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Transcript of Chapter 2 Electrostatics - Komaba Particle Theory...
Chapter 2
Electrostatics
2.1 Coulomb’s law
Coulomb’s law1 (1785) (Cavendish 1772: unpublished )
The force exerted on q by q′
~F = kqq′
R2R , (2.1)
~R = ~r − ~r′ (2.2)
R =~R
R2, R = |~R| =
√~R · ~R (2.3)
q′ q
O
~r~r′
~R
2 Dimension、system of units、constant of proportionality k :
Dimension: Every physical quantity has a dimension, made up of a com-
bination of of M =mass, L =length and T =time. We denote the dimension
of A as [A].
Examples: velocity [v] = L/T、acceleration [a] = L/T 2、force [F ] = [ma] =
ML/T 2、energy [E] = [mv2] = ML2/T 2, etc.
Unit: Unit is the standard by which one measures a quantity having a
definite dimension.
1Columbus in italian. (The real name is Colombo.)
1
Examples: Unit of length L : m, cm, etc.
MKS system of units: In this system one meaures (L,M, T ) by (m,Kg, s)
Remark: In the case of E & M, there is more freedom in the choice of units
compared to the case of mechanics.
Dimensions of k and q: From the Coulomb’s law, one has
[kqq′] = [kq2] = [FR2] =ML3
T 2. (2.4)
From this relation we can only fix the dimension and the unit of the com-
bination kq2. Those of individual k and q are not uniquely determined. In
the following, we will use the so-called MKSA system .
Introduce independent unit for current
In the MKS system, first we define the unit of electric current as “Ampere”
and then fix the unit of charge called “Coulomb”. Concretely, this is done in
the following way:
• Prepare two parallel currents 1m apart and let the same strength of
currents go through as in the figure. Then if the Lorentz force2 exerted
on the 1m segment is 2× 10−7N , we define the amount of the current
as 1A (Ampere)3
d = 1m
l = 1m
~F1A 1A
• Unit of electric charge C(Coulomb) is defined as
C = A · s (2.5)
2We will explain the Lorentz force in detail later.3The general formula is
F = km2I1I2
dl, km =
k
c2= 10−7
2
• In this case the unit of k becomes [k] = Nm2/C2. As for the actual
value of k, experiments tell us that
k = 10`7c2 = 8.988× 109 (2.6)
c = speed of light = 2.998× 108 m/s (2.7)
The reason why the speed of light appears here will be clarified when
we deal with the electromagnetic wave.
Further we denote k defined this way as
k =1
4πε0
(2.8)
ε0 = dielectric constant of the vacuum (2.9)
To really understand the meaning of this constant, we must learn the
physics of dielectric material. This is an advanced topic, which we will
not be covered in this course.
Exercise 2.1 Actually the“Coulomb” is a tremendously large unit. (The
amount discharged in a lightening is on the order of a few Coulombs )
When two bodies with mass 1 Kg and charge 1C each are placed L meters
apart, the electrostatic force exerted on one of them turned out to be equal
to the gravitational force exereted on the surface of the Earth. What is the
value of L ?
Exercise 2.2 What is the ratio of the gravitational force and the Coulomb
force for a proton ? The mass, the the charge of the proton and the Newton’s
gravitational constant are given by
mp = 1.672× 10−27 kg , ep = 1.602× 10−19 C (2.10)
G = 6.67× 10−11 Nm2/Kg2 (2.11)
3
2.2 Concept of electric field and superposi-
tion principle: Coulomb’s law for contin-
uous distribution of charges
Consider the situation where a charge q is receiving Coulomb forces due to
other charges. Take the point of view that the charges first create an electric
field at the position of q and then it exerts the force on q. This is expressed
as
~F (~r) = q ~E(~r) . (2.12)
Then since we know that the forces can be superposed, we must conclude
that the electric fields can also be superposed. This is called the superpo-
sition principle for electric fieds. If we apply this to the fields created
by each individual charge, we obtain the formula
q
qN
q2
q1
~E
~E(~r) =1
4πε0
N∑i=1
qi
R3i
~Ri (2.13)
~Ri ≡ ~r − ~ri (2.14)
2 Dimension and unit of E :
From F = qE, we find that the unit of E is [E] = N/C
Using E = (1/4πε0)q/R2, the dimension of E must be
4
[E] =1
[ε0]
[Q
L2
](2.15)
This formula is extremely useful in checking the answer for all the problems
for computing the electric field. If E is not of this form, one must have made
a mistake !
2 Electric field due to continuous distribution of charges :
Since the charges are carried by electrons and protons, classically the distri-
bution of charges is strictly speaking discrete. However, for a macroscopic
body with an enormous number (∼ 6 × 1023) of atoms it is possible and
preferable to regard the distribution as continuous.
ρ(~r) = electric charge density (2.16)
ρ(~r)dV = amount of charge in the volume dV
in the vicinity of ~r (2.17)
dV = dxdydz = d3r (2.18)
From this we see that to deal with a continuous distribution, all we have
to do is to make the following replacements in the formulas for the discrete
distribution
qi ⇒ ρ(~r′)d3r′ (2.19)∑
i
⇒∫
(2.20)
In this way we easily obtain the formula
~E(~r) =1
4πε0
∫ρ(~r′)
|~r − ~r′|3 (~r − ~r′)d3r′ (2.21)
5
2.3 Simple applications of Coulomb’s law
Example 1. Electric field due to uniform distribution of charges
on a plane
Consider the situation where charges are uniformly distributed on the y-z
plane with the area density = σ. (Area density = charges per unit area).
From the uniformity, without loss of generality we can choose the position
at which to measure the electric field to be on the x axis.
First we compute the electric field created by the charges in the infinitesimal
area dy′dz′, namely σdy′dz′, located at ~r′. (See the figure in the next page.)
2 Symmetry and dimensional analysis :
It is always recommended to guess the answer.
For this purpose, consideration of the symmetry of the problem and the
dimensional analysis is of utmost importance, as we shall see.
From the symmetry, the only non-vanishing component is Ex perpendicular
to the plane.
