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Electrostatics of Continuous Media
COULOMB’S LAW
Reference -- Jackson, "Classical Electrodynamics," Wiley, 1962, pp1-14.
In 1785, the French physicist Charles Augustin Coulomb measured the
force between two charged bodies and determined that:
F
1 Q1Q2
40 r 2
Time Out: units
Actually, Coulomb established the proportionality between force, charge and distance. The
proportionality constant depends on the units used for charge. The expression above, in which
the 4 has been factored out, is said to employ rationalized SI units:
0  8.854187817  10-12
Coul2
N-m2
By factoring the 4 out here, we avoid the appearance of 4 later on (in Gauss’s law and
elsewhere); hence, the constant has been “rationalized”.
Some authors still employ
nonrationalized units, which we will denote by adding a prime to the notation:
1
1
N-m2

 7.15207  108
40 0
Coul2
A third set of units, called the electrostatic system of units (esu’s), is also sometimes used.
Instead of coulombs, another unit of charge is defined (called the statcoulomb) such that when
two point charges, each having one statcoulomb of charge, are separated by one centimeter, they
produce an electrostatic force equal to one dyne:
1
1 statcoul  1 dyne-cm 2   109 Coul
3
In this system of units, the permittivity is unity (dimensionless):
1
1
dyne-cm2

1
1
40 0
statcoul2
By defining the unit of charge in this way, we have reduced the number of fundamental units
required from 4 in SI units (mass, length, time and charge) to 3 in esu’s (mass, length and time).
Copyright© 2016 by Dennis C. Prieve
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Engineers have encounted such a reduction before. Recall Newton’s second law: in the English
engineering units (lbm, lbf, ft and sec), it is written using a proportionality constant (gc):
F
1
Ma
gc
and
gc  32.2
ft-lbm
lbf -sec2
In the metric system (SI), the unit of force (N) is defined as the force required to give an
acceleration of 1 m/sec2 to a mass of 1 Kg:
1N 1
Kg-m
sec 2
for which the proportionality constant in Newton’s law becomes unity:
1
N-sec2
1
1
gc
Kg-m
and we need only 3 fundament units (Kg, m and sec) such the unit of force is derived.
Time In: now that we have discussed the various systems of units, let’s return to Coulomb’s law.
Suppose our two charges are located at x1 and x2. The direction of the force on the charge at x2
(say) is pointing away from Q1 if Q1Q2>0 and toward Q1 if Q1Q2<0. One vector pointing away
from 1 in the direction of 2 is:
x2 - x1
Converting this into a unit vector and multiplying it by the
magnitude of the force as given by Coulomb's law, yields:
F
Q1Q2 x 2  x1 Q1Q2 x2  x1
1

40 x  x 2 x 2  x1
40 x  x 3
2
1
2
1
Now this is the vectorial force on the charge located at point 2. A
second part of Coulomb's law is superposition: in the case of multiple point charges, the net
force is the linear sum of the forces due to each point charge:
F
x  xi
Q n
Qi

40 i 1 x  x 3
i
is the net force on test charge Q located at x. The force per unit charge is called the electric
field:
Copyright© 2016 by Dennis C. Prieve
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E x 
x  xi
F
1 n

Qi

Q 40 i 1 x  x 3
i
(1)
Indeed, this is the electric field at location x due to the other charges.
GAUSS'S LAW: INTEGRAL FORM
To obtain a more useful form of Coulomb’s Law, reconsider the electric field due to a single
point charge, q, which we will locate at the origin of a spherical coordinate system (i.e. x' = 0).
Coulomb's law then predicts the electric field will depend only on the radial distance from the
point charge and the force will be directed away from the point. In spherical coordinates, this
direction is er:
E r  
q
40
er
r2
Now suppose we integrate the normal component of
this vector over some arbitrary closed surface. First
let's consider the contribution from one element of
surface having
n
da
r
da = area of element
d
n = unit outward normal
E.n da 
q er.n da
q

d
40 r 2
40
n
The second fraction on the right-hand-side of the first equation turns
out to be the element of solid angle, d, subtended by the surface da.
To see this, refer to the sketch at right, which is a edge view of the
surface element da, whose normal n makes an angle of  with the unit
vector er. Thus
n.er = cos

da

r
da cos
and n.er da is the area of a spherical cap of radius r subtended by
d. Integration over any closed surface is now quite easy:

A
E.n da 
q
40

d 
A
q
0
(2)
4
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where there are 4 solid radians (steradians - a contraction of stereo and radian) in a sphere, just
like there are 2 radians in a circle.* Note that the position of the charge inside the surface is not
important to the value of this integral. Due to linear superposition (a consequence of Coulomb's
law), I can easily write the result for any number of point charges, distributed in any way through
the interior of the surface A:

q
Qi
all point charges
inside surface A
where the sum is over all charges inside the closed surface A. Charges located outside the
surface do not contribute (as it turns out) to the integral. This result is called Gauss's Law.
COULOMB’S LAW FOR CONTINUA
When the number of charges n is very large, then we might consider describing the
distribution of charges using (x) instead of Qi and xi.
q
  x   lim  
V  0 V 
which we call the charge density. It’s like the more familiar mass density (also commonly
denoted by ) except that the numerator is the total charge inside V instead of the total mass.
Making the following substitutions
xi  x'
Qi    x dV 
and
the sum in (1) can be replaced by the integral
E x 
1
40

x  x
universe x  x
3
  x  dV 
(3)
Equation (3) is equation (1.5) from Jackson (p3), which is the direct distributed counterpart to
Coulomb’s law.
Example #1: Electric Field Near a Charged Plate
Assuming all the free charges are confined to the plane z=0, (3) yields
* This is the origin of the factor of 4, which distinguishes “rationalized” units from “non-rationalized” units. By
writing Coulomb’s law with 4  in the denominator, we avoid 4 in Gauss’s law. The corresponding value of 0 is
in “rationalized” units.
Copyright© 2016 by Dennis C. Prieve
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E x 
1
x  x
1
x  x
  x 
dV  

da 


3
3
4
4



xx
xx
Ez  r , , z  
Dotting both sides by ez:
(4)

z
da 

4 x  x 3
(5)
where, in cylindrical coordinates, we have
x = rer() + zez
x' = r'er(')
and
On the axis of the disk, r=0 and we have
2
x  x   r 2  z 2
(5) becomes


2
2
R
R1
r dr
2z
z 2 d r  z
2r  dr  

32
32
32
4  2
2  2
2
2
2
0 r  z
0 r z
0 r  z2
R

Ez  0, , z  
4 
z





2
2
R
rR
z d  r  z  z
2
z  1
1








3
2
1
2
1
2
1
2
4
4 2
2
2
2
2
2
0 r2  z2 

 r  z  r  0   z   R  z  


 
z
 
z
1

 O 


2
R
R 2  z 2  2

In the limit that the test charge is placed very close to the plate (i.e. z <<
R), the normal component of the electric field becomes

En 
2
En

En
(6)
Since the z-axis is defined to be pointing away from the disk, this electric field also points away
from the disk (assuming >0).
Copyright© 2016 by Dennis C. Prieve
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Example #2: Electric Field Near 2-Sided Disk
Let’s now suppose that the disk has two different sides.
Suppose that the charge density of these two sides is different. In
principle, Coulomb’s law is additive and we can just sum the
separate effects of the two uniformly charged surfaces. But what
value of  is used in (6)? For the region between the two charged
plates, there is no ambiguity because a single material having 2
separates all the charges, including the test charge:
1
E1
2
3
E2
  12
E2  23
22
12
23
E3
(7)
where E2 is positive when directed upward. Notice that if the two surfaces have equal charges
(i.e. 12 = 23), there is zero electric field inside the disk.
To determine E1, we apply the general electrostatic boundary condition
 1E1  2E2 .n1  12
where n1 is the unit vector pointing normal to the interface into material “1” (i.e. upward).
Keeping in mind the sense defined for E2, we note that E2.n1 = +E2 and this b.c. becomes
1E1  2 E2  12
Substituting E2 from (7):
  12
1 E1   2 E2  12  23
2
23 12
2
  12
E1  23
21
or
(8)
The b.c. at the second interface becomes
 3E3  2E2 .n3  23
where n3 is the unit vector pointing normal to the interface into material “3” (i.e. downward).
Keeping in mind the sense defined for E2, we note that E2.n3 = –E2 and this b.c. becomes
3 E3   2 E2  23
23 12
2
Copyright© 2016 by Dennis C. Prieve
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  12
E3  23
23
or
(9)
Comparing (8) and (9), we see that the magnitude of the electric fields on either side of the 2sided disk are the same (assuming 1 = 3), but their directions are opposite.
This is Coulomb’s law for continua. Now we proceed one step further by showing that this
electric field is derivable from a potential. This requires the following equation
 1
  
