Electrostatics
1
Introduction
Electrostatics is the study of the properties of stationary charges, where the charges do
not change with time. It also assumes that there is no magnetic field. The study of
electrostatics is critical to understand the fundamental principles of electromagnetics.
Many phenomena such as lightening, corona, applications such as oscilloscope, ink-jet
printer, xerography have their working based on electrostatics.
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Coulomb’s Law and Electric Field Intensity
Coulomb’s law deals with the force exerted by a point charge on another point charge.
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−
The law states that the force F between two point charges with charge Q1 and Q2 is 1. along the line joining them,
2. directly proportional to the product, Q1 Q2 , of the charges and
3. inversely proportional to the square of the distance, R, between them.
Mathematically the force experienced by Q2 due to Q1 is represented as,
→
−
kQ1 Q2
F =
âr
R2
where k is a proportionality constant whose value is 9 × 109 m/F, âr is the unit vector
along the direction from Q1 to Q2 and R is the distance between the charges. Force
between like charges is repulsive, while that between unlike charges is attractive.
Thus, a stationary charge creates a force-field around it. The strength of the electric
field is quantified through electric field intensity. The electric field intensity is the force
experienced by a small stationary test charge when placed in an electric field. The test
charge should be small such that its presence doesn’t change the electric field. If q is the
test charge, then the electric field intensity is given by
→
−
→
−
F
E = lim
q→0 q
→
−
The electric field intensity is measured in V/m. The force, F experienced by a stationary
charge in an electric field is
→
−
→
−
F = qE,
1
3
Fundamental Postulates of Electrostatics
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−
E is a vector function that satisfies the following two relations - these are the fundamental
postulates of electrostatics.
→
− →
−
ρ
∇. E =
0
(1)
→
− →
−
∇ × E = 0,
(2)
and
where ρ is the volume density of free charges and 0 is the permittivity of free space.
These postulates hold good at every point in space; these are also referred to as ’point
form’ or ’differential form’, since they involve divergence and curl operators.
1
× 10−9 F/m is the permittivity of free space. Eqn.(1) relates the divergence
0 = 36π
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−
of electric field intensity to the charge density and eqn.(2) implies that E is irrotational.
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Electric Flux and Gauss’ Law
The total field due to a distribution of charges can be obtained from an integral form of
eqn.(1). Taking a volume integral of eqn.(1) on both sides,
Z
Z
→
− →
−
1
∇. E dV =
ρdV
0 V
V
. Using divergence theorem,
Z
→
− →
−
∇. E dV =
I
→
→
− −
E .dS.
V
Also,
Z
ρdV = Q
V
where Q is the total charge contained in the volume V enclosed by the surface S. Thus,
I
→ Q
→
− −
E .dS =
(3)
0
S
Eqn (3) is referred to as Gauss’ law which states that the net outward flux of an
electric field over any closed surface in free space is equal to the total charge enclosed by
the surface divided by 0 . Eqn.(1) represents the point form of Gauss’ law. Note that
the surface represented in eqn (3) need not be a physical surface.
2
5
Coulomb’s Law from Gauss’ Law
Coulomb’s law can be derived from the fundamental postulates. Consider a point charge
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−
q, at rest, in free space. To find E due to the charge, draw a hypothetical sphere of
radius R centered at q. This surface is chosen based on the symmetry of the problem.
Since a point charge has no preferred direction, its electric field must be radial and it
should have the same intensity at all points on the spherical surface. Using the point
form of Gauss law to this problem,
I
→
→
− −
q
E .dS =
0
S
−
→
→
−
E = Er âr ; dS = dsâr
I
Z
−
→
− →
q
⇒
E .ds = Er dS = Er (4πR2 ) =
0
S
→
−
1 q
E =
âr V /m
4π0 R2
So the force experienced by a unit charge is proportional to the charge and inversely
proportional to the square of the distance from the charge. When a point charge q2 is
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−
placed in the field of another point charge q1 , the force experienced by q2 due to E due
to q1 is
→
−
q2 E =
1 q1 q2
âr
4π0 R2
This is Coulomb’s law. Thus, Coulomb’s law and Gauss law are equivalent.
