UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lect. Notes 6
Prof. Steven Errede
LECTURE NOTES 6
THE METHOD OF IMAGES
•
A useful technique for solving (i.e. finding) E ( r ) and / or V ( r ) for a certain class /
special classes of electrostatic (and magnetostatic) problems that have some (or high)
degree of mirror-reflection symmetry. ← Exploit “awesome” power of symmetry
intrinsic to the problem, if present.
•
Idea is to convert a (seemingly) difficult electrostatic problem involving spatiallyextended charged objects (e.g. charged conductors) and then replace them with a finite
number of carefully, intelligently chosen / well-placed discrete point charges!!
•
Solving the simpler point-charge problem is the solution for the original, more
complicated problem!!!
•
Can replace e.g. a charged surface of a conductor (which is at constant potential – an
equipotential) by an equivalent / identical equipotential surface (at same potential) due to
one (or more) such / so-called “image charges”.
→ By doing this replacement, the original boundary conditions associated with original
problem are retained / conserved.
∴ E -fields and potentials V of the original and “surrogate” problems must be the same /
identical!!
•
The method of images is best learned by example…
©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
Example 1:
Fall Semester, 2007
Lect. Notes 6
Prof. Steven Errede
Point Charge q Located Near An Infinite, Grounded Conducting Plane:
n.b. A grounded conductor is a special type of equipotential: infinite amounts of electrical
charge (± Q) can flow from / to ground to / from the conducting surface so as to maintain
electrostatic potential V = 0 (Volts) at all times. (In reality / real life, ∃ no such thing. e.g.
especially / particularly for AC / time-varying electromagnetic fields for frequencies f 1 MHz,
due to inductance effects (magnetic analog of capacitance)).
What are:
E ( r ) , V ( r ) and σ ( x, y, z = 0 )
ẑ
( σ ( x, y, z = 0 ) = induced surface
Source
charge density on ∞-plane)
r ≡ r − r′
Point, S
q
d
E ( r ) @ Field Point, P
r
r′
r
∞-conducting plane
@ V ( x, y , z = 0 ) = 0
induced surface charge
density σ ( x, y, z = 0 )
Ο
ŷ
wire
Earth ground
Symbol
2
x̂
©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 6
Prof. Steven Errede
Side View of Problem:
ẑ
Lines of E :
* point radially outward @ +q
* are ⊥ to surface of ∞-conducting
plane @ z = 0 ∀ ( x, y )
+q
d
E (r )
σ ( x, y , z = 0 )
∞-conducting plane
V ( x, y, 0) = 0
ŷ
x̂
( x̂ points outward from paper)
Now mirror-reflect above problem:
i.e. Let z → − z and simultaneously let + q ( z ) → − q ( − z )
(more generally let r → − r for objects in problem)
ẑ
+q
Grounded,
∞-conducting
plane
d
E (r )
σ ( x, y , z = 0 )
x̂
ŷ
V ( x, y, z = 0) = 0
d
−q
Now (mentally) remove the grounded, ∞-conducting plane:
©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
Lect. Notes 6
Prof. Steven Errede
Side View:
ẑ
Original Charge
@ (0,0,+d)
+q
z = +d
Field / Observation
Point P
r
d
Ο x
ŷ
d
z = −d
Image Charge
@ (0,0, −d)
−q
We have replaced ∞-conducting grounded plane (equipotential, V ( x, y, z = 0) = 0 ) with an image
point charge –q located at ( x, y, z ) = ( 0, 0, −d ) .
⇒ The method of images is highly analogous to mirror-type optics problem!!
− here we have point “object” +q at ( x, y, z ) = ( 0, 0, + d ) and plane mirror at ( x, y , z ) = ( x, y, 0 ) .
An image of point object is formed as a point image a distance d behind the mirror; point
image –q is located at ( x, y, z ) = ( 0, 0, −d ) .
Optical mirrors are essentially equipotentials – optics works for electrostatic problems!!!
Mathematical constraint (boundary condition) on V: V ( x, y, z = 0 ) = 0 ∀ ( x, y, z = 0 ) .
(i.e. everywhere on ∞-conducting grounded plane.)
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 6
Prof. Steven Errede
3-D View of Image Charge Problem for Grounded Infinite Conducting Plane:
ẑ
Source Point, S1
+q (x,y,z) = (0,0,+d)
We
already
know how
to solve /
do this
problem!!
r1 = r − r1′
r
r1′
d
Field / Observation
Point, P at (xp,yp,zp)
Ο
ŷ
r2 = r − r2′
r2′
d
x̂
−q
Source Point, S2
(x,y,z) = (0,0,−d)
Use Principle of Superposition to solve / determine total potential at observation / field point r :
VTOT ( r ) = ?
