Download Lecture Notes 03: Electrostatic Potential, Poisson and Laplace Equation, Boundary Conditions

UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 3
Prof. Steven Errede
LECTURE NOTES 3
The Scalar Electric Potential, V ( r )
The electric field E ( r ) is a very special type of vector point function/vector field, which for
electrostatics, the CURL of E ( r ) = zero, i.e. ∇ × E ( r ) = 0 . The physical meaning of the curl of
a vector field is as follows:
For an arbitrary vector field A ( r ) , if ∇ × A ( r ) ≠ 0 for all (or some) points in space, r then the
vector A ( r ) rotates/circulates/ swirls, or shears in some manner in that region of space − e.g. the
velocity field of a whirlpool, vwp ( r ) , or that associated e.g. with a wind shear field, vshear ( r ) :
Curl of Whirlpool Field ∇ × vwp ( r ) ≠ 0 :
Curl of Shear Field ∇ × vshear ( r ) ≠ 0 :
In Cartesian Coordinates: v ( r ) = vx ( r ) xˆ + v y ( r ) yˆ + vz ( r ) zˆ
⎛ ∂v ( r ) ∂v y ( r ) ⎞
⎛ ∂v y ( r ) ∂vx ( r ) ⎞
⎛ ∂vx ( r ) ∂vz ( r ) ⎞
Curl: ∇ × v ( r ) = ⎜ z
−
−
−
⎟ xˆ + ⎜
⎟ zˆ
⎟ yˆ + ⎜
∂z ⎠
∂x ⎠
∂y ⎠
⎝ ∂z
⎝ ∂y
⎝ ∂x
∇ × v ( r ) is a vector quantity. Compare this with
Divergence: ∇iv ( r ) (= measure of how much the vector v spreads out, or diverges in space):
In Cartesian Coordinates: ∇iv ( r ) =
∂vx ( r ) ∂v y ( r ) ∂vz ( r )
+
+
∂x
∂y
∂z
∇iv ( r ) is a scalar quantity (i.e. number).
∇ iv ≠ 0
(c)
©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
Lecture Notes 3
Prof. Steven Errede
We saw in the previous lecture (P435 Lect. Notes 2, p.15) by use of Stokes’ Theorem,
(for electrostatics) that:
∫ ∇ × E ( r )idA = ∫ E ( r )id = 0
S
C
(closed surface) (closed contour)
from which there are two implications (assuming E (r ) ≠ 0 everywhere):
1.) ∇ × E (r ) = 0 everywhere (for arbitrary closed surface S).
2.)
∫
C
E (r )id = 0 implies path independence of this (arbitrary) closed contour, C.
i.e./e.g. taking path ( i ) in figure below gives identical result as taking path ( ii ) in the figure
below:
point b
path (i)
∫
•
b
a
b
b
a
a
E ( r )i d = ∫ E ( r ) i d = ∫ E ( r ) i d
path (i )
path (ii )
any path
point a •
path (ii)
because
b
∫ E ( r )i d
is independent of the path taken from point a → b.
a
We now define a scalar point function, V ( r ) known as the electric potential, as follows:
Electric Potential
(integral version)
V ( r ) ≡ −∫
r
Ο ref
E ( r )id
By convention, the point r = Ο ref is taken to be a
standard reference point of electric potential,
V ( r ) where V ( r = Ο ref ) = 0 (usually r = ∞) .
SI Units of Electric Potential = Volts
If V ( r = Ο ref ) = 0 @ the reference point, Ο ref then V ( r ) depends only on point r .
