

PHYSICS11
Chapter 29  Electrostatics
Definations:
1. Electric field strength: It is defined as the field or space
upto which force due to electric charges is exert.
Mathematically, electric field strength is electrostatic force
per unit charge.
= /q
= (1/4πε_{0})(q.q_{1}/r^{2}.q)
therefore, E= (1/4πε_{0})(q_{1}/r^{2})
if test charge is very small then,
E= q
0 (F/q)
2. Superposition of electric field intensity: If there are
various electric field intensity due to multiple charges then
total intensity can be calculated
by the vector sum of various intensity.
Let, E_{1}, E_{2}, E_{3}, ..... are
various electric field acting at a source charge then total
intensity is
=
_{1}
+ _{2} +
_{3}
+ _{4} +
.....
Electric flux
It is defined as electric lines of force/electric field
passing through a surface area held perpendicular to the
direciton of
electric field.
Let, 'E' be an electric field due to any charge at a point
then if it passes through a surface area A which is
perpendicular to the
field then electric flux (ɸ) is defined as,
ɸ= E.A < (i)
Electric flux due to a point charge:
Consider a point charge Q with electric field E which
exerts electric force uniformly in all direction. Then
from definition of electric
flux,
(1/4πε_{0})(q_{1}/r^{2})= E.A
ɸ= E x 4πr^{2}
< (ii)
Also we have,
Electric field intensity, E=
(1/4πε_{0})(q/r^{2}) < (ii)
From equation (ii) and equation (iii),
ɸ= (1.q/4πε_{0}r^{2})
ɸ= q/ε_{0}
which is the electric flux due to the point
charge in the space.
Gauss law
The total flux passing through any closed surface is
equal
to 1/ε times of the total charge enclosed in the
surface.
Mathematically, it is the ratio of total charge present
in
the given surface to permittivity.
ɸ= Q/ε (for medium)
ɸ= Q/ε_{0} (for air/space)
Application of Gauss law:
1. Electric field due to point charge at surface of
sphere.
Consider a sphere having radius 'r' and 'q' be the
charge
present at any point of the sphere (+ve charge) here we
are
going to find
the electric field intensity at the surface of this
sphere
due to charge 'Q'.
From the defination of electric flux,
ɸ= E.A
ɸ= E.4πR^{2}
< (i) [where Area=4πr^{2}]
Again, from the definition of gauss law,
ɸ= Q/ε_{0}
< (ii)
from equation (i) and equation (ii)
Q/ε_{0} = E.4πR^{2}
E= (Q/ε_{0}4πR^{2})
therefore, E=
(1/4πε_{0}).(Q/R^{2})
< (iii)
2. Electric field due to point charge
outside
the sphere:
Consider a sphere of radius R. Let, r be the
distance from center of sphere to the point
outside the sphere where electric field is
to be
calculated.
From the definition of electric flux,
ɸ= E.A [where E is electric field]
A= area of sphere upto the distance 'r'
outside
the surface of sphere
ɸ= E.4πr^{2}
< (i)
Also, from gauss law,
ɸ= Q/ε_{0}
< (ii)
from equation (i) and equation (ii)
Q/ε_{0} = E.4πr^{2}
therefore, E=
(Q/4πr^{2}ε_{0})
Conclusion: This equation indicates
that
the charge is uniformly distributed
over
the surface of sphere. It behaves as
if
the whole charge is concentrated
as the centre of sphere.
3. Electric field inside the
sphere:
Consider a sphere of radius 'r'.
Let, Q
be the uniformly distributed charge
over
the sphere. Now we have to calculate
electric field intensity inside the
sphere.
From the definition of electric
flux,
ɸ= E.A [where E is electric
field]
A= area of sphere and ɸ is electric
flux.
ɸ= E.4πR^{2}
< (i)
Also, from gauss law,
ɸ= Q/ε_{0}
< (ii)
from equation (i) and
equation
(ii)
Q/ε_{0} =
E.4πR^{2}
E=0 then, Q=0
No charges are present
inside
the sphere. Therefore total
electric field inside the
sphere
is always 0.
4. Electric field outside
plane charged sheet:
Consider a plane charged
sheet,
where a circle is drawn
outside
where the electric field is
to
be calculated. The electric
field is uniform in all part
of
this circle.
Since, electric field is
perpendicualr to the charged
sheet but parallel to the
region
where electric field
intensity
is to be calculated.
Let, ɸ be an electric flux
due
to charges then,
ɸ= E.A < (i)
where A is area of
cylinder
Again from gauss law of
electrostatics,
ɸ= Q/ε_{0}
< (ii) [where Q is charge present and ε_{0} is
permittivity in
space.]
