Download Lecture Notes 04: Work and Electrostatic Energy

UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lect. Notes 4
Prof. Steven Errede
LECTURE NOTES 4
Work & Electrostatic Energy:
Consider the following situation:
A stationary / fixed configuration of source charges is used to generate a (net) electric
field E ( r ) .
A test charge QT is moved from point a to point b in this electric field E ( r ) . How much
mechanical work W is done on the test charge QT in moving it (slowly) from point a to point b?
At any point r along the path a → b, the electrostatic force acting on QT is: FC ( r ) = QT E ( r )
b
q1 ·
q2 ·
The mechanical force required to
counter/balance the electrostatic force is:
E (r )
q3 ·
a
· →
·
· →
·
· →
·
· →
·
· →
qN ·
Fmech ( r ) = − FC ( r ) = −QT E ( r )
QT @ P
ẑ
r
Fmech
E (r )
FC
Ο
(Net E-field)
ŷ
x̂
The mechanical work done on the test charge QT along the path a → b is:
b
b
a
a
W = ∫ Fmech ( r )idl = −QT ∫ E ( r )idl
b
But: E ( r ) = −∇V ( r ) OR: ΔVab ≡ V ( b ) − V ( a ) = − ∫ E ( r )idl
a
∴ W = QT ⎡⎣V ( b ) − V ( a ) ⎤⎦ = QT ΔVab
The mechanical work done on the test charge QT along the path a → b:
Fmech ( r ) = − FC ( r ) = −QT E ( r )
b
b
b
W = QT ΔVab = QT ⎡⎣V ( b ) − V ( a ) ⎤⎦ = ∫ Fmech ( r )idl = − ∫ FC ( r )idl = −QT ∫ E ( r )idl
a
a
a
Now, if point a is the reference point ra = ∞ where V ( ra ) = V ( ∞ ) = 0 and point rb = r
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
1
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lect. Notes 4
Prof. Steven Errede
Then: W = QT ⎡V ( r ) − V ( ∞ ) ⎤ = QT V ( r )
⎣
⎦
i.e. the test charge QT is (slowly) mechanically moved from the point ra = ∞ to the point rb = r .
Then W = QT V ( r ) is the mechanical work done on the charge QT in moving it along the path a
→ b from ra = ∞ to rb = r .
Thus, now we also see that W = work done on charge QT is also equal to the POTENTIAL
ENERGY, P.E. = W = QTV ( r )
P.E. is linearly proportional to the potential V(r) (when
referenced to V (ra = ∞) ).
POTENTIAL ENERGY = amount of work W it takes to create the system (Joules).
POTENTIAL DIFFERENCE = work W it takes to create the system per unit charge
(Joules/Coulomb = Volts).
UNITS OF WORK, W = UNITS OF POTENTIAL ENERGY, P.E. = JOULES (SI UNITS)
W = P.E.= QT V ( r ) = Coulomb-Volts = Joules
i.e. 1 Coulomb x 1 Volt = 1 Joule
Fundamental unit of electric charge:
1e = 1.602×10−19 Coulombs = 1.602×10−19 C
∴ 1 electron volt = 1.602×10−19 Joules ← energy conversion factor for eV ⇔ Joules
= eV
Example 1:
An electron is initially infinitely far away from a proton, and is initially at rest (as is the proton).
Assume the proton is fixed / rigidly attached to the head of a pin at the origin
Ο( x, y, z ) = (0, 0, 0) .
The electron is “released” at zero velocity and is attracted to the proton. What is its kinetic
energy and speed when it is a distance of 1 meter from the proton?
Use Energy Conservation:
INIT
FINAL
ETOT
= ETOT
=0
INIT
ETOT
= KEeINIT + PEeINIT = 0
FINAL
ETOT
= KEeFINAL + PEeFINAL = 0
2
KEeINIT = 12 me vinit
=0
KEeFINAL = 12 me v 2final ≥ 0
vinit = 0
2
v final = ??
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Prof. Steven Errede
PEeFINAL = QeV p ( rb = 1m ) where V p ( r ) =
PEeINIT = QeV p ( ra = ∞ )
Qe=−e
Lect. Notes 4
= −eV p ( ra = ∞ )
= −eV p ( rb = 1m )
1 e2
=−
=0
4πε o ∞
=−
So:
INIT
FINAL
ETOT
= 0 = ETOT
= KEeFINAL + PEeFINAL = 0
Or:
KEeFINAL = − PEeFINAL = +
1 Qp
,
4πε o r
with Qp=+e.