The electric potential difference between two points a & b is thus:
⎛ b
V ( b ) − V ( a ) = ⎜ − ∫Ο
⎝
b
= − ∫Ο
b
= − ∫Ο
ref
2
⎞ − ⎛ − a E ( r )i d ⎞
⎟ ⎜ ∫Ο ref
⎟
⎠ ⎝
⎠
a
E ( r )i d + ∫Ο
ref
Ο
ref
Ο
= − ∫a
ref
E ( r )i d
ref
E ( r )i d − ∫a
b
E ( r )i d − ∫Ο
ref
ref
E ( r )i d
E ( r )i d
E ( r )i d
©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
Lecture Notes 3
Prof. Steven Errede
Thus:
b
ΔVab ≡ V ( r = b ) − V ( r = a ) = − ∫ E ( r )id
Integral Version
a
b
path Ο ref → b
path a → b
Ο ref
( e.g.@ r = ∞ )
a
path Ο ref → a
The fundamental theorem for gradients states that:
b
b
a
a
Potential difference: ΔVab ≡ V ( r = b ) − V ( r = a ) = ∫ ∇V ( r )id = − ∫ E ( r )id
This is true for any end-points a & b (and any contour from a → b). Thus the two integrands
must be equal:
E ( r ) = −∇V ( r ) Differential Version
Now (for electrostatics): ∇ × E ( r ) = 0, Thus: ∇ × E ( r ) = ∇ × ( −∇V ( r ) ) = −∇ × ∇V ( r ) = 0
See inside front cover of Griffiths, ∇ × ∇f ( r ) = 0 , where f ( r ) is a scalar point function.
For Electrostatic problems, ∇ × E ( r ) = 0 will always be true. For such problems, this means
that (both) F ( r ) = QT E ( r ) and E ( r ) can be expressed as the (negative) gradient of a scalar
point function, V ( r ) ,
i.e.
E ( r ) = −∇V ( r )
F ( r ) = −QT ∇ ( r )
A scalar point function ( V ( r ) here) is one which is a scalar quantity (not a vector quantity)
who’s numerical value depends on position in space, r – e.g. a continuous/well-behaved
function which is mathematically defined at every point r = xxˆ + yyˆ + zzˆ.
⇒ Knowing V ( r ) enables you to specify/calculate E ( r ) !!
©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
Lecture Notes 3
Prof. Steven Errede
What is the physical meaning of the electric potential, V ( r ) ?
Newton-meters
Also
Coulomb
Newtons ⎛ F ( r ) ⎞
SI units of E ( r ) =
⎜=
⎟ but SI units of E ( r ) = Volts meter ( since E = −V ( r ) )
Coulomb ⎝ QT ⎠
SI units of V ( r ) = Volts =
The electric field, E ( r ) is the (negative) spatial gradient of electric potential, V ( r )
In Cartesian coordinates, ∇V ( r ) =
∂V ( r )
∂V ( r )
∂V ( r )
xˆ +
yˆ +
zˆ
∂x
∂y
∂z
Why is E ( r ) specified as negative gradient of scalar quantity, the electric potential??
Because of the way we define (by convention) the reference point for absolute voltage/potential,
when r = ∞ .
Consider our point charge problem (again) {n.b. choose local origin @ the point charge}:
1 ⎛q⎞
⎜ ⎟
4πε o ⎝ r ⎠
⎛ 1 ⎛ q ⎞⎞
1
⎛1⎞
E ( r ) = −∇V ( r ) = −∇ ⎜
q∇ ⎜ ⎟
⎜ ⎟⎟ = −
4πε o
⎝r⎠
⎝ 4πε o ⎝ r ⎠ ⎠
n.b. V ( r ) for a point charge has no θ or ϕ -dependence
V (r ) =
In spherical-polar coordinates: ∇ =
∂
1 ∂ ˆ
1
∂
rˆ +
θ+
ϕˆ
∂r
r ∂θ
r sin θ ∂ϕ
1 ∂ ˆ
1
∂ ⎫⎛ 1 ⎞
⎛1⎞ ⎧ ∂
∴ ∇ ⎜ ⎟ = ⎨ rˆ +
θ+
ϕˆ ⎬ ⎜ ⎟
r ∂θ
r sin θ ∂ϕ ⎭ ⎝ r ⎠
⎝ r ⎠ ⎩ ∂r
=0
=0
=
1
∂ ⎛1⎞
1 ∂ ⎛1⎞ ˆ
∂ ⎛1⎞
⎜ ⎟ ϕˆ
⎜ ⎟ rˆ +
⎜ ⎟θ +
r ∂θ ⎝ r ⎠
r sin θ ∂ϕ ⎝ r ⎠
∂r ⎝ r ⎠
∂ ⎛1⎞
1
⎜ ⎟=− 2
∂r ⎝ r ⎠
r
∴ E ( r ) = −∇V ( r ) = +
∂ ⎛1⎞
⎜ ⎟=0
∂θ ⎝ r ⎠
∂ ⎛1⎞
⎜ ⎟=0
∂ϕ ⎝ r ⎠
1 ⎛q⎞
⎜ ⎟ rˆ
4πε o ⎝ r 2 ⎠
1
1
and thus both E ( r ) , FC ( r ) ~ 2 for point electric charge. The electrostatic potential
r
r
V ( r ) associated with a point charge q is a central potential; it varies as ~ 1/r.
