01 Electric Fieeld and charges Notes.pdf

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Electrostatics deals with the study of forces, fields and potentials arising from static
charges. OR The branch deals with study of charges at rest.
ELECTRICITY: The flow of charges is called as electricity. The name electricity is coined
from the Greek word "electron " meaning amber.
ELECTRIC CHARGE: An basic entity (particle) of each matter which experience force of
attraction or repulsion is called the electric charge. Hence there were only two kinds of
charges were named as positive and negative by the American scientist Benjamin
Franklin.
If an object possesses an electric charge, it is said to be electrified or charged. When
it has no charge it is said to be neutral. Hence process of making objects charged is called as
electrification or charging.
GOLD-LEAF ELECTROSCOPE
A simple apparatus to detect charge on a body is the "gold-leaf electroscope."
Construction and working: It consists of a vertical metal rod housed in a box, with
two thin gold leaves attached to its bottom end. When a charged object 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 measures the amount of charge.


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.

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CONDUCTORS AND INSULATORS
The substances 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.
Example- Metals, human and animal bodies and earth etc.
The substances which do not allow electricity to pass through them easily are called
Insulators. There are no free electric charges inside material.
Examples- Glass, porcelain, plastic, nylon, wood etc.

CHARGING METHODS
The object can be acquired a charge by following three process.
1)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. Here both bodies get oppositely charged.
2)Charging by contact: When a charged body brought in contact with uncharged body
then both the bodies acquire the same charge.
3)Charging by induction: When charging a conductor by induction, a charged object is
brought close to but does not touch the neutral conductor. In the end the conductor acquire
charge of opposite sign as the charge on the object.


 NOTE: The process of sharing the charges with the earth is called grounding or
earthing. Earthing provides a safety measure for electrical circuits and appliances.
 How to make two bodies charged oppositely by method of induction?

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r
BASIC PROPERTIES OF ELECTRIC CHARGE
1. Additivity of charges: The total charge of the system is the algebraically sum of all
the charges within the system. If a system contains 'n' charges q1, q2, q3, ..., qn, then
the total charge of the system is q1 + q2 + q3 + ... + qn .

2. Charge is conserved: Charges are neither created nor destroyed but only transfer
from one body to another such that total charge of the isolated system is always
conserved (constant).

3. Quantisation of charge: The charges are integral multiples of a basic unit of charge
denoted by e. Thus charge Q on a body is always given by
Q = ±ne
where n is any integer, positive or negative & e =1.602×10
-19C
 Note: The quantisation of charge was first experimentally demonstrated by Millikan
in 1912.
>SI Unit of charge is called a 'Coulomb' and is denoted by the symbol C.
>One coulomb: Charge flowing through a wire in 1 s if the current is 1 A.
“Thus, there are about 6 × 10
18 electrons in a charge of –1C”.

COULOMB’S LAW OF ELECTROSTATIC
"Coulomb measured the force between two point charges and found that it varied
inversely as the square of the distance between the charges and was directly proportional
to the product of the magnitude of the two charges and acted along the line joining the two
charges.”
If two point charges ??????
5 , ??????
6 are separated by a distance 'r' in vacuum, then
magnitude of the force (F) between them is given by

F=
??????
??????????????????
??????
??????
????????????
??????
??????
??????

Where ??????=
5
8
,
= 9 × 10
9 Nm
2C
-2
And ??????
4=permittivity of free space = 8.854×10
-12 C
2N
-1m
-2
When ??????
5=??????
6= 1?????? and r = 1m then F = 9 × 10
9N
Define One coulomb:
1C is the charge that when placed at a distance of 1 m from another charge of the same
magnitude in vacuum experiences an electrical force of repulsion of magnitude 9 × 10
9 N.

??????
5
??????
6

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q<0 q>0
FORCES BETWEEN MULTIPLE CHARGES
Force on any charge due to a number of other charges is the vector sum of all the
forces on that charge due to the other charges, taken one at a time. The individual forces
are unaffected due to the presence of other charges. This is termed as t
superposition

ELECTRIC FIELD: The electric field due to a charge Q at a point in space may be defined
as the force that experience a unit positive test charge if placed at that point.

