I was trying to find some nice articles on method of images and fou...
Approaching Method of Images and It’s relation
with real Images
by K.A.Rousan
kaziaburousan@gmail.com
Student of Sripatsingh College, India
10th march, 2019
Abstract
We know that It is invented by noticing the results. But what is the
proof that the results of different problems will give you same result, can
we really apply them without hampering the basic physics & what is the
general Approach to get the charges and distances of image charges?, and
also Is there any relationship between the real mirror images and image
Charges.This paper is written for that purpose only to Ensure that Indeed
we can use the Metho d of Images for some cases and a general approach
to it and to notice that Real mirror images and this image method is
equivalent.
1 Introduction
From the very beginning of The ElectroStatics we just have 3 objective, we want
to find the charge density, or Electric-Field or Potential. we have read many
topics but most of it was done only to find those 3 easily. We have a direct
formula for calculating E, φ ,
E =
Z
l
λ(r)ˆr
kr − r
0
k
2
dl +
Z
s
σ(r)ˆr
kr − r
0
k
2
ds +
Z
v
ρ(r)ˆr
kr − r
0
k
2
dv (1)
and the potential
φ(r) =
Z
l
λ(r)
kr − r
0
k
dl +
Z
s
σ(r)
kr − r
0
k
ds +
Z
v
ρ(r)
kr − r
0
k
dv (2)
remember here the linear, surface and volume charge densities are function of
source charge or source charge element.I have not used the primed notation just
to make things look simple
1
Method of Images
But they are not sufficient enough for finding charges.For Giving the potential
at some surfaces (v.g., metalic plates connected to batteries, induced charges...),
find the total potential everywhere. This is a bit trickier using Direct method
and Superposition principle, so we then use Laplace and Poisson’s Equation
and we see those problems are then easy.
So Now we are Happy, but not for too long. We see solving the partial Dif-
ferential Equation of Laplace(which are easier than Poisson’s ) can give you
nightmares even using most suitable Boundary conditions. At this point when
all hopes are lost(I mean it’s like we have to solve those monsters), Lord Kelvin
Came into picture. He noticed something which no-one before him notice. We
can use the knowledge of Equipotential Surface which we have gotten when solv-
ing previous problems and The Uniqueness Theorem to solve the problems
for some cases.And This is the thought that changed everything and how we
looked towards the problems. we should remember that Method of images have
not developed like other Solving tools. It is developed by seeing the results,
i.e., It is invented by first using the result and then thinking that how can we
configure out system to get same result but with logic if it was done randomly
then nothing would have been gotten from that. I mean same kind of problems
can be solved is the proof that the method is correct.
Figure 1: Lord Kelvin , he is the one invented the method of images
To begin the discussion of method of images first we need to know the
Boundary Conditions and The Uniqueness Theorem.You may say why?
, The cause will be clear once we understand the Theorem.
2 K.A.Rousan
Method of Images
2 Developing The Charge and Image
2.1 Uniqueness Theorem
Boundary Condition and Uniqueness Theorem: I have we all know the
Poisson Equation,
∇
2
φ = −
ρ
0
(3)
where ρ is the charge density of the region we are dealing with and
0
is the
permittivity of the free space. Now when the charge density of our interested
region is zero (by zero I don’t mean the total charge of the whole region be zero.
I just mean the region we are looking at ...only there the total charge maybe
zero) then The Poisson’s Equation goes down to Laplace’s Equation
and we get
∇
2
φ = 0 (4)
Now for those 2 equations to get a solution for a special case i.e., for problem
of interest we have to get the boundary condition. As with out those we can
have infinity many solution of a single problem and without that we can’t de-
cide which solution have to use. Also if we have a region where one part of
it satisfies eq
n
-(3)(the function which is the solution here is φ
1
) and other-one
satisfies eq
n
-(4)(here solution is φ
2
) then φ
1
+φ
2
is also a solution.so for many
more reason Boundary condition is necessary.
Uniqueness Theorem:The solution to Laplace’s Equation (also Poisson’s Equa-
tion) in some volume V is uniquely determined if φ is specified on the boundary
surface S. so The The boundary condition is that the φ must be known
on the boundary and after that From the uniqueness Theorem tells us that
There is only one function for us.
