Electric charges and fields are ok .pptx

ELECTRIC CHARGES
AND FIELDS
REVIKUMAR. R
CHAPTER ONE
CLASS XII
PGT PHYSICS
JNV KOLLAM

CONTENTS
 Introduction
 Electric Charges
 Conductors and Insulators
 Charging by Induction
 Basic Properties of Electric Charge
 Coulomb’s Law
 Forces between Multiple Charges
 Electric Field
 Electric Field Lines
 Electric Flux
 Electric Dipole
 Dipole in a Uniform External Field
 Continuous Charge Distribution
 Gauss’s Law
 Application of Gauss’s Law

Electrostatics
Electrostatics is the branch of Physics
which deals with the study forces,
fields, and potentials arising from static
charges.

INTRODUCTION -ELECTROSTATICS
• The word electricity is derived from the
Greek word elektron which means amber.
• In 600 BC, the Greek philosopher Thales
of Miletus observed that when amber is
rubbed with wool it attracts light objects.

ACTIVITY USING COMB AND PAPER
BITS

ACTIVITY USING PLASTIC SCALE AND PAPER BITS

Methods of charging
There are three methods:
a) Rubbing (charging by friction)
b) Charging by Conduction
c) Charging by Induction

Rubbing (charging by friction)

Rubbing (charging by friction)
When glass rod is rubbed with silk, glass
acquires positive charge and silk acquires
negative charge.
When plastic is rubbed with fur, plastic
acquires negative charge and fur acquires
positive charge.
Electricity developed on bodies, when two
suitable bodies rubbed with each other is
called frictional electricity or static electricity.

Charging by friction
When we rub a glass rod with silk, some of the
electrons from the rod are transferred to the
silk cloth. Thus the rod gets positively charged
and the silk gets negatively charged.
Similarly, when we rub a plastic rod with fur,
some of the electrons from fur are transferred
to the plastic rod. Thus the fur gets positively
charged and the plastic rod gets negatively
charged.

ELECTRON THEORY OF
ELECTRIFICATION
To electrify a neutral body, we need to add or
remove one kind of charge. When we say that
a body is charged, we always refer to this
excess charge or deficit of charge. In solids,
some of the electrons, being less tightly bound
in the atom, are the charges which are
transferred from one body to the other. A body
can thus be charged positively by losing some
of its electrons. Similarly, a body can be
charged negatively by gaining electrons.

a) It was observed that if two glass rods rubbed
with wool or silk cloth are brought close to
each other, they repel each other.
b) Similarly, two plastic rods rubbed with cat’s
fur repelled each other.
c) On the other hand, the plastic rod attracts the
glass rod.

• If a plastic rod rubbed with fur is made to touch two
small pith balls suspended by silk or nylon thread,
then the balls repel each other.
• A similar effect is found if the pith balls are touched
with a glass rod rubbed with silk .
• A pith ball touched with glass rod attracts another pith
ball touched with plastic rod.
Like charges repel and unlike charges attract
each other.

Like charges repel while unlike charges
attract each other.

ELECTROSCOPE
A simple apparatus to detect
charge on a body is the gold-
leaf electroscope. It consists
of a vertical metal rod housed
in a box, with two thin gold
leaves attached to its bottom
end. When a charged touches
the metal knob at the top of
the rod, charge flows on to
the leaves and they diverge.
The degree of divergence is
an indicator of the amount of
charge.

Conductors and insulators
Some substances readily allow passage of
electricity through them, others do not. Those
which allow electricity to pass through them easily
are called conductors. They have electric charges
(electrons) that are comparatively free to move
inside the material. Metals, human and animal
bodies and earth are conductors. Most of the non-
metals like glass, porcelain, plastic, nylon, wood
offer high resistance to the passage of electricity
through them. They are called insulators.

Charging by conduction
When a charged body is brought in to
contact with an uncharged conductor,
charge flows from the charged body to the
uncharged body.
+
+
+
+
+
+
+
+
+
+
+ +
+

I. Bring two metal spheres, A and B, supported on
insulating stands,in contact as shown in Fig.(a)
II. Bring a positively charged rod near one of the
spheres, say A. The free electrons in the
spheres are attracted towards the rod. This
leaves an excess of positive charge on the rear
surface of sphere B. The left surface of sphere
A, has an excess of negative charge and the
right surface of sphere B, has an excess of
positive charge.
The process is called induction of charge and
happens almost instantly. The accumulated
charges remain on the surface, as shown, till
the glass rod is held near the sphere.
Charging by Induction

iii) Separate the spheres by a small distance
while the glass rod is still held near sphere A,
as shown in Fig. (c). The two spheres are
found to be oppositely charged and attract
each other.
(iv) Remove the rod. The charges on spheres
rearrange themselves as shown in Fig. (d).
(v)Now, separate the spheres quite apart.
The charges on them get uniformly
distributed over them, as shown in Fig. (e).
In this process, the metal spheres will each
be equal and oppositely charged. This is
charging by induction.