1 ⎛ e2 ⎞
⎜ ⎟<0
4πε o ⎝ 1 ⎠
1 ⎛ e2 ⎞
⎜ ⎟
4πε o ⎝ 1 ⎠
r = 1 meter
−19
e = 1.602x10 coulomb
ε o = 8.85x10−12 Farads/meter
1.6 ×10−19 )
(
1
=
Joules
4π ∗ 8.85 × 10−12
1
2
∴ KEeFINAL
KEeFINAL = 2.3 × 10−28 Joules
But: 1 eV = 1.6 ×10−19 Joules
1 eV
1.6 ×10−19
1 Joule = 6.242×1018 eV
OR: 1 Joule =
∴ KEeFINAL = 1.437 × 10−9 eV
Now KEeFINAL =
1
me ve2 FINAL = 2.3 ×10−28 Joules
2
But me = 9.11×10−31 kg
1 Joule = 1 Newton-meter
b
W = ∫ Fmech idl
a
veFINAL =
veFINAL
2 KEeFINAL
2 ∗ 2.3 ×10−28 Joules
2
=
= 504.94 ( m / s )
−31
me
9.11×10 kg
22.5 m/sec
at
r = 1m
22.5 m / s
speed
of light
NOTE: veFINAL ( r = 1m ) c
(22.5 m/s)
(c = 3×108 m/s)
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
3
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
KEe(r) > 0
1
= me ve2 ( r )
2
Lect. Notes 4
Prof. Steven Errede
ETOT ( r ) = KEe ( r ) + PEe ( r ) = 0
i.e. KEe ( r ) = − PEe ( r )
ETOT=0
r
1
1 ⎛ e2 ⎞
2
me ve ( r ) = +
⎜ ⎟
2
4πε o ⎝ r ⎠
PEe(r) < 0
=−
1 ⎛ e2 ⎞
⎜ ⎟
4πε o ⎝ r ⎠
1 ⎛ e2 ⎞
⎜
⎟
2πε o ⎝ me r ⎠
ve ( r ) =
Example 2:
An electron is located at x = − ∞ and has an initial velocity v = 22.5 m / s xˆ and is initially
infinitely far away from another electron, which is rigidly fixed at the origin Ο( x, y, z ) = ( 0, 0, 0 )
at rest.
veINIT
= 22.5 m / s
1
e1
e2
x = −∞
Ο = ( 0, 0, 0 )
x̂
What is the distance of closest approach of e1 to e2? Why is there a minimum distance?
TOT
TOT
We know that EeINIT
= EeFINAL
(Energy is conserved in this process)
1
1
TOT
EeINIT
= KEeINIT
+ PEeINIT
1
1
1
TOT
EeFINAL
= KEeFINAL
+ PEeFINAL
1
1
1
1
1 e2
2
me vINIT
+ 0 = 2.3 ×10−28 Joules
= 0+
2
4πε o r 2
22.5 m/s
repulsive
closest approach
Solve for rclosest :
=
approach
(1.6 ×10 ) = 1 meter !
1 ⎛ e2 ⎞
1
=
=
⎜
⎟
2
2πε o ⎝ me vinit
2π ∗ 8.85 × 10−12 9.11× 10−31 × ( 22.5 )2
⎠
−19 2
∴ rclosest
approach
4
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lect. Notes 4
Prof. Steven Errede
INIT
FINAL
ETOT ( r ) = ETOT
= ETOT
= 2.3 × 10−28 Joules = constant =
KE(r)+PE(r)
2.3×10−28
Joules
KE(r)
ETOT(r)
KE(r)
or:
PE(r)
PE(r)
0m
0 Joules
r = 1m
r
Classical turning point
What happens:
e1 comes in from r = ∞ , is decelerated and stopped at r =1m (= distance of closest approach),
and then e1 is accelerated back out to r = ∞ . vefinal
= 22.5 m / s at r = ∞ again.
1
How much mechanical work is done in taking a charge QT around a closed path / closed
contour C in an external electric field E ( r ) ?
∫
a
a
= −QT
∫
W=
∫ ( − F ( r ) )idl = − ∫
Fmech ( r )idl =
a
a
C
C
C
a
a
a
a
FC ( r )idl
C
E ( r )idl = 0
because:
C
∫ E ( r )idl = 0
NO work is done on closed contour, because the electrostatic Coulomb Force, FC ( r ) is a
conservative force.