Potential @ point ( r )
Potential @ point ( r )
due to point charge
+q @ point ( r1′)
due to point charge
−q @ point ( r2′ )
VTOT ( r ) = V1 ( r ) + V2 ( r )
=
=
1 ⎛ +q ⎞
1 ⎛ −q ⎞
⎜
⎟+
⎜
⎟
4πε o ⎝ r1 ⎠ 4πε o ⎝ r2 ⎠
q ⎛1 1⎞
⎜ − ⎟
4πε o ⎝ r1 r2 ⎠
separation distances
c
What are r1 & r2 ??
a
Use law of cosines c = a + b − 2ab cos θ
to obtain expressions for r1 + r2 .
2
b
2
2
θ
©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 6
Prof. Steven Errede
Here it is easier simply to use basic definition of separation distance, i.e.
Let: θ = θ1
+q
d
0
r ≡ Δx 2 + Δy 2 + Δz 2
and: θ2 = π – θ1 = π – θ
z
θ1
r1 =
r
θ2
(x
(x
r1 ≡
P
r1
p
− x1 ) + ( y p − y1 ) + ( z p − z1 )
p
− 0) + ( y p − 0) + ( z p − d )
2
2
2
r1 = x 2p + y 2p + ( z p − d )
r2
2
2
2
2
d
(x
(x
p
− x2 ) + ( y p − y2 ) + ( z p − z2 )
p
− 0) + ( y p − 0) + ( z p + d )
2
Law of cosines:
r2 ≡
r12 = d 2 + r 2 − 2dr cos θ
r2 =
r22 = d 2 + r 2 − 2dr cos (π − θ )
r2 = x 2p + y 2p + ( z p + d )
−q
cos (π − θ ) = cos π cos θ + sin π
=0
2
2
2
2
2
2
sin θ
= − cos θ
∴ r = d 2 + r 2 + 2dr cos θ
2
2
Then : VTOT ( r ) =
q ⎛1 1⎞
⎜ − ⎟
4πε o ⎝ r1 r2 ⎠
Drop “p” on field point subscript:
⎧
q ⎪
1
−
VTOT ( r ) =
⎨
4πε o ⎪ x 2 + y 2 + ( z − d )2
⎩
1
x2 + y 2 + ( z + d )
2
⎫
⎪
⎬
⎪⎭
Note that:
π
⎛
⎞
1. VTOT ( r @ z = 0 ) = 0 i.e. VTOT ⎜ θ1 = θ 2 = θ = = 90o ⎟ = 0
2
⎝
⎠
2. VTOT ( r → ∞ ) = 0
⎧
q ⎪
1
−
VTOT ( r ) =
⎨
4πε o ⎪ x 2 + y 2 + ( z − d )2
⎩
1
x2 + y 2 + ( z + d )
2
⎫
q
⎪
⎬=
⎪⎭ 4πε o
⎧1 1 ⎫
⎨ − ⎬ = V1 ( r ) + V2 ( r )
⎩ r1 r2 ⎭
We can now determine ETOT ( r ) from either:
(a.) ETOT ( r ) = E1 ( r ) + E2 ( r ) =
or:
q ⎧ rˆ1 rˆ2 ⎫
⎨ − ⎬ (using the Principle of Superposition)
4πε o ⎩ r12 r22 ⎭
What are r̂1 and r̂2 ?