V (r ) ~
Note that V ( r ) has no θ- and/or φ-dependence.
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
Lecture Notes 3
Prof. Steven Errede
⇒ The Coulomb force is a central force (as is e.g. the gravitational force). Thus, the Coulomb
force is a conservative force, like gravity, because F ( r ) = −QT ∇V ( r ) can be written as the
(negative) gradient of scalar point function, V ( r ) .
Let’s plot V ( r ) for a point charge Q. For definiteness’ sake, we will plot V ( r ) for Q = + e
and Q = −e (e = magnitude of charge of an electron or proton, i.e. 1.602 x 10−19 Coulombs).
n.b. again, we choose the local origin to be located at the point charge.
Electric Potentials & Fields:
+e ⎛ 1 ⎞
V ⊕ (r ) =
⎜ ⎟
4πε o ⎝ r ⎠
V ( r ) =
E ⊕ ( r ) = −∇V ⊕ ( r )
E  ( r ) = −∇V  ( r )
=+
e ⎛1⎞
⎜ ⎟ rˆ
4πε o ⎝ r 2 ⎠
Radial outward
Lines of E ( r )
−e ⎛ 1 ⎞
⎜ ⎟
4πε o ⎝ r ⎠
=−
e ⎛1⎞
⎜ ⎟ rˆ
4πε o ⎝ r 2 ⎠
Radial inward
Lines of E ( r )
Thus, we see that by defining E ( r ) as the negative gradient, this also simultaneously defines the
convention that lines of E point outward from ⊕ charge, and point inward for  charge.
Graph of the electrostatic potential V ( r ) for ⊕ and charges:
V ( r ) = +∞
SI UNITS
of V(r):
Volts
V (r ) = 0
V ⊕ (r ) ~ +
1
(Volts )
r
·
V ⊕ (r = ∞) = 0
r
V  (r ) ∼ −
1
(Volts )
r
V  (r = ∞) = 0
V ( r ) = −∞
r = ∞ is the reference point, where V ( r = ∞ ) = 0
©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
Lecture Notes 3
Prof. Steven Errede
Equipotentials:
1 ⎛q⎞
⎜ ⎟ (i.e. potential for a point charge, q) that for r = constant,
4πε o ⎝ r ⎠
e. g. r = R, then V ( r = R ) = constant.
Note that from e.g. V ( r ) =
Thus, for a point charge, q, there exist “imaginary” surfaces – concentric spheres of varying radii
r = R1 < R2 < R3 < … RN whose spherical surfaces are surfaces of constant potential (i.e.
potential = same value, in Volts everywhere on one of these surfaces, RN, where N = 1, 2, 3, …)
These “imaginary” surfaces of constant potential are known as equipotential surfaces –
projecting them onto a 2-D surface, contours of constant potential can be seen.
Note that the equipotentials/contours of constant electrostatic potential V ( r ) are everywhere
perpendicular to lines of E ( r ) ! e. g. for a +ve point charge, +q looks like a contour map! (It
is!!!)
1 ⎛ +q ⎞
V (r ) =
⎜
⎟
4πε o ⎝ r ⎠
E ( r ) = −∇V ( r )
1 ⎛ +q ⎞
=
⎜
⎟ rˆ
4πε o ⎝ r 2 ⎠
6
Equipotential Surfaces:
r = R1 V1 ( R1 ) = +100V
r = R2 V2 ( R2 ) = +80V
r = R3 V3 ( R3 ) = + 65V
r = R4 V4 ( R4 ) = + 55V
r = R5 V5 ( R5 ) = + 50V
©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
Lecture Notes 3
Prof. Steven Errede
Equipotential surfaces exist for (arbitrary) electrostatic charge distributions – e.g. a charged lump
of metal:
e.g.