The dimension of σ is [σ] = Q/L2. This is precisely the dimension of ε0E
Therefore we should have
Ex = cσ
ε0
(2.22)
where c is dimensionless. Since x is the only quantity having the dimension
of length, c can only be a constant (one cannot form a ratio of x with
another length). Consequently, Ex cannot depend on x! Thus, all we have
to compute is the value of the constant c.
2 Actual computation :
6
Parametrize the positions as
~r′ = (0, y′, z′) , ~r = (x, 0, 0) ,
q qq ~r − ~r′ = (x,−y′,−z′) , (~r − ~r′)2 = x2 + y′2 + z′2 (2.23)
~r
z
y
σdy′dz′
~r′
x
Thus from the basic formula for the electric field,
Ex
Ey
Ez
=
σ
4πε0
∫dy′dz′
1
(x2 + y′2 + z′2)3/2
x−y′
−z′
The components with y′ and z′ multiplied vanish since they are integrals
over odd functions from −∞ to ∞. (This was obvious from the symmetry
consideration.) Therefore we are left with one integral to perform:
Ey = Ez = 0
Ex =σx
4πε0
∫dy′dz′
1
(x2 + y′2 + z′2)3/2
2 Idea of Scaling :
To compute this type of integral with ease, it is extremely useful to use the
scaling consideration and extract the dependence on x. Since x is the
only quantity having the dimension of length, we can express y′ and z′ in
units of x. Then the x dependence is trivially extracted and the rest is just
a dimensionless integral giving a number.
y′ = xu , z′ = xv∫dy′dz′
1
(x2 + y′2 + z′2)3/2=
C
x
where C =
∫dudv
(1 + u2 + v2)3/2= number
7
C can be computed easily with the use of the polar coordinate:
u
v
a
da
(u, v)
φ
u = a cos φ , v = a sin φ
C =
∫ ∞
0
2πada
(1 + a2)3/2
= π
∫ ∞
0
db
(1 + b)3/2(b ≡ a2)
= 2π (2.24)
In this way, we obtain the simple result
Ex(~r) =σx
4πε0
· 2π
x=
σ
2ε0
= constant (2.25)
Ey = Ez = 0 (2.26)
We will later rederive this result using a beautiful and powerful formula
called the Gauss’ law.
Exercise 2.3 We wish to find the electric field produced by the charge
uniformly distributed along the z-axis with the line density (charge per unit
length) τ .
(1) First guess the form of the answer from the symmetry consideration and
the dimensional analysis.
(2) Find the electric field by actual computation
8
Er(r)r
Example 2. Electric field produced by the uniform charge distribu-
tion on the surface of a sphere
Radius of the sphere = a. Total amout of charge = Q.
Then the area density is σ = Q/(4πa2).
From the symmetry we only have the normal component En only 。From the previous consideration, the form of En must be
En =1
ε0
Q
L2
where L2 = r2, a2, ar,etc. Noticing that the dimensionless ratio a/r is ava-
ialbe, in general it can be of the form
En =1
4πε0
Q
r2f(a/r)
where f(x) is an arbitrary function. However, since the sphere appears point-
like when looked at from infinity ( r → ∞), the function must have the
property f(0) = 1.
The direct calculation using the Coulomb’s law is rather complicated and we
will not discuss it in this course. However, the result is astonishingly simple:
En(~r) =
Q
4πε0r2 r > a
0 r < a(2.27)
9
(discontinuous at r = a)
For r > a, the answer is the same as for the case where a point
charge Q is located at the origin. On the other hand, inside the
sphere, the field cancels out completely.
This feature will be very easily deduced from the powerful Gauss’ law, to be
explained in the next section.
2.4 Gauss’ law and Maxwell’s first equation
As we have seen, the Coulomb’s law is applicable to general distribution of
charges but its direct application often requires rather involved calculations.
Also it is expressed as a law at a distance. In order to capture the essence of
the Coulomb’s law in the form of the Maxwell’s first equation for the electric
field, we shall now develop a new perspective.
2 Basic picture :
When constructing his theory of E & M, Maxwell made essential use of the
analogy with the fluid dynamics. One such viewpoint is to regard the
charge as the source of the fluid and focus on the conservation of the
fluid as it flows across a closed surface. More specifically, we can derive
the extremely powerful Gauss’ law by considering the following situation:
q
q
2.4.1 Gauss’ law
2 Case where the charge is inside a closed surface:
10
We define an infinitesimal electric flux as the product of the infinitesimal
area dS and the normal component of the electric field En = ~E · n at that
point. The total flux is then given by
Φe =
∫EndS =
∫~E · ndS . (2.28)
q
dΩ
~E
En
n
r
r2dΩ
Now consider a unit sphere around the charge q, and
define the solid angle dΩ as the infinitesimal area cut
out from this sphere by the cone as in the figure. (Note
that the ordinary angle is measured by the length of the
arc.)
θ
θ radian1
From this definition we get
∫dΩ = 4π
= surface area of a unit sphere
dS
r2dΩ
θ
Also, from the figure it is easy to see that
r2dΩ = dS cos θ → dS =r2dΩ
cos θ(2.29)
Since En = E cos θ ( E ≡ | ~E|), we can write
Φe =
∫E(~r)r2dΩ (2.30)
Now apply the Coulomb’s law. We get
Φe =
∫q
4πε0r2r2dΩ =
q
4πε0
∫dΩ
=q
4πε0
4π =q
ε0
q qq∫
~E · ndS =q
ε0
(2.31)
11
This says that the total flux is proportional to the charge inside. Note that
for this conclusion the inverse square law E ∼ 1/r2 is of crucial im-
portance.
2 Case where the charge is outside the closed surface :
dS1
~E(1)
~E(2)
dS2
q
Remembering that n is defined to be the outward normal, the flux is com-
puted as
∫EndS =
∫E(1)
n dS1 −∫
E(2)n dS2
= 0 ← integrals over the same solid angle
For the general distribution of charges, we can apply the superposition principle
and obtain the following result:
Gauss’ law= Integral form of the Maxwell’s first equation
∫
∂V
~E · ndS =1
ε0
∑i
qi =1
ε0
(total charge inside V )
=1
ε0
∫
V
ρ(~r)d3r
∂V = the boundary of V
12
2.4.2 Calculation of the electric field using Gauss’ law
Not only is the Gauss’ law beautiful in form and easy to understand phys-
ically, it can be an extremely powerful tool for computing the electric field
itself when the system has an appropriate symmetry.