⎧∂
∂
∂ ⎫
(b.) ETOT ( r ) = −∇VTOT ( r ) == − ⎨ xˆ +
yˆ + zˆ ⎬VTOT ( r ) (e.g. in Cartesian coordinates)
∂y
∂z ⎭
⎩ ∂x
6
©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 6
Prof. Steven Errede
Because we will see this same problem again (in the near future) from a different perspective, let
us rewrite the problem in spherical polar coordinates, using the law of cosine results:
{r =
r 2 + d 2 − 2rd cos θ
1
and
r2 = r 2 + d 2 + 2rd cos θ
}
⎫
q ⎧
1
1
−
⎨ 2
⎬
4πε o ⎩ r + d 2 − 2rd cos θ
r 2 + d 2 + 2rd cos θ ⎭
⎧∂
1 ∂
1
∂ ⎫
ETOT ( r ) = −∇VTOT ( r ) = − ⎨ rˆ +
θ+
ϕˆ ⎬VTOT ( r )
r ∂θ
r sin θ ∂ϕ ⎭
⎩ ∂r
VTOT ( r ) =
= ErTOT rˆ + EθTOT θ + EϕTOT ϕˆ
1
1
2
3
⎫
∂
q ∂ ⎧
1
1
−
VTOT ( r ) = −
⎨ 2
⎬
2
2
2
4πε o ∂r ⎩ r + d − 2rd cos θ
∂r
r + d + 2rd cos θ ⎭
⎧
⎫
2r − 2d cos θ )
2r + 2d cos θ )
(
(
q ⎪ ⎛1⎞
⎪
⎛1⎞
=−
+⎜ ⎟
⎨− ⎜ ⎟
3
3 ⎬
4πε o ⎪ ⎝ 2 ⎠ ⎡ r 2 + d 2 − 2rd cos θ ⎤ 2 ⎝ 2 ⎠ ⎡ r 2 + d 2 + 2rd cos θ ⎤ 2 ⎪
⎣
⎦
⎣
⎦ ⎭
⎩
ErTOT = −
⎧
⎫
r − d cos θ )
r + d cos θ )
(
(
q ⎪
⎪
=−
−
⎨
3
3 ⎬
4πε o ⎪ ⎡ r 2 + d 2 − 2rd cos θ ⎤ 2
⎡⎣ r 2 + d 2 + 2rd cos θ ⎤⎦ 2 ⎪
⎦
⎩⎣
⎭
2
⎧∂
⎫
n.b. ⎨ cos θ = − sin θ ⎬
⎩ ∂θ
⎭
⎫
1 ∂
q ∂ ⎧
1
1
−
VTOT ( r ) = −
⎨ 2
⎬
r ∂θ
4πε o r ∂θ ⎩ r + d 2 − 2rd cos θ
r 2 + d 2 + 2rd cos θ ⎭
⎧
⎫
+2rd sin θ )
−2rd sin θ )
(
(
q ⎛ 1 ⎞⎪ ⎛ 1 ⎞
⎪
⎛1⎞
=−
+⎜ ⎟
⎜ ⎟ ⎨− ⎜ ⎟ 2
3
3 ⎬
4πε o ⎝ r ⎠ ⎪ ⎝ 2 ⎠ ⎡ r + d 2 − 2rd cos θ ⎤ 2 ⎝ 2 ⎠ ⎡ r 2 + d 2 + 2rd cos θ ⎤ 2 ⎪
⎣
⎦
⎣
⎦ ⎭
⎩
EθTOT = −
⎧
q ⎛ 1 ⎞⎪
=+
⎨ r d sin θ
4πε o ⎜⎝ r ⎟⎠ ⎪
⎩
(
)
⎡
⎤⎫
1
1
⎢
⎥⎪
+
3
3
⎢ 2
⎥⎬
2
2
⎡⎣ r 2 + d 2 + 2rd cos θ ⎤⎦ 2 ⎥ ⎪
⎢⎣ ⎡⎣ r + d − 2rd cos θ ⎤⎦
⎦⎭
⎧
⎫
qd sin θ ⎪
1
1
⎪
=+
+
⎨
3
3 ⎬
4πε o ⎪ ⎡ r 2 + d 2 − 2rd cos θ ⎤ 2
⎡⎣ r 2 + d 2 + 2rd cos θ ⎤⎦ 2 ⎪
⎦
⎩⎣
⎭
3
i.e.
1
∂
VTOT ( r ) = 0 because VTOT ( r ) has no ϕ -dependence.
r sin θ ∂ϕ
VTOT ( r ) and ETOT ( r ) are azimuthally symmetric (invariant under ϕ → ϕ ′ rotations)
EϕTOT =
because original charge distribution is azimuthally symmetric – no ϕ -dependence.
©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 6
Prof. Steven Errede
=0
Thus, ETOT ( r ) = ErTOT rˆ + EθTOT θ + EϕTOT ϕ̂ , or:
ẑ
⎡⎧
⎫
r − d cos θ )
r + d cos θ )
(
(
q ⎢⎪
⎪
−
ETOT ( r ) = +
⎨
⎬r
3
3
2
2
2
4πε o ⎢ ⎪ ⎡ r 2 + d 2 − 2rd cos θ ⎤ 2
⎡
⎤
⎢⎣ ⎩ ⎣
⎦
⎣ r + d + 2rd cos θ ⎦ ⎪⎭
+q
d
2d
d
−q
with:
⎧
⎫ ⎤
d sin θ )
d sin θ )
(
(
⎪
⎪ ⎥
+⎨
+
θ
3
3 ⎬ ⎥
⎪ ⎡⎣ r 2 + d 2 − 2rd cos θ ⎤⎦ 2
⎡⎣ r 2 + d 2 + 2rd cos θ ⎤⎦ 2 ⎪ ⎥
⎩
⎭ ⎦
⎫
q ⎧
1
1
−
VTOT ( r ) =
⎨ 2
⎬
4πε o ⎩ r + d 2 − 2rd cos θ
r 2 + d 2 + 2rd cos θ ⎭
The above expressions are potential and electric field associated with a spatially-extended
electric dipole, with electric dipole moment p ≡ + qr1 − qr2 = qΔr21 (SI Units: Coulomb-meters)
with separation distance Δr21 ≡ r1 − r2 = dzˆ − d ( − zˆ ) = dzˆ + dzˆ = 2dzˆ. Here (i.e. in this problem),
the separation distance Δr21 = 2d .