V1 ( R1 ) = + 100V
V2 ( R2 ) = + 80V
V3 ( R3 ) = + 65V
V4 ( R4 ) = + 55V
etc.
Close in to the actual conducting surface (itself an equipotential!), the equipotentials outside
the charge distribution “follow” the shape of the conducting surface. However, note that the
further away from the conducting surface that the equipotential surface is, it becomes
“smoother”/less bumpy/less wiggly. Very far away, the equipotential surface is nearly spherical
in shape, independent of the shape of the actual object, at least for reasonable/small h x w x l
aspect ratios.
⇒ This has ramifications for the ability to measure/infer shape of object
by mapping out equipotential surfaces. Lose detailed information about the
geometrical shape of the object, the farther away one gets!!
•
•
•
•
Neglecting internal stress/strains, plate tectonics, tidal effects, earth’s rotation, etc. The
earth’s surface is an equipotential of the earth’s gravito-electric field!
Indeed, sea level (if you also don’t think too much about details of this) is also an
equipotential of the earth’s gravito-electric field! (neglecting tidal effects, Global
warming, ice ages, El Nino, La Nina…)
We define atmospheric pressure = 1 ATM (@ T = 20°C) and use this pressure as our
“reference standard” to which other pressures are related, defined by pressure differences
from sea level pressure (can also do this relative to zero pressure also, i.e. absolute
pressure) because we have developed technology of vacuum pumps.
Similarly, in the “real world” we find that only potential differences have practical
meaning. We cannot physically measure V ( r = ∞ ) = 0 because it’s in outer space
(somewhere)…
©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
Lecture Notes 3
Prof. Steven Errede
•
For convenience, scientists have defined the earth’s surface (everywhere) as our (local)
electrostatic zero of potential, Vearth = 0. (But in reality, this is also not true….see
comment below…)
•
In practice, people drive a ~ 6' long copper-coated steel rod nearly entirely into the
ground – that rod is defined as Vearth = 0, from which (relative) potential difference
(voltage difference) measurements can be made.
•
In reality, ∃ (there can/sometimes do exist) huge ground currents in the earth (from
magnetic storms/ solar flares) – “earth ground” potential = 0 is not ideally so at every
point on the planet at all times!
•
Also, what is the electrostatic potential difference between: (earth – moon)? (earth –
mars)? (earth – venus)? (earth – sun)?? We know these potential differences are in fact
huge, because powerful electric fields exist in/throughout the solar system, e.g. driven by
the sun’s solar wind & solar flares (!)
ΔVsun −earth = Vearth − Vsun = − ∫
earth
sun
If Esun ( r ) ~ 100 V m
E
sun
( r )i d
′
d sun −earth
1.5 × 1011 m
Then ΔVsun −earth ~ 1.5 × 1013 Volts = 15 Tera-Volts (a lot!!)
The electrostatic potential (i.e. “voltage”) is analogous e.g. to the pressure of a gas:
Electrostatic potential differences between two points in space, ΔVab (due to gradients in
electrostatic potential) create electric fields, E ( r ) which in turn can accelerate charges
( F ( r ) = QE ( r ) = ma ( r ) ) causing them to move – in turn producing electric currents,
I=
dQ
dt
( Amperes = Coulombs/sec )
Likewise, pressure differences/pressure gradients can cause mass flow. In a gas (or a fluid,
more generally speaking) – mass currents = mass flow:
“I m =
8
dm
”
dt
©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
Lecture Notes 3
Prof. Steven Errede
The Electrostatic Potential V ( r ) and the Superposition Principle
We have seen that the Coulomb Force, Fc ( r ) =
1 QT QS
rˆ and the electrostatic field,
4πε 0 r 2
E ( r ) obey the principle of superposition:
F
C NET
N
(r ) = ∑ F ( r ) = F ( r ) + F
i =1
ENET ( r ) =
F
Ci
C NET
C1
(r )
C 2
( r ) + FC 3 ( r ) + … FC N (r )
N
QT
= ∑ Ei ( r ) = E1 ( r ) + E2 ( r ) + … EN ( r )
i =1
The above relations hold/are valid for any arbitrary electrostatic charge distributions: qi (ri),
λ ( r ) , σ ( r ) , ρ ( r ) , etc.