Condition:
Symmetry of the system allows one to take the closed surface ∂V
in such a way that En on it is constant. In such a case, En can be
taken outside the integral and can be computed.
Specifically,
∫
∂V
~E · ndS = En
∫
∂V
dS = EnS =Q
ε0
q qq En =Q
ε0S(2.32)
Example 1. Electric field due to the uniform charge on the surface
of a sphere
We already said that we can compute the electric field for this case using the
Coulomb’s law. However, the computation can be enourmously simplified if
we use the Gauss’ law. From symmetry, only the radial component can be
non-vanishing and moreover it depends only on r not on the angle. Denoting
this as E(r), Gauss’ law tells us
∫
S
EndS = E(r)4πr2
=
Q/ε0 if r > a0 if r < a
q qq E(r) =
Q
4πε0r2 if r > a
0 if r < a
Qa
r
~E(~r)
Exercise 2.4 Find the electric field produced by the uniformly charged
sphere ( including the interior). In particular what is the field inside ?
13
Example 2. Electric field produced by the uniformly distributed
charge along an infinitely long straight line
Let the line-density be τ . In the right figure, E points
along the radial direction by symmetry and therefore
there is no contributions from the upper and the lower
surfaces. Therefore we only need to consider the integral
over the side of the cylinder. Gauss’ law then dictates
ε0
∫EndS = ε0E(r)2πrl = lτ
q qq E(r) =1
2πε0
τ
r(2.33)
電荷 τ l
rEr
l
So the electric field decreases only as 1/r. The intuitive reason for
this behavior is as follows. Since the system is uniform in the z direction,
this direction can be ignored and effectively the problem becomes that of
computing the field due to a point charge at the origin on a plane. ( Indeed
the form of the answer does not contain the length l along z direction.) Thus
the integral for the Gauss’ law gives us the circumference 2πr instead of the
area. This shows that due to the inherently two-dimensional nature, the
electric flux can diverge only on a plane and hence the field does not die
away as quickly as in 3 dimensions.
Exercise 2.5 Using the Gauss’ law, find the electric field produced by
the charge distributed uniformly on the y-z plane.
Exercise 2.6 Consider the situation where the electric charge is uni-
formly distributed, with area density σ, over the side of an infinitely long
cyliner of radius R. Compute the electric field produced (inside and outside
the cylinder) using the Gauss’ law.
14
2.4.3 Gauss’ theorem
Gauss’ law gives a relation between an area integral and a volume integra.
We will now derive the differential form of the Maxwell’s first equation
by extracting a law which holds locally at every point. What allows us to
do this is the Gauss’ theorem, a mathematical identity. (Later, we will
encouter a similar theorem called “Stokes’ theorem”. ) The essence of the
theorem is the conversion of the area integral into a volume integral.
2 Gauss’ theorem :
Let ~E(~x) be a vector field, V be a finite 3-dimensional region and ∂V be its
boundary. Then, the following identity holds:
∫
∂V
~E · ndS =
∫
V
~∇ · ~~E d3x (2.34)
~∇ · ~~E =∂Ex
∂x+
∂Ey
∂y+
∂Ez
∂z
=∑
i
∂iEi
Proof:
(x, y, z)
Ex(x + ∆x, y′, z′)
Ex(x, y′, z′)
∆z
∆y
∆x
S1
S2
Decompose the region into small rectangular parallelepiped’s
Focus on the component in the x direction. The flux going out in the x
15
direction from this parallelopiped is
Φx =
∫
S′1
Ex(x + ∆x, y′, z′)dy′dz′
−∫
S1
Ex(x, y′, z′)dy′dz′
Expanding y′, z′ around y, z and keeping only the largest contribution
Φx ' [Ex(x + ∆x, y, z)− Ex(x, y, z)] ∆y∆z
=∂Ex
∂x(x, y, z)∆x∆y∆z (2.35)
Similarly, we add the flux going out in the y, z directions. Then we get, for
this parallelepipied,∫
~E · ndS =
(∂Ex
∂x+
∂Ey
∂y+
∂Ez
∂z
)∆V (2.36)
Thus the theorem is proved for this small parallelepiped.
Now we put together the contributions from these infinitesimal paral-
lelpiped to recover the entire region. Then, except for the contrigution
of the surface of the entire region, all the contributions on the left hand
side cancel with each other. Thus the Gauss’ theorem holds for arbitrary V .
//
2.4.4 Maxwell’s first equation: differential form
We now apply the Gauss’ theorem to the LHS of the Gauss’ law. This gives∫
∂V
~E · ndS =
∫
V
~∇ · ~~Ed3x =1
ε0
∫
V
ρd3x (2.37)
Since this equality is valid for arbitrary V , the integrand must be equal.
In this way, we obtain the Maxwell’s first equation:
~∇ · ~E =ρ
ε0
(2.38)
16
Remember that this law came from the Coulomb’s law. Conversely, we
will show, after introducing the concept of electrostatic potential, that the
Coulomb’s law can be obtained as the solution of this equation regarded as
the differential equation for ~E.
Coulomb
Gauss Maxwell 1st eq
2.5 Potential energy and the electric poten-
tial
2.5.1 Reviews of the “work” and the potential energy
2 work and conservative force :
Consider the electric Lorentz force ~F = q ~E acting on a charge q. Then the
work done by this force in bringing the charge from P0 to P along a path is
given by
W =
∫ P
P0
~F · d~x =
∫ t
t0
~F (~x(t)) · d~x
dtdt
(2.39)
If this work does not depend on the path, ~F is said to
be a conservative force. In such a case, by fixing the
starting point P0 once and for all W becomes a single-
valued function of the other end point P .
P0
P~F
d~x
This quantity is written as −U(P ), where U(P ) is called the potential en-
17
ergy . If we denote the position of P as ~x, we have thus
U(~x) =
∫ ~x
(−~F ) · d~x (2.40)
Now we want to solve ~F in terms of U . To do this, take the time derivative
of both sides. The one for the LHS is
dU
dt=
∂U
∂x
dx
dt+
∂U
∂y
dy
dt+
∂U
∂z
dz
dt
= ~∇U · d~x
dt(2.41)
On the other hand the one for the RHS is
d
dt
∫ t
(−~F (~x(t))) · d~x
dtdt = −~F · d~x
dt(2.42)
Since the path of the integration is arbitrary, d~x/dt is an arbitrary function.
