CHAPTER
1
Electrostatics: Electric
Field and Potential
1.1 ELECTRIC CHARGE
Electric charges possess two basic properties—(i) charge is quantized and
(ii) charge is conserved. The quantisation of charge is a universal law of nature.
The fundamental smallest charge found in nature is denoted by e. Its value is
1.602 × 10–19 C. All other charges, no matter what their origin is, are integral
multiples of e. The electron has a charge –e, while the proton charge is +e.
The charge conservation law states that the total electric charge in an
‘isolated’ system never changes. This is an empirical law supported by various
experimental observations, e.g., those connected with pair-creation and
annihilation processes, radioactive decays, nuclear reactions, etc.
1.2 COULOMB’S LAW
Electrostatics is the study of the physics of stationary electric charges. The
interaction between such charges is described by Coulomb’s law.
Coulomb’s law states that “two stationary point charges repel or attract one
another with a force proportional to the product of the magnitude of the charges
and inversely proportional to the square of the distance between them”. This
force also depends on the nature of the medium in which the charges are placed.
Mathematically, Coulomb’s law can be stated as follows:
qq
F= k 1 2
r2
where k is a positive constant and r is the distance between the point charges
q1 and q2. The force F acts along the line joining the two charges and is directed
away from each charge (repulsion) for like charges. If the charges have opposite
signs, the force is directed towards each charge (attraction).
2
ELECTRICITY AND MAGNETISM
In SI units, the distance r is in meters (m), the force F is in newtons (N),
and the unit of charge is the Coulomb (C). In practical work, the microcoulomb
(1 µ C = 10–6 C) and the nanocoulomb (1 nC = 10–19 C) are frequently employed.
In rationalised SI units, the coulomb constant k for charges in free space (or
vacuum) is given by
k=
1
= 9 × 109 Nm2C–2
4 π ∈0
where ∈0 = 8.85 × 10–12 C2N–1 m–2 is called the permittivity of free space. An
equivalent unit for ∈0 is farads per meter (F/m). Thus Coulomb’s law becomes
(for vacuum)
F=
1 q1 q2
...(1.1)
4 π ∈0 r 2
In vector form, F2, the force on charge q2 due to charge q1, is given by.
F2= k
q1 q2
r2
rˆ =
1 q1 q2
r ,...(1.2)
4 π ∈0 r 3
where r is the vector displacement from q1 to q2 and rˆ. =
r
is the unit vector in
r
the direction of r. The force is Newtonian and F2 = – F1
When the surrounding medium is not a vacuum, ∈0 in Coulomb’s law must
be replaced by ∈ = K∈0, where ∈ is called the permittivity of the material and
K the dielectric constant. Now, Eq. (1.1) becomes
F=
1 q1 q2
...(1.3)
4π ∈ r 2
For vacuum, K = 1 and for air, K = 1.0006. Since K = ∈/∈0 represents the
ratio of the permittivity of the medium to the permittivity of free space, it is
often called the relative permittivity of the medium (∈r). Thus K = ∈r = ∈/∈0, or
∈ = ∈0∈r.
The total force exerted upon a point charge qj by a set of N point charges qi
is, from the principle of superposition,
N
Fj= ∑ Fij =
i=1
qj
4 π ∈0
N
∑
i =1
qi
rij3
rig
(i ≠ j) ...(1.4)
1.3 THE ELECTRIC FIELD
A charged body modifies the state of affairs in the space around it and produces
what is known as an electric field. Suppose, we have some arrangement of
ELECTROSTATICS: ELECTRIC FIELD AND POTENTIAL
3
charges, q1, q2, ..., qN, fixed in space and we are interested in their effect on a
test charge q0 which is located at a point (x, y, z) in the vicinity of the above
charge configuration. The force on the charge q0 is:
F0=
1
4 π ∈0
N
∑
q0 q j
3
j =1 r0 j
r0 j =
q0
4 π ∈0
N
qj
∑ r3
j =1 0 j
r0 j ...(1.5)
where r0j is the vector from the jth charge to the point (x, y, z). The charges q1,
..., qN are called sources of the field.
The electric field intensity E at the point (x, y, z) due to the above source is
a vector defined by the equation.
E (x, y, z)=
F0
1
=
q0 4 π ∈0
N
qj
∑ r3
j =1 0 j
r0 j ...(1.6)
The electric field intensity at a given point in an electric field is the force per
unit charge, in magnitude and direction, that a positive test charge q0 would
experience at that point in the field.
It is assumed in the above definition that the test charge is small enough
to ensure that it does not disturb the electric field due to the existing charge
configuration. This is expressed by writing E in the form:
E= lim
q0 → 0
F0
...(1.7)
q0
The SI units of E are newtons per coulomb (N/C) or the equivalent, volts
per meter (V/m). If a charge q is placed at a point where the electric field due to
other charges is E, the charge will experience a force F given by
F=qE...(1.8)
The electric field at a distance r from a point charge q is obtained by using
Coulomb’s law:
E=
1
q
...(1.9)
4 π ∈0 r 2
I f the source of electric field is a continuous charge distribution rather
than point charges, we must replace the sum in Eq. (1.6) with the appropriate
integral. The electric field at (x, y, z) is now given by (Fig. 1.1)
E (x, y, z)=
1
4 π ∈0
∫
ρ( x ′, y′, z′) r dx ′ dy′ dz′
...(1.10)
r3
This is a volume integral ρ is called the volume charge-density function and
4
ELECTRICITY AND MAGNETISM
ρ(x′, y′, z′) dx′ dy′ dz′ is the charge contained in the little box of volume dx′ dy′
dz′ located at the point (x′, y′, z′). ρ is measured in units of C/m3.
z
Field point
(x, y, z)
r
dxdydz
(x, y, z)
Charge
distribution
x
o
y
Fig. 1.1
1.4 ELECTRIC FLUX
Flux is a property of any vector field. The electric flux can be used to express
the relation between the electric field and its sources.
Figure 1.2 shows a closed surface S of any shape in an electric field. The
whole surface is divided into a number of small patches, each considered as
a vector quantity. For the patch number j, we have a vector aj giving its area
and orientation. Let Ej be the electric field vector at the location of the patch
number j. The scalar product Ej . aj is a number, called the electric flux through
the patch j. The electric flux through the entire surface is a scalar quantity
denoted by Φ:
Φ=
∑ E j . a j ...(1.11)
all j
If the patches are infinitesimally small in area and very large in number,
we can replace the sum in Eq. (1.11) by a surface integral:
Fig. 1.2
ELECTROSTATICS: ELECTRIC FIELD AND POTENTIAL
5
Φ= ∫s E.da ...(1.12)
Thus electric flux is the surface integral of the electric field vector.
1.5 GAUSS’S LAW
Gauss’s law is another basic law of electrostatic interactions. The integral form
of Gauss’s law can be stated, in SI units, as follows:
“The total outward electric flux through any closed surface is equal to
the total charge contained within that surface divided by the free-space
permittivity.”
For a closed surface S enclosing a charge q (Fig. 1.3), Gauss’s law may be
stated mathematically as:
Fig. 1.3
q
∫s E . dS = ∫s En dS = ∈0 ,...(1.13)
where En represents the normal component of the electric field at the position
of the surface element ds. En is positive wherever E emerges from the surface
and negative wherever E enters the surface. The closed surface S on which
Gauss’s law is applied is known as the Gaussian surface. If the Gaussian
surface encloses both positive and negative charges, their algebraic sum must
be taken to apply Gauss’s law.
In the case of continuous distribution of charges enclosed by a surface,
Gauss’s law may be written as:
1
∫s E . dS = ∈0 ∫v ρdv, .
where ρ is the volume charge density. Applying divergence theorem:
∫s E . dS = ∫v ∇. E dv
6
ELECTRICITY AND MAGNETISM
we have
1
∫v ∇. E dv = ∈0 ∫ ρdv,
v
or,