= Vb − Va
Since E ( r ) = −∇V ( r ) or
b
Δ V ( r ) = − ∫ E ( r )i d
a
ab
If we integrate from a common reference point, a = Ο ref (It doesn’t matter which point is
taken as the common reference point, because VΟref ( r = Ο ref ) = 0 will be the same in each
expression (as we saw above)…
Thus, we can show that electrostatic potential also obeys the principle of superposition:
ΔVNET ≡ VNET ( r ) − VΟref ( r = Οref )
(
N
) (
ΔVNET = ∑ ΔVi = V1 ( r ) − VΟref ( Οref ) + V2 ( r ) − VΟref ( Ο ref
i =1
(
+ … VN ( r ) − VΟref ( Οref )
))
)
Adding VΟref ( Ο ref ) = 0 to LHS and RHS, we see that:
N
VNET ( r ) = ∑ Vi ( r ) = V1 ( r ) + V2 ( r ) + V3 ( r ) + …VN ( r )
i =1
Note that this is a scalar (i.e. ordinary) numerical sum, not a vector sum!
©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
Fall Semester, 2007
Lecture Notes 3
Prof. Steven Errede
Griffiths Example 2.7
Electrostatic Potential, V ( r ) and Electric Field E ( r ) of a uniformly charged spherical
(conducting) shell of radius, R:
V (r ) =
1
4πε o
σ ( r ′ ) dA′
∫
sphere
r2
=
1
4πε o
σ dA′
∫
sphere
and r = r − r ′, r = r = r − r ′
r2
σ ( r ′ = R ) = σ = constant on sphere.
ẑ
field point/oberservation point, P
r
θ
R
Ο
x̂
QToT = σ Asphere = 4πσ R 2
R
dθ ′
R
Note that we can calculate V ( r ) from V ( r ) = ∫
r′
ŷ
Charged spherical
conducting shell of
radius R with uniform
surface charge, σ
Coulombs ⎞
⎛
⎜ SI Units:
⎟
meter 2 ⎠
⎝
i
sphere
dV ′ ( r ) where dV ′ ( r ) = potential @ P due
( m ) and annular area dA′ = 2π R sin θ ′dθ (← note
to infinitesimal annular charged strip σ C
that V (r ) has no ϕ-dependence)
2
2
dQ = σ dA′
dV ′ ( r ) =
i.e.
=
1
dQ
4πε o r
1 σ dA′
4πε o r
2
1 σ ( 2π R sin θ ′dθ ′ )
=
r
4πε o
10
©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
Lecture Notes 3
Now, use law of cosines, c 2 = a 2 + b 2 − 2ab cosθ to find r:
r 2 = z 2 + R 2 − 2 zR cosθ ′
Prof. Steven Errede
c
a
θ
·
b
r = r − r ′ = z 2 + R 2 − 2 zR cos θ ′
R sin θ ′
z=r
R
r' = R
θ′
·
∴ V (r ) =
1
( 2πσ R ) ∫
2
4πε o
sin θ ′dθ ′
θ ′=π
θ ′= 0
z + R 2 − 2 zR cos θ ′
2
Recall that d cos θ ′ = sin θ ′dθ ′
Make a change of variables: define u = cos θ ′
If u ≡ cos θ ′, then when θ ′ = 0, cos θ ′ = 1, u = 1
when θ ′ = π , cos θ ′ = −1, u = −1
Note also: du = d ( cos θ ′ )
Then: V ( r ) =
σ R2
2ε o
d ( cos θ ′ )
θ ′=π
∫θ
z 2 + R 2 − 2 zR cos θ ′
′= 0
σ R2
becomes: V ( r ) =
2ε o
∫
u =−1
u =+1
σ R2
=−
2ε o
z 2 + R 2 − 2 zRu
du
∫
dU
u =+1
u =−1
z + R 2 − 2 zRU
2
−1
1
dx
2
2
2
2
=
−
=
−
=
a
bx
dx
a
bx
a − bx
[
]
[
]
∫ a − bx ∫
b
b
where: a = z 2 + R 2 and b = 2 zR.