>>Unit of electric field - NC
-1
>>It is a vector quantity.






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FORCES BETWEEN MULTIPLE CHARGES (Principle of superposition)
Force on any charge due to a number of other charges is the vector sum of all the
forces on that charge due to the other charges, taken one at a time. The individual forces
are unaffected due to the presence of other charges. This is termed as t


The electric field due to a charge Q at a point in space may be defined
as the force that experience a unit positive test charge if placed at that point.
??????=
??????
??????

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rinciple of superposition)
Force on any charge due to a number of other charges is the vector sum of all the
forces on that charge due to the other charges, taken one at a time. The individual forces
are unaffected due to the presence of other charges. This is termed as the principle of
The electric field due to a charge Q at a point in space may be defined
as the force that experience a unit positive test charge if placed at that point.

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Properties of electric field lines:
1. Electric the field lines directed radially outwards from the positive charge.
2. Electric the field lines directed radially inward for negative the charge.
3. Electric field line are continuous smooth curve starts from positive charge and ends
at negative charge.
4. Electric field line never form close loop.
5. Electric field lines never cross each other.
6. They are perpendicular to the surface of charge.
7. Tangent at any point on electric field lines gives direction.

ELECTRIC FLUX (
??????)
The number of electric field lines passing through a certain cross-section area.
∅
??????= EA ????????????????????????
Where E is the magnitude of the Electric field (NC
-1), A is the area of the surface(m
2),
and θ is the angle between the electric field lines and the normal ( perpendicular ) to A.
 Unit of Electric flux is NC
–1m
2
 It is a scalar quantity.
Electric field due to a isolated single point charge (3 marks)





Consider a positive point charge is placed at 'A'. Let the point P is at distance r from the
charge +q, where electric field is to be estimated. Let us imagine a test charge qo placed at
point 'P' .The electric field due to charge +q exerts a force on test charge hence according to
Gauss law,

5
8"
?
??
?
?
.
-----------------(1)
but by definition of electric field is the force per unit charge.
E=
?
?
?
------------------(2)
From (1) and (2) we get



q0 (test charge)
A
+Q
P
r
F
?
6

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ELECTRIC DIPOLE :
An electric dipole is a pair of equal and opposite point charges
distance 2a.
The total charge of the electric dipole is zero. But the electric field of the electric dipole
is not zero.
Electric dipole moment (P): It is defined as the product of
distance between the charges.

 The direction of the dipole
 It is an vector quantity.
 Unit is ‘Cm’
ELECTRIC FIELD DUE TO DIPOLE
i)The field of an electric dipole
Let the point 'P' be at distance '
+q, as shown in Figure below.

The electric field at point P due to +q charge and
respectively. Then

>?

??
Where ??????̂ is the unit vector along the axis of dipole.
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An electric dipole is a pair of equal and opposite point charges q and
The total charge of the electric dipole is zero. But the electric field of the electric dipole
It is defined as the product of either charge on dipole or


The direction of the dipole moment is from negative charge to positive charge.
ELECTRIC FIELD DUE TO DIPOLE
The field of an electric dipole when the point is on the dipole axis
'P' be at distance 'r' from the centre of the dipole on the side of the charge

The electric field at point P due to +q charge and -q charge is represented by E
?
8
,(???)
.
----------------(1)
?
8
,(?>?)
.
---------------(2)
is the unit vector along the axis of dipole.
P=2qa
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and –q, separated by a
The total charge of the electric dipole is zero. But the electric field of the electric dipole

either charge on dipole or the
is from negative charge to positive charge.
when the point is on the dipole axis.(5Marks)
the dipole on the side of the charge

q charge is represented by E+q and E-q
(1)