∴ Mechanical work done on test charge QT is independent of the path:
E (r )
E (r )
E (r )
b
QT
path1
path2
a
Note:
∫
C
a
a
QT
=∫
b
a
+∫
E ( r ) is produced by some (arbitrary)
source charge distribution.
a
b
path1 path2
path1 + path2 are arbitrary because FC ( r ) is conservative (Coulomb Pot’l is 1/r, central pot’l no θ , ϕ dependence.)
FC ( r ) = −QT ∇V ( r )
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
5
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lect. Notes 4
Prof. Steven Errede
ELECTROSTATIC ENERGY OF ASSEMBLY
OF A POINT CHARGE DISTRIBUTION
How much work does it take to assemble a collection of point charges – bringing them in
from infinity, one by one? Bringing in the first charge q1 takes NO work (W1 =0), since there is
no electric field present, initially.
Now bring in the 2nd charge q2 from infinity. The work done in bringing in q2 from infinity is:
W2 =
q2 ⎛ q1 ⎞
⎜ ⎟ = q2V1 ( r2 )
4πε o ⎝ r12 ⎠
with
V1 ( r2 ) =
1 ⎛ q1 ⎞
⎜ ⎟
4πε o ⎝ r12 ⎠
where r12 = r2 − r1 and
r12 = r12 = r2 − r1 = separation distance between r2 & r1 (i.e. between charges q1 & q2)
step 2:
∞
step 1:
∞
ẑ
q2
r12
q1
r2
r1
Ο
ŷ
x̂
Now bring in the 3rd charge from infinity. This requires work W3 = q3V1,2 ( r3 ) where V1,2 ( r3 ) is
the (total) potential due to charges q1 & q2:
1 ⎛ q1 q2 ⎞
V1,2 ( r3 ) = V1 ( r3 ) + V2 ( r3 ) =
⎜ + ⎟
4πε o ⎝ r13 r23 ⎠
∴ W3 = q3V1,2 ( r3 ) =
q3 ⎛ q1 q2 ⎞
⎜ + ⎟
4πε o ⎝ r13 r23 ⎠
Similarly for the 4th charge: V1,2,3 ( r4 ) = V1 ( r4 ) + V2 ( r4 ) + V3 ( r4 ) =
∴ W4 = q4V1,2,3 ( r4 ) =
6
1 ⎛ q1 q2 q2 ⎞
+ ⎟
⎜ +
4πε o ⎝ r14 r24 r34 ⎠
q4 ⎛ q1 q2 q3 ⎞
+ ⎟
⎜ +
4πε o ⎝ r14 r24 r34 ⎠
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lect. Notes 4
Prof. Steven Errede
The total work necessary to assemble the first 4 charges is thus:
WTOT = W1 + W2 + W3 + W4 = 0 + q2V1 ( r2 ) + q3V1,2 ( r3 ) + q4V1,2,3 ( r4 )
=
1 ⎛ q1q2 q1q3 q1q4 q2 q3 q2 q4 q3 q4 ⎞
+
+
+
+
+
⎜
⎟
4πε o ⎝ r12
r13
r14
r23
r24
r34 ⎠
We can generalize this relation for N charges as:
WTOT =
1
4πε o
N
∑
i =1
⎛ qi q j
⎜
∑
⎜ rij
j =1
j ≠i ⎝
N
⎞
1
⎟=
⎟ 8πε o
⎠
j >i
N
∑
i =1
⎛ qi q j
⎜
∑
⎜ rij
j =1
j ≠i ⎝
N
so we don’t
count same
pair twice!
WTOT
⎛
1 N ⎜ N 1 ⎛ qj
⎜
= ∑ qi ∑
2 i =1 ⎜⎜ j =1 4πε o ⎜⎝ rij
⎝ j ≠i
⎞
⎟
⎟
⎠
double-counts pairs but factor of 8 (vs. 4)
takes care of this!