From equation (i)
and
equation (ii)
E.A= Q/ε_{0}
E= Q/ε_{0}A
<(iii)
From the
definition
of surface
charge
density
σ= Q/A
A=Q/σ
therefore,
E=σ/ε_{0}
5. Electric
field
due to one
dimensional
(linear)
charged
rod:
Consider an
infinite
charged
conducting
rod of linear
charge
density (λ).
Let,
'p' be any point
'r'
distance away
from
conducting rod.
By symmetry at
any
all other point
at
same distance
from
conducting rod,
electric field
intensity is
uniform. So, we
have
considered a
cylinder
of length 'L' as
shown in figure
above.
Now applying
Gauss
law,
ɸ=
Q/ε_{0}
< (i)
Also from
definition
of
electric
flux
ɸ= E.A
From
equation
(i) and
equation
(ii)
E.A=
Q/ε_{0}
therefore,
E=
λ/2πε_{0}r
6.
Electric
field
intensity
due to
infinite
plane
sheet
of
charge:
Consider a
plane
sheet of
surface
charge
density
'σ'. Let,
'P' be
any point at
distance 'r'
from sheet
where
electric
field
intensity is
to
be
calculated.
A
cylinder of
area
'A' is
constructed
as
shown in
figure.
By symmetry
magnitude of
electric
field
at distance
'r'
in opposite
side
of
'p' is same
as
at point P.
Applying
Gauss
theorem,
electric
flux
(ɸ)=
Q/ε_{0}
<(i)
Again
from
definition
of
electric
flux,
ɸ= E2A
< (ii) [2A for considering both side of
charged sheet]
From
equation
(i)
and
equation
(ii)
E2A=
Q/ε_{0}
[Q=
σ.A]
E=
Q/2Aε_{0}
[σ=
Q/A]
E=
σ/2ε_{0}
7.
Force
experienced
by
charged
particle
in
electric
field:
A
negatively
charged
oil
drop
is
placed
in
between
two
parallel
plates
A
and
B as
shown
in
figure
above.
Electric
potential
'V'
is
supplied
is
placed
in
plates
A
and
B so
force
is
experienced
by
charged
oil
drop
in
between
the
parallel
plates.
Let
'Q'
be
the
charged
of
supplied
electric
field
then
force
experienced
by
charged
oil
drop
due
to
electric
field,
F=
Q.E
< (i)
Since,
the
weight
of
charged
oil
drop
is
acting
vertically
downward.
therefore,
downward
acting
force
(W)=
m.g
< (ii)
For
the
charged
drop
remain
to
be
stationery
in
between
the
parallel
plates
upward
acting
force
must
balance
downward
acting
force.
therefore,
QE=
mg
Q.(V/d)=
volume(v)
x
ρ
x
g
[E=V/d]
and
[m=
v
x
ρ]
Q.(V/d)=
4/3(πr^{3}.ρ.g)
Q=
4/3V(πr^{3}.ρ.d.g)
< (iii)
if
'n'
be
the
number
of
electron
in
given
potential
then,
Q=
ne
ne=
4/3V(πr^{3}.ρ.d.g)
therefore,
n=
4/3eV(πr^{3}.ρ.d.g)
COULUMB'S
LAW
1.
Columb's
law
of
force
of
attraction
or
force
of
repulsion
between
two
charges:
The
force
of
attraction
or
repulsion
between
two
charges
is
defined
by
columb's
law
considered
q_{1}
and
q_{2}
be
two
charges
'r'
distance
apart
from
each
other.
F
∝
(q_{1}.q_{2})/r^{2}
F=
K(q_{1}.q_{2})/r^{2}
[where
'k'
is
proportionality
constant]
K=
(1/4πε_{0})
[where
'ε_{0}'
is
permittivity
of
free
space]
Therefore,
F=
(1/4πε0)
(q_{1}.q_{2})/r^{2}
2.
Permittivity:
The
permittivity
explains
the
respnose
of
given
medium
to
the
charges
present
on
it.
The
unit
of
permittivity
is
ε_{0}=
N^{1}m^{2}C^{2}
ε_{0}=
C^{2}/Nm^{2}
ε_{0}=
C^{2}N^{1}m^{2}
And
it's
value
in
air
is
ε0=
8.85
x
10^{2}
C^{2}N^{1}m^{2}
3.