How can you charge a metal sphere positively without
touching it?

 Figure (a) shows an uncharged metallic sphere on an
insulating metal stand.
 Bring a negatively charged rod close to the
metallic sphere, as shown in Fig.(b).
As the rod is brought close to the sphere, the free
electrons in the sphere move away due to repulsion
and start piling up at the farther end. The near end
becomes positively charged due to deficit of
electrons.
 Connect the sphere to the ground by a
conducting wire. The electrons will flow to the
ground while the positive charges at the near end
will remain held there due to the attractive force of
the negative charges on the rod, as shown in Fig.(c).
 Disconnect the sphere from the ground. The
positive charge continues to be held at the near end
[Fig. (d)].
 Remove the electrified rod. The positive charge
will spread uniformly over the sphere as shown in
Fig.(e).

BASIC PROPERTIES OF ELECTRIC CHARGE
Additivity of charges
The total charge of an isolated system is
equal to an algebraic sum of individual
charges of the system.
For example, the total charge of a system
containing five charges +1, +2, –3, +4 and
– 5,in some arbitrary unit, is
(+1) + (+2) + (–3) + (+4) + (–5) = –1 in the
same unit.

Conservation of charge
The total charge of an isolated system is
always conserved that means charge can
neither be created nor be destroyed but can
be transferred from one body to another.
When bodies are charged by rubbing, there
is transfer of electrons from one body to the
other; no new charges are either created or
destroyed.

Quantisation of charge
Any charged body has a total charge ± ne where
‘n` is an integer (n =0,1,2,3………..). This
experimental fact is called quantization of charge.
q = ± ne, where n is an integer
and e = 1.6 × 10 -19
C
By convention, the charge on an electron is taken
to be negative; therefore charge on an electron is
written as –e and that on a proton as +e.
The SI unit of charge is Coulomb and is denoted
by the symbol C.
1 C=10
𝞵 -6
C

How many electrons constitute one coulomb
of charge?
q = ne
q=1C e = 1.6 × 10 -19
C
n=6.25 x 1018
electrons

Why can one ignore quantisation of electric charge when
dealing with macroscopic i.e., large scale charges?
At the macroscopic level, one deals with charges that
are enormous compared to the magnitude of charge e. A
charge of magnitude, say 1 μC, contains something like
1013
times the electronic charge. At this scale, the fact that
charge can increase or decrease only in units of e. Thus,
at the macroscopic level, the quantisation of charge has
no practical consequence and can be ignored.
At the microscopic level, where the charges involved are
of the order of a few tens or hundreds of e, i.e.,they can
be counted and quantisation of charge cannot be ignored.

Charles Augustin de Coulomb
(1736 – 1806)

COULOMB’S LAW
F α
The force of attraction or repulsion between
two charges is directly proportional to product
of their charges and inversely proportional to
square of the distance between them.
r
q1 q 2

COULOMB’S LAW
F α
F= k
k=
is called permittivity of air or free space.
(absolute permittivity)
F=
Value of = 8.85x10 -12

= 9x109
F=
If the charges are placed in a medium of
permittivity 𝟄
F=
F = 9x109

Relative permittivity Or Dielectric constant
( K Or 𝟄r)
It is the ratio of permittivity of a medium to
the permittivity of free space.
𝟄r =
𝟄= 𝟄0 𝟄r
F=
F= OR F=

1.How does the force between two charges change if the
a) distance is doubled?
b) distance is halved?
c) dielectric constant of the medium is increased?
d) charges are immersed in water (K=81)?
2.The force between two charges placed in air at a
distance r apart is F. What must be the distance between
two charges so that the force become
e) 3F
f) F/3
3.What is the force between two small charged spheres
having charges of 2x10-7
C and 3x 10-7
C placed 30cm apart
in the air?