ẑ
+q
n.b. The vector Δr21 ≡ r1 − r2 points from –q to +q ← important convention!!!
p = 2qdzˆ
2d
−q
8
©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 6
Prof. Steven Errede
Electric Field E ( r ) and Equipotentials of an Electric Dipole with Electric Dipole Moment, p = pzˆ :
ẑ
electric field lines
equipotentials ( ⊥ to E everywhere)
+q
ŷ
x̂
−q
We can now determine the surface free charge density σ free ( x, y, z = 0 ) on the infinite,
grounded conducting plane e.g. via two methods:
METHOD 1: σ free ( x, y, z = 0 ) = −ε o
i.e. here, σ free ( x, y, z = 0 ) = −ε o
∂VTOT ( r )
∂ ⎡ gradient normal to surface ⎤ ∂
= ⎢
where
⎥ = ∂z ( here ) .
of
∞
-conducting
plane
∂n
∂
n
⎣
⎦
surface
∂VTOT ( r )
∂z
z =0
⎧
q ⎪
1
In Cartesian coordinates: VTOT ( x, y, z ) =
−
⎨
4πε o ⎪ x 2 + y 2 + ( z − d )2
⎩
Thus: σ free ( x, y, z = 0 ) = −
⎧
q ∂ ⎪
1
−
⎨
4π ∂z ⎪ x 2 + y 2 + ( z − d )2
⎩
1
x2 + y 2 + ( z + d )
1
x2 + y2 + ( z + d )
2
2
⎫
⎪
⎬
⎭⎪ z =0
⎫
⎪
⎬
⎪⎭
z =0
©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
Lect. Notes 6
Prof. Steven Errede
⎧
⎫
2( z − d )
2( z + d )
⎪⎛ 1 ⎞
⎪
−
⎨⎜ − ⎟
3
3 ⎬
⎪⎝ 2 ⎠ ⎡ x 2 + y 2 + ( z − d )2 ⎤ 2
⎡ x 2 + y 2 + ( z + d )2 ⎤ 2 ⎪
⎣
⎦
⎣
⎦ ⎭ z =0
⎩
⎧
⎫
q ⎪
d
−d
⎪
=+
−
⎨
3
3 ⎬
4π ⎪ ⎡ x 2 + y 2 + d 2 ⎤ 2
⎡⎣ x 2 + y 2 + d 2 ⎤⎦ 2 ⎪
⎦
⎩⎣
⎭
2qd
1
p
1
=−
= −
3
4π ⎡ x 2 + y 2 + d 2 ⎤ 2
4π ⎡ x 2 + y 2 + d 2 ⎤ 3 2
⎣
⎦
⎣
⎦
q
=−
4π
Electric dipole moment p = 2qd = q ( 2d ) where 2d = charge separation distance.
→ Note that the sign of the induced surface free charge on ∞-conducting plane is opposite to
that of original charge q.
→ Note also that σ free ( x, y, z = 0 ) is greatest at ( x = 0, y = 0, z = 0 ) - directly underneath
free
=−
original charge, q. i.e. σ max
2qd ⎛ 1 ⎞
2q ⎛ 1 ⎞
⎜ 3⎟=−
⎜ ⎟
4π ⎝ d ⎠
4π ⎝ d 2 ⎠
⇒ See plot of σ free ( x, y, z = 0 ) below (on p. 12).
METHOD 2: Use Gauss’ Law:
∫
S
E idA =
free
Qencl
εo
Use “shrunken” Gaussian Pillbox of height δh centered on / around ∞-conducting plane:
induced free charge σ ( x, y, 0 ) on
zˆ, nˆ1
surface
S1
E
½ δh
½ δh
n̂2
E = 0 inside conducting plane (z < 0)
S2
S3
nˆ3 = − zˆ
On the conducting surface (@ z = 0), θ =
10
E ≠ 0 above conducting plane (z > 0)
⎛π ⎞
⎛π ⎞
⇒ sin θ = sin ⎜ ⎟ = 1 and cos θ = cos ⎜ ⎟ = 0 .