With this in minde, compare (2.41) with (2.42). We must then have
~F (~x) = −~∇U(~x) (2.43)
Conversely, when ~F (~x) can be expresed in this form, W does not indeed
depend on the path. The work needed to move a body from ~x0 to ~x against
the conservative force ~F is given by
∫ ~x
~x0
(−~F ) · d~x = U(~x)− U(~x0) (2.44)
This amount of energy is stored as the potential energy.
A simple example: Gravitational potential energy of a body at height h as
measured from the surface of the Earth is mg × h = mgh.
2.5.2 Potential energy and electrostatic potential dueto Coulomb force
The electric field produced at ~x by a charge q located at ~x′ is
~E(~x) =1
4πε0
q
|~x− ~x′|3 (~x− ~x′) (2.45)
18
This can be interpreted as the force exerted on a unit charge at ~x. The
fact that this force is conservative can be easily checked by the calculation
∂
∂xi
1
|~x− ~x′| =∂
∂xi
[(~x− ~x′)2]−1/2 = −1
2[(~x− ~x′)2]−3/2 ∂
∂xi
∑j
(xj − x′j)2
= −1
2[(~x− ~x′)2]−3/2 × 2(xi − x′i) = − xi − x′i
|~x− ~x′|3 (2.46)
So we have the formula
~∇ 1
|~x− ~x′| = − ~x− ~x′
|~x− ~x′|3 (2.47)
This formula will be important later.
Comparing this with the expression for ~E, we immediately see that
~E(~x) = −~∇φ(~x) (2.48)
φ(~x) =1
4πε0
q
|~x− ~x′| + const (2.49)
φ(~x) is called the electrostatic potential. When it is multiplied by a
charge, it acquires the dimension of energy. The constant in the above ex-
pression can be set to zero by defining φ(~x = ∞) = 0. Then, we can write
φ(~x) =1
4πε0
q
|~x− ~x′|
= −∫ ~x
∞~E(~y) · d~y (2.50)
19
This expresses the work done on a unit charge by moving it from ∞ to ~x
against the electric field ~E. Electrostatic potential has the following impor-
tant properties:
1. Superposition of fields −→ superposition (addition) of the potential
2. The electric field is a vector field with 3 components. In contrast, the
electrostatic potential is a “scalar” having only one component and
hence is easier to deal with. Thus it is advantageous to compute the
electrostatic potential first and then compute ~E from ~E = −~∇φ. This
is what we will do from now on.
2 electrostatic potential due to general distribution of charges :
From the superposition principle we have
φ(~x) =1
4πε0
∑i
qi
|~x− ~xi| (2.51)
φ(~x) =1
4πε0
∫ρ(~x′)d3x′
|~x− ~x′| (2.52)
Exercise 2.7 (1) Compute the electrostatic potential due to the charges
uniformly distributed, with line-density τ , along an infinitely long straight
20
line (say z-axis). Take the zero of the potential to be at the point r = a,
where r is the perpendicular distance from the line.
Remark:To avoid a divergence (infinity) in the intermediate steps, take the
length of the straight line to be finite (for example 2l) and then take the limit
l →∞ at the end of the calculation.
(2) Compute the electric field from the electrostatic potential and confirm
that the answer is the same as we got before.
2 Equipotential surface and the direction of the electric field :
If we know the equipotential surface, we can immedi-
ately learn the directions of the electric field by a simple
geometrical consideration.
equipotential surface ⇐⇒ φ(~x) = C = const
q qq 0 = φ(~x + ∆~x)− φ(~x)
= ∆~x · ~∇φ(~x)
⇒ ∆~x · ~E(~x) = 0
(2.53)
∆~x
~x ~x + ∆~x
φ = const
Thus, the electric field is perpendicular to the equipotential surface.
The equipotential surface (or line) corresponds to the isobaric surface (or
line) in the weather chart, and ~E corresponds to the wind velocity vector.
~Eφ = const
21
2.5.3 The equation satisfied by the electrostatic poten-tial: The Poisson equation
As we have seen, one can compute the electric field by first obtaining the
electrostatic potential and then differentiating it. Now if we combine the two
equations we already know, i.e.
~∇ · ~~E =ρ
ε0
(2.54)
~E = −~∇φ (2.55)
we can eliminate the electric field and obtain an equation for the potential φ
~∇ · ~~E = −~∇ · ~∇φ
= −∇2φ =ρ
ε0
(2.56)
This is the Poisson equation
∇2φ = − ρ
ε0
(2.57)
∇2 =∂2
∂x2+
∂2
∂y2+
∂2
∂z2= Laplacian (2.58)
In particular, when ρ = 0, it is called the Laplace equation.
2.5.4 General solution of the Poisson equation
Since we can make use of our knowledge of the electrostatic potential due
to the Coulomb force, it is not so difficult to obtain the general solution of
the Poisson equation. Suppose we find, by some means, one solution of the
Poisson equation (usually called a special solution) φ0(x). If we denote the
general solution by φ(x) we have
∇2φ = − ρ
ε0
∇2φ0 = − ρ
ε0
22
We then find that the difference satisfies the Laplace equation and hence
∇2(φ− φ0) = 0
q qq φ(~x) = φ0(~x) + f(~x) (2.59)
f(x) = general solution of the Laplace equation
(2.60)
When an appropriate boundary condition is specified, the solution of the
Laplace equation can be obtained, for simple cases, by methods to be de-
scribed later.
Since the electrostatic potential due to the Coulomb’s force
must be a solution of the Poisson equation the following theorem holds:
Theorem
General solution of the Poisson equation φ(x)
φ(~x) =1
4πε0
∫d3x′
ρ(~x′)
|~x− ~x′| + f(~x) (2.61)
∇2f(~x) = 0 (2.62)
The function f(x) depends on the specific situation of the problem (to be
elaborated later).