 x  x
3
x  x

x  x



(10)
Proof: to convince yourself that this last equation is correct, consider a change of variables: let r
= x – x', where r = rer is the position vector in spherical coordinates and er() is the unit vector
pointing in to the r-direction (direction of unit vector depends on spherical coordinates  and ).
This is equivalent to locating the origin of spherical coordinates at point x'. Then the above
equation can be written as
x  x
x  x
3
r

r
3

re r
r3

 1
1
er       
 x  x
r
r2

1



(10) into (3), then interchanging order of integration (with respect to x') and differentiation (with
respect to x), yields
E  x   
where
  x  
1
40

universe
(11)
  x 
dV 
x  x
which is called the electrostatic potential. To the extent that this integral converges, we have
just proven that E is derivable from a potential. From Theorem III, we also know that E is
irrotational:
 E  0
To calculate the total charge inside a macroscopic surface (like A in (2)), we just integrate the
charge density over the region inside. Thus (2) becomes

A
1
E.n da 
0

 dV
V
q
Copyright© 2016 by Dennis C. Prieve
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Using the Divergence Theorem, the left-hand side can also be expressed as a volume integral
over the same domain:

V
1
.E dV 
0

 dV
V
Combining the two volume integrals:
 .
 
  E   dV  0
0 
V 

Since this integral must vanishe for all choices of the region V, the integrand must vanish at
every point in the region; thus
.E 

0
(12)
which is also called Gauss’s Law (differential form). Substituting (11) into (12)
2  

0
we obtain a single scalar equation in the electrostatic potential.
Equation.
This is called Poisson’s
BOUNDARY CONDITIONS AT A CHARGED INTERFACE
Consider a charged interface between two different dielectric
media. To obtain an appropriate boundary condition for such an
interface, we apply Gauss's law to a “pill-box” straddling the
interface. For a dielectric medium, the displacement vector D [
E + (1/0)P] replaces E in the integral form of Gauss's law given
by (2). After substituting Error! Reference source not found.,
we have D = rE and (2) becomes:
  E.n da  q
S
Let the area of the faces be Af and the thickness of the pill-box be
l (lI in Phase I and lII in Phase II). Applying the Mean Value
Theorem, we can write the integrals as
Copyright© 2016 by Dennis C. Prieve
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I
II
side
I
II
 E.n A f   E.n A f   E.n
Cl  A f   A f l I   A f l II
  E.n da
q
S
where the carets
denote the appropriate “mean” values. The surface integral was
decomposed into three contributions: the two plane faces and the curved side wall of the
cylinder; C is the perimeter of the face and Cl is the area of the side wall. Next, we will let the
thickness l of the pill box shrink to zero. Three of the five terms vanish (those proportional to l),
leaving
I
II
 E.n   E.n  
l  0:
after dividing out Af. If we now let Af be sufficiently small, we can replace the mean values by
the point values. Finally noting that nI = -nII  n, we obtain
(1E1  2 E2 ).n1  
Af  0:
where n1 is the unit normal pointing into medium 1. This implies that a discontinuity in the
normal component of E may occur at an interface.
To get a relation involving the tangential components of E, we can exploit the conservative
nature of the electric field. The existence of a scalar field, (x), such that:
E = -


 E.dx  0
means that
C
for all closed contours, C. For example, let's choose the contour
to be a thin rectangle straddling the interface and lying in the xy
plane, where the yz plane is the charged interface. On the vertical
paths dx = +eydy on the right side (direction is upward) and dx =
-eydy on the left side (direction is downward). On the horizontal
path, dx = -exdx on the top (direction is toward left) and dx =
+exdx on the bottom (direction is toward right). Applying the
Mean Value Theorem to evaluating the contour integral:



II
II
I
E.e y
y  E. e x  x II  E. e x  x I  E. e y

I
y 
I
II
 E.e x x I  E.e x x II  0
Letting x 0:
Copyright© 2016 by Dennis C. Prieve


I
II
E.e y
 E. e y
0
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after the y has been cancelled out.
Letting y 0:
E2y = E1y
Similiarly, we could have choosen our contour to lie in the xz plane, and show that:
E2z = E1z
Thus we have proved that the tangential components are continuous across a charged interface.
We can write this more compactly as a single vector equation:
(E1  E2 )  n1  0
These two equations relating the normal and tangential components of the electric field on either
side of a charged interface serve as boundary conditions in problems on electrostatics
Copyright© 2016 by Dennis C. Prieve
printed July 11, 2016 at 1:44 PM