2
⎝2⎠
⎝2⎠
π
©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
surf
TOT
E
Fall Semester, 2007
Lect. Notes 6
Prof. Steven Errede
⎡⎧
⎫ ⎧
⎫ ⎤
q ⎢⎪
r
r
d
d
⎪ ⎪
⎪ ⎥
−
+
θ
r+⎨
( x, y , z = 0 ) =
⎨
3
3 ⎬
3
3 ⎬ ⎥
⎢
4πε o ⎪ ⎡ r 2 + d 2 ⎤ 2 ⎡ r 2 + d 2 ⎤ 2 ⎪ ⎪ ⎡ r 2 + d 2 ⎤ 2 ⎡ r 2 + d 2 ⎤ 2 ⎪
⎢⎣ ⎩ ⎣
⎦
⎣
⎦ ⎭ ⎩⎣
⎦
⎣
⎦ ⎭ ⎥⎦
⎡
⎤
q ⎢
2d
π
0rˆ +
θ⎥
=
When θ = = 90o
3
⎥
4πε o ⎢
2
⎡⎣ r 2 + d 2 ⎤⎦ 2 ⎥
⎢⎣
⎦
Now zˆ = cos θ rˆ − sin θθ
Please remember / derive this!!
ẑ
ϕ̂
r̂
Consider rˆ, r
θ
to lie in yˆ − zˆ
plane:
Ο
θ
ϕ
ŷ
x̂
ẑ
ẑ
r̂
zˆ = rˆ cos θ
θ
θ
r̂
zˆ = θ cos α α
π
⎛
⎞ π
α = ⎜π − −θ ⎟ = −θ
2
2
⎝
⎠
zˆ = θ sin θ
θ
π
π
⎛π
⎞
cos α = cos ⎜ − θ ⎟ = cos cos θ + sin sin θ = sin θ
2
2
⎝2
⎠
Thus, when θ = 90o =
⎛π ⎞
⎛π ⎞
, zˆ = cos θ rˆ − sin θθ = cos ⎜ ⎟ rˆ − sin ⎜ ⎟ θ = −θ
2
⎝2⎠
⎝2⎠
π
q ( 2d )
1
surf
∴ ETOT
zˆ
( x, y , z = 0 ) = −
4πε o ⎡ r 2 + d 2 ⎤ 3 2
⎣
⎦
q ( 2d )
1
surf
ETOT
zˆ
( x, y , z = 0 ) = −
4πε o ⎡ x 2 + y 2 + d 2 ⎤ 3 2
⎣
⎦
On conducting plane z = 0
r 2 = x2 + y2 + z 2
r 2 = x 2 + y 2 on conducting plane
©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
Gaussian Pillbox Surface:
∫
S
Fall Semester, 2007
E idA = ∫ E1 idA1 +
S1
∫
S2
Lect. Notes 6
=0
E2 idA2 +
∫
E3 idA3
E3 = 0
inside
conductor
E just ε above surface
S1
σ free ∼ 1
=0
S3
E2 ⊥ dA2
S2 shrinks → 0
anyway
= ∫ E1 idA1
Prof. Steven Errede
r3
xˆ, yˆ
area of surface S1
Gauss Law:
q ( 2d )
1
zˆ i A1 zˆ
4πε o ⎡ x 2 + y 2 + d 2 ⎤ 3 2
⎣
⎦
free
σ ( x, y, z = 0 ) A1
Q
= encl = free
∫ E idA = ∫ E1 idA1 = −
S1
εo
∴ σ free ( x, y, z = 0 ) = −
εo
q ( 2d )
1
p
1
=−
3
4π ⎡ x 2 + y 2 + d 2 ⎤ 2
4π ⎡ x 2 + y 2 + d 2 ⎤ 3 2
⎣
⎦
⎣
⎦
Electric dipole moment p = q ( 2d ) ( Coulomb-meters ) where 2d = charge separation distance
⇒ Same answer as obtained in Method 1.
plane
TOT
NOTE: Q
p
= ∫ σ free ( x, y, 0 ) dA = −
plane
4π
∞
2π
0
0
∫ ∫
1
qd
rdrdϕ =
2
2
r2 + d 2
⎣⎡ r + d ⎦⎤
∞
= −q
0
The net force of attraction of charge +q to ∞-conducting plane is just that of force of charge +q
attracted to its image charge, −q a separation distance Δr21 ≡ r1 − r2 = r = 2d away!!!
