrong:
• Ohm’s law includes signs of V and I
V
I
I = V / R, so
slope = 1/ R
I
R
+ V −
Analyzing a reducible circuit: how arethe 600 ohm and 560 ohm resistor
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the 600 ohm and 560 ohm resistor
related?
A In series
B In parallel
C neither D both
E not enough information
1 2 V
600 Ω
+
800 Ω
400 Ω
5 6 0 Ω
Analyzing a reducible circuit
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y g
• A circuit is reducible if we can use simple
resistor/capacitor combinations to reduce the circuitto a single source and a single resistor and/orcapacitor.
• Usually (but not always), a circuit with only one source is
reducible
1 2 V
600 Ω
+
800 Ω
400 Ω
1 2 V +
800 Ω
240 Ω
5
6 0 Ω
5
6 0 Ω
1 2 V +
800 Ω
800 Ω
1 2 V +
800 Ω 8 0 0 Ω
1 2 V +
400 Ω
Space
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• The space station has
charge dissipators• Xenon ions carry away
built-up charge
NASA says there is a
health risk if the ISS
charges to more than 40
volts, and a 2002 NASAreport says that voltages
as low as 60 volts could
cause cardiac arrest in a
spacewalking astronaut.
Sparks
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p
• Dielectric strength of air ~3e6
V/m• At ~100 V => 3e-5 m ~ 0.03 mm
electrical switch
“Drawing” current
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• Nothing “draws” current
• Devices allow current to flow if pushed by a
voltage (potential difference)
g
I
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Analyzing a 2-loop circuit: which
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resistors can we combine?
A 300 and 100B 300 and 200
C 100 and 200
D All 3 at onceE None of them
19 V
300 Ω
+200 Ω
12 V +
100 Ω
I 1 I 2
Analyzing an irreducible 2-loop circuit
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• Can we reduce it?
• How many unknowns are there?• What about the unlabeled current?
19 V
300 Ω
+200 Ω
12 V +
100 Ω
I 1 I 2
? I 1 - I 2
1 1 219 300 100 12 0 I I I
1 2 212 100 200 0 I I I
1 1 2 1 2
1 2 2 1 2
1 2
1 1
2 2
7 300 100 400 100
12 100 200 100 300
21 1200 300
33 1100 0.03 A
12 3 300 0.05 A
I I I I I
I I I I
I I
I I
I I
1 2
1 2
1
Or:
19 300 200 0
19 300 200
33 1100
I I
I I
I
Current-voltage relation for acapacitor: i(t) ? v(t)
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capacitor: i(t ) ? v(t )
• Note reference polarities
• What is relation between q and i?
• What is relation between q and v?
• Finally, what is relation between i and v?
i
C
+ v −
q(t )
( )( )
dq t i t
dt
( )( ) ( ) ( )
dv t q t Cv t i t C
dt
What is the current?
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