⎞⎞ 1 N
⎟ ⎟ = ∑ qiV ( ri )
⎟ ⎟⎟ 2 i =1
⎠⎠
= V ( ri )
Notice WTOT can be < 0 or can be > 0, depending on signs of qi & qj
e.g.
if all qiqj terms > 0, WTOT > 0 (for pair-wise repulsive forces)
if all qiqj terms < 0, WTOT < 0 (for pair-wise attractive forces)
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
7
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lect. Notes 4
Prof. Steven Errede
ELECTROSTATIC ENERGY OF A
CONTINUOUS CHARGE DISTRIBUTION
For a discrete / discretized charge distribution:
WTOT
⎛
1 N
1 N ⎜ N 1 ⎛ qi
= ∑ qiV ( ri ) = ∑ qi ∑
⎜
2 i =1
2 i =1 ⎜⎜ j =1 4πε o ⎜⎝ rij
⎝ j ≠i
⎞ ⎞⎟
⎟⎟ ⎟
⎠ ⎟⎠
N
∑ q V (r ) ⇒ ∫
For a continuum volume charge density, ρ ( r ) :
i =1
WTOT =
i
i
V
dqV ( r ) = ∫ ρ ( r ) V ( r ) dτ
V
1
ρ ( r ) V ( r ) dτ
2 V∫
For a continuum line charge density, λ ( r ) :
WTOT =
1
λ ( r ) V ( r ) dl
2 C∫
()
For a continuum surface charge density, σ r :
WTOT =
1
σ ( r ) V ( r ) dA
2 ∫S
Using Gauss’ Law:
ρ ( r ) = ε o ∇i E
WTOT =
ε
1
ρ ( r ) V ( r ) dτ = o
∫
2 V
2
∫ ( ∇i E )V ( r ) dτ
V
Now use integration by parts to transfer the derivative from E to V (and also use the divergence
theorem):
∫ ∇i( f A) dτ = ∫ f ( ∇i A) dτ + ∫ Ai( ∇ f ) dτ = ∫
Or: ∫ f ( ∇i A ) dτ = − ∫ Ai( ∇ f ) dτ + ∫ fAida
V
V
V
Thus: WTOT =
8
V
V
S
fAida
(see Griffith’s eqn. 1.59, p. 37)
S
εo ⎡
− ∫ E ( r )i( ∇V ( r ) ) dτ + ∫ V ( r ) E ( r )idA⎤
2 ⎣
V
S
⎦
= −E ( r )
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lect. Notes 4
Prof. Steven Errede
εo ⎡
− ∫ E ( r )i( ∇V ( r ) ) dτ + ∫ V ( r ) E ( r )idA⎤
WTOT =
2 ⎣
V
⎦
S
=
εo ⎡
=
εo ⎡
2
∫ E ( r ) dτ + ∫ V ( r ) E ( r )idA⎤
2 ⎣ ∫V
2 ⎣
( E ( r )i E ( r ) ) dτ + ∫ V ( r ) E ( r )idA⎤⎦
S
V
⎦
S
Now integration volume V and enclosing surface of integration S are arbitrary, as long as all
charges are contained within the volume V and enclosed by the surface S.
⎛S⎞
So we can take ⎜ ⎟ → all space (i.e. ∞ - volume & surface) without any loss of generality.
⎝V ⎠
Then the surface integral over all space ∫ V ( r ) E ( r )idA vanishes, because for a localized
S
All Space
charge distribution i.e. one which has finite spatial extent {with characteristic size ~ d}, far away
from the localized charge distribution, if there is a net electric charge associated with the
localized charge distribution, then V ( r d ) ~ 1 r and E ( r d ) = −∇V ( r d ) ~ 1 r 2 .
If the localized charge distribution has no net electric charge, then for r d the potential
V ( r d ) and hence the electric field E ( r d ) will fall of more rapidly than 1 r and 1 r 2 ,
respectively. {Please see/read P435 Lecture notes 8 (and/or Griffiths 3.4, p. 146-159) for further
details/understanding of these large-r dependencies.} Thus, we see that both V ( r = ∞ ) = 0
and E ( r = ∞ ) = 0 , hence
∴ WTOT =
εo
2
∫
∫
S
All Space
E 2 ( r ) dτ
V ( r ) E ( r )idA = 0 and thus:
(n.b. ≥ 0 always!)
V
all
space
However, note that for a single point charge, since E ( r ) =
q
Then: WTOT
=
εo
q2
2 ( 4πε o
) ∫
2
V
all
space
1 2
q2
r
dr
sin
d
d
θ
θ
ϕ
=
r4
8πε o
∫
∞
0
q
1
rˆ
4πε o r 2
1
dr = ∞ !!!
r2
The electrostatic energy associated with a point electric charge is infinite (i.e. singular)!