ρ
∫v  ∇.E − ∈0  dv =0
Since dv is arbitrary, we have:
or
∇. E −
ρ
= 0,
∈0
∇. E=
ρ
...(1.14)
∈0
Equation (1.14) is known as the differential form of Gauss’s law.
Gauss’s law can be used to calculate electric field in situations where the
symmetry of the charge distribution is high.
(a) Field due to a uniformly charged sphere:
This is a case of spherically symmetrical distribution of charge. Consider a
sphere of radius R having a uniform volume charge density ρ. The electric field
outside the sphere at a distance r from the center is given by
E=
ρR3
3 ∈0 r 2
(r > R) ...(1.15)
The electric field inside the sphere at a distance r from the center is given
by
E=
ρr
(r < R) ...(1.16)
3 ∈0
(b) Field due to a uniformly charged infinite cylinder:
This is a case of linear charge distribution. Consider an infinite cylinder of
radius a having a uniform linear charge density λ. The electric field outside the
cylinder at a distance r from its axis is
E=
λ
(r > a) ...(1.17)
2πr ∈0
The electric field inside the cylinder at a distance r from its axis is
E=
λr
(r < a) ...(1.18)
2πa2 ∈0
ELECTROSTATICS: ELECTRIC FIELD AND POTENTIAL
7
(c) Field due to an infinite plane sheet of charge:
Electric charge distributed uniformly in a thin sheet is called a surface charge
distribution. Consider a flat sheet infinite in extent, with the constant surface
charge density σ. The electric field is independent of the distance from the
sheet. It is constant on, both sides of the sheet, but oppositely directed. The
field is
E=
σ
...(1.19)
2 ∈0
(d) Field close to a charged insulated conductor:
The field strength at a point close to the surface of a charged conductor has a
constant value:
E=σ/∈0...(1.20)
where σ is the surface density of charge on the conductor. Eq. (1.20) is often
called Coulomb’s theorem.
(e) Electrical stress on the surface of a charged conductor:
The electrical stress on the surface of a charged conductor of surface density σ
is given by
stress=σ2/2∈0...(1.21)
1.6 ELECTRIC POTENTIAL
The electric potential at a point in an electric field is defined as the amount of
work done against electric force in carrying a unit positive charge from a given
reference point (usually infinity) to that point. The electric potential is thus
the electric potential energy, with respect to infinity, of a unit positive charge
and is a scalar quantity. The potential at a point P at a distance r from a point
charge q in free space is given by
V=
1 q
...(1.22)
4 π ∈0 r
The potential difference between two points in a given electric field is the
work done against electrical force in moving a unit positive charge from point
1 to point 2. If W is the work done in carrying a charge q from point 1 to point
2, then
W=q(V2 – V1) = qV...(1.23)
where V is the potential difference.
The SI unit of potential or potential difference is volt.
8
ELECTRICITY AND MAGNETISM
1 volt=1 joule/coulomb
= 1
or,
newton × meter
coulomb
1V=1 JC–1 = 1 NmC–1
Relation between electric field and potential
The electrostatic field is the negative gradient of the potential function:
 ∂φ
∂φ ˆ ∂φ
j+
E= −  iˆ +
 ∂x
∂y
∂z