NET
+q
+q
F
( r ) = + qE− q ( r = 2d ) =
1
+ q ( −q )
4πε o
( 2d )
2
zˆ = −
1
q2
4πε o ( 2d )2
zˆ
r = 2d
−q
E− q
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
Lect. Notes 6
Prof. Steven Errede
We can also obtain the net force of attraction of the charge +q and grounded, infinite conducting
plane by adding up all of the individual contributions qdE ( r = ( 0, 0, d ) ) due to σ free ( x, y, z = 0 ) :
q ( 2d )
1
2qd
1
=−
3
4π ⎡ x 2 + y 2 + d 2 ⎤ 2
4π ⎡ x 2 + y 2 + d 2 ⎤ 3 2
⎣
⎦
⎣
⎦
1
⎛d ⎞
⎛1⎞
σ free ( x, y, 0 ) ⎜ 2 ⎟ rˆ dA where dA = 2π rdr and rˆ = cos θ zˆ = ⎜ ⎟ zˆ
4πε o
⎝r ⎠
⎝r⎠
σ free ( x, y, z = 0 ) = −
F+NET
( r ) = q ∫ dETOT = q
q
∫
S
plane
q 2 ( 2d ) ⎛ 1 ⎞ ∞
1
d
⎛ 1 ⎞
∗⎜ 2
∗
∗ 2π rdr zˆ
⎜
⎟ ∫0
3
2 ⎟
4π ⎝ 4πε o ⎠ ⎡ r 2 + d 2 ⎤ 2 ⎝ r + d ⎠ r 2 + d 2
⎣
⎦
2 2
qd ∞
rdr
1
q2
ˆ
=−
=
−
z
zˆ
4πε o ∫0 ⎡ r 2 + d 2 ⎤ 3
4πε o ( 2d )2
⎣
⎦
F+NET
(r ) = −
q
Integration over the conducting plane:
r̂
ẑ
θ
q
θ
r 2 = x2 + y 2 + d 2
d
dr
ŷ
r
x̂
The work done to assemble the Image Charge Problem (i.e. put +q first at (x,y,z = d) and then
bring in –q at (x,y,z = –d) from ∞) is:
1
2d
WICP = ∫ Fmech idl = Fmech ∗ ( 2d ) = −
∞
Also:
WICP =
εo
2
∫
all
space
E 2 dτ =
εo
2
∫
all
space
q2
4πε o ( 2d )
E i E dτ =
εo
2
∗ 2d = −
2
∫ ( E iE
r
r
q2
(Joules)
4πε o ( 2d )
1
+ Eθ i Eθ ) dτ
all
space
Note that this integral includes both the z > 0 and z < 0 regions for the image charge problem.
©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
Lect. Notes 6
Prof. Steven Errede
However, for the actual problem, i.e. the charge +q above grounded ∞-conducting plane,
there is NO ELECTRIC FIELD in the z < 0 region!
1
Thus Wactual = WICP
2
1 ⎛ 1 ⎞ q2
i.e. Wactual = − ⎜
⎟
2 ⎝ 4πε o ⎠ ( 2d )
For the actual problem we can obtain Wactual by calculating the work required to bring +q in
from infinity to a distance d above the grounded ∞-conducting plane. The mechanical force
required to oppose the electrical force of attraction is:
Fmech = − FE =
1
q2
4πε o ( 2 z )2
zˆ (along ẑ axis)
d
1
∞
4πε o
Wactual = ∫ Fmech idl =
d
q2
∫ ( 2z )
∞
dz =
2
q2
16πε o
∫
d
∞
dz
z2
d
q 2 ⎛ −1 ⎞
q2
=
⎜ ⎟ =−
16πε o ⎝ z ⎠ ∞
16πε o d
1 ⎛ 1 ⎞ q2
Wactual = − ⎜
⎟
2 ⎝ 4πε o ⎠ ( 2d )
Same answer as that obtained above!
IMPORTANT NOTES / COMMENTS ON IMAGE CHARGE PROBLEMS
1.) Image charges are always located outside of regions(s) where V ( r ) and E ( r ) are to be
calculated!!
→ Image charges cannot / must not be located inside region where V ( r ) and E ( r ) are to be
calculated (no longer the same problem!!)
2.) WICP (all space) = 2×Wactual (half space).
→ In general, this is not true ∀ image charge problems. Be careful here! Depends on detailed
geometry of conducting surfaces.
3.) Depending on nature of problem, image charge(s) may or may not be opposite charge sign!!
4.) Depending on nature of problem, image charge(s) may or may not be same strength as
original charge Q.
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
Lect. Notes 6
Prof. Steven Errede
Image Charge Problem
Example 2: (Griffiths Example 3.2 p. 124-126)
Point charge +q situated a distance a away from the center of a grounded conducting sphere of
radius R < a. Find the potential outside the sphere.
ẑ
V(r = R) = 0
on sphere
(equipotential)
R
+q
Ο
ŷ
a
Take origin of coordinates to be @ center of
sphere
x̂
From spherical and ŷ axial symmetry (rotational invariance) of problem, if solution for image
charge q′ is to exist, it must be:
1.) inside spherical conductor (r < R)
2.) q′ image charge must lie along ŷ axis (i.e. along line from charge +q to center of
sphere).
3.) because V (r = R) = 0 on sphere, q′ must be opposite charge sign of +q.
