Emt Asea
Emt Asea
Emt Asea
by
Or
The charge element dQ and the total charge Q due to these charge
distributions can be obtained by
The electric field intensity due to each charge distribution ρL, ρS and
ρV may be given by the summation of the field contributed by the
numerous point charges making up the charge distribution.
Electric Field Intensity due to Line Charge
Consider a line charge with uniform charge density ρL extending from
A to B along the z-axis.
and
Electric Field Intensity due to Surface Charge
Consider an infinite sheet of charge in the xy plane with uniform
charge density ρS. The charge associated with an elemental area dS
is
The contribution to
Electric Field at Point P
(0, 0, h) by the elemental
surface is
Electric Field Intensity due to Surface Charge
So E has only z-
component
Electric Field Intensity due to Surface Charge
D oE
The electric flux ψ in terms of D can be defined as
The vector field D is called the electric flux density and is measured in
coulombs per square meter.
Electric Flux Density
Qenc
(i)
Using Divergence Theorem
(ii)
Hence
Applications of Gauss Law (Infinite Line charge)
Consider the infinite line of uniform charge ρL C/m lies along the z-
axis.
To determine D at a point P, consider a cylindrical surface centered at
Origin.
D is constant and normal to the cylindrical Gaussian surface.
Here
Applications of Gauss Law (Infinite Line charge)
Hence
Applications of Gauss Law (Infinite Sheet of charge)
Hence
or
Electric Potential
Electric Field intensity, E due to a charge distribution can be obtained
from Coulomb’s Law.
or using Gauss Law when the charge distribution is symmetric.
We can obtain E without involving vectors by using the electric scalar
potential V.
From Coulomb’s Law the force on point
charge Q is
F QE
The work done in displacing the charge
by length dl is
dW F .dl Q E.dl
The positive sign indicates that the work is being done by an external agent.
The total work done or the potential energy required in moving the
point charge Q from A to B is B
W Q E.dl
A
Dividing the above equation by Q gives the potential energy per unit
charge.
B
W
E.dl V AB
Q A
V AB is known as the potential difference between points A and B.
1. If V AB is negative, there is loss in potential energy in moving Q from
A to B (work is being done by the field), if V AB is positive, there is a
gain in potential energy in the movement (an external agent does the
work).
2. It is independent of the path taken. It is measured in Joules per
Coulomb referred as Volt.
The potential at any point due to a point charge Q located at the origin is
Q
V
4 o r
The potential at any point is the potential difference between that point
and a chosen point at which the potential is zero.
Assuming zero potential at infinity, the potential at a distance r from
the point charge is the work done per unit charge by an external agent
in transferring a test charge from infinity to that point.
r
V E.d l
If the point charge Q is not at origin but at a point whose position
vector is r ' , the potential V ( r ' ) at r ' becomes
Q
V (r )
4 o | r r ' |
For n point charges Q1, Q2, Q3…..Qn located at points with position
vectors r 1 , r 2 , r 3 .....r n the potential at r is
n
1 Qk
V (r )
4 o
| r r
k 1 |
k
If there is continuous charge distribution instead of point charges then
the potential at r becomes
E.d l 0 (i)
It means that the line integral of E along a closed path must be zero.
Physically it means that no net work is done in moving a charge along
a closed path in an electrostatic field.
Applying Stokes’s theorem to equation (i)
E.d l ( E).d S 0
E 0 (ii)
Equation (i) and (ii) are known as Maxwell’s equation for static
electric fields.
Equation (i) is in integral form while equation (ii) is in differential
form, both depicting conservative nature of an electrostatic field.
Also
E V
It means Electric Field Intensity is the gradient of V.
The negative sign shows that the direction of E is opposite to the
direction in which V increases.
Electric Dipole
If r >> d, r2 - r1 = d cosθ
and r1r2 = r2 then
But d cos d .ar where d d a z
(i)
= -
Put this in (i) and integrate over the entire volume v’ of the dielectric
(ii)
where an’ is the outward unit normal to the surface dS’ of the dielectric
The two terms in (ii) denote the potential due to surface and volume
charge distributions with densities
where ρps and ρpv are the bound surface and volume charge densities.
Bound charges are those which are not free to move in the dielectric
material.
Equation (ii) says that where polarization occurs, an equivalent
volume charge density, ρpv is formed throughout the dielectric while
an equivalent surface charge density, ρps is formed over the surface of
dielectric.
The total positive bound charge on surface S bounding the dielectric is
Total charge =
Hence
Where
The effect of the dielectric on the electric field E is to increase D inside
it by an amount P .
The polarization would vary directly as the applied electric field.
We know that
and
Thus
or
where
o r
and
(i)
But
Equation (i) now becomes
or (ii)
or
where
and
Boundary Conditions (Between two different dielectrics)
But
Thus the tangential components of E are the same on the two sides of
the boundary. E is continuous across the boundary.
But
Thus
or
Putting Δh 0 gives
or
(surface current)
(volume current)
Ampere’s circuit Law
The line integral of the tangential component of H around a close path
is the same as the net current Iinc enclosed by the path.
But
Comparing we get
(i)
(ii)
where Ho is to be determined.
Evaluating the line integral of H along the closed path
(iii)
Comparing (i) and (iii), we get
(iv)
where an is a unit normal vector directed from the current sheet to the
point of interest.
Magnetic Flux Density
where the magnetic flux ψ is in webers (Wb) and the magnetic flux
density is in weber/ square meter or Teslas.
Magnetic flux lines due to a straight
wire with current coming out of the
page
Each magnetic flux line is closed
with no beginning and no end and
are also not crossing each other.
or
Now we define
The magnetic dipole moment is the product of current and area of the
loop and its direction is normal to the loop.
We can apply this equation for loop of any arbitrary shape. The only
limitation is that the magnetic field should be uniform.
Faraday’s law
According to Faraday a time varying magnetic field produces an
induced voltage (called electromotive force or emf) in a closed circuit,
which causes a flow of current.
The induced emf (Vemf) in any closed circuit is equal to the time rate of
change of the magnetic flux linkage by the circuit. This is Faraday’s
Law and can be expressed as
where N is the number of turns in the circuit and ψ is the flux through
each turn.
The negative sign shows that the induced voltage acts in such a way to
oppose the flux producing in it. This is known as Lenz’s Law.
Transformer and Motional EMF
(i)
Thus
also
Displacement Current
For static EM fields
(i)
But the divergence of the curl of a vector field is zero. So
(ii)
But the continuity of current requires
(iii)
or (vi)
Putting (vi) in (iv), we get