Gauss’s law:
0 . q s d E Here it is important to note that any closed surface, whatever its shape, would trap the same number of field lines, regardless of its size. So the flux through any surface enclosing the charge “q” is q/0. Now suppose instead of a single charge at origin, we have number of charges scattered about. According to principle of superposition, the total
field is simply
n i i E E 1
; Then the flux through any surface enclosing all the charges
n i n i o i i q s d E s d E 1 1 . . Where Qenc is the total charge enclosed within the surface. This is Gauss’s law in “Integral form”.
Gauss’s divergence theorem
surface volume
dv E s
d
E. . ...(2)
and Q dv ...(3)
volume
enc
Where is the volume charge density. Substituting eqn.2 and 3 in 1, we get Gauss’s
law in differential form
volume volume d d E 0 . Electric potential:
Since the integral of E “a” to “b” is independent of path, so we can define a function V depending only on the point P by
b a l d E PV( ) . ...(1)
1 1 ...(2)
4 . 0
ba rb ra
q l d E
. ( ) ( ) ( ) ( )
. ...(3)b a b a l d E a V b V a V b V l d
E
Using fundamental theorem of gradient
. ...(4)) ( )
(b V a V dl
V b a
b
a b a l d E l d
V
.
. EV ... (5)
From Gauss’s law
02 0 0 / . .
E V V
0 .
E
surface enc Q s dE. ...(1)
0
This is known as Poisson’s equation. In the regions where there is no charge, =0, Poisson’s equation reduces to 2V 0
. Which is called as Laplace’s equation.
Energy density in Electrostatic field:
To determine the energy present in assembly of charges, Let’s find the work done to assemble them. Let us first bring a charge Q1 from infinity to point P1 the work done will be W1 =0. Now for bringing the charge Q2 from infinity to point P2 the work done will be W2 = Q2 V21
Similarly W3 = Q3 (V32 - V31) (for Q3 charge) So the total work done in the assembly of these three charges will be
) 1 ( ... )
(
0 2 21 3 32 31 3
2
1 W W Q V Q V V
W
WE
If the charges were positional in reverse order, then
) 2 ( ... )
(
0 2 23 1 12 13 1
2
3 W W Q V Q V V
W
WE
Adding eq. (1) & (2)
2WE Q1(V12V13)Q2(V21V23)Q3(V32V31)
0r 2WE Q1V1Q2V2 Q3V3
where V1, V2 & V3 are the total potentials at point P1, P2 & P3, respectively. So
n k
k k
E Q V in Jules
W
1
) 4 ( ... 2
1
Instead of point charges, the region has continuous charge distribution
L
E Vdl line ch e
W ( arg )
2 1
S
E Vds surface ch e
W ( arg )
2 1
v
E Vd volume ch e
W ( arg ) ... (5)
2
1
From Gauss theorem we have
0 .
E .(0E) case,
tic electrosta
For Let us define D 0E
.D ... (6)
From (5) & (6)
v
E D Vd
W ( . ) ... (7) 2
1
From vector properties .(A)A.(.A)
or
.A .
A A. Using this property above eqn. will reduce to
v v
E VD d D V d
W ( . ) ... (8)
2 1 ) .( 2 1
v s
E VD ds D V d
W ( . ) ... (9)
2 1 ).
( 2
1
For a point charge 2
/ 1 &
/
1 r D or E r
V V.D 1/r3
and ds r2 this implies that in the eqn. (9) the first term will tend towards zero when r; means surface becomes very large.
) (
) 10 ( 2
1 .
2 1
sin )
. ( 2 1 )
. ( 2 1
0 2
0E d D E
d E D W
V E
ce d
E D d
V D W
v v
E
v v
E
Electrostatic energy per unit volume will be
0 2 2 0
2 2
1 . 2 1 .
D E E
D d
dW E
U E
ng
el
Note:
(1) Above study of electrostatic fields is carried out in free space or space that has no materials in it. It is known as “vacuum field theory”.