2 Check:
Let us check that the Coulomb potential actually satisfies the Poisson equa-
23
tion. If we apply ∇2 and compute naively, we find
~∇ 1
|~x− ~x′| = − ~x− ~x′
|~x− ~x′|3 (2.63)
q qq ~∇ · ~∇ 1
|~x− ~x′| =−~∇ · (~x− ~x′)
|~x− ~x′|3 − (~x− ~x′) · ~∇ 1
|~x− ~x′|3
= − 3
|~x− ~x′|3 − (~x− ~x′) · (−3)(~x− ~x′)
|~x− ~x′|5= 0! (2.64)
So it appears that one cannot account for the source term of the Poisson
equaiton. Although very subtle, there is actually an implicit assumption
made in the calculation above. Since the denominator involves the factor
|~x − ~x′| which vanishes for ~x′ = ~x, we have assumed ~x 6= ~x′. However, since
we have to integrate over ~x′ over the entire space to get φ(x), there must be
an instance where ~x′ coincides with ~x. At that point the calculation above
does not apply.
Therefore we have to consider carefully how we should perform the inte-
gral over ~x′ in the small neighborhood of ~x.
So consider a small sphere around ~x of radius ε and try to perform an
integral inside this region. But we wish to avoid integrating over the point
~x = ~x′. How can we achieve such a thing ? There is a clever trick. Recall the
Gauss’ theorem which converts between the volume integral and
the surface integral. Since the surface of the sphere never touches the
center, we can avoid the problem by replacing the integral inside the sphere
by an integral over its surface.
What we want to compute is
∇2φ(~x) =1
4πε0
∫
|~x′−~x|≤ε
d3x′ρ(~x′)∇2 1
|~x′ − ~x|(2.65)
=1
4πε0
∫
|~x′−~x|≤ε
d3x′ρ(~x′)~∇ · ~∇ 1
|~x− ~x′|(2.66)
ε
~x
~x′n
If we take ε to be sufficiently small, we can approximate ρ(~x′) ' ρ(~x). Also,
24
we can write ~∇· ~∇(1/|~x− ~x′|) = ~∇′ · ~∇′(1/|~x− ~x′|). By using the formula we
derived before (which is completely valid since ~x′ 6= ~x here) i.e. ~∇′(1/|~x −~x′|) = (~x − ~x′)/|~x − ~x′|3, we can compute the integral in the following way
using the Gauss’ theorem:
∇2φ(~x) =ρ(~x)
4πε0
∫
|~x′−~x|≤ε
d3x′~∇′ ·(
~x− ~x′
|~x− ~x′|3
)
=ρ(~x)
4πε0
∫
|~x′−~x|=ε
ε2dΩn · (~x− ~x′)
ε3← Gauss’ theorem
= −ρ(~x)
4πε0
∫dΩ = −ρ(~x)
ε0
(2.67)
Thus, we do recover the source term and check that the Coulomb potential
is indeed a solution of the Poisson bracket.
Exercise 2.8 Suppose the electrostatic potential is of the so-called Yukawa
form
φ(r) =e−kr
r,
(k > 0 , r =
√x2 + y2 + z2
)
(i) Find the electric field ~E(~x).
(ii) Using the Gauss’ law, find the total amount of charge inside the sphere
of radius R, including the origin.
(iii) Find the charge inside an infinitesimal neighborhood of the origin. Also
find the total charge present in the entire space.
(iv) Find the charge density outside the origin .
(v) Find the total amount of charge in the entire space excluding the origin.
(Use the formula∫∞
0drre−kr = 1/k2 if needed)Compare the result with that
of (iii) and discuss the consistency.
25
2.6 The electrostatic potential and the elec-
tric field produced an electric dipole
The superposition of the electric field and the potential is one of the cru-
cial principles of E & M. It in turn means that one can decompose the
general electric field and the potential into a sum of some “basic
configurations of charges”. What then should be considered as such ba-
sic configurations ? Let us begin with some intuitively plausible answer and
then justify it more logically.
2 1. A point charge at the origin:
This should obviously be fundamental. The electrostatic potential it pro-
duces is
φ0(~r) =1
4πε0
q
r(2.68)
2 2. Electric dipole :
Consider a pair of charges of equal magnitude with opposite sign closely
located to each other near the origin.
∆r = d cos θ
r
r + ∆rθ
q
−q
d
If we assume d ¿ r、hence ∆r ¿ r, the electrostatic potential is given by
φ1(r) =1
4πε0
(q
r− q
r + ∆r
)
=1
4πε0
q∆r
r(r + ∆r)
' 1
4πε0
qd cos θ
r2(2.69)
Now if we take the limit d → 0 with p ≡ qd held fixed the system becomes
the so-called an electric dipole. Its potential is given by
26
φ1(~r) =1
4πε0
p cos θ
r2
=1
4πε0
~p · rr2
∼ 1
r2
Remarks: Here θ is the angle betwee ~p and ~r. ( Not the angle from the
y-axis.)
In the second expression, ~p is a vector of magnitude p in the direction from
−q to q and is called the electric dipole moment (vector).
2 Electric field produced by an electric dipole:
To get the profile of the electric field, we should first draw the equipotential
surface (↔ r2 ∝ cos θ) and then draw the electric field perpendicular to this
surface. Since there is an obvious axial symmetry, it suffices to consider the
projection to the z-y plane.
z
y
θ = 0 line
θ = π line
~p
cos θ > 0
cos θ < 0
Precise expression of the electric field becomes
~E1(~r) =1
4πε0
1
r3(3(~p · r)r − ~p) (2.70)
Exercise 2.9 Compute the field ~E1 from the potential φ1.
27
2.7 System of conductors and the charge dis-
tribution, the electrostatic potential and
the electric field produced by them
Matter can be classified according to its electrical conductivity roughly into
two categories
• Electric conductor: Example metal(conduction electrons)、Elec-
trolytic solution(positive and negative ions)
• Insulator, or dielectric media (no freely moving charge carriers
inside)
Electrostatiscs of conductors is quite important practically as well as theo-
retically.
2.7.1 Static properties of a conductor
Static = no flow of electric current
Characteristic properties
1. No electric field in a conductor:← If there is, then a current must
flow and it is not static.
2. No charge density inside a conductor:← If there is a charge
distribution, then from the Gauss’ law it produces an electric field.
3. Therefore charge and electric field can exist only onthe surface
of a conductor .