Figure 2: Here is a charged region (yellow colour) covered by red curve
3 K.A.Rousan
Method of Images
Proof: We will prove for the Poisson’s Equation as it is more general than
Laplace’s one.
Suppose,In the region there are 2 solutions of the eq
n
-(3) then,
∇
2
φ
1
= −
ρ
0
and
∇
2
φ
2
= −
ρ
0
Now consider the Difference φ
3
=φ
1
-φ
2
So, Now operating the ∇
2
in this equa-
tion we get, ∇
2
φ
3
=∇
2
φ
1
-∇
2
φ
2
or ∇
2
φ
3
=-
ρ
0
+
ρ
0
=0 So we see φ
3
satisfies Laplace’s Equation and it is zero on
the boundary as as on the surface potential is specified and so φ
1
= φ
2
. Now we
know The solution of the Laplace’s Equation takes maximum value on
Surface and on surface it is zero. so it is zero all over the Volume. And hence
φ
3
= 0 all over and
φ
1
= φ
2
. So we see that If we know the Potential of any Bounded region at the surface
, then it’s potential all over the region can be specified by only one function.
So Now we are ready to Go for the main topic.
2.2 Surfaces and Equivalent conductors
So We know the Uniqueness Theorem. But so what ?, I have told you before that
The method of images developed by remembering previous results
and then using them to find important for different configurations.
How can we guarantee that both will have same Potential or field for
2 different methods and more important things ,Can we apply them
and if we can for which cases?
Remember one thing, If we can show somehow that for 2 different cases the
potential function (for other functions we need) are same, i.e., unique then we
can show that our argument can be proven.. There are many types of uniqueness
theorems and main component of this theorem is that Dirichlet’s Condition and
Neumann Condition but for our purpose it is not that much necessary to told
them.
4 K.A.Rousan
Method of Images
Figure 3: upper one is the image of a surface with charges distributed like it is
shown and later one is the image of a conductor which is kept on the surface
without hampering the charges except the inner ones.
To see why it is the case and to see how powerful The Uniqueness Theorem
is Let’s consider, A charge distribution as in Fig-3•1 with the Equipotential
surface γ covering it. As we know the distribution and position so we also know
Charge-density(ρ) on the surface and also the Field(E) and φ everywhere in our
region of interest (our region of interest is the region outside the surface γ i.e.,
The region between the Surface and the surface at Infinity).
Now we take a Conductor with same geometry as The surface and placed
it on the same place as the surface with out disturbing the charges. So the 2
system are identical. We know the charges and potential on The Curve(γ), Let’s
suppose The potential on the surface of it is φ
1
and field outside is E. Now as
the outside charges are same. So if we can make the Potential and Field on
the surface same then Boom!!..we are done. And it’s a easy task. Suppose we
have to place Q to the conductor to make it’s Potential φ
1
. Then making so
we will know the potential at every-single position in our region of interest (As
we know potential at the surface so Due to the uniqueness Theorem we will get
only a single function everywhere). As the charge distribution outside is not
changed and as We have the same potential on surface of the Conductor so,
By uniqueness Theorem the potential and by it the field on both cases will be
same. See I told you we can relate two separate cases. Now what is the amount
of charge needed to do this all work? i.e., what is the value of Q? By just using
the Gauss’s Law we can see that
Q =
n
X
i=1
Q
i
(5)
where the Q
i
is the i
th
charge in the surface γ. so using eq
n
(5) we can say for
5 K.A.Rousan
Method of Images
our Fig•3 the
Q = Q
1
+ Q
2
+ Q
3
.
2.3 Method of Images
We are now fully ready to know the Method of images.What we learn in the
previous section will help us develop the method itself. Let’s start by assuming
A charge pair of charge as shown in Fig•(4-1). A q charge is kept on the x-axis
and assuming it as the origin and the other one q´ is kept at a distance d toward
the left i.e., it is on (-d,0,0). Let the points are on Q(0,0,0) and P(-d,0,0) and
we want to find the Equipotential Surface(already told why).
