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PREPARED BY
RAHUL SINHA-130280109107
INTRODUCTION
MAGNITUDE OF ELECTRIC FIELD INTENSITY
ELECTRIC FIELD DUE TO POINT CHARGE
PRINCIPLE OF SUPERPOSITION
ELECTRIC FIELD DUE TO N CHARGES
ELECTRIC FIELD LINES RULES & PATTERNS
GAUSS’S LAW
ELECTRIC FIELD DUE TO INFINITE LINE CHARGE
ELECTRIC FIELD DUE TO INFINITE SURFACE
CHARGE
In physics, the space
surrounding an electric charge
has a property called an
electric field.
The electric field exerts a force
on other electrically charged
objects.
The concept of an electric field
was introduced by Michel
Faraday.Michael Faraday
(1791-1867)
The magnitude of electric field intensity at any point in
electric field is given by force that would be
experienced by a unit positive charge placed at that
point.
UNIT: N/C OR V/M
FE
Q
The electric field at a point P due to a charge q is the
force acting on a test charge q0 at that point P,
divided by the charge q0 :
For a point charge q
0
0
q
FE p
q
ˆ( )4 ^ 2
q
qE p r
r
The total field at a point is the vector sum of the
individual component fields at the point.
The field intensity of the charge Q1 at the point P is E1
and field due to charge Q2 is E2.
The total field at P due to both charges is the vector sum
of E1 and E2.
When there are n number of charges,the field due to
each point charge is given by,
4 ^ 2r
QE a
r
By the principle of superposition electric field due to n
number of charges is,
1 2
1 21 22 2 2
1 2
21
...
...4 4 4
1
4
N
nr r n
n
ni
ri
i i
E E E E
QQ QE a a a
r r r
QE a
r
For a positive charge electric field lines radiate
outwards.
For a negative charge electric field lines point inwards.
Gauss’s law: The electric field at any point in
space is proportional to the line density at that point.
Line density s
DN
N
As
DD
Consider the field near a positive charge q.
Then imagine a surface of radius r surrounding the
charge q.
E is proportional tos
0
E
NE
A
NE
A
s
D
D
D
D
Gaussian Surface
Radius r
r
The proportionality constant for line density is known
as the permittivity and is given by
Recalling the relationship with line density we have:
212
0 2
18.85 10
4 .
C
k N m
0
0
NE
A
N E A
D
D
D D
Summing over the entire area A gives the total lines as
If we represent q as net charge then
0N EA
Gauss’s law: The net number of electric field
lines crossing any closed surface in outward
direction is numerically equal to the net total
charge within that surface.
0
qEA
0N EA q
Consider the infinite line
charge is placed along the
z-axis.
Consider the line charge as
axis of the cylinder in
cylindrical co ordinates
The point where we desire
the field is in xy plane at
point P.
Consider a small charge dQ on z-axis at point Q.
So,Vector from point P to Q
So , unit vector
l ldQ dl dz
zR a za
^ 2 ^ 2
z
R
a zaa
z
Thus ,the diffrential electric field due to dQ is
04 ^ 2R
dQdE a
R
04 ^ 2 ^ 2 ^ 2
za zadQdE
R z
04 ( ^ 2 ^ 2) ^ 2 ^ 2
zla zadz
dEz z
By solving above equation we get the equation of
electric field intensity due to infinite line charge
04 ( ^ 2 ^ 2) ^ (3 / 2)
l dzdE a
z
02
lRE a
R
Consider the infinite sheet
with charge density given in
the xy-plane.
The point where we desire the
electric field is on z-axis refer
fig given.
The vector from Q to P and its unit vector are given by,
The electric field due to small charge dQ on the sheet is
given by,
^ 2 ^ 2
z
z
R
R a za
a zaa
z
04 ^ 2R
dQdE a
R
Putting the value of R we get the equation
The total electric field is obtained by integrating dE we
get,
04 ( ^ 2 ^ 2) ^ 2 ^ 2
zsa zad d
dEz z
2
00 0
( )4 ( ^ 2 ^ 2) ^ 3 / 2
sz
d dE za
z
By solving the above equation we get the electric field
intensity due to infinite sheet charge,
02
snE a
Magnitude of electric field
Electric field due to point charge
ˆ( )4 ^ 2
q
qE p r
r
FE
Q
GAUSS’S LAW
Electric field due to infinite line charge
Electric field due to infinite surface charge
02
snE a
02
lRE a
R
0
qEA
http://en.wikipedia.org/wiki/Electric_field
http://en.wikipedia.org/wiki/Gauss%27s_law
http://en.wikipedia.org/wiki/Field_line
THANK YOU
