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MTE1 results
Scores available at Learn@UW, your TAs have examsIf your score is an F or a D, talk to us and your TAs for suggestions on how to improve
Mean 75% = 90/120
From last times
Electric charges and forceElectric Field and ways to calculate itMotion of charges in E-fieldGauss’ Law
Today:
More on Gauss’ law and conductors in electrostatic equilibrium
Work, potential energy Electric potential
PHYS208, SPRING 2008
What is the Electric Flux?
(Component of E-field ⊥ to surface) x (surface area)
So you need to understand how the E-field is directed respect to each piece of surface through which you have to calculate it
Why do we need to calculate it?to use Gauss’ law to determine E
PHYS208, SPRING 2008
How to calculate Electric Flux?
Electric flux through a surface: (component of E-field ⊥ to surface) x (surface area)
!
"E
!
ˆ s
Ecos! is component of E-fieldperpendicular to surface
PHYS208, SPRING 2008
GAUSS’ LAW for any closed surface
Flux thru closed surface depends ONLY on the charge enclosed by surface!!
But you need to calculate the Flux to determine EThe Flux depends on the normal component of E that
crosses the surfacebecause of the scalar product in the surface integral
PHYS208, SPRING 2008
Quiz 1
Rank fluxes through the blue dashed spherical surfaces for the 3 cases:
+Q
+Q
+Q+2Q
2)3)
1)
A) 1)<2)=3)B) 3)<1)<2)C) 1) = 3) < 2)D) 1) < 2) < 3)
!
Qr3
a3"0
<Q
"0
<2Q
"0
Flux thru spheres:
!
4"r2E(r)
PHYS208, SPRING 2008
Conductors in equilibrium electrostatic equilibrium
In a conductor in electrostatic equilibrium there is no net motion of charge
Property 1: E=0 everywhere inside the conductor
Conductor slab in an external field E:
• if E-field not null inside the conductor, free electrons would be accelerated
• These electrons would not be in equilibrium.
• When the external field is applied, the electrons redistribute until the magnitude of the internal field equals the magnitude of the external field
•The total field inside the conductor is zero
Etot = E+Ein= 0
Ein
Etot =0
PHYS208, SPRING 2008
Property 2: Charge on only conductor surface
• chose a gaussian surface inside the conductor (as close to the surface as desired)
•There is no net flux through the gaussian surface (since E=0)
•Any net charge must reside on the surface (cannot be inside!)
E=0
PHYS208, SPRING 2008
Magnitude and direction of E-Field
Field lines
E-field perpendicular to the surface:parallel component to E would put force on chargescharges would accelerate along the surface NO equilibrium
Applying Gauss’s law
Remember from P207
• 1st Newton’s law: An object at rest will remain at rest and a moving object will continue to move at constant velocity on a straight line if F=0.
• 2nd Newton’s law(superposition principle applies):
• A moving particle has kinetic energy
• Work:
• Total work and work-energy theorem:
!
r F
net= m
r a
!
F
ds
>0: Force is in direction of motion
<0: Force is opposite to direction of motion
=0: Force is perpendicular to direction of motion
!
dW =r F " d
r s = Fdscos#
Infinitesimal displacement
!
W = F " dsA
B
# = KB$K
A
!
K=1
2mv
2
Potential energy
Fg = mg
F
yf
yi
Work to lift the block done by external agent (F): W = mgh = mg(yf-yi) > 0 (the agent has to make an effort to move the block)Work done by gravitational field: Wg = -mgh
y
h
Conservative Forces: work done by the force is independent on the path and depends only on the starting and ending locations (if the initial and final point coincide => W = 0)
It is possible to define the potential energy U
Wconservative = -! U = Uinitial - Ufinal = !K
"Ug = Ug - U0 = -mghUg = U0 + mgy for gravitational force and U0 =U(y=0) = 0
Electric force workConsider bringing two positive charges together
They repel each other Force is conservative
Pushing them together requires work
Stop after some distance
How much work was done?
+ +
!
r F " d
r x =
xinitial
x final
# ke
Q1Q2
R $ x( )2
dx =xinitial
x final
# ke
Q1Q2
R $ x xinitial
x final
=
= ke
Q1Q2
R $ x final
$ ke
Q1Q2
R $ xinitial
== ke
Q1Q2
rfinal
$ ke
Q1Q2
rinitial
> 0
Calculating the work! E.g. Keep Q2 fixed, push Q1 at constant velocity
! Net force on Q1 ?
! Force from hand on Q1 ?
Zero
!