Now:
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
11
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 3
σ R2 ⎛ 2 ⎞ 2
u =+1
2
⎜
⎟ z + R − 2 zRu u =−1
2ε o ⎝ 2 zR ⎠
σ R2
=−
z 2 + R 2 − 2 zR − z 2 + R 2 + 2 zR
2ε o z
σR
=+
z 2 + R 2 + 2 zR − z 2 + R 2 − 2 zR
2ε o z
σR
=
z 2 + 2 zR + R 2 − z 2 − 2 zR + R 2
2ε o z
Prof. Steven Errede
Thus: V ( r ) = −
{
{
{
}
}
}
⎧
⎫
( z + R )( z + R ) − ( ± ( z − R ) ) ∗ ( ± ( z − R ) ) ⎪⎬
⎪⎭
⎩
always positive
must take positive root:
If z > R : ( z − R ) * ( z − R )
σR ⎪
=
⎨
2ε o z ⎪
If z < R :
Vzoutside
( r = zzˆ ) =
>R
σR
σ R2
+
−
−
=
z
R
z
R
{( ) ( )} ε z
2ε o z
o
ˆ) =
V(inside
z < R ) ( r = zz
σR
σR
( z + R ) − ( R − z )} =
{
2ε o z
εo
( − ( z − R )) ∗ ( − ( z − R ))
( z > R)
( z < R)
ˆ ) = constant!!!
⇐ V(inside
z < R ) ( r = zz
( ≠ 0!!)
n.b. surface of charged sphere is an equipotential: ⇒ V ( r = R ) =
Let r = z, then:
Vsphere ( r )
σR
εo
constant
~1
σR
≠ 0!
ε0
r
r
r=R
(surface of sphere)
12
©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
Lecture Notes 3
Prof. Steven Errede
We calculated that the total electric charge on the surface of the sphere is: Qsphere = 4πσ R 2
⎧ outside
σ R 2 4π σ R 2 4πσ R 2 Qsphere
ˆ
V
r
zz
=
=
=
=
=
(
)
⎪ ( z>R)
ε o z 4π ε o z
4πε o z 4πε 0 z
⎪
Then: ⎨
2
⎪V inside ( r = zzˆ ) = σ R = 4π R σ R = 4πσ R = Qsphere = constant ≠ 0
⎪⎩ z < R
ε o 4π R ε o
4πε o R 4πε o R
Now since the sphere has rotational invariance, then more generally, we can replace z with r
(= radial distance of field point, P from the center of the sphere, then V(r) will have only
r-dependence, no θ- or ϕ-dependence)
⎧ outside
σ R 2 4πσ R 2 Qsphere
=
=
=
V
r
(
)
⎪ ( r >R )
εor
4πε o r 4πε o r
⎪
Then: ⎨
2
⎪V inside ( r ) = σ R = 4πσ R = Qsphere = constant ≠ 0
⎪⎩ ( r < R )
εo
4πε o R 4πε o R
n.b. for r > R, this is same V(r) as
for point charge, with q = Qsphere!!!
Then electric field, E ( r ) = −∇V ( r )
In spherical coordinates, ∇ =
Then: E(outside
r =
r >R) ( )
∂
1 ∂ ˆ
1
∂ ˆ
θ+
θ
rˆ +
∂r
r ∂θ
r sin θ ∂ϕ
1 Qsphere
rˆ
4πε o r 2
same as for E ( r ) for point charge, q = Qsphere
inside
E(inside
r = 0 because: V(inside
r < R ) ( r ) = constant , i.e. no gradient of V( r < R ) ( r ) for r < R!!!
r <R) ( )
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
13
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 3
Prof. Steven Errede
POISSON’S EQUATION & LAPLACE’S EQUATION
Since the (electrostatic) electric field E ( r ) can be written as the negative gradient of a scalar
point function – the electrostatic potential, V ( r ) , i.e. E ( r ) = −∇V ( r )
Then with ∇i E ( r ) =
ρencl ( r )
and ∇ × E ( r ) = 0
εo
We see that: ∇ × E ( r ) = −∇ × ∇V ( r ) = 0
∇i E ( r ) = −∇i∇V ( r ) = −∇ 2V ( r ) =
Or:
∇ 2V ( r ) = −
ρencl ( r )
εo
ρencl ( r )
⇐ Poisson’s Equation
εo
Laplacian Operator = ∇ 2 = ∇i∇ ⇐ n.b. scalar quantity!