(2) For study of Electrostatic phenomena in material medium, most of the previous formulas are valid with some modifications.
(3) Materials can be classified in terms of their electrical properties as conductor, semiconductor or non – conductor (insulators or dielectrics).
(4) In broad sense, the materials may be classified in terms of conductivity , in mhos per meters or siemens/meter
metals high conductivity ( 1) insulators low conductivity ( 1) semi conductors in between
(5) Conductivity increases as the temperature decreases. At T00 K, some conductors exhibits infinite conductivity and they are called as supper conductors, e.g., Pb (40K 1020 mhos/m), Al
Electrostatic field in matter:
Dielectrics (or insulators) and Polarization: In a dielectric all charges are attached to specific atoms or molecules. They are not able to move freely. Finite forces find them and we may certainly expect a displacement when an external force is applied.
Let an dielectric atom with charge +q in nucleus is placed in an electrical field E. The positive charge is displaced from its equilibrium position in the direction of
E by force F qE while the negative charge is displaced in the opposite direction by the force F qE. “A dipole result from the displacement of charges
and the dielectric is said to be polarized”. A similar picture can be obtained for a
dielectric molecule. One can treat the nuclei is the molecules as a point charges and electronic.
Structure as a single cloud of negative charge: in the polarized state, applied field E
distorts the electron cloud. This distorted charge distribution is equivalent, by the principle of superposition, to the original distribution and a dipole whose moment is
E p or E p also d
q
p
where is called atomic polarizability and d is the distance between the two charges. If there are N dipoles in a volume of the dielectric, the total dipole moment due to the electric field is
n
i i i n
nd qd
q d
q d q d q
1 3
3 2 2 1 1
As a measure of intensity of the polarization, we define polarization P(coulombs/m2) as the dipole moment per unit volume of the dielectric as
As a conclusion we can say that the major effect of the electric field Eon a dielectric is the creation of dipole moments that align themselves in the direction of E. This type of dielectrics are called as non-polar, e.g., Hydrogen, Oxygen , Nitrogen and the rare gases . Note:
1. Non-polar dielectric molecules do not passes dipoles until the application of electric field.
2. There are certain molecules, which have built in permanent dipoles (randomly oriented). These are called polar dielectrics, e.g., H2O, S, dioxide, HCl etc.
3. When an electric field Eis applied to a polar molecule, the permanent dipole experiences a torque tending to align its dipole moment parallel withE.
The Electric displacement (
Gauss’s Law in the presence of Dielectrics
):
The effect of polarization produces accumulation of bound charge b .P (within thedielectric) and b Pn
.
(on the surface). The free charge might consist of electrons on a conductor or ions embedded in the dielectric materials or whatever (in other words the charge that is not result of polarization). So, within the dielectric, the total charge density can be written as
b f
and In total field E the Gauss’s law reads
E b f0 . .Pf
EP
f. 0
p d
d A Q d q P
n
i i
i
. ' lim 0 1
.D f
Where parameter D 0EP is known as the dielectric displacement. Hence in term of electron displacement, the Gauss’s law is
.D f
or, in integral form
surface
fenc
Q s d D.
where Qfenc denotes the total free charge enclosed in the volume . It is interesting to note that this eqn. Deals with only free charges so it becomes easy to use these formulas in case of dielectric
Linear Dielectrics (Susceptibility, Permittivity, Dielectric constant):
We know that the polarization of a dielectric ordinarily results from an electric field, which lines up the atomic or molecular dipoles. In many substances, in fact, the polarization is proportional to the field,
E P
E
P 0e ………(1)
The constant of proportionality, e is called the “electric susceptibility” of the medium. (Note: A factor 0has been extracted to make e depends on the microscopic structure of the substance in use. The materials that obey eqn.(1) is called as “linear dielectrics”. In linear media, we have
E E
P E
D 0 00e
D 0(1e)E ………(2)
So, not only PE but also DE
Hence D E ………(3)
Where 0(1e) ………(4)
is called the permittivity of the material. In a vacuum, where there is no matter to polarize, the susceptibility is zero, and the permittivity is 0. That is why we called 0 as
the permittivity of free space.