Moreover, in order to avoid a current flow, the electric field must
be perpendicular to the surface. According to Gauss’ law, its
magnitude is
En =σ
ε0
(2.71)
28
En
dS
EndS = σε0dS
Since the electric field is produce only on oneside (i.e. outside of the
conductor), its magnitude is twice as large compared to the case where
it exists on both sides.
The charges distribute themselves so that ~E = 0 inside the conductor.
Put it another way, the distribution is such that the electrostatic
potential is constant inside and on the surface.
4. When on places a conductor in an external electric field, even if the
conductor does not have an overall charge, some charge distribution get
induced on the surface. This phenomenon is called the electrostatic
induction.
⊕+ + +
− − −=
+ + +
− − −~E = 0
We can summarize the above three properties in one figure
En = σε0
~E = 0, ρ = 0φ = const
φ = const
The essential characteristic is “conductor=equipotential body”
29
2.7.2 How to compute the electrostatic field around aconductor
Special feature of a conductor:
So far, we have been computing the electrostatic potential φ(~r) and the
electric field from a given charge density ρ(~r). However, in the case involving
a conductor, we have to devise a new method since the charge distribution
on the surface of the conductor is not apriori known.
2 How to set up the problem :
The electrostatic field around a conductor is determined from the following
principle.
1. We know the answer inside the conductor, namely
ρ = ~E = −~∇φ = 0
→ φ(~r) = c = const.
2. On the boundary of the conductor the potential is constant, called the
Dirichlet boundary condition
φ = c = const. (2.72)
3. We then must solve the Poisson equation for the potential in the outside
region with this boundary condition
−∇2φ =ρ
ε0
(2.73)
(ρ denotes the charge density outside)
In summary, what we have to do is to solve the so-called boundary value
problem restricted to the outer region. In this regard, we can forget
about the fact that we have a conductor in the inner region.
30
Now recall that we already have the general solution of the Poisson equation
(2.74):
φ(~x) =1
4πε0
∫d3x′
ρ(~x′)
|~x− ~x′| + f(~x) (2.74)
∇2f(~x) = 0 (2.75)
The problem here is how to determine the fucntion f(~x). As we empha-
sized already, the crucial requirement is that φ(~x) = c on the boundary. So
we must choose f(~x) in such a way that this is realized. Moreover we must
make sure that f(~x) satisfies the Laplace equation. Apparently this requires
the knowledge of the general solution of the Laplace equation, which is in
general a difficult problem.
Fortunately, however, when the shape of the conductor is simple enough,
we can easily find the solution by an ingeneous method called Kelvin’s
method of images without invoking the detailed knowledge of the general
solution of the Laplace equation.
2.7.3 Kelvin’s method of images
Let us explain this method by way of examples.
Example 1.
Consider the situation where a charge q is placed
at a certain distance from the infinite plane, say
y-z plane, which forms the surface of a conduc-
tor and is grounded, as in the figure. We wish
to find the electrostatic potential at an arbitrary
point P (x, y, z). Here grounding means that the
potential is set equal to the one for the earth, which
can be regarded as an “infinitely large conductor”
with φ = 0.
y-z plane P
q−q
rr′
φ = 0
image charge
grounded
Clearly, one obvious solution of the Poisson equation in the outside region is
the Coulomb potential given by
φ0(P ) =1
4πε0
q
r
31
However, clearly it does not satisfy the boundary condition that φ = 0 on
the surface of the conductor, i.e. the y-z plane. To remedy this situation,
we place a virtual so-called image charge −q as in the figure. Then the
potential created by this charge is
φ1 =1
4πε0
−q
r′(2.76)
Adding this to the Coulomb potential, we obtain
φ =1
4πε0
q
(1
r− 1
r′
)
Evidently, this satisfies the condition that φ = 0 on the boundary.
Why are we allowed to put such an additional contribution ?
Note that the additional potential (2.76) is of Coulombic in form and,
since the virtual charge is placed inside the conductor, r0 never vanishes
outside. In such a situation、as we checked before, it satisfies the Laplace
equation. So we have obtained the appropriate solution f(~x) = φ1(~x) of the
Laplace equation by the trick of representing it as the potential produced by
a virtual image charge.
2 Charge distribution on the surface of the conductor :
Having obtained the electrostatic potential in the outside region, we can
easily compute the charge distribution on the surface of the conductor σ(y, z).
We already derived the relevant formula, namely Ex = σ/ε0. So all we have
to do is to compute Ex = −∂φext/∂x. Let us coordinatize the problem as
follows
R2 ≡ y2 + z2
r =√
(x− d)2 + R2
r′ =√
(x + d)2 + R2
φ =q
4πε0
(1√
(x− d)2 + R2− 1√
(x + d)2 + R2
)(for x ≥ 0)
φ = 0 (for x < 0)
32
Performing the differetiation,
Ex = − ∂
∂xφext
= − q
4πε0
( −x + d
((x− d)2 + R2)3/2+
x + d
((x + d)2 + R2)3/2
)
Setting x = 0, we get the electric field on the surface as
Ex(0, y, z) = − q
4πε0
2d
(d2 + R2)3/2
=σ(y, z)
ε0
Thus the surface charge density takes the form
σ(y, z) =−q
2π
d
(d2 + R2)3/2
Further, let us find the total charge induced on the surface:
Qind =
∫dy dz σ(y, z)
= 2π
(−qd
2π
) ∫ ∞
0
RdR
(d2 + R2)3/2
= −qd1
d
∫ ∞
0
udu
(u2 + 1)3/2
︸ ︷︷ ︸1
← R = ud
= −q (2.77)
Thus we find that the total induced charge is precisely equal to that of the
image charge.
Exercise 2.10 Derive this simple fact by using the Gauss’ law.
Exercise 2.11 Consider the configuration where a charge is placed at
(b, 0, a) in the empty region surrounded by the surfaces of a conductor as in
the figure, namely the planes given by z = 0, x ≥ 0 and x = 0, z ≥ 0, as in
the figure.
33
b
ax
zq
1. Applying the method of images, find the electrostatic potential at an
arbitrary point (x, y, z) inside the empty region x ≥ 0, z ≥ 0.
