Oct. 2, 2008 1
From last time(s)…
Finish conductors in electrostatic equilibrium Work, energy, and (electric) potential Electric potential and charge Electric potential and electric field.
Gauss’ law Conductors in electrostatic equilibrium
Today…
Oct. 2, 2008 2
Exam 1 Scores
Class average = 76%
(This is 84/110)
Curve: B / BC boundary is 76%
Your score postedat learn@uw
Oct. 2, 2008 3
Conductor in Electrostatic Equilibrium
In a conductor in electrostatic equilibrium there is no net motion
of charge E=0 everywhere inside the conductor
Conductor slab in an external field E: if E 0 free electrons would be accelerated
These electrons would not be in equilibrium
When the external field is applied, the electrons redistribute until they generate a field in the conductor that exactly cancels the applied field.
Etot = E+Ein= 0
Ein
Etot =0
Oct. 2, 2008 4
Conductors: charge on surface only
Choose a gaussian surface inside (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
Oct. 2, 2008 5
E-Field Magnitude and Direction
E-field always surface: Parallel component of E would put force on
charges Charges would accelerate This is not equilibrium
Apply Gauss’s law at surface
€
ΦETot = EA
€
Qencl = ηA
€
⇒ E = η /εo
€
rE || this surface
€
⇒ ΦE = 0
€
rE ⊥this surface
€
⇒ ΦE = EA
€
rE = 0
this surface
Oct. 2, 2008 6
Summary of conductors everywhere inside a conductor
Charge in conductor is only on the surface
surface of conductor
€
rE = 0
€
rE ⊥
----
-
-
+ +++++
Oct. 2, 2008 7
Electric forces, work, and energy
Consider positive particle charge q, mass m at rest in uniform electric field E Force on particle from field Opposite force on particle from hand
Let particle go - it moves a distance d How much work was done on particle? How fast is particle moving?
+ +v=0 v>0€
W = Fd = qEd
€
ΔK .E . =1
2mv 2 = W = qEd
Oct. 2, 2008 8
Work and kinetic energy Work-energy theorem:
Change in kinetic energy of isolated particle = work done
€
dW =r F ⋅d
r s = Fdscosθ
Total work
€
ΔK = dWstart
end
∫ =r F • d
r s
start
end
∫
In our case,
€
rF = q
r E
Oct. 2, 2008 9
Electric forces, work, and energy
Same particle, but don’t let go How much force does hand apply?
Move particle distance d, keep speed ~0 How much work is done by hand on particle? What is change in K.E. of particle?
++
€
F = qE
€
W = Fd = qEd
€
ΔK .E . = 0
Conservation of energy? W stored in field as potential energy
Oct. 2, 2008 10
Work, KE, and potential energy
If particle is not isolated,
€
Wexternal = ΔK + ΔUWork done on system
Change in kinetic energy
Change in electric potential energy
Works for constant electric field if
Only electric potential energy difference Sometimes a reference point is chosen
E.g. Then for uniform electric field
€
ΔU = −qr E ⋅Δ
r r
€
Ur r ( ) = 0 at
r r = (0,0,0)
€
Ur r ( ) = −q
r E ⋅
r r
Oct. 2, 2008 11
Electric potential V
Electric potential difference ΔV is the electric potential energy / unit charge = ΔU/q
For uniform electric field,
€
ΔVr r ( ) =
ΔUr r ( )
q=
−qr E ⋅Δ
r r
q= −
r E ⋅Δ
r r
This is only valid for a uniform electric field
Oct. 2, 2008 12
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 +100µC charge from A to B?
A. +2 mJ
B. -20 mJ
C. +8 mJ
D. +100 mJ
E. -100 mJ
Oct. 2, 2008 13
Check for uniform E-field
++
Push particle against E-field, or across E-field
Which requires work? Constant electric potential in this direction
Decreasing electric potential in this direction
Increasing electric potential in this direction
Oct. 2, 2008 14
Potential from electric field
Potential changes largest in direction of E-field.
Smallest (zero) perpendicular to E-field
€
dV = −r E • d
r l
€
dV = −r E • d
r l
€
dr l
€
rE
V=Vo
€
V = Vo −r E d
r l
€
V = Vo +r E d
r l
€
dr l
€
dr l
€
V = Vo
Oct. 2, 2008 15
Electric potential: general
Electric field usually created by some charge distribution. V(r) is electric potential of that charge distribution
V has units of Joules / Coulomb = Volts
€
ΔU =r F Coulomb • d
r s ∫ = q
r E • d
r s ∫ = q
r E • d
r s ∫
Electric potential energy difference ΔU proportional to charge q that work is done on
€
ΔU /q ≡ ΔV = Electric potential difference
Depends only on charges that create E-fields
€
= r
E • dr s ∫
Oct. 2, 2008 16
Electric potential of point charge
Electric field from point charge Q is
What is the electric potential difference?
€
rE =
kQ
r2ˆ r
€
ΔV =r E • d
r s
start
end
∫ = kQ
r2dx
rinitial
rfinal
∫
= −kQ
r rinitial
rfinal
= kQ
rinital
− kQ
rfinal
Define
€
V r( ) = kQ
rfor point charge
€
V r = ∞( ) = 0 Then
Oct. 2, 2008 17
Electric Potential of point charge
Potential from a point charge Every point in space has a
numerical value for the electric potential
y
x
€
V =kQ
r
Distance from ‘source’ charge +Q
+Q
Oct. 2, 2008 18
Potential energy, forces, work
U=qoV Point B has greater potential
energy than point A Means that work must be done
to move the test charge qo
from A to B. This is exactly the work to
overcome the Coulomb repulsive force.
Ele
ctric
pot
entia
l ene
rgy=
qoV
A
B
qo > 0Work done = qoVB-qoVA =
€
− r
F Coulomb( ) • dr l
A
B
∫
Differential form:
€
qodV = −r F Coulomb • d
r l
Oct. 2, 2008 19
V(r) from multiple charges
Work done to move single charge near charge distribution. Other charges provide the force, q is charge of interest.
q
q1
q2
q3
€
U =r F • d
r s ∫ =
r F q1
+r F q 2 +
r F q3( ) • d
r s ∫
= Uq1r1( ) + Uq 2 r2( ) + Uq 3 r3( )
= kq1q
r+ k
q2q
r+ k
q3q
r
= q kq1
r+ k
q2
r+ k
q3
r
⎛
⎝ ⎜
⎞
⎠ ⎟
= q Vq1 r( ) + Vq2 r( ) + Vq 3 r( )( )Superposition of individual electric potentials
€
V r( ) = Vq1 r( ) + Vq 2 r( ) + Vq 3 r( )
Oct. 2, 2008 20
Quick Quiz 1 At what point is the electric potential zero for this
electric dipole?
+Q -Q
x=+ax=-aA
B
A. A
B. B
C. Both A and B
D. Neither of them
Oct. 2, 2008 21
Superposition: the dipole electric potential
+Q -Q
x=+ax=-aSuperposition of• potential from +Q• potential from -Q
Superposition of• potential from +Q• potential from -Q
+ =
V in plane