From eqn. (4) e K
1 0
………(5) Where K is called the dielectric constant.
Electrostatic boundary condition
:
Usually in electrostatic problem we have to find out the electric field E, produced by source charge distribution . When we deal with inhomogeneous medium (dependent on media), boundary conditions become important. The conditions that the field must satisfy at the interface reporting the media are called boundary conditions. These conditions are helpful in determining the field on one side of the boundary if the field on other side is known. In practice these types of situations occurs in case of reflections of waves at an
air dielectric interface, multiple dielectric capacitor and guiding of waves in metallic wave-guide. When field vector crosses the boundary surface between different media it is characterized by change in its magnitude and direction.
To determine the boundary condition, we need to use and (or D) relations (Maxwell’s electrostatic equations)
E.dl0 ………(1)
Dds Qfencand . ………(2)
Also we need to decompose Einto two components (orthogonal components)
n t E
E
E
Where Etand Enare the tangential and normal components of E to the interface of interest. A similar decomposition can be done for electric field density D.
Dielectric – dielectric boundary conditions: Consider the Efield, existing in a Region consisting of two different dielectrics is characterized by 1 0r1 and 2 0r2, denoted by medium (1)
& (2). So, E1
and E2
in these mediums can be decomposed as
n t E
E
E1 1 1 ………(3)
n t E
E
E2 2 2 ………(4)
Let us apply eqn. (1) to the closed path “a b c d a” (a cross section normal to the interface separating two media of different permittivity). We know the line integral along a closed path is zero
0
.
Edl
.
.
.
. 0a b c d
l d E l d E l d E l d
E
0 ) 2 2
( )
2 2
( 1 2 2 2 1
1
w E h E h E w E h E h
Et n n t n n ………(5)
Where Et Et
and En En
as h0, eqn. (5) will become
t t E
E1 2 ………(6)
Thus the tangential components of electric field are continuous on both sides of boundary. In other words there is no changes in the tangential part of electric field on the boundary.
Also D = Dt + Dn = E D1 = D1t + D1n = 1E1 & D2 = D2t + D2n = 2E2
n t n t
E E D D
E 1 1
1 1 1 1
1 ; 1 1 1 1 1 1
1 t n
n t
E E D D
E
& t n
n t
E E D D
E 2 2
2 2 2 2
1 1 ; 1
1 1 1
1 t n
n t
E E D D
E
& t n
n t
E E D D
E 2 2
2 2 2 2
2 ………(7)
Hence, from (6) & (7)
t t t t
D D
D D
2 1 1
2 2
2 1
1
………(8)
It indicates that Dt undergoes some change across the interface. Hence Dt is said to be discontinuous across the interface.
Similarly we will apply eqn.(2) to the pillbox (Gaussion surface)
Let us see first the Qfenc in the pillbox Qfenc = f s; with the condition h0. Here f is the free charge density placed deliberately at the boundary. Let us assume that E is directed from region2 to region1.
Therefore, on the pillbox
s
n n s D S
D s d
D. 1 2
Hence from Gauss’s law
s
fenc
Q s d D.
S S
D S
D1n 2n s D1n D2n s ………(9)
Where S is the surface charge density on the interface? Above equation does not
include the outflow of flux of D through the side S, because this flow can be made negligible by letting h0, while at the same time the terms in eqn. (9) remain unaffected. For a dielectric S 0, unless a surface charge is actually not placed at the interface
D1n = D2n ………(10)
Thus the normal component ofD is continuous across the interface. Since D E
n
n E
E1 2 2
1
………(11)
Showing that the normal component of Eis discontinuous at the boundary. Equation 6, 8, 10 & 11, are collectively called as boundary condition at the boundary separating two different dielectrics.