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The total electric field at point P is the vector sum of field due to +q and -q charge.
??????= (??????
>?−??????
??)??????̂
??????=
??????
4????????????
4(??????+??????)
6
−
??????
4????????????
4(??????−??????)
6
??????̂
??????=
??????
4????????????
4
d
1
(??????+??????)
6
−
1
(??????−??????)
6
h??????̂
??????=
??????
4????????????
4
d
4????????????
(??????
6
−??????
6
)
6
h??????̂
??????=
6??
8
,
B
6?
(?
.
??
.
)
.
C??????̂ But P=2qa
??????=
2????????????
4????????????
4 (??????
6
−??????
6
)
6
??????̂
The electric field of a dipole at large distances along the axis
For r>> a,
6??????
8
, ?
/

ii)The field of an electric dipole when the point is on the equatorial plane.(5marks)
Electric field of a dipole at a point on the equatorial plane of the dipole at point 'P'.
Having the dipole moment vector of magnitude p = q × 2a and directed from –q to q.

The electric filed at P point due to +q and -q charge get resolved into its horizontal
and vertical components. Clearly, the components normal to the dipole axis cancel away.
Let r be the distance from the center of the dipole to point P. The components along
the dipole axis add up. The total electric field is opposite to electric dipole moment(P).
Components of electric field
??????
>??????



??????
???????

??????
>?????????????????????????
??????
??????????????????????????
??????
??
???????????????????????? ??????
>?
????????????????????????
θ
θ
EP
E
>o

E
?o
θ
θ

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Electric field due to +q and –q charge at point P is given by,

>?
?
8
,(?
.
>?
.
)
&
??
?
8
,(?
.
>?
.
)


The net electric field is vector sum of the electric field due to both charges,
??????= −k??????
>?+??????
?? ocos?????? ??????̂

Where ??????̂ is the unit vector along dipole moment.
Negative sign indicates that Net E and dipole moment P are in opposite direction.

??????= −l
??????
4??????????????????(??????
2
+??????
2
)
+
??????
4??????????????????(??????
2
+??????
2
)
pcos?????? ??????̂

But from figure cos??????=
??????
???????
2
+??????
2

??????= −l
2??????
4??????????????????(??????
2
+??????
2
)
p
??????
???????
2
+??????
2
???????
By simplifying the above equation we get,
??????= −L
??????????????????
????????????????????????(??????
??????
+??????
??????
)
??????
??????
M???????
For r>>a,
The electric field of dipole at large distances at point on equatorial plane is
??????= −l
??????
4????????????
0
??????
3
p Where P = 2qa
DIPOLE IN A UNIFORM EXTERNAL FIELD(Torque on a dipole)
Consider a electric dipole placed in a uniform Electric field E. The axis
of the dipole makes an angle 'θ' with the direction of electric field as shown in figure below.







The force acting on the charge +q is, F=+qE in the direction of 'E'
The force acting on the charge -q is, F=-qE in the opposite direction of 'E'
+qE
-qE
θ
θ
2a
A
B
C
??????
→
2a sinθ

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These two opposite and parallel non-linear forces acting on dipole forms a couple
that exerts a torque and rotates the dipole in clockwise direction.
Torque( ??????) =(magnitude of one of the force×Normal distance between two forces,AB)
??????=(????????????×2??????sin??????) sin??????=
?????? ????????
??
=
??
6?

Also P= 2qa AB=2aSinθ
??????=????????????????????????????????????
Vector form of Torque,
 Torque is maximum when, ??????=????????????
??????

 Torque is minimum when, ??????=??????????????????
??????
???????????? ??????
??????
(parallel or antiparallel)
Torque is a measure of the force that can cause an object to rotate about an axis.
Just as force which causes an object to accelerate in linear kinematics, torque is force that
causes an object to acquire angular acceleration. Torque is a vector quantity.
CONTINUOUS CHARGE DISTRIBUTION
The continuous charge distribution system is a system in which the charge is
uniformly distributed over the conductor. There are three types of the continuous charge
distribution system.
 Linear Charge Distribution
 Surface Charge Distribution
 Volume Charge Distribution

 Linear Charge Distribution
When the charge is non-uniformly distributed over the length of a conductor, it is called
linear charge distribution. It is also called linear charge density and is denoted by the
symbol λ (Lambda).

∆?
∆?