Suppose we Have a point I(x,y,z) where we wish to find the potential. So
potential at I is
φ
I
=
q
4π
0
p
x
2
+ y
2
+ z
2
+
q
0
4π
0
p
(x + d)
2
+ y
2
+ z
2
(6)
Now for simplicity we want the equipotential surface in which φ=0 ,
Note: We can do for any φ , just to reduce our effort we use 0. So Now making
φ
I
=0 in equation (6) we get,
q
p
(x + d)
2
+ y
2
+ z
2
= −q
0
p
x
2
+ y
2
+ z
2
(7)
Now square both sides of eq
n
(7) we get
q
2
((x + d)
2
+ y
2
+ z
2
) = (q
0
)
2
(x
2
+ y
2
+ z
2
) (8)
Notice the information of negative sign is lost due to squaring, so One thing
we should remember if one charge is positive then other one is negative.
or x
2
(q
2
− q
02
) + y
2
(q
2
− q
02
) + z
2
(q
2
− q
02
) + 2xdq
2
+ d
2
q
2
= 0
x
2
+ y
2
+ z
2
+ 2
dq
2
x
(q
2
− q
02
)
+
d
2
q
2
(q
2
− q
02
)
(9)
Now we know for an equation like
x
2
+ y
2
+ z
2
+ 2gx + 2fy + 2hz + c = 0
The solutions are on a Sphere of Radius given by
r =
p
g
2
+ f
2
+ h
2
− c
and center at (-g,-f,-h).
6 K.A.Rousan
Method of Images
Figure 4: upper one is the image of 2 charges kept at Q and P and I=(x,y,z) and
lower one is the locus of the point I for which we get a Equipotential Surface
with potential 0.
Now notice eq
n
(9) is the Equation of the form of Sphere with
Center = O = (
dq
2
(q
2
− q
02
)
, 0, 0)
And the radius
r = −
s
(
dq
2
(q
2
− q
02
)
2
)
2
−
d
2
q
2
(q
2
− q
02
)
why negative?, because I told you before that if one charge is positive another
one is negative. so
radius = r = −
qq
0
d
q
2
− q
02
Now we Kept the center at O in the fig-4•2 , So by doing so we get
P Q = d
i.e., d is the distance between two charges.
r = radius
which is between P and Q ,
OP = b
i.e., distance between the center and The charge q’ and finally
OQ = s
7 K.A.Rousan
Method of Images
distance between the center and the charge q. So
s =
q
2
d
q
2
− q
02
(10)
Now to get the value of b
b = s − d =
q
2
d
q
2
− q
02
− d =
q
2
d − q
2
d + q
02
d
q
2
− q
02
b =
q
02
d
q
2
− q
02
(11)
and last the radius
r = −
qq
0
d
q
2
− q
02
(12)
Now by diving eq
n
(10) and by eq
n
(11) we get
s
r
= −
q
q
0
(13)
or
q
0
= −q
r
s
(14)
also
q
0
= −q
r
b
s
(15)
Equation (14) and Equation (15) tells something very important. we will
see later. Now from (10) and (11) we get
b =
r
2
s
(16)
I suggest Remember and praise the beauty of Equation (14) and (16).
So Now we have all pieces of the puzzle. As we have seen in the previous subsec-
tion Surfaces and Conductors we can now Put a Conductor of Spherical
Shape on the place of spherical Equipotential surface of potential = 0. To show
it is zero we grounded it.
So Now we have a Spherical grounded conductor with Center at O(0,0,0) and
Radius given by eq
n
(12). To make the potential at the surface same for both
cases we have to put the charge
q
0
= −q
r
s
as given by eq
n
(14). So by doing all this we can calculate E, φ and the surface
charge density(σ) which is due to the charge q and induced charge q’. So What
is method of images?
8 K.A.Rousan
Method of Images
2.4 calculation by Method of Image:
Let’s see some definition:
Image charge:The fictitious Point charges, placed in the region where the field
is not required and producing the same field in the desired region(already told
you about the region) as with the actual electrification of the surface ,are defined
as the Electrical Images. why they are called image, well we will see later that
the expression here are similar to the expression for Mirror images.
Steps of using It:
1.For simplicity and elegantisity of this method we only use it for Equipotential
Surface for zero potential.