1
4"#o
Q1Q2
R2
! Total work done by hand
Force in
direction of motion
+ +
Q1 Q2R
xinitial xfinal
+
rfinalrinitial
!
rfinal < rinitial "1
rfinal>
1
rinitial for like charges
!
r F
!
r F
Coulomb
x
Potential energy of 2 point charges
The external agent makes positive work (because charges repel) and changes the potential energy of the system
Wext = !U = Ufinal - Uinitial >0 => U increases.
The energy is stored in the electric field as electric potential energy.
We can set: Uinitial = U(!) = 0 (at infinite distance force becomes null). Hence, electric potential energy of 2 charges at distance r:
!
Wext = keQ1Q2
rfinal" ke
Q1Q2
rinitial= ke
Q1Q2
r12
Q1
Q2
If we initially put Q1 very far away (rinitial = !) and move Q1 at distance rfinal =r12
from Q2:
!
Uelec = keQ1Q2
r
More about U of 2 charges
! Like charges ! U > 0 and work must be done to bringthe charges together since they repel (W>0)
! Unlike charges ! U < 0 and work is done to keep thecharges apart since the attract one the other (W<0)
Quick QuizTwo balls of equal mass and equal charge are held fixed a distance r12 apart, then suddenly released. They fly away from each other, each ending up moving at some constant speed.If the initial distance between them is reduced by a factor of four, their final speeds are
A. Two times bigger
B. Four times bigger
C. Two times smaller
D. Four times smaller
E. None of the above
!
Wel = "#U =Uinitial "Ufinal = keQ1Q2
r12
!
Wel = "K = K final #Kinitial = K final =1
2mv
2= ke
Q1Q2
r12
=0 very far away
=0 initially fixed
if r’12 = 1/4 r12 =>K’final = 4 Kfinal
and v’final = 2 vfinal
U with multiple charges! If there are more than two
charges, then find U foreach pair of charges andadd them
! For three charges:
Quick QuizHow much work would it take YOU to assemble 3 negative charges?
A. W = +19.8 mJ
B. W = 0 mJ
C. W = -19.8 mJ
!1µC
!3µC
!2µC5 m
5 m5 m
Electric Potential
!
"U /q #V = Electric potential
Work done by E-field produced by some source charge on a test charge q to move it along a path:
!
Wel = "#U =r F Coulomb $ d
r s = q
r E $%% d
r s = q
r E $ d
r s %
Define:
V scalar quantity and units volts 1 V = 1 J/C V created by some source charge or charge distribution and
independent on test charge V can be used to determine potential energy of charge q and
source charge system V connected to E due to source charge (more in this week
laboratory)
Equipotential lines
Lines of constant potentialIn 3D, surfaces of constant potential(your Lab)
Quick QuizIn the Figure, q1 is a negative source charge and q2 is a test charge. If q2 is initially positive and then is changed to a negative charge of the same magnitude, the potential at the position of q2 due to q1
(A.) increases (B.) decreases (C.) remains the same
+ ! -
Answer: (c). The potential is established only by the source
charge and is independent of the test charge.
Electric potential at point charge
Consider one charge as ‘creating’ electric potential, the other charge as ‘experiencing’ it
q
Q
!
UQq r( ) = keQq
r
!
Vq r( ) =UQq r( )q
= keQ
r
Electric Potential of point charge
Ele
ctr
ic p
ote
ntial energ
y=qoV
A
B
qo > 0
Every point in space has a numerical value for the electric potential
Potential energy: U = q0V (q0 test charge)VB > VA
Means that work must be done to move the test charge q0 from A to B to overcome the Coulomb repulsive force.
y
x+Q
!
V =kQ
r
Distance from ‘source’ charge +Q
Work done = qoV
B-q
oV
A =
!
"r F
Coulomb( ) • dr l
A
B
#
Differential form:
!
qodV = "r F Coulomb • d
r l
Quick QuizTwo points in space A and B have electric potential VA=20 volts and VB=100 volts. How much work does it take to move a +100"C charge from A to B?
A. +2 mJ
B. -20 mJ
C. +8 mJ
D. +100 mJ
E. -100 mJ
Superposition: electric potential of dipole
+Q -Q
x=+ax=-a
Superposition of
• potential from +Q
• potential from -Q
+ =
y
Eg: V(y=0) = kQ/a-kQ/a=0In general: for a group of point charges
!
V = keqi
rii
"
Electric Potential of a continuous charge distribution
Consider a small charge element dq. The potential at some point due to this charge element is
Total potential:
This value of V used the reference that V = 0 at infinite distance
Van de Graaff generator
!V
• The high-voltage electrode is a hollow metal dome mounted on an insulated column
• Charge is delivered continuously to dome by a moving belt of insulating material
•Large potentials can be developed by repeated trips of the belt
•Protons can be accelerated through such large potentials
•Sparks: electrons from rod to dome excite air molecules that emit light
!
"U = q"V = #"K = K f