The unit of linear charge density is Cm
-1(Coulomb per meter)

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 Surface Charge Distribution
When the charge is uniformly distributed over the surface of the conductor, it is
called Surface Charge Density or Surface Charge Distribution. It is denoted by the
symbol σ (sigma) symbol.

∆?
∆?

The unit is C m
-2(Coulomb per meter square)
 Volume Charge Distribution
When the charge is distributed over a volume of the conductor, it is called Volume
Charge Distribution. In other words charge per unit volume is called Volume Charge
Density. Mathematically, volume charge density. It is denoted by symbol ρ (rho).
??????=
∆??????
∆??????

Its unit is Cm
-3 (Coulomb per meter cube)



GAUSS’S LAW OF ELCTROSTATIC
Electric flux through a closed surface ‘S’ is equal to @
5

,
A times charge enclosed by the
surface.
?
?

,

Significance of Gauss’s law:
 Gauss’s law is true for any closed surface, no matter what its shape or size.
 The charges may be located anywhere inside the surface.
 The surface that we choose for the application of Gauss’s law is called the Gaussian
surface. You may choose any Gaussian surface and apply Gauss’s law.
 Gauss’s law is based on the inverse square dependence on distance contained in the
Coulomb’s law
 Gauss’s law is useful for calculation of the electrostatic field when the system has
some symmetry.(Sphere, ring, rod, cylinder)

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APPLICATIONS OF GAUSS’S LAW
I) FIELD DUE TO AN INFINITELY LONG STRAIGHT UNIFORMLY CHARGED WIRE.

Consider an infinitely long thin straight wire with uniform linear charge density λ.


















Figure: The Gaussian surface for a long thin wire of uniform linear charge density
Since the wire is infinite, electric field does not depend on the position of P along
the length of the wire. Hence the electric field is everywhere radially outward through the
Gaussian surface and its magnitude depends only on the radial distance r.
The electrical flux through the bottom and top circular plane area is Zero, because E
and Area vector A area perpendicular to each other i.e cos90 = 0 hence ∅
I= 0
The electrical flux is passing through only curved surface radially symmetric. ‘E’ is
normal to the surface at every point, and its magnitude is constant. The surface area of the
curved part is 2πrl, where l is the length of the cylinder.
The Electrical flux is passing through only curved surface is,
∅
??????= EA ????????????????????????
Where A=2πrl and cos??????= 1 (i.e. E parallel to A )

∅
I= E2πrl -------------------------------(1)
According to Gauss law,
∅
I=@
?

,
A -----------------------(2)
Equating equations (1) and (2) we get and also q= λ l
l
r
r
P
E
l= length of cylinder.
r = Radius of Gaussian surface
around charged wire
P= Point where electric field is to
be estimated.
E=Electric field through a point ‘P’

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E2πrl =
λ ??????

,



II.FIELD DUE TO A UNIFORMLY CHARGED INFINITE PLANE SHEET.
Let σ be the uniform surface charge density of an infinite plane sheet. Let P be the
point at distance r from the sheet.















From eq.(1) and (2) we get
2EA=
????????????
??????
0





Hence electric field due to plane sheet independent on the distance from the plate.
NOTE: Where ??????? is a unit vector normal to the plane and going away from it. 'E' is directed
away from the plate if ?????? is positive and toward the plate if ?????? is negative.

III). FIELD DUE TO A UNIFORMLY CHARGED THIN SPHERICAL SHELL
Let ?????? be the uniform surface charge density of a thin spherical shell of radius R.
(i) Field outside the shell: Consider a point P outside the shell with radius vector r. To
calculate Electric field at P, we take the Gaussian surface to be a sphere of radius r
and with centre O, passing through P.


r
??????
→
??????
→
??????
←
??????
←
We can take the Gaussian surface to be a
cylindrical parallel piped of cross-sectional area
A. From figure, electrical flux passing through
only the two faces S1 and S2 will the rest curved
surface do not contribute to the total flux.
Therefore
∅
I= EA + EA =2EA ------(1)
Therefore the net flux through the Gaussian
surface is 2EA. The charge enclosed by the
closed surface is q=σA. Therefore by Gauss’s law,
∅
I=@
?