2.The image charge should be such that Laplace’s Equation is satisfied in our
region of interest with the boundary conditions but not on the point where we
have placed the charge.
3.The image charge should be taken in the region which is not of our interest.
The image charge forms in the position of the inverse reflection of the Real
charge. For plane conductor the inverse reflection = Normal reflection.
Figure 5: This is the sphere which we have given and we wish to find potential
at L(r,θ) with O is the origin and using polar coordinates and the Radius of
the sphere is ”a”
The classical Image problem,Grounded Conducting Sphere: Now, Let’s
Suppose we have Spherical Grounded Conductor and a charge +q is placed at a
distance s from it’s center. Now, as the conductor is grounded we can see that
the surface is at Zero potential. Now if we need the surface charge density
or Potential(φ
L
) at any point L, or Electric-Field at any point outside
, then what can we do?, we can’t find it directly as we don’t know the charge
induced but then we remember From previous subsection that we can remove
9 K.A.Rousan
Method of Images
the sphere and can place a charge q
0
given by eq
n
(14) at a distance b from the
center O given by eq
n
(15) and we will get similar result.
Figure 6: This is the equivalent figure of Fig:5 with center at O and q’ at P
So Now we can use Fig:6 to find potential at L. So the potential at L is
φ
L
=
1
4π
0
[
q
√
r
2
+ s
2
− 2rscosθ
+
q
0
√
r
2
+ b
2
− 2rbcosθ
] (17)
As
(QL)
2
= (OQ)
2
+ (OL)
2
− 2(OQ)(OL)cosθ
and
(P L)
2
= (OL)
2
+ (OL)
2
− 2(OL)(OL)cosθ
from ∆OLQ and ∆OLP respectively. Now putting the values of q’ and b in
eq
n
(17) we get
φ
L
=
q
4π
0
[
1
√
r
2
+ s
2
− 2rscosθ
−
a
√
r
2
s
2
+ a
4
− 2rsa
2
cosθ
] (18)
So this is the general Expression for potential at any point outside the sphere,i.e.,
region of interest. Now for the r-component of E we partially differentiate
eq
n
(18) w.r.t r and get
E = −
∂φ
L
∂r
=
q
4π
0
[
r − scosθ
(r
2
+ s
2
− 2rscosθ)
3
2
−
as(rs −a
2
cosθ)
(r
2
s
2
+ a
4
− 2rsa
2
cosθ)
3
2
] (19)
So This is the general Expression for Field outside the sphere at any point L.
Now as E is perpendicularly outward for conductors so that anypoint on the
surface like I the field is
E
n
=
q
4π
0
l
3
(a −
s
2
a
) (20)
10 K.A.Rousan
Method of Images
where
QI = l = (a
2
+ s
2
− 2ascosθ)
1
2
Notice one thing maximum value of l = (s+a) when cosθ=1 (θ=0) and minimum
value l= (s-a) when cosθ=-1(θ=π).
Now surface charge density
σ =
0
E
n
And Putting the values from eq
n
(20) we get
σ = −
q
4πl
3
(
s
2
− a
2
a
) (21)
. Notice for l greater than a σ is negative as there is only q’. Now you may
cross-check the total induced charge on the sphere i.e., q’, using σ but I don’t
find any point on that. The force on +q due to the Sphere is the same as
between +q and −q(
a
s
).
Analogy with Real Convex Mirror: After all this we notice something
extraordinary. We know that
1
u
+
1
v
=
1
f
where u is the distance of object from Surface of Mirror , v is the distance of
image from surface of the mirror and f is the focal length. We also know u is
positive and v,f are negative. And also as u → ∞ ; v becomes f i.e.,
lim
u→∞
v = f
Now we shift the origin of Fig:5 at T point then doing so The coordinate of
+q becomes (s-a), The coordinate of q’ becomes (a −
a
2
s
).