4.) want to replace grounded conducting sphere with equipotential V (r = R) = 0 by use
of image charge q′ at distance b away from center of sphere:
R
b
q′
+q
ŷ
Ο
a
©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
Lect. Notes 6
Prof. Steven Errede
NOTE: Two points on the surface of sphere where the potential VTOT (r = R) = 0 is easy to
calculate - is on the ŷ axis at the field points P1 and P2:
r1 = r − r1 and r2 = r − r2
V (r = R) = 0
P3
P (General Field Point)
r
R
P1
Ο
b q′
r2
r2
r1
P2
r1
+q
ŷ
a
In general:
At point P1:
VP1 ( r = R ) = 0
1 ⎛ q q′ ⎞
⎜ + ⎟
4πε o ⎝ r1 r2 ⎠
r1 = ayˆ , r = R ( − yˆ ) = − Ryˆ , r1 = r − r1 = − Ryˆ − ayˆ = − ( R + a ) yˆ
VTOT ( r ) = V1 ( r ) + V2 ( r ) =
r1 = r1 = ( R + a )
r2 = byˆ , r = R ( − yˆ ) = − Ryˆ , r2 = r − r2 = − Ryˆ − byˆ = − ( R + b ) yˆ
r2 = r2 = ( R + b )
1 ⎛ q q′ ⎞
⎜ + ⎟
4πε o ⎝ r1 r2 ⎠
1 ⎛ q
q′ ⎞
=
+
⎜⎜
⎟=0 ⇒
4πε o ⎝ ( R + a ) ( R + b ) ⎟⎠
VP1 ( r = R ) =
q
q′
=−
( R + a) ( R + b)
Relation #1
At point P2:
VP2 ( r = R ) = 0
r1 = ayˆ , r = R ( + yˆ ) = + Ryˆ , r1 = r − r1 = Ryˆ − ayˆ = ( R − a ) yˆ
r1 = r1 = ( a − R )
(a > R) !!
r2 = byˆ , r = R ( + yˆ ) = + Ryˆ , r2 = r − r2 = Ryˆ − byˆ = ( R − b ) yˆ
r2 = r2 = ( R − b )
VP2 ( r = R ) =
1 ⎛ q
q′ ⎞
+
⎜⎜
⎟=0 ⇒
4πε o ⎝ ( R − a ) ( R − b ) ⎟⎠
q
q′
=−
(a − R) ( R − b)
Relation #2
16
©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 6
Prof. Steven Errede
We now have two equations (Relations # 1 & 2), and we have two unknowns: q′ and b.
Solve equations simultaneously!
- First, we eliminate q′ :
⎡R+b⎤
From Relation #1 we have: q ' = − ⎢
q
⎣ R + a ⎥⎦
⎡R−b⎤
q
From Relation #2 we have: q ' = − ⎢
⎣ a − R ⎥⎦
⎡R+b⎤ ⎡R−b⎤
∴ ⎢
=
OR: ( R + b )( a − R ) = ( R + a )( R − b )
⎣ R + a ⎥⎦ ⎢⎣ a − R ⎥⎦
− R 2 + aR + ab − bR = R 2 + aR − bR − ab
−2 R 2 + 2ab = 0
OR: ab=R2
2
b=R
OR:
a
⎡R+b⎤
Then: q ' = − ⎢
q
⎣ R + a ⎥⎦
⎡ R + R2 ⎤
⎡1 + R ⎤
a ⎡1 + R ⎤
a⎦ q
a ⎥ q = −⎛ R ⎞ ⎣
a ⎥ q = −R ⎢
= −⎢
⎜
⎟
⎢ R+a ⎥
⎢ R+a ⎥
⎝ a ⎠ [ R + a]
⎣
⎦
⎣
⎦
⎛R⎞
⎛ R ⎞ [a + R]
⎛ R ⎞ [ R + a]
q = −⎜ ⎟q
q = −⎜ ⎟
= −⎜ ⎟
⎝a⎠
⎝ a ⎠ [ R + a]
⎝ a ⎠ [ R + a]
⎛R⎞
Thus: q ' = − ⎜ ⎟ q
⎝a⎠
©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
Lect. Notes 6
Prof. Steven Errede
CHECK:
⎛ R2 ⎞
⎛R⎞
Does q ' = − ⎜ ⎟ q , located at r2 = byˆ = ⎜ ⎟ yˆ satisfy the B.C. that V ( r = R ) = 0 for any r = R?
⎝a⎠
⎝ a ⎠
VTOT ( r ) = V1 ( r ) + V2 ( r ) =
1 ⎧ q q′ ⎫
⎨ + ⎬
4πε o ⎩ r1 r2 ⎭
At an arbitrary field point P3 anywhere on the surface of sphere, r = R:
r1 = a 2 + R 2 − 2aR cos θ
and
r2 = b 2 + R 2 − 2bR cos θ
Note that we changed to ẑ axis here in order
to define (& use) the polar angle, θ !!!