a) F=
3F=
Ans:r1= r/
b) F=
F/3=
Ans:r2= r

UNIT OF CHARGE- COULOMB
F=
If q1=q2= 1C and r=1m
F = 9x109
N
1 Coulomb is that charge when placed in air
or vacuum at a distance of 1m from an equal
and similar charge experiences a force of
9x109
N.
F = 9x109

Question
Identify the given ratio ke2
/G memp as
dimensionless quantity or not. From the table of
Physical Constants determine the value of this
ratio. Also, mention the signification of the ratio.
ke2
/G memp
=
= 1 (dimensionless)

Coulomb’s law in vector form
q1 r q2
Force on q2 due to q1, =
is the unit vector pointing from q1 to q2
=
=

q1 r q2
Force on q1 due to q2, =
is the unit vector pointing from q2 to q1
=
=
= -
= -
ie Coulomb’s law obey Newton’s third law

Force between multiple charges
(Principle of superposition)
q1
q2
q3
q4
F43
F42
𝐹 41
= +

Principle of superposition
=
=
= +
In case of n charges
= +………………….

Principle of superposition
Total force on any charge due to number of
charges is the vector sum of all the forces due to
the other charges.

What is the force acting on a charge Q placed at
the centroid of the triangle?
F1 = F2 = F3 = F
R =
R =
R =
R =
R =F
Net force acting
on the charge at
the centroid of
the triangle is
zero

Four point charges qA = 2 μC, qB = –5 μC, qC = 2 μC, and qD = –5 μC
are located at the corners of a square ABCD of side 10 cm. What is the
force on a charge of 1 μC placed at the centre of the square?
The repulsive force between
the charges at A and at the
centre O is same in magnitude
with the repulsive force by the
corner C to the centre O, but
these forces are opposite in
direction. Hence, these forces
will cancel each other.
Similarly attractive force
between charges at D and O is
cancelled by the attractive
force between the charges at
B and O. Therefore, the net
force on 1 μC at the centre is
zero..

ELECTRIC FIELD
The electric field is defined as the region or
space around a charge where an electric force
of attraction or repulsion can be experienced.

Electric field Intensity
Consider a charge Q placed in vacuum. If we place
another point charge q at a point P, then the charge
Q will exert a force on q as per Coulomb’s law.
Let be the force experienced
by the charge q.
Force experienced by unit charge ,
=
q
q
P
The electric field or field intensity at a point is defined as
the force experienced by unit positive charge placed at
that point.
E=lim
𝑞→𝑜
F
q

Significance of is that the test charge q
should be negligibly small so that the source
charge Q remain at its original position.
Unit Electric field Intensity is N/C
or
V/m
Electric field Intensity is a vector quantity.
The force acting on the charge q is
= q

Electric field intensity due to a point charge
+q r P E
+1
Consider a point P at a distance r from a point
charge +q.
Electric field intensity at the point P,
E=

Electric dipole and Electric Dipole moment
-q 2a +q
An electric dipole is a pair of equal and opposite point charges q
and –q, separated by a short distance 2a.
Electric Dipole moment is the product of one of the charges and
distance between them.
= q
The unit of electric dipole moment is coulomb –meter (C-m)
Electric dipole moment is a vector quantity and by convention, its
direction is from –q to +q.
+
-

A system has two charges qA = 2.5 × 10–7
C and qB = –2.5 × 10–7
C
located at points A: (0, 0, –15 cm) and B: (0,0, +15 cm), respectively.
What are the total charge and electric dipole moment of the system?
Total charge
= 2.5 × 10–7
C –2.5 × 10–7
C
= 0
q = 2.5 × 10–7
C
2a = 15Cm + 15Cm = 30Cm
= 0.3 m
p =2a q
p =0.3 x 2.5 × 10–7
p =7.5 × 10–8
Cm

Electric field of a dipole (On axial line)
A o B E2 P E1
-q 2a +q +1
Consider a dipole of charge q and length 2a. Let P be a
point at a distance x from the centre of the electric dipole.
Electric field intensity at the point P due to the charge +q ,
Electric field intensity at the point P due to the charge -q ,
E2= along PA
The total electric field at P due to the dipole is
E = E1- E2
E1= along BP
x

E = -
E = ( - )
E = ()
E = a<<x (a2
can be neglected)
E =
E = E =
E = =

Electric field of a dipole (On equatorial line )
The electric field at P due to the
charge +q,
E1 Can be resolved into two
components E1 Cos𝞱 and E1 Sin𝞱.
r
r x r
E1 Cos𝞱
E2 Cos𝞱
�
�
�
�
�
�
�
�
E1= along BP