kˆ  = −∇φ, ...(1.24)

where φ is the potential function. The SI unit of E is thus Vm–1. The components
of E in different coordinate systems are:
(i) Cartesian coordinates (x, y, z):
Ex= −
∂φ
∂φ
∂φ
...(1.25)
, E y = − , Ez = −
∂x
∂y
∂z
(ii) Spherical polar coordinates (r, θ, ϕ):
Er= −
∂φ
∂φ
1 ∂φ
1
, Eθ =
, Eϕ = −
...(1.26)
∂r
r ∂θ
r sin θ ∂ϕ
(iii) Cylindrical polar coordinates (r, θ, z):
Er= −
∂φ
1 ∂φ
∂φ
, Eθ =
, Ez = −
...(1.27)
∂r
r ∂θ
∂z
Potential of a charge distribution
The potential at any point, in the field of an isolated point charge q, is given by
Eq. (1.22). Because potential is a scalar quantity, the absolute potential at a
point due to several point charges is given by, using the principle superposition,
V=
1
4 π ∈0
N
q
∑ ri , ...(1.28)
i =1 i
where ri is the distance of the charge qi from that point.
If the charge distribution is continuous, the potential is specified by the
scalar integral (See Fig. 1.1):
V(x, y, z)=
1
4 π ∈0
∫all sources
ρ( x ′, y′, z′) dx ′ dy′ dz′
...(1.29)
r
where r is the distance from the volume element dx′ dy′ dz′ to the field point
(x, y, z) at which the potential is being evaluated:
ELECTROSTATICS: ELECTRIC FIELD AND POTENTIAL
9
1/ 2
r= ( x − x ′)2 + ( y − y′)2 + ( z − z′)2 


Equation (1.29) can be applied to calculate the electric potential and
field around a uniformly charged disk. Let a be the radius of the disk and σ
the uniform surface charge density. The potential at a point P on the axis of
symmetry of the disk is given by
V=
σ  2
a + r 2 − r  ,...(1.30)

2 ∈0 
where r is the distance of the point P from the center of the disk. By symmetry,
the electric field E at P will be directed along the axis of the disk and is given by
E=
σ
2 ∈0

1 −



 ...(1.31)
a2 + r 2 
r
1.7 ENERGY OF A SYSTEM OF CHARGES
The electric potential energy of a system of point charges is defined as the work
required to assemble this system of charges by bringing them in from infinity.
It is assumed that the charges are all at rest when they are infinitely separated,
i.e., they have no initial kinetic energy.
q1
q2
r12
r12
q1
q2
r23
r31
q3
Fig. 1.4
Consider two charges q1 and q2, a distance r12 apart (Fig. 1.4). The electric
potential energy U of the system (q1 + q2) is given by
U=
1 q1 q2
...(1.32)
4 π ∈0 r12
For systems containing more than two charges the procedure is to compute
the potential energy for every pair of charges separately and to add the results
algebraically. Thus for the system of three ch~arges q1, q2, q3 in the specified
10
ELECTRICITY AND MAGNETISM
configuration, the electric potential energy is given by
U=
1
4 π ∈0
 q1 q2 q2 q3 q3 q1 
+
+

 ...(1.33)
r23
r31 
 r12
In general,
U=
1 N
∑
2 j =1
∑
The double-sum notation,
q j qk
k≠ j
rjk
.
1
...(1.34)
4 π ∈0
N
∑ ∑ , says: Take j = 1 and sum over k = 2, 3,
j =1 k≠ j
4,..., N; then take j = 2 and sum over k = 1, 3, 4, ... N; and so on, through j = N.
Clearly this includes every pair twice, and to correct for that we put in front
the factor
1
.
2
1.8 ELECTRIC DIPOLE
Two equal and opposite charges separated by an infinitesimally small distance
constitute an electric dipole.
Consider a dipole AB with +q charge at A and –q charge at B, separated by
a distance 2a (Fig. 1.5).
Fig. 1.5
The electric dipole moment p is a vector quantity defined by
p=2 a q
and points from the negative charge to the positive charge.
The electric potential at a point P (r, θ) in space due to the dipole is given by
V=
1
p cos θ
3 ...(1.35)
.
4 π ∈0
r2