Refraction of the electric field across the interface: Let us consider vector E1and E2
making angle 1 and 2 with normal to the interface as illustrated in fig. From first boundary condition
E1 sin1 = E2sin2 ---(12) 1 E1 cos1 = 2 cos2 ---(13)
2 0
1 0 2
1 2
1 tan tan
r r
2 1 2
1 tan tan
r r
This is the law of refraction of the electric field at a boundary free of charge (since f = 0, is assumed at the interface). Thus, an interface between two dielectric produces bonding of the flux lines as a result of unequal polarization charges, which accumulate on the sides of the interface.
Magneto static Fields
In simple words, a magneto-static field is produced by a constant current flow (or direct current). This current flow may be due to magnetization currents as in permanent magnets, electron beam currents as in vacuum tubes, or conduction currents as in currents carrying curves.
Analogs: Most of the eqn. derived for electric field may be readily correlated to obtained corresponding equations for magnetic fields for equivalent emitted analogous quantities.
Term Electric Magnetic
Basic law : r
r q q
F 2
0 2 1 4
0 2
4 r r l Id B
d
(Coulomb’s law) (Biot-Savart’s law)
:
D.dsQfenc
H.dlIfenc(Gauss’s law) (Amper’s law)
Force law :F qE F IdlB Field Intensity :
m V l V
E
m A l I
H
Flux density :
2m c S
D
2
m S
B b
Relationship between fields :D 0E B0H
Flux :
D.ds
B.ds:
dt dV c
I
dt dI L V
Potential :
r dl V
0 4
r Idl A
4 0
Energy density :WE D.E
2 1
Wm B.H
2 1 Electrostatics : Source charge is at rest.
Magnetostatics : Charges are moving with constant velocity.
Hall Effect
When a current carrying thin conductor is placed in cross magnetic field, the magnetic field exerts a transverse force on the moving charge carriers. This force tends to push charges to one side of the conductor. This buildup of charge at the sides of the conductors will balance this magnetic influence, producing a measurable voltage between the two sides of the conductor. The presence of this measurable transverse voltage is called the Hall effect.
Let a magnetic field is applied perpendicular to the direction in which holes drift in a p – type bar. The total force on a single hole due to electric and magnetic fields is
E v B
q
F Fy q
Ey vxBz
Thus a +ve charge carrier experiences a net force qvxBz in y – direction due to magnetic field, unless an electric field Ey is established in y – direction and the holes are shifted slightly in -y direction. Therefore to maintain a steady state flow of holes along the length of the bar the net force Ey must be zero,
Ey vxBz.
The establishment of the electric field Ey is known as Hall effect, and the resulting voltage in –y direction Vy = Eyw is called the Hall voltage. Substituting the value of drift velocity of holes (vx = Jx/qp0) in above expression
z x H z x
y R J B
qn B J
E
0
Where RH 1(qn0) is called the Hall coefficient. A measurement of the Hall voltage for
a known current and magnetic field yields a value for the charge concentration n0.
y z x
y z x
y z x
H qtV
B I w qV
B wt I qE
B J qR n
/ / 1
0
Where w and t are the width and the thickness of the conductor, respectively. Thus measurement of Hall coefficient provide the following information
1. The sign of charge carrier. 2. The carrier concentration.
3. The mobility of charge carrier is measured directly.
4. Type of given material whether it is a conductor, semiconductor or insulator. 5. If RH is known then by measuring VH the unknown magnetic field can be
measured.
Ampere’s Law:
Now, suppose we have a bundle of straight wires. Each wire that passes through the loop would contribute 0I and those out side would contribute. The line integral then would be
………..(2)
Where Ienc is the total current enclosed by the integration path. This is known as Integral
form of Ampere’s law. If the flow of the charge is represented by a volume current density J, the enclosed current is
S enc Jds
I . ………..(3)
Now,from Stoke’s theorem
S l
l d B s d
B
.