,
A=
?

,
--------------------(2)
S1
S2
??????
????????????
??????

E
??????
6?
,p

Vector form
??????
????????????
??????

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The electric field at each point of the Gaussian surface, therefore, has the same
magnitude E the flux through the through the small area ∆S is,
∅
I= E∆S
The total flux through the Gaussian surface is,
∅
I= E×total area of Gaussian surface(A)
∅
I= E×4πr
6
---------------------------- --(1)

From Gauss law , ∅
I=@
?

,
A ----------------------------(2)

From equation (1) & (2) we get, E×4πr
6
=@
??????
??????
0
A
but ??????=??????4????????????
6

E×4πr
6
=m
??????4????????????
2
??????
0
q


OR E =@
??????
4????????????
0??????
2
A

NOTE: The electric field is directed outward if q > 0 and inward if q < 0. This, however, is
exactly the field produced by a charge q placed at the centre O. Thus for points outside the
shell, the field due to a uniformly charged shell is as if the entire charge of the shell is
concentrated at its centre.

Surface charge
density ??????=
?
?

Gaussian
surface
P
R
r
O
E
??????=
????????????
??????
??????
????????????
??????

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(ii) Field inside the shell: In Figure the point P is inside the shell. The Gaussian surface is
again a sphere through P centered at O.












(iii) Field on the shell: When the point ‘P’ just lies on the shell then the Gaussian surface
just enclose the charged spherical shell. Hence applying the Gauss theorem
∅
I=
??????
??????
0
and ∅
I=EA=E×4πR
2

E×4πR
6
=
?

,
but ??????=????????????????????????
??????


E×4πR
6
=
??????4????????????
6
??????
4

E =
??????
??????
4


Variation of electric field due to charged spherical shell from the center of the shell















R
P
r
O
The Gaussian surface encloses no charge (q=0)
then according to Gauss’s law,
∅
I=@
?

,
A and ∅
I= E×4πr
6

E×4πr
6
=
?

,
but q=0



The field due to a uniformly charged thin shell
is zero at all points inside the shell.
Gaussian
surface
Surface charge
density ??????=
?
?

E=0
o
E
r
R
E=0

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IMPORTANT

Physical Quantity Symbol Dimensions Unit
Vector area element ∆S [??????
??????
] m
2
Electric field E [??????????????????
???????
??????
???????
] NC
-1 or Vm
-1
Electric flux ∅
I [????????????
??????
??????
???????
??????
???????
] NC
-1m
2 or Vm
Dipole moment P [??????????????????] Cm
Linear Charge density λ [??????
???????
????????????] Cm
-1
Surface Charge density ?????? [??????
???????
????????????] Cm
-2
Volume Charge density

[??????
???????
????????????] Cm
-3

1 MARK QUESTIONS
1) What is additivity of charges?
2)What is conservation of charges ?
3) What is quantisation of charges?
4) What is electrification?
5) What is charging by induction?
6) Define one coulomb of charge.
7) Define electric field at a point.
8) Write the S.I. unit of electric field.
9) Define electric flux.
10) Write the S.I. unit of electric flux.
11) What is an electric dipole?
12) What is the net charge of an electric dipole?
13) What is the direction of electric dipole moment?
14) What is a Gaussian surface?
15) How many types of charges exist in nature?
16) Name the SI unit of charge.
17) Name the instrument which detects the charge on an object.
18) What is the nature of force between like charges?
19) What is the nature of force between unlike charges?
20) What is gold leaf electroscope?
21) What is earthning?
22)How many electron are there in body whose charge is -1C?
23) How does the electrostatic force between two point charges change when a dielectric
medium is introduced between them?
24) What is the value of electric field inside spherical shell?