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Electric Potential
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−
We now know two ways to find the electric field intensity E , namely through Coulomb’s
law and Gauss’ law. It is convenient to use Gauss’ law where the charge distribution
is symmetrical; Coulomb’s law is used otherwise. Another way of obtaining the electric
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−
field E is from the electric scalar potential,V. The scalar potential V is defined such that,
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−
E = −∇V . Scalar functions are easier to handle than vector functionsl so if we are able
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−
to find V, E can be found from a gradient operation. The physical significance of electric
potential is defined as follows.
Suppose we want to find the work done to move a point charge Q from point A to
point B in an electric field. The work done against the field is,
Z B
−
→
− →
W =−
F . dl
A
Z B
−
→
− →
= −Q
E . dl
A
3
Figure 1: Work done in an electric field
If we divide W by Q, we get the potential energy per unit charge.
Z B
Z B
−
→
−
→
− →
E . dl =
(∇V ). dl
−
A
A
Z B
=
dV = VB − VA
= VAB
A
This quantity denoted by VAB is known as the potential difference between points A and
B. Thus,
Z B
−
→
− →
W
VAB =
=−
E . dl
Q
A
VAB can be regarded as the potential at B with respect to A. In case of point charges,
the reference A is usually taken to be infinity where the potential is zero. Thus, the
potential at any point is the potential difference between that point and a chosen point(or
reference point) at which the potential is zero.
From the integral, we can see that the potential doesn’t depend on the path taken
between A and B. Such fields are known as conservative fields.
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Conductors in a Static Field
Conductors are characterized by the presence of a large number of charges, free to move.
When an external field is applied, the positive free charges are pushed in the direction
of applied field while the negative charges move in the opposite direction. This charge
migration happens very quickly, due to which,
(i) they accumulate on the surface of the conductor thus forming an induced surface
charge and
(ii) they set up an internal field which is equal and opposite to the external applied
field.
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Thus, a perfect conductor (with infinite conductivity) cannot contain an electrostatic
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−
field within it. According to Gauss’ law, if E = 0, the charge density ρ should be zero,
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−
inside the conductor. Also, E = −∇V ; and hence the conductor is an equipotential
body - where the potential is the same everywhere on the surface.
Figure 2: Isolated conductor in the presence of applied field; the net field inside the
conductor is zero under static equilibrium
→
−
When a potential difference is maintained at the two ends of a conductor, then E 6= 0
inside the conductor. The external source of potential forces the charges to move in a
specific direction. The damping force experienced by the electrons as they move in the
conductor is the reason for resistance.
Figure 3: Conductor of uniform cross-section under an applied field E due to a potential
difference V
Conduction current is the rate of flow of charges, defined as
I=
dQ
dt
If ∆I is the current through a planar surface, ∆S, the current density is, J =
assuming the current density is perpendicular to the surface.
Z
→
→
− −→
→
− −
∆I = J .∆S; I =
J .dS
S
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∆I
,
∆S
→
−
→
−
→
−
Thus, I is the flux of the current density, J . It can be shown that, J = σ E , and this
relation is Ohm’s law, obeyed by all conduction currents.
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Dielectrics in Static Field
Dielectrics have bound charges and an electric field causes a polarization of these bound
charges, thus creating a electric dipole. Though electric dipoles can exist in a dielectric
media even in the absence of external field, they are randomly oriented. The volume
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−
density of electric dipole moment is defined as polarization, P . These dipole moments
inside the dielectric produces an electrostatic potential - contributed from the bound
surface charge and volume charge distribution, inside the dielectric. It can be shown
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−
→
− →
−
that, the bound volume charge density ρv is related to P as, ρv = − ∇. P .
Applying Gauss’ Law,
→
−
→
−
∇.(0 E ) = (ρ + ρv )
→
− →
−
= (ρ − ∇. P )
→
− →
−
→
−
∇.(0 E + P ) = ρ
→
− →
−
∇. D = ρ
→
−
→
−
→
−
Here ρ is the volume density of free charges. D = 0 E + P is defined as the electric
displacement or electric flux density. Thus, the effect of dielectric on electric field is to
→
−
→
−
→
−
increase the flux density inside the dielectric. In free space, P = 0, so D = 0 E . If
→
−
→
−
the dielectric medium is linear and isotropic, then P = 0 χ E where χ is the electric
susceptibility.
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