P3
R
r2
θ q′
b
r1
+q
ẑ
a
Then: VTOT ( r ) =
1 ⎧ q q′ ⎫
⎨ + ⎬ with:
4πε o ⎩ r1 r2 ⎭
⎛R⎞
q ' = −⎜ ⎟q
⎝a⎠
⎛ R2 ⎞
b=⎜ ⎟
⎝ a ⎠
⎧
q ⎪
1
VTOT ( r = R ) =
−
⎨ 2
2
4πε o ⎪ a + R − 2aR cos θ
⎩
=
=
18
q
4πε o
q
4πε o
( R a)
⎫
⎪
⎬
2
2
b + R − 2bR cos θ ⎪
⎭
⎧
⎫
R
⎪⎪
⎪⎪
1
a
−
⎨ 2
⎬
2
2
2
⎪ a + R − 2aR cos θ
R2
+ R2 − 2 R
R cos θ ⎪
a
a
⎪⎩
⎪⎭
⎧
⎪⎪
1
1
−
⎨ 2
2
2
3
⎪ a + R − 2aR cos θ
a
R2
+ R2 − 2 R
cos θ
R
a
a
⎪⎩
( )
( ) (
( )
( )
)
( )
⎫
⎪⎪
⎬
⎪
⎪⎭
©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
⎧
q ⎪⎪
1
VTOT ( r = R ) =
−
⎨ 2
4πε o ⎪ a + R 2 − 2aR cos θ
⎪⎩
q ⎧
1
=
−
⎨ 2
4πε o ⎩ a + R 2 − 2aR cos θ
= 0 ∀ θ ,ϕ
(@ r = R )
Lect. Notes 6
Prof. Steven Errede
⎫
⎪⎪
⎬
2
a
R 4 2 + a 2 − 2aR cos θ ⎪
R
a
⎪⎭
⎫
1
⎬
a 2 + R 2 − 2aR cos θ ⎭
1
)
( )(
YES!!!
The scalar potential for an arbitrary point outside the grounded, conducting sphere (r > R) is:
⎧
1
q ⎪⎪
VTOT ( r ) =
−
⎨ 2 2
4πε o ⎪ a + r − 2ar cos θ
⎩⎪
Then: ETOT ( r ) = −∇VTOT ( r ) and thus:
And thus: σ free ( r = R ) = −ε o
⎫
⎪⎪
⎬
2
2
2
R2
R
r cos θ ⎪
+r −2
a
a
⎭⎪
( )
( R a)
( )
Er ( r ) = −
∂VTOT ( r )
∂r
∂VTOT ( r )
∂V ( r )
= −ε o TOT
= +ε o Er ( r )
∂n
∂r
r =R
r=R
total
Total charge on surface of the conducting sphere:
Q sphere
=
free
∫
sphere
r=R
⎛R⎞
⎟q
⎝a⎠
σ free dA = − ⎜
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
19
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
R
Image Charge Problem
Lect. Notes 6
q′
+q
Prof. Steven Errede
ŷ
a
Example #3:
Point charge +q near a charged conducting sphere of radius R.
(variation on image charge Example #2)
→ Use the superposition principle for image charges!!
( a ) q located at
Step 1: Replace the conducting sphere by an image charge q ' = − R
( a ) yˆ
2
rq′ = byˆ = R
(same as in Example #2)
→ This makes surface of sphere an equipotential surface V (r = R) = 0.
Step 2: Add a second image charge q′′ at center of sphere to raise potential on surface of
sphere to achieve required potential V (r = R) = V (positive or negative constant
potential on sphere)
Note: q′′ is also on same axis ( ŷ ) as q and q ' .
′
′′
Then: σ TOT
free = σ free ( q ) + σ free ( q )
Surface free charge
Density due to q′
surface free charge
density due to q′′
Then: VTOT ( r ) = Vq ( r ) + Vq ' ( r ) + Vq '' ( r ) =
( a) q
But: q ' = − R
1 ⎧⎪ q q′ q′′ ⎫⎪
⎨ + + ⎬
4πε o ⎩⎪ rq rq' rq'' ⎭⎪
=r
and
Qsphere = q′ + q′′
Then: ETOT ( r ) = −∇VTOT ( r ) and
σ TOT
free = −ε o
∂VTOT ( r )
∂V ( r )
= −ε o TOT
= +ε o Er ( r )
∂n
∂
r
r =R
r=R
Since: Er ( r ) = −
20
r =R
∂VTOT ( r )
∂r
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.