. ………..(4)
so,from(2) & (4)
J B or
s d J s
d B
S S
0 0 . .
This is the differential form of the Ampere’s law and gives the Curl of magnetic field B. It states that the line integral of the tangential component of B around a closed path is the same as the net current 0Ienc enclosed by the path.
Magnetic field strength H:
The magnetic fields generated by currents are calculated from Ampere’s law or Bio – Savart Law. These fields are characterized by B(Tesla). But when generated fields are passes through magnetic materials which themselves contribute internal magnetic fields, another magnetic field quantity is characterized. It is usually called the magnetic field strength and designated by
M B B
H
0 0 0
B 0(H M)
where M is the magnetization induced in the magnetic material in applied magnetic field.
Magnetic Susceptibility and Permeability: In paramagnetism and diamagnetism, the field maintains the magnetization in a material; when Bis removed, M disappears. Actually, for most substances the magnetization is proportional to the field, provided the field is not too great.
MH
or M mH (1)
The constant of proportionality m is called the magnetic susceptibility. It is a dimensionless quantity that varies from one substance to another. It is positive for paramagnets and negative for diamagnets.
A material that obeys eqn. (1) is called a linear medium.
Since 0( ) (2)
0
B M B H M
H
From (1) & (2) B0(HmH)
H B H
B
H
B m
0(1 )
Where 0
1m
is called the permeability of the material. In vacuum 0 and 0 is called as the permeability of free space.We can also write as B0rH ; Where r 1m /0is called as the relative
permeability of the material. Above relationship holds only for the linear and for isotropic (non- crystalline) mediums (materials).
Classification of Magnetic materials:
Magnetic Materials
Linear Non-linear
Superconductor Diamagnetics Paramagnetics Ferromagnetics
m 0, r 1 m 0, r 1 m r
Temperature ~ 00k Bi, Pb, Cu, Si, Nacl Air, Pt, Sn, K iron, steel, Ni r Perfect diamagnetism m ~ -110-5 m ~ 10-5to 10-3 varies in a big
m = -1, r=0 range and is very
And B=0 sensitive to temp.
Boundary conditions:
To determine the boundary condition for the normal components of the magnetic flux density at the interface between the two regions, let us construct a Gaussian pillbox with vanishingly small thickness, as shown in Figure. Since the magnetic flux lines are continuous, we have
S
S d B. 0
where S is the entire surface of the pillbox. Neglecting the flux that flows through the vanishingly small thickness of the pillbox, this equation becomes
1
2 1
2
0 .
.
S S S
n n
S
ds B ds B S d B S d
B Bn1 Bn2 1Hn1 2Hn2
Thus normal component of B is continuous while for H is discontinuous across the boundary.
To obtain the boundary condition for the tangential components of the H field, consider the closed path shown in Figure. Applying Ampere’s law to the closed path, we obtain
C
I l d H. Applying Stoke’s law
0 ). (
.
.
S S
S d J H S
d J S d
H
[I
J.dS]where I is the total current enclosed by the closed path c. The path c2 and c4 are each of vanishingly small thickness, w0, and their contributions to the total mmf drop can be neglected. Thus, dropping these integrals, we have
1
2 1
2
. .
C S S
t t
C
I dl H dl H l d H l d
H
2 2 1
1 2
1
t t t
t
B B H
H
which states that the tangential components of the field at the boundary are discontinuous.
Refraction of the Magnetic field across the interface: Let us consider vector B1
and
2
B making angle 1 and 2 with normal to the interface as illustrated in fig. From first
boundary condition
B1 cos1 = B2 cos ---(12) 1 B1 sin1 = 2 B2 sin2 ---(13)
2 1 2
1 tan tan
2 1 2
1 tan tan
r r