The origin r = 0 cannot truly be included here – this formula breaks down – because it
mathematically describes only classical physics – and e.g. we know that quantum mechanics
takes over at distance scales of:
r≤
e
=
c
me c 2
197 MeV − fm
0.511
385.5 fm 385.5 ×10−15 m
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
9
UIUC Physics 435 EM Fields & Sources I
e
=
λe
h
, c 197 MeV − fm
2π
me c 2 = 511 keV = 0.511 MeV
=
and
Lect. Notes 4
Prof. Steven Errede
c
= “reduced” Compton wave length of electron, here h = Planck’s constant,
me c 2
=
2π
Fall Semester, 2007
(1 MeV = 10 eV ) , electron rest mass energy,
6
And 1 fm = 1 femto − meter = 1 fermi = 10−15 m
Actually, we know that for atoms & molecules, quantum mechanics takes over at distance scales of
~ Bohr radius (H-atom): ao = 0.53Þ= 0.53 ×10−10 m = 0.053 nm
385.5 ×10−15 m
e = 385.5 fm
c
q
The classical formula for WTOT
breaks down / is invalid for point charges, for r ≤ e =
385.5 fm
me c 2
At an even shorter distance scale, that of the so-called Planck distance LP =
with corresponding Planck time t P = LP c =
GN c 5
GN c 3
1.6 × 10−35 m ,
5.4 ×10−44 s , the “classical”, continuous
nature of space-time itself becomes “foam-like” – thus, we certainly have no reason to believe in the
validity of Coulomb’s law below the Planck distance scale.
WTOT =
=
εo
2
εo
2
∫ E ( r )i E ( r ) dτ
(Joules)
This formula works fine/is valid for
continuous charge distributions:
λ ( r ) ,σ ( r ) , ρ ( r )
V
all
space
∫
E 2 ( r ) dτ > 0 (always)
V
all
space
E ( r ) is total/net electric field arising from charge distribution, ρ ( r ) .
Define: uE ( r ) ≡
Joules/meter3
WTOT =
∫
εo
ε
E ( r )i E ( r ) ) = E
(
2
2
o
u E ( r ) dτ =
V
all
space
=
1
2
εo
2
∫
2
E 2 ( r ) dτ
1
( r ) = ρ ( r )V ( r )
2
= electrostatic energy density =
Physically, says that the energy density, uE ( r ) = ε2o E 2 ( r )
V
all
space
is associated entirely with the electric field, E ( r ) .
∫ ρ ( r )V ( r ) dτ
Physically, says that the energy density, u E ( r ) = 12 ρ ( r ) V ( r )
V
all
space
is due to product (i.e mix) of volume charge density, ρ ( r ) and
electrostatic potential, V ( r ) .
Electrostatic energy density: uE ( r ) =
10
εo
2
E2 (r ) =
1
ρ ( r ) V ( r ) (Joules/meter3)
2
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lect. Notes 4
Prof. Steven Errede
ELECTROSTATIC ENERGY AND THE SUPERPOSITION PRINCIPLE
Note that since W (work) / P.E. (potential energy) is quadratic in the electric field, i.e.
W=
εo
2
∫
E 2 ( r ) dτ =
εo
∫ E ( r )i E ( r ) dτ
2
V
all
space
V
all
space
then for example, if we double E , i.e. E → 2 E , the work needed / done to assemble the nowdoubled charge distribution / stored potential energy in the electrostatic field quadruples (i.e
increases 4x).
Thus, work done to assemble the charge distribution / potential energy stored in the electrostatic
field of the charge distribution does NOT obey the principle of linear superposition.
This is not surprising, as there exist many other physical examples where the superposition
principle does not hold – e.g. the superposition of two (or more) acoustic / sound waves, or e.g.
the superposition of two (or more) light waves – sound/light overall intensities (proportional to
square of overall/total amplitude) result from interference effects at the amplitude level.
Here, e.g. let ETOT ( r ) = E1 ( r ) + E2 ( r )
Then: WTOT = P.E. =
=
=
εo
2
∫ ( E ( r )i E ( r ) ) dτ
TOT
εo
2
∫ ( E ( r ) + E ( r ) )i( E ( r ) + E ( r ) ) dτ
1
2
1
2
V
all
space
εo
2
TOT
V
all
space
∫ ( E ( r ) + 2 E ( r )i E ( r ) + E ( r ) ) dτ
2
1
1
2
2
2
V
all
space
WTOT = P.E. = W1 + W2 + ε o
∫ ( E ( r )i E ( r ) ) dτ
1
2
V
all
space
Clearly, the principle of linear superposition is not (always) obeyed for WTOT – there exists an
additional cross-term involving the product the two electric fields: E1 ( r )i E2 ( r ) .
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
11