The electric field at P due to the
charge -q,
E2= along PA
E2 Can also be resolved into two
components E2 Cos and
𝞱 E2 Sin .
𝞱
Here Sin components are equal
𝞱
and opposite, therefore they cancel
out. But Cos components are in
𝞱
the same direction, they can be
added up.
r
r x r
E1 Cos𝞱
E2 Cos𝞱
�
�
�
�
�
�
�
�

The total electric field at P due
to the dipole,
E = E1 Cos𝞱 + E2 Cos𝞱
E = 2 Cos𝞱
Cos𝞱= a/r
E = 2
E =
E = 2aq=p
E = if a is very small r ≃ x
E =
r
r x r
E1 Cos𝞱
E2 Cos𝞱
�
�
�
�
�
�
�
�
E = Cos𝞱 + Cos𝞱
a

Torque acting on a dipole in an electric field
-q
+q
A
B
N
+qE
- qE
2a �
�
�
�
E
Consider a dipole of charge q and length 2a placed in a
uniform electric field makes an angle with the direction of the
𝞱
electric field.

Torque acting on a dipole in an electric field
The charges +q and –q experience
forces + qE and –qE respectively.
These two equal and unlike forces
constitute a couple.
Torque(τ) = force x
Perpendicular distance
τ = qE x BN
Sin𝞱=
BN= 2a Sin𝞱
τ = qE x 2a Sin𝞱
τ = 2aqE Sin𝞱
2aq=p
τ = PE Sin𝞱
= X
Torque is perpendicular toand
2a
2aSin
𝞱
+
The net force acting on
the dipole + qE- qE = 0

Draw the orientations of a dipole in an electric field corresponding
to:
1) 𝞽 =0
2) 𝞽 = max
3) 𝞽 = max
Ans:
1) a.
b.
-q +q
E
2a
Parallel
+q -q
2a
E Anti parallel

2) 𝞽 = max
3) 𝞽 = max
E
+q
- q
2a
-q
+q
E
30o

An electric dipole with dipole moment 4 × 10–9
C m is aligned
at 30° with the direction of a uniform electric field of
magnitude 5 × 104
NC–1
. Calculate the magnitude of the
torque acting on the dipole.
p = 4 × 10–9
C m
E = 5 × 104
NC–1
𝞱 = 30°
τ = PE Sin𝞱
τ = 4 × 10–9
X 5 × 104
Sin 30°
τ = 4 × 10–9
X 5 × 104
x
τ = 10–4
Nm.

Electric field lines
The electric field lines are imaginary lines
drawn in such a way that the tangent to
which at any point gives the direction of
the electric field at that point.

Electric field lines of a single positive Charge

Electric field lines of a single negative Charge

The field lines of a single positive
charge and a single negative
The field lines of a single positive charge are radially
outward while those of a single negative charge are
radially inward.

Field lines around the system of two positive
charges

Field lines around the system of two positive charges
gives a different picture and describe the mutual
repulsion between them.

Field lines around a system of a positive
and negative charge (Electric dipole)

Field lines around a system of a positive
and negative charge clearly shows the
mutual attraction between them.

Field lines around the system of two negative
charges

Electric charges and fields are ok .pptx

ELECTRIC FIELD LINES IN A UNIFORM
ELECTRIC FIELD

Electric field corresponding to a negative
charge is placed with in the vicinity of a
metal plate

Properties of Electric field lines.
Electric field lines start from +ve charge and end in –ve
charge.
 Electric field lines do not form any closed loop.
 Electric field lines never intersect each other.
If two lines intersect at a point, it means that at the
point of intersection electric field can be two directions
and hence they never intersect each other.
 If the field lines are crowded, then the field is strong and
if the field lines are not crowded, then field is weak.
 The electric field lines are always normal to the surface
of the charge body.
 Electric field lines can be taken to be continuous curves
without any breaks.