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th, PU I & II (Sci. & Commerce), NTSE, Olympiad, KVPY, NDA, CET, NEET, JEE. Call: 9663320948 Page 16

2 MARKS QUESTIONS
1) State and explain Coulomb’s law in electrostatics.
2) Write Coulomb’s law in vector forms and explains the terms.
3) How the coulombs force between the two charges changes when the medium is
replaced with dielectric?
4) Define electric dipole moment and write its S.I. unit
5) When does an electric dipole placed in a uniform electric field experience maximum
and minimum torque?
6) Define linear density of charge. Write its S.I. unit.
7) Define surface density of charge. Write its S.I. unit.
8) Define volume density of charge. Write its S.I. unit.
9) State and explain Gauss’s law in electrostatics.
10)State principle of superposition of forces between multiple charges.
3 MARKS QUESTIONS
1) Mention the three basic properties of an electric charge?
2) Derive an expression for electric field due to an point charge.
3) Mention the three methods of electrification/charging the body?
4) Mention any three properties of electric field lines.
5) Draw electric field lines around A i) single positive charge(q>0) ii) Two equal positive
charge ii) An electric dipole iv) single Negative charge(q<0)
6) Derive an expression for torque on a dipole placed in a uniform electric field.
7) Write three significance of Gauss’s law of electrostatic.
8) What is ratio of electric force and gravitational force between a proton and an
electron?

5 MARKS QUESTIONS
1) Derive an expression for electric field due to an electric dipole at a point on an
equatorial line.
2) Derive an expression for electric field due to an electric dipole at a point on the axial
line.
3) State Gauss’s law in electrostatics. Using the law derive an expression for electric field
due to a uniformly charged thin spherical shell at a point outside and inside the shell.
4) State Gauss’s law in electrostatics. Using the law derive an expression for electric field
at a point due to a uniformly charged infinite plane sheet.
5) State Gauss’s law in electrostatics. Using the law derive an expression for electric field
at a point due to an infinitely long, charged straight conductor using Gauss’s law.

17. 1.Electric charges and Fields Xplore Learning Centre

Coaching for 8
-10
th, PU I & II (Sci. & Commerce), NTSE, Olympiad, KVPY, NDA, CET, NEET, JEE. Call: 9663320948 Page 17

NUMERICAL PROBLEMS
1. Two point charges 20 µC and 10µC are separated by 0.05m in free space. Find the force
between them. Also calculate the force when a dielectric medium of dielectric constant
3 is introduced between them.[720N, 240N]

2. Two point charges 16nC and 8nC are situated at the corners B and C of an equilateral
triangle of side 0.03m. Find the magnitude and direction of the resultant electric field at
the vertex A of the triangle.[21.2X10
4NC
-1 , 40
053’ angle with E1]
3. Two pith balls of mass 10mg each are suspended by two threads from the same support
are charged identically. They move apart by 0.08m and threads make an angle 60
0 with
each other. Find the charge on each pith ball.[6.33nC ]

4. Two identically oppositely charged metallic spheres placed 0.5m apart attract each
other with a force of 0.108N., when they are connected to each other by a copper wire
for a short while, they begin to repel with a force of 0.036N. Calculate the initial charges
on the spheres.[+3µC, -1µC]

5. Two positively charged particles each of mass 1.7X10
-27Kg, carrying a charge of
1.6X10
-19C are kept at a certain distance in air. If each charge experiences a repulsive
force equal to its weight, find the distance of separation between the charges. [0.117m]

6. The electrostatic force on a metal sphere of charge 0.4µC due to another identical metal
sphere of charge -0.8µC in air is 0.2N. Find the distance between the same two spheres
and also find the force between the same two spheres when they are brought in to
contact and then replaced in their initial positions.

7. Three charges each equal to +4nC are placed at the three corners of a square of side 2cm.
find the electric field at the fourth corner. (M-18)

8. Two point charges of 3μC and -3μC are located 20cm apart in vacuum.
(a) What is the electric field at the midpoint ‘O’ of the line joining the two charges?
(b) If a negative test charge of magnitude 1.5nC is placed at that point, what is the force
experienced by the test charge?
9. Two point charges qA=3μC and qB=-3μC are located 20cm apart in vacuum.
a) What is the electric field at the midpoint O on the line AB joining the two charges?
b) If a negative test charge of magnitude 1.5X10
-9C is placed at this point, what is the
force experienced by the test charge? (M-17)