Relative
permittivity
(ε_{r}):
Relative
permittivity
is
the
ratio
of
permittivity
of
any
medium
to
permittivity
in
its
space
and
it
is
called
as
dielectric
constant.
Mathematically,
Dielectric
constant
is
a
ratio
of
the
columb's
force
between
two
charges
in
air
to
columb
force
between
the
same
charges
in
any
other
medium.
Dielectric
constant
(k)=
(1/4πε_{0})[q_{1}.q_{2})/r^{2}]
=
{(1/4πε_{0})[q_{1}.q_{2
r}>)/r^{2}]}
/
(1/4πε_{0})[q_{1}.q_{2})/r^{2}]
therefore,
K=
ε_{r}
Electrostatic
Induction:
THe
temporary
electrification
of
conduction
by
bringing
a
charged
body
by
closed
to
it
is
called
Electrostatic
Induction.
For
eg:
charging
a
body
by
friction,
by
conduction,
or
by
induction.
(a.)
By
friction:
In
this
method,
two
bodies
are
rubbed
against
each
other
due
to
which
there
is
transfer
of
charge(electrons)
between
these
two
bodies
so
that,
one
becomes
positively
charged
and
another
becomes
negatively
charged.
In
this
case,
nature
of
the
charge
depends
upon
the
materials
used.
(b.)
By
conduction:
The
process
of
charging
a
body
by
touching
it
with
a
charged
body
is
called
charging
by
conduction.
In
this
method,
the
charge
produce
on
the
conductor
is
of
the
same
nature
as
is
on
the
charging
body
as
shown
in
figure:
Conductor
A
is
positively
charged.
So,
it
has
a
deficiency
of
electron.
When
it
is
brought
in
contact
with
a
neutral
conductor
B,
the
free
electrons
of
B
will
flow
to
A
to
makeup
for
its
deficiency.
Due
to
loss
of
electron
in
B,
it
becomes
positively
charged.
On
the
other
hand,
if
B
is
to
be
negatively
charged,
a
negatively
charged
conductor
A
is
made
to
touch
it.
In
this
case
transfer
of
electron
takes
place
from
A
to
B.
(c.)
By
induction:
Charging
a
body
positively
by
induction:
An
uncharged
body
can
be
charged
positively
by
following
ways:
i.
Considered
as
uncharged
body
(sphere)
having
two
sides
A
and
B
as
shown
in
figure:
Step1:
A
charge
rubber
rod
kept
near
to
side
A
of
the
sphere,
all
positively
charges
are
attracted
towards
the
rod
as
shown
in
figure
above.
Here
side
A
is
filled
with
positive
charges
and
side
B
is
filled
with
negative
charges.
Step2:
With
the
help
of
a
conducting
wire
side
B
is
grounded.
So,
all
the
negative
charges
move
towards
earth.
Hence,
only
positive
charges
remaining
in
the
sphere
are
being
attracted
towards
the
rubber
rod
which
is
shown
in
figure
above.
Step3:
Now
the
conducting
wire
is
remove
from
the
sphere,
only
positive
charges
remained
on
a
sphere
attracting
towards
the
rubber
rod
as
shown
in
figure
above.
Step4:
Finally,
rubber
rod
is
removed
from
side
A
of
the
sphere.
Now,
positive
charges
spread
all
over
the
sphere
as
shown
in
figure
above.
Linear
charge
distribution
It
is
defined
as
total
charge
present
on
distributed
in
1
dimensional
body
per
unit
length.
Therefore,
linear
charge
distribution
is
equal
to:
Total
charge
present/length
Q/L
columb
per
metre
Surface
charge
distribution
It
is
defined
as
total
charge
present
in
two
dimensional
body
per
unit
area
of
body.
Therefore
surface
charge
distribution
is
equals
to:
(total
charge
present)
/
area
Q/A
C/m^{2}
(columb
per
meter
square)
Volume
charge
distribution
It
is
defined
as
total
charge
present
in
three
dimensional
body
per
unit
volume
of
a
body.
Therefore
volume
charge
distribution
is
equals
to
:
total
charge
present/volume
Q/V
C/m^{3}
(Columb
per
meter
cube)
The
distribution
of
charge
depends
on
instructure
and
shape
of
a
body.
It
is
also
affected
by
a
conduction
present
near
to
it.
The
greater
the
curvature
at
any
point
the
greater
will
be
the
accumulation
of
the
charge
at
that
point.
In
case
or
rectangular
shape
of
body,
the
charge
density
is
more
at
corners.
Similarly,
in
case
of
conical
charge
body
maximum
charges
are
present
at
apex.,
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