2. When a = b, find the charge density induced on the surface of the con-
ductor. Also find the total induced charge. Use the following formulas
as needed.∫
dx
(x2 + c2)3/2=
1
c2
x√x2 + c2∫
dx
x2 + c2=
1
ctan−1 x
c
2.8 Concept of the energy of the electric field
In this subsection we compute the electrostatic potential energy of the basic
charge distributions and learn that they can be interpreted as the energy
stored in the electric field itself.
2.8.1 Electrostatic energy
As we already mentioned, the product of the charge and the electrostatic
potential has the meaning of the potential energy. Below we compute the
concrete form for simple charge distributions.
34
1. Energy of the system of two point charges:
U =1
4πε0
q1q2
r12
(2.78)
2. The case of three charges
U =1
4πε0
(q1q2
r12
+q2q3
r23
+q3q1
r31
)
=1
4πε0
1
2
∑
i6=j
qiqj
rij
(2.79)
Here the factor of 1/2 corrects the double-counting
of qiqj/rij and qjqi/rji.
r12
r12
r23
r31
q1
q1
q2
q2
q3
3. The case of n charges: Generalizing the case of 3 charges, we imme-
diately get
U =1
4πε0
1
2
∑
i 6=j
qiqj
rij
(2.80)
4. Expression in terms of the electrostatic potential produced by qj (j 6=i) at the position of qi : Since the electrostatic potential is
φi =1
4πε0
∑
j 6=i
qj
rij
U above can be reexpressed as
U =1
2
n∑i=1
qiφi (2.81)
5. The case of continuous distribution of charges:
qi −→ ρ(~x)d3x
φi −→ φ(~x)∑
i
−→∫
d3x
35
This leads to
U =1
2
∫d3xρ(~x)φ(~x) (2.82)
6. When the charges are on the surface of conductors:Since φ =const on each conductor, the expression
for the electrostatic energy simplifies to
U =1
2
∑i
∫
Vi
d3x ρ(~x)φi
=1
2
∑φi
∫
Vi
d3x ρ(~x)
=1
2
∑Qiφi (2.83)
Here Vi = volume of the i-th conductor
Qi = total charge on the i-th conductor
This result shows that we can effectively think of
each conductor as a charged particle。
Q1
Q2
Q3
φ2
φ3
φ1
V1
V2
V3
36
2.8.2 Electrostatic energy of simple systems
1. Conducting sphere: From Gauss’ law, the electric
field outside is exactly the same as for a point particle
at the center with total charge Q. Therefore
electrostatic potential φ =1
4πε0
Q
a
Therefore U =1
2Qφ =
1
8πε0
Q2
a
2. Uniformly charged sphere: Denote the constant
charge density as ρ:
ρ =Q
43πa3
= const.
First, from the Gauss’ law, the electric field can be com-
puted as
Q
a
Q
a
Outside (r ≥ a) En(r) =1
4πε0
Q
r2=
ρa3
3ε0
1
r2
Inside (r ≤ a) En(r) =ρ
4πε0
(4/3)πr3
r2=
ρr
3ε0
(2.84)
From these result, the electrostatic potential is obtained as
φ(r) = −∫ r
∞En(r′)dr′ = −
∫ a
∞
ρa3
3ε0
dr
r2−
∫ r
a
ρr
3ε0
dr
=ρa3
3ε0a− ρ
6ε0
(r2 − a2) =ρ
6ε0
(3a2 − r2) (2.85)
a
φ
r0
ρa2
3ε0
ρa2
2ε0
37
The total energy is
U =1
2
∫ρφd3x =
ρ2
12ε0
∫ a
0
(3a2 − r2)4πr2dr
=4πρ2a5
15ε0
(2.86)
Dimensional analysis:
φ ∼ EL ∼ Q
ε0L(2.87)
U ∼ Qφ ∼ Q2
ε0L(2.88)
On the other hand ρ2 ∼(
Q
L3
)2
=Q2
L6
q qq U ∼ 1
ε0
ρ2L5 (2.89)
Comment:Very roughly one can regard a nuleus as a uniformly charged
sphere where ρ is constant and independent of the radius. Then
Mass M ∝ a3
U ∝ a5
q qq U ∝ M5/3 (2.90)
This shows that the electrostatic energy increases as M5/3. This ex-
plains the fact that the nucleus of heavy element is easy to undergo
fission.
38
2.8.3 Interpretation of the electrostatic energy as theenergy of the electric field
Let us try to express the electrostatic energy U directly in terms of the
electric field ~E:
U =1
2
∫d3xρφ
Substitute ρ = ε0~∇ · ~E
Then U =ε0
2
∫d3xφ~∇ · ~E
=ε0
2
∫d3x
(~∇ · ( ~Eφ)− ~E · ~∇φ
)
=ε0
2
∫
∞dSEnφ +
ε0
2
∫d3x~E · ~E
In the first integral, En and φ behave at infinity at most like
En ∼ 1
r2, φ ∼ 1
r
Therefore this interal vanishes. Therefore
U =ε0
2
∫d3x| ~E|2
u(~x) = Energy density of the electric field =ε0
2| ~E|2
2.9 electrostatic potential of a system of con-
ductors and the concept of electric capac-
ity
As we stressed, a conductor can be characterized as an equipotential body.
We now point out that, due to the superposition principle, there is a simple
39
relation between the electrostatic potential of a conductor and the charge
on the conductor.
2 Case of a single conductor :
Let us derive a general relation between the electrostatic potential φ and the
charge Q of a conductor.
For this purpose we recall how φ is determined. When
there is no true charge outside, the properties of φ for
the conductor are
(i) In the space outside ∇2φ = 0
(ii) Boundary condition at ∞ (convention) φ = 0
(iii) Boundary on the surface φ = φc = const
φC
These conditions determine φ uniquely. From such φ, we can compute the
electric field En on the surface and this in turn will dictate the total charge
on the surface in the following way:
En = n · (−~∇φ) (2.91)
Q = ε0
∫dSEn (2.92)
The crucial point is that all these relations are linear. This is due to
the superposition principle of electric field and the electrostatic potential.
Specifically this means that if φ is a solution of the Laplace equation, then
λφ, where λ is an arbitrary constant, is also a solution. In this case, the
boundary condition on the surface is scaled by λ. Then clearly the
charge on the conductor is scaled by the same factor λ. That is
φ −→ λφ
Q −→ λQ
Evidently this means that φ and Q must be proportional to each other. This
is exprssed as
40
Q = Cφc (2.93)
The constant of proportionality C is called the electric capacity of the
conductor.