Cartesian Coordinates:
∂2
∂2
∂2
∇ = 2+ 2+ 2
∂x ∂y
∂z
2
Cylindrical Coordinates: ∇ 2 =
Spherical Coordinates:
∇2 =
1 ∂ ⎛ ∂ ⎞ 1 ∂2
∂2
r
+
+
⎜
⎟
r ∂r ⎝ ∂r ⎠ r 2 ∂ϕ 2 ∂z 2
1 ∂ ⎛ 2 ∂ ⎞
1
∂ ⎛
∂ ⎞
1
∂2
θ
r
sin
+
+
⎜
⎟
⎜
⎟
∂θ ⎠ r 2 sin 2 θ ∂ϕ 2
r 2 ∂r ⎝ ∂r ⎠ r 2 sin θ ∂θ ⎝
Poissons’ Equation is a linear, inhomogeneous 2nd order differential equation.
In regions of space where the volume charge density, ρ ( r ) = 0 , then Poisson’s equation ⇒
Laplace’s Equation ∇ 2V ( r ) = 0 ⇐ linear homogenous 2nd order differential equation.
We will discuss and use these two differential equations (much) more in the near future….
Usually in an electrostatics problem, for example:
1) A charge distribution q ( r ) , ∑ qi ( r ), λ ( r ) , σ ( r ) and or ρ ( r ) is specified (i.e.
given) afore-hand, and you are asked to calculate e.g. E ( r ) . Generally speaking, it’s
best (i.e. easiest) to calculate V ( r ) first (as an intermediate step), and then calculate
E ( r ) = −∇V ( r )
14
©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
Lecture Notes 3
Prof. Steven Errede
THUS:
⎧ 1 q
⎫
1 N qi
,
, or
∑
⎪
⎪
4πε o r 4πε o i =1 ri
⎪
⎪
⎧charge distribution ⎫
⎪
′
′
′
′
r
d
r
dA
λ
σ
(
)
(
)
1
1
⎪ N
⎪
⎪
⎪⎪
,
⎨
⎬ ⇒ V (r ) = ⎨
⎬ ⇒ E ( r ) = −∇V ( r )
∫
∫
C
S
q
q
λ
σ
ρ
,
,
,
,
πε
πε
r
r
4
4
∑
i
o
o
⎪
⎪
⎪
⎪
⎩ i =1
⎭
⎪ 1
⎪
′
′
ρ (r ) d r
⎪
⎪
∫ r
⎪⎩ 4πε o V
⎪⎭
OR:
⎧ 1 q
⎫
1 N qi
,
rˆ or
∑
⎪
⎪
2
2
⎪ 4πε o r 4πε o i =1 ri
⎪
⎪⎪ 1
λ ( r ′ ) rˆ d ′ 1
σ ( r ′ ) rˆ dA′ ⎪⎪
,
E (r ) = ⎨
⎬
∫C r 2
r2
4πε o ∫S
⎪ 4πε o
⎪
⎪ 1
⎪
ρ ( r ′ ) rˆ dτ ′
⎪
⎪
∫V r 2
⎩⎪ 4πε o
⎭⎪
2) On the other hand, if V ( r ) is specified (i.e. given). then we can use Poisson’s
equation ∇ 2V ( r ) = −
ρ (r )
to find ρ ( r ) .
εo
3) If E ( r ) is given/specified, then use ΔV ( r ) = − ∫ E ( r )id ′ to find V ( r ) and then
c
use ∇i E ( r ) = ρ ( r ) ε o to find ρ ( r ) .
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
15
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 3
Prof. Steven Errede
ELECTROSTATIC BOUNDARY CONDITIONS
Consider a 2-dimensional infinite-planar surface/sheet charge distribution – then place e.g. a
cylindrical Gaussian pillbox centered on the charged planar surface as shown in the figure
below:
n.b. Due to symmetry
of this problem, E (r )
can have no x- or ydependence!