Now for our case real charge is at Q and for our case imaginary charge i.e.,
image charge is at P so,
u = T Q = (s −a)
v = T P = (
a
s
)(s − a)
We now notice
lim
s→∞
u = ∞
lim
s→∞
v = a
so we can say now
lim
u→∞
v = a (22)
And also
1
s − a
+
1
−(
a
s
)(s − a)
=
a − s
a(s − a)
(23)
1
s − a
+
1
−(
a
s
)(s − a)
= −
1
a
(24)
11 K.A.Rousan
Method of Images
The negative sign are used as u=positive, v= negative , f= negative. and also
we see that
magnification =
q
0
q
= −
image
0
sposition
objectposition
= −
a(s − a)
s(s − a)
= −
a
s
Then we get
q
0
= −
a
s
q (25)
So we see this is the same as we get in eq
n
(14).So this also Satisfies Magnification
property. So the information we get by This method not only help us in
this but also Satisfies the Properties of real Images which we get by
real mirrors and hence The name ”Method of Images”.
The classical Image Problem,Infinite grounded Conducting Plane:
As same as before we can’t Find any information we like in Electrostatics di-
rectly as we don’t know the inducted charge. So how we should approach?, Well
Notice a thing You know the basic Apollonius Theorem:Given 2 points A
and B and a number α locus of P if (
AP
BP
)=α is circle And If α= 1 ,
then It’s perpendicular Bisector of AB.
Figure 7: This is the same Fig:5 just the origin is shifted from O to T and
(s-a)=d
Now shifting the origin to T we get Q = ((s −a), 0) , P = ((−(
a
s
)(s −a)), 0).
Now let, I is a point with coordinate (x, y). So Now if α=1 then
P I = IQ
[x + (
a
s
)(s − a)]
2
+ y
2
= y
2
+ [x − (s − a)]
2
12 K.A.Rousan
Method of Images
or,
2(
a
s
)(s − a)x + (
a
s
)
2
(s − a)
2
= −2x(s − a) + (s − a)
2
or
2x(s − a)(
a
s
+ 1) = (s − a)
2
[1 − (
a
s
)
2
]
or
x =
s − a
2s
Now If we put s − a = d then get
x =
d
2(d + a)
Now we see as a→∞ , x→0 and only y value exist for I. So we get a perpendicular
plane for a→∞. Then
q
0
= −
a
s
q = −
a
d + a
q = −
1
1 +
d
a
Now as a→∞ we see
q
0
= −q (26)
and if v be the distance of the image from the plane i.e., from T, them by similar
approach
v = −d (27)
So in this case we have a plane i.e., similar to a mirror and it is
working as a plane mirror(we will see more later). Let’s find E, φ and
σ for this case.
Figure 8: We see a charge at (0,0,d) and it’s image charge at (0,0,d’) i.e., (0,0,d)
and the plane is the conductor which is earthed.
13 K.A.Rousan
Method of Images
You can even prove that for this case it will be a plane as we did for Sphere
and get sphere’s equation. So now we wish to find potential at a point D in
our region of interest (remember image will not form in the region of
interest).
φ
D
=
1
4π
0
[
q
p
x
2
+ y
2
+ (z − d)
2
+
q
0
p
x
2
+ y
2
+ (z + d)
2
] (28)
φ
D
=
q
4π
0
[
1
p
x
2
+ y
2
+ (z − d)
2
−
1
p
x
2
+ y
2
+ (z + d)
2
] (29)
Now is it satisfies the basic Boundary Condition of Potential ? As as φ=0 as
z=0 and φ→ 0 as z→∞. Now perpendicular component of E (we are just using
perpendicular component but remember other components also exist).
E
n
= −
δφ
δz
Then we get
E
n
=
q
4π
0
[
z − d
(x
2
+ y
2
+ (z − d))
3
2
−
z + d
(x
2
+ y
2
+ (z + d))
3
2
] (30)
So Now we have Potential and Field. Then
Surfacecharge = σ =
0
E
n
(atz = 0) (31)
σ = −
qd
2π(x
2
+ y
2
+ d
2
)
3
2
=
−qd
2π(r
2
+ d
2
)
3
2
(32)
,where r
2
= x
2
+ y
2
and
areaelement = ds = rdrdφ
Now you can check the total charge to be -q using σ by this equation
Q =
Z
2π
0
Z
∞
0
σds
The force between the conductor and the charge q is also same as the force
between q and -q but The energy stored is not same as for 2 charges q
and q’, the energy is half of that. This is the case because in half of the
region there are not any field so half of the energy is missing as shown in Fig:9.
so For 2 charges the energy is
E
q
=
1
4π
0
q
2
2d
and for the conductor and a charge it is
E
c
=
1
4π
0
q
2
4d
14 K.A.Rousan
Method of Images
The whole thing can be proven by simple integration and is very common to
find in book.