Continuous Charge Distribution:
A system of closely spaced charges is said to form a
continuous charge distribution.
(i) Linear charge density
If the charge is distributed over a line then the distribution
is called ‘linear charge distribution’.
Linear charge density is the charge per unit length. Its SI
unit is C / m.
=
dq
dl
+ + + + + + + + + + + +
dq
dl
or
𝝀=
𝒒
𝒍

ii) Surface Charge Density ( σ ):
o
r
If the charge is distributed over a surface, then the
distribution is called ‘surface charge distribution’.
Surface charge density is the charge per unit area.
Its SI unit is C / m2
.
+ + + + + + + + + + +
+ + + + + + + + + + +
+ + + + + + + + + + +
+ + + + + + + + + + +
+ + + +
dS
dq
𝝈 =
𝒒
𝒔
𝜎 =
𝑑𝑞
𝑑𝑠

(iii) Volume Charge Density ( ρ ):
If the charge is distributed over a volume, then the
distribution is called ‘volume charge distribution’.
Volume charge density is the charge per unit volume. Its
SI unit is C / m3
.
dq
dV
or

Electric flux (ϕ)
The electric flux is
defined as the measure
of total number of
electric field lines
passing normally
through a given surface.
If the surface is
perpendicular to the
field, then the flux
through an area ΔS is
Δϕ = E ΔS

Electric flux (ϕ)
If the normal to the surface makes
an angle with the electric field ,
𝞱
Flux through the surface
Δϕ = E ΔS Cos 𝞱
Δϕ = E . ΔS
Total Flux through a given surface
ϕ =𝞢 E . ΔS
OR
ϕ = E . S
Unit of electric flux is Nm2
/C
�
�

GAUSS’S LAW
Gauss‘s law state that
the total electric flux or
total number of field
lines passing through
any closed surface is
equal to times the
charges enclosed by
the surface.
ϕ =
E . S =

The electric flux through the
surface = E . S
The electric field intensity
at P
E=
Surface area of the spherical
surface , S = 4𝞹r2
The electric flux through the
surface = E . S = x 4𝞹r2
E S =
ϕ =
ie Gauss’s Law
P

APPLICATIONS OF GAUSS’S LAW
Field due to an infinitely long straight uniformly
charged wire.
Consider an infinitely long thin
straight wire with uniform linear
charge density λ. Let P be a point at
a distance r from the straight wire.
The electric field lines are radially
outward. To find the electric field
intensity at P, imagine a Gaussian
surface of radius r and length l. The
electric flux through two flat
surfaces is zero because the
electric field lines are radially
outward and the area vector is
purpendiculat to E
�
�

The electric flux through
though the curved surface,
E . S =
E S =
q = 𝞴l
S = 2𝞹rl
E 2𝞹rl =
E =
=
where is the radial unit vector plane
normal to the wire.
Electric Flux
ϕ =
ϕ =
�
�

Field due to a uniformly charged thin spherical shell
(i) Field outside the shell
Consider a spherical shell of
radius R with uniform surface
charge density . Let P be a
𝞼
point at a distance r from the
center of the spherical shell.
Here the electric field lines are
radially outward. To find the
electric field intensity at P
imagine a Gaussian surface of
radius r.

Electric flux through the
surface,
E . S =
E S =
q = 4𝞹R2
𝞼
S = 4𝞹r2
E 4𝞹r2
=
E =
If the point is on the surface of the
charged spherical shell, r =R
E =

(ii)Field inside the shell
If the point P is inside the Shell,
the Gaussian surface is again a
sphere through P centred at O.
The flux through the Gaussian
surface,
E . S =
Here the charge enclosed by
the Gaussian surface is zero.
(q = 0)
E . S = 0
E =0
The electric field inside a
Charged spherical shell is zero.

Variation of electric field with distance
from the centre of the spherical shell
E =
E =
E
=0

Field due to a uniformly charged infinite
plane sheet
x x
ΔS
�
�

plane sheet
Consider an infinite plane
sheet of charge with
uniform charge density .
𝞼
To find the electric field
intensity at P, imagine a
Gaussian cylinder of cross
sectional area A normal to
the plane of the sheet.
Since the electric field lines
are parallel to the curved
surface, the flux through
this surface is zero.
x
x
�
�
ΔS

plane sheet
The flux through two flat
surfaces,
E . S =
E S =
S = 2ΔS
q =𝞼 ΔS
E 2ΔS =
E =
E =
E is independent of x
x
x
�
�
ΔS

Electric field between two parallel plates
+𝞼 -𝞼
I III
II
E = (
⎻ =0 E = (=0
E = (
E = (
E =
E =
Electric field
intensity between
two parallel plates E
=

Electric field between two parallel plates

Electric charges and fields are ok .pptx

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