The unit of C: [C] = Coulomb/Volt = F (Farad). Here, Volt = [E/Q] =
Joule/Coulomb. Just like the unit of charge Coulomb, Farad is actually too
large for practical use. What are commonly used are
µF = 10−6F micro Farad
pF = 10−12F pico Farad
2 How C is determined :
The electric capacity C is determined by the size and the shape of the
conductor. To actually compute C, we should either
(i) compute φc for a given value of Q,
(ii) or instead, solve the Laplace equation with a prescribed φc as the bound-
ary condition and then compute Q.
In this way on can find the relation between Q and φc and hence C.
2 Example: :
Let us compute the electric capacity of a spherical conductor with radius
a = 1m. If the total charge on the conductor is Q, the electrostatic potential
and the capacity are given by
φ =1
4πε0
Q
a
q qq C = 4πε0a =107
c2= 1.1× 10−10F = 110 pF
(2.94)
2 System of two conductors =Condenser :
41
We can generalize the concept of the electric capacity to a system of more
than one conductors.
For simplicity, consider a system made of two conductors. Endow charges Q1
and Q2 to conductors 1 and 2 respectively. Then the electrostatic potential
produced outside is, according to the superposition priniciple, the sum
of the potential due to Q1 and that due to Q2.
Q1 Q2
φ2
φ1
For the case Q2 = 0, φ1, φ2 must be proportional to Q1, which is the only
charge. Therefore
φ1 = D11Q1 , φ2 = D21Q1
Similarly, when Q1 is zero, we have
φ1 = D12Q2 , φ2 = D22Q2
Then the potential for the general case with both charges non-vanishing is
obtained by the superposition of these two results, i.e.
φ1 = D11Q1 + D12Q2
φ2 = D21Q1 + D22Q2
(2.95)
We can summarize the result as
φi =∑
j
DijQj (2.96)
42
The coefficients Dij are sometimes called the coefficients of electrostatic
potential . Solving these equations for Qi, we get
Qi =∑
j
Cijφj (2.97)
Cij are called the coefficients of capacity4. Once Dij or Cij are obtained
we can compute φi from Qi or vice versa.
2 Condenser and its capacity :
In particular, when the two conductors carry
charges related as Q1 = Q = −Q2, the system is
called a condenser. In this case, φ1 and φ2 are
both proportional to Q and hence the potential
difference V = φ1 − φ2 is also proportional
to Q. Therefore one has the well-known relation
Q −Q
φ2
φ1
Q = CV (2.98)
C is called the capacity of the condenser. Note that the potential dif-
ference and hence the capacity of a condenser do not depend on what
conventinal value we assign to the electrostatic potential at ∞.
Exercise 2.12 Express the capacity of a condenser C in terms of the
coefficients of electrostatic potential Dij.4In some cases, the diagonal elements Cii are called the capacities while the off-diagonal
elements Cij , i 6= j are called the coefficients of induction.
43
2 Examples of condensers :
Example 1: Parallel plate condenser
Q
−Q
~E d
Area = S
From the figure
σ = Q/S = Surface charge density = finite
E = σ/ε0
V = Ed = σd/ε0 =d
ε0SQ
Therefore we obtain
C = ε0S
d∼ ε0
area
length(2.99)
Dimensional analysis:
Q = Cφc ∼ CQ
ε0L
q qq C ∼ ε0L ∼ ε0L2
L(2.100)
So the smaller the distance between the plates the smaller the potential
difference and the larger the capacity. (But if we take the distance too small,
there occurs an electric discharge ! )
Example 2: Concentric spherical condenser
44
−Q
Qa
b
From Gauss’ law, there is no electric field outside since the total charge
enclosed is Q + (−Q) = 0. This clearly shows that the electric field is non-
zero only in the intermediate region. The field and the potential difference
there is
E(r) =1
4πε0
Q
r2
V =
∫ a
b
(−E(r))dr =Q
4πε0
(1
a− 1
b
)=
Q
4πε0
b− a
ab
Therefore the capacity is
C =4πε0ab
b− a(2.101)
Again, apart from 1/ε0, the dimension is area/distance.
Exercise 2.13 Compute the coefficients of capacity Cij for the concen-
tric spherical condenser as above and show that they have the following
properties:
Cij = Cji i 6= j
Cii ≥ 0
Cij ≤ 0
2.10 Force acting on conductors
2 Example: The force acting on the parallel plate condenser :
45
Consider a parallel plate condenser as in the fig-
ure. Since the two plates are charged oppositely,
Coulomb force acts to attract them to each other.
What is the magnitude of the force ?
One way is to honestly compute the Coulomb force
exerted on each small part of the plates and add
them up. This computation is rather cumbersome.
There is , however, a clever and simpler method of
wide applicability. It is the method of virtual
work invented by D’Alembert in 1743.
Q
−Q
x~Fext
~F~E
If we let the Coulomb force act freely, of course the two plates get pulled
toward each other. To keep the plates still as in the figure, we must exert
an external force ~Fext which is opposite in direction and exactly the same in
magnitude as the Coulomb force, i.e. ~Fext = −~F , so that they balance.
Now imagine that we increase the external force infinitesimally by δFext
and move the plates a small distance δx away from each other. The work5
done by the external force in this process is (Fext + δFext)δx ' Fextδx and
the potential energy of the condenser must increase by this amount. This is
expressed by
Fextδx = −Fδx = δU (2.102)
Dividing both sides by δx we get
F = −δU
δx(2.103)
Note that this relation is exactly the same as the one between the conservative
force and the potential energy.
Let us apply this formula to the system above and compute the force.
5Since we are “imagining” this ideal process to take place, it is called the “virtualwork”.
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We get
U =1
2
∑i
Qiφi =1
2Q(φ1 − φ2) =
1
2QV
V = Ex =σ
ε0
x , σ =Q
S
q qq U =1
2QEx
q qq δU =1
2QEδx
q qq F = −1
2QE = − Q2
2ε0S(2.104)
The negative sign means that the direction of the force is such that the
distance between the plates decreases, i.e. attractive.
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