So it is not given.
Figure 9: Here you see the fields from q to -q but as -q doesn’t exist in reality
the field lines terminate on the surface and the dotted lines shown there actually
doesn’t exist so the Energy stored there also don’t exist.
Analogy with real Planar Mirror We see that like normal mirror For this
case the image form at the distance −d and also for solving Laplace equation
and to make potential zero at the surface we see that the property of charges
are also reverse but opposite in magnitude. Now suppose we have two mirrors
with angle between them θ then we know Number of images(n) is
n = b
360
θ
c − 1
for n = even and
n = b
360
θ
− 1c
for n = odd
15 K.A.Rousan
Method of Images
Figure 10: Here green lines represent 2 conductor plane with φ = 0 and the blue
lines are imaginary image lines and c is the point here the +q charge is kept
and here θ=α=β, and J is the image of C with charge -q
So Now suppose we have 2 conductor infinite plane which are making angle
θ between them. Then there is a charge kept between the region. Now what will
be the force between them?, well See that for satisfying the Laplace equation
along with making the potential at the conducting surface zero we need
n = b
360
θ
c − 1
for n = even and
n = b
360
θ
− 1c
for n = odd .
For the proof let At C +q charge is kept then Assuming OB and OA as
two independent Conducting plates. Then the image of +q at C is formed at
J and to make OB potential zero , the charge at J is -q. And similar for AO.
Now But taking the charge at J there changes the potential at AO. So we have
to take another surface OH and there we have to take another charge +q and
it will go till the whole 360 is completed with the region similar to θ.
Then for θ even 360 is divisible by θ. So number of θ needed to cover full circle
is (
360
θ
). So number of images = total charges -1.
n = (
360
θ
) − 1
Now if θ is odd then
b
360
θ
c ≤
360
θ
≤ b
360
θ
c + 1
So finally we get
n = b
360
θ
− 1c
16 K.A.Rousan
Method of Images
for n = odd So we see that this is similar to the number of images formed for
similar arrangement of mirrors. So, The method gives similar results like that of
Mirrors (Finding the charge using magnification like we did for spherical Case is
similar and very straightforward for this case, so it is left). After all this finally
we can see that the method of images is valid and very powerful and also it is
related nicely with Real images. But there is so much more to see about this
beautiful method like,
This is similar to image and real convex mirror and not just any image it is the
Circular Inversion. As
(OP )(OQ) =
a
2
s
s = a
2
which is the exact result for Circular Inversion in complex number. Noticing
this we can notice something very beautiful from eq
n
(18) that the first term is
potential due to charge q and second term is potential due to charge q’. Then
the potential at r due to q’ and q are related by
φ
0
q
0
(r, θ) = (
a
r
)φ
q
(
a
2
r
, θ)
where φ
0
is potential Function due to q’ and φ is potential function for q.
This is called Method of Inversion and it can be generalized for any number of
charges. There are so many properties and similarities that I can’t even end
them in a whole book. So let’s stop here for now and praise the beauty which
we learnt.
Hope you understand it and feel the absolute beauty of this method.If you
for any question feel free to contact.
References
[1] D.Chattopadhyay and P.C.Rakshit. Electricity and Magnetism [with Elec-
tromagnetic Theory and Special Theory of Relativity]. New Central Book
Agency (P) Ltd, 2018.
[2] David J.Griffiths. Introduction to Electrodynamics(Fourth Edition). Pearson,
2019.
[3] Richard Fitzpatrick: Method of Images,
http://farside.ph.utexas.edu/teaching/em/lectures/node64.html
[4] Method of Images, COURSE AVAILABLE FROM 31-DECEMBER-2009 ;
COURSE CO-ORDINATED BY : IIT GUWAHATI
https://nptel.ac.in/courses/117103065/18
17 K.A.Rousan
I was trying to find some nice articles on method of images and found it . It is really nice but the language is quite poor.