2
Gauss’ Law:
∫
E idA′ =
Gaussian
surface, S
Qencl
ε0
Now shrink height of cylindrical Gaussian pillbox to be infinitesimally above/below charged
sheet,
i.e. let hpillbox → 0 .
vanishes for hpillbox → 0
Then:
∫
Gaussian
surface, S
E idA′ = ∫ E idA1 + ∫ E idA2 + ∫ E idA3
S1
S2
S3
dA1 = π R 2 nˆ1 = π R 2 zˆ
dA2 = π R 2 nˆ2 = −π R 2 zˆ = − dA1
Eabove = Ezˆ
Ebelow = − Ezˆ
⎞ Qencl π R 2σ
⎞
2 ⎛
2 ⎛
′
ˆ
ˆ
ˆ
ˆ
i
π
π
E
dA
R
E
z
z
R
E
z
z
=
⋅
+
−
⋅
−
=
⎜
⎟=
⎜
⎟
∫
εo
⎝ =1 ⎠
⎝ =1 ⎠ ε o
Gausian
surface
Or: E =
σ
, as we obtained previously.
2ε o
However, what we really want to point out is that E is discontinuous across a charged interface.
For the “shrunken” Gaussian pillbox, we can write Gauss’ Law as:
∫ Eabove idA1 + ∫ Ebelow idA2 = ∫ Eabove idA1 − ∫ Ebe; pw idA1
= ∫ ( Eabove − Ebelow )idA = ∫ ( Eabove − Ebelow )inˆ1dA
⊥
⊥
Now: ( Eabove − Ebelow )inˆ1 = Eabove
− Ebelow
=σ
εo
i.e. the perpendicular (normal) components of E are discontinuous across a charged
⊥
⊥
surface/interface (with surface charge, σ ) by an amount: Eabove − Ebelow = σ ε
16
o
©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
Lecture Notes 3
Prof. Steven Errede
What about the tangential components of E across a charged surface/interface?
∫
We know that:
c
E (r )idl ′ = 0
Shrink contour, C such that height h of vertical ( 2 , 4 ) portions shrink to infinitesimal size,
i.e. h → 0.
∫
C
E id = ∫ Eabove id
1
1
+ ∫ E2 id
2
= ∫ Eabove id
1
1
1
3
+ ∫ Ebelow id
+ ∫ E4 id
− ∫ Ebelow id
= ∫ ( Eabove − Ebelow = 0 )id
1
= Eabove d
Eabove − Ebelow = 0
1
OR:
4
=0
4
3
d
1
= dy yˆ
d
3
= −dy yˆ = −d
1
1
3
1
∴
3
3
= ∫ Eabove id
Now: Eabove id
+ ∫ Ebelow id
2
1
=0
(E
⇒
and Ebelow id
1
above
= Ebelow d
− Ebelow )id
1
=0
1
Eabove = Ebelow
i.e. the tangential component of E is
always continuous across an interface
b
⇒ Potential ΔV = Vabove − Vbelow = − ∫ E ( r )id = 0
a
Thus, V is (also) always continuous across an interface: Vabove = Vbelow
point a is located infinitesimally below the interface
point b is located infinitesimally above the interface
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
17
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 3
Prof. Steven Errede
⊥
⊥
⎧ Eabove
− Ebelow
= σ ⎫⎪
σ
⎪
εo ⇒ E
Since: ⎨
nˆ where n̂ points from “below” to
⎬
above − Ebelow =
εo
⎪⎩ Eabove − Ebelow = 0 ⎪⎭
“above”.
But: E = −∇V , thus: ∇Vabove − ∇Vbelow = −
σ
nˆ
εo
Or, more specifically, if n̂ is the unit outward normal of interface, at/on the interface,
⎛σ ⎞
∂Vabove
∂V
⎛σ ⎞
− below
= −⎜ ⎟
Then: ∇Vabove − ∇Vbelow = − ⎜ ⎟ nˆ can be written as:
∂n interface
∂n interface
⎝ εo ⎠
⎝ εo ⎠
∂V ( r )
Where:
= ∇V ( r )inˆ
= normal derivative of the potential, V(r) on the interface.
interface
∂n interface
= spatial gradient in the direction perpendicular (normal) to the interface, on the interface.
18
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.