Physics; presentation electrostat; -harsh kumar;- xii science; -roll no 08

Presentation
ELECRTOSTATS
XII Science.
Harsh Kumar
Presents
Physics
on
of

Electrostatics
The study of all electrical
properties of system having
charge, when the charges are in
state of rest is known as
electrostatics.

Charge
It is the basic property of a body
which explain electrical
properties of the body.
Charge on a body
define by number of free
electrons transacted
from a body
•Unit of charge= C(Coulomb)
•1µC=10¯⁶C
Q =±n e

Types of charge:
•Positive (+)ve
•Negative (-)ve
Charge of electron = -1.6×10¯¹⁹C
Charge on body = 1C
Q = n e
n=1/1.6×10¯¹⁹ = 6.25×10¹⁸

Properties of charge:
Charge can be produced by three ways:
I. Friction.
II. Conduction.
III. Induction.
Friction – When two charged particles are rubbed each
other then one get (+)vely charged and other gets (-)vely
charged.
E.g.: When we rub glass rod by woolen cloth then glass rod gets (+)vely charged
and woolen cloth gets (-)vely charged
Conduction- When two conducting substance attached to
each other then charges overlap on one another.
Induction- When a (-)vely charged particle is kept beside
(+)vely charged particle then opposite charges are induced on
each other.

 Charge are additive in nature.
It is a scalar quantity.
SI unit of charge is Coulomb (C).
CGS unit of charge is “Stat Coulomb”.
1 farad=96500C.
Moving charge with constant velocity produces
electric field and magnetic field.
Static charge produces electric field only.
When charge is moving with accelerated motion
then it produces electric field, magnetic field and
radiates electro magnetic wave.

Coulombs law
The force of attraction between two point
charges is directly proportional to magnitude
of product of both charges and inversely
proportional to square of distance between
them.
F  q₁ q₂ ———①
F  1/r² ———②
From ①&②
where ‘K’ is
proportionality constant
F=K q₁ q₂/r²

K = 1 (in CGS unit)
K = 1/4  Є˳ Єг (in SI unit)
K = 9×10⁹ (in air)
Where Є˳ is absolute
permittivity in free space and
Єг is relative permittivity of
medium.
For air Єг = 1
For water Єг = 81Coulombs law is also called
inverse square law.
The force of attraction is a
central force, it is added by

Charge Density
It is a scalar quantity, there are three
types of charge density.
 Linear charge density:- Charge per unit length is called linear charge density.
It is denoted by .
=Q/l
 Surface charge density:- Charge per unit surface area is called surface charge
density. It is denoted by .
=Q/A
 Volume charge density:- charge per unit volume is called volume charge
density. It is denoted by .
=Q/V

Electric field
Electric field is the field where a force acts on any
charged particle.
Electric field intensity:- Electric field intensity at a
point is the force experienced by unit (+)ve charge at
that point.
It is denoted by E.
It is a vector quantity.
It’s unit is N/C.
It is added by vector law of addition.
→
E=F/q
→

Electric field lines or Electric lines of force:-
Electric lines of force is a curved imaginary lines which
denotes electric field.Properties of electric field lines:-
Electric lines of force never interact each other.
It emerges normally to a conductor.
Number of lines of force per unit area gives strength of electric field.
Tangent on lines of force gives direction of electric field.
Electric lines of force emerges normally from (+)ve charge and terminates to (-)ve charge.
E E’
E > E’
E
→

Electric lines of force never made a close loop.
Electric lines of force per unit surface area is called electric flux.

Electric field intensity due to unit (+)ve charge in an electric field
Let us conceder a point ‘P’ at distance ‘r’ from a test charge ‘q’.
A unit (+)ve charge ‘q˳’ is kept at ‘P’, we have to found
the electric field intensity due to unit (+)ve charge.
(q) (q˳)
A r P
F = (q.q˳/4  Є˳ Єг r²)
We know that Єг=1 for air
E=F/q˳
̥˚˳ E = (q.q˳/q˳ 4  Є˳ r²)
E=(q/4  Є˳ r²)

Electric flux
Lines of force in a unit surface area is called electric flux.
It is dot product of electric field intensity and
surface area. It is denoted by Ø.
Ø=E.ds
→ →
Ø=E ds cosƟ
Flux comes out from a closed
cube is q/є˳
Where ‘q’ is charge

Dipole
A system of two same but opposite charges placed at some
distance is known as Dipole.
It is a combination of two charges of same magnitude but
opposite nature kept at some distance apart.
-q +q
2L
LL
P
→
Dipole Moment:- It is the product of 1 charge and distance between two charges.
It is a vector quantity.
It is denoted by P .
It’s direction is from (-)ve to (+)ve charge.
→

Torque acting on a electric dipole in an electric field
ᅳᅳᅳᅳᅳᅳᅳᅳᅳᅳᅳᅳ
Ɵ
Ɵ
B
E
E
+q
-q
2L
A
E
O
C
→
→
→
Let us consider a dipole of dipole moment P in an
electric field E due to opposite nature of charge on
dipole it experiences electric force in opposite
direction which is converted into Torque and dipole
starts rotating
→ →
=Force × ʳ distance

= F × (BC)
= q E × 2L sinƟ sinƟ=BC/AB
̥˚˳ BC=2L sinƟ
=P E sinƟ ———————①
=P × E
→ →

Work done by electric dipole in an electric field
Ɵ
Ɵ
B
E
E
+q
-q
2L
A
E
O
C
→
→
→
Work done of electric
dipole can be given
as :
dw =  × dƟ
Putting value of ‘’ from eqᵑ ①.
dw = P E sinƟ dƟ
Integrating both side with taking proper limit.
 dw = P E  sinƟ dƟ
w = P E [-cosƟ]
w= P E [-cosƟ₂ - (-cosƟ₁)]
w = P E [cosƟ₁ - cosƟ₂]
0
w
Ɵ₁
Ɵ₂
Ɵ₂
Ɵ₁
If Ɵ₁ = 0˚ ; Ɵ₂ = Ɵ
w = P E [1-cosƟ]

Gauss’s law
Flux linked with
closed surface is
1/є˳ times of a
charge kept in
that close
surface.
i.e. Ø= q/є˳

Proof:
dω
ds
E
→
→
q
dØ = E . ds
→→
Consider a charge ‘q’
enclosed in a Gaussian
sphere of radius ‘r’. Let an
small area ‘ds’ at any point
on sphere; whose direction
is along the direction of
electric field intensity.
Then Flux linked with this
small area :

dØ = E . ds
dØ = E ds cos
dØ = E ds cos0˚ [ =0˚]
dØ = E ds
dØ = (q/4  Є˳ r² ) ds
Integrating along closed surface
∮dØ = ∮ds
 Ø = (q/4Є˳ r²) (4r² - 0)
 Ø = q/Є˳
dω
ds
E
→
→
q

Presentation Title Goes Here
XII Science.
Harsh Kumar
Presented
by
of
08Roll
No.

Physics; presentation electrostat; -harsh kumar;- xii science; -roll no 08

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