1 electrostatic 09

1.1 Electrostatic:-Study of electric charge at rest is called electrostatics or static electricity. i.e. Charge is not
allowed to flow from one body to another body.
The physical phenomenon which involves electric charge & its effect is called electricity.
Thalse a Greek philosopher discovered friction electricity in 600 BC. He observed that when a piece of amber (a
kind of resin) is rubbed with fur it acquires a property of attracting small pieces of dry leaves & dust. In 1600, Sir
Dr. William Gilbert (physician to Queen Elizabeth-I) found that some other bodies also acquire the same property.
He gave name electricity for this phenomenon from Greek word electrum for Amber.
The electricity produce in a body due to friction between the two bodies is called friction electricity.
OR - Charging of an object by rubbing it with another object is called friction electricity.
The agency, which gives the attractive power, is called electricity.
The body, which acquires the attracting power, is called electrified or charged.
According to Gilbert charge can be classified in two classes. Charge on the amber is called resinous because amber
is a resin. Charge developed on the wool is called vitreous.
An American scientist Benjamin Franklin (1706 to 1790) introduced a sign convention according to which charge
on amber is −ve & charge on the wool is + ve. These conventions followed till today.
Positive Charge Wool Glass rod Fur or woollen cloth Dry hair Cat skin Fur
Negative Charge Plastic Silk cloth Ebonite or amber Comb Rubber Cat skin
Application of electrostatic:- (1)Prevention of pollution of atmosphere by electrostatic precipitation of fly ash. (2)
In designing the electrostatic generator e.g. Van de Graff generator. (3) In electrostatic spraying of paint & powder.
(4) In designing the cathode ray tube for radar & TV etc. (5) In electrostatic loud speaker microphone.
1.2 Two Type of Charge:-Two glass rod
rubbed with silk , one is suspended by a
thread & another similarly charged rod
bring closed to first. The two-glass rod
repels each other. Similarly two ebonite rods
rubbed with fur also repel each other. But
the glass rod attracts the ebonite rod.
Since charge produce on the glass (ebonite)
rods is of same kind, so same type of charge repels each other. While charge on glass rod is different from charge
of ebonite so, different kind of charge attract each other.
Hence charge is of two types- (a) Positive charge (b) Negative charge.
1.3 Modern Electronic Theory:-According to this theory an atom consist of a heavy central core called nucleus.
Whole mass and total positive charge of atom lies in this small region of nucleus. The Magnitude of charge of
proton is equal to magnitude of charge of electron 1.6×10−19
Coulomb.
Electrons of outermost orbit are loosely bound. When two bodies are rubbed with each other there is transfer of
electron from one body to another body because the friction provides the necessary energy for ionisation. The body
looses electron become positively charged (glass rod) due to deficiency of electrons (mass of the body decreases).
The body gaining the electron becomes negatively charged (silk) due to excess of electrons (mass of the body
increases). However mass of electron is very small
Conservation of Charge: -Individual charges can neither be created nor destroyed it can be transferred, so we say
the total charge of a system (e.g. a rubber rod and a piece of wool) is conserved.
However, it is possible to separate positive and negative charges by transferring electrons from one material to
another. When the rubber rod is rubbed against the piece of wool, the rubber pulls electrons away from the atoms on
the surface of the wool. This leaves the wool with less electrons (becomes positively charged) and the rubber rod
with more electron (become negatively charged).
We call this state an electrified state in which the electrical balance is upset with either some electrons being
removed from or added to an electronic orbit. In this electrification process the total negative charge is equal to the
total positive charge in the system.
Quantisation of Charge: - Minimum Charge posses by a body is equal to the charge of one electron. We denote
charge by symbol "Q" and measure it in coulombs (C). The charge of a proton or an electron is the elementary
charge (e.c.) of all the matter. One coulomb is equal to the total charge of 6.25 × 1018
electrons.
According to Millikan−” charge posses by a body is equal to integral multiple of charge of electron (basic charge ‘e’)”.
Thus charge on a body exist in terms of discreet packets or quanta of `e` this property is known as quantisation of
charge. Because fraction of independent charge is not observed yet. Charge on a body is Q = ± n e
Where n = 1,2,3, - - - - - - - - - & e = 1.6 × 10– 19
Coulomb.

Recently particles of charge e/3 & 2e/3 have been observed known as quarks. But no quark has been detected in
experiments, although there are indirect evidences in favour of quarks.
1.4 Properties & difference between Mass & Charge:-
Conductors are materials through which allow electrons to move freely. Metals are the best conductors because the
outermost electrons of the atoms are loosely bounded and so can move freely between atoms. e.g.
All Metals
Insulators are materials that do not allow the electrons to move through freely. e.g. water, human body, moist air.
Most non-metals are poor conductors of electricity because they have few, or no, free electrons.
Semiconductors are materials through which very few electrons are able to move freely. e.g. Ge, Si
1.5 Types of Electrification :-(i) Charging Conductors by Conduction:- A neutral body can be charged by
touching a charged body. This is called charging by conduction. After the process, the charges on both bodies have
the same polarity.
(ii)Charging Conductors by Induction: - To charge a conductors equally and oppositely by bringing a charged
conductor closed to it (without touching) is called electric induction.
(a) A positively charged rod (inducing body) is brought near to an insulated metal conductor without touching.
Charge induced on the either side of conductor due to electric induction.
Due to the electric force, the electrons will be attracted and move closer to the rod & it become negative it is called
bound charge. Due to deficiency of electron the opposite side of conductor become positively charged it is called
free charge.
(b) Keeping the rod in position if we touch the conductor by our fingertip, the free charge (+ve) passes through our
body to the earth. (c) Now first remove the finger. (d) And then remove the rod. The bound (−ve) on
conductor distributed uniformly over its surface and the body become negatively charged.
Laws of Electrostatic induction:- (I) Inducing charge is equal to each of the induced charge.
(II) In electric induction dissimilar kind of charge induced on the body Bound Charge & Free charge.
(iii) Charging Conductors by Friction: By rubbing two bodies together, both positive and negative charges in equal
amounts appear simultaneously due to transfer of electrons from one body to the other.
(a) When a glass rod is rubbed with silk, the rod becomes positively charged while the silk becomes negatively
charged. The decrease in the mass of glass rod is equal to the total mass of electrons lost by it.
(b) Ebonite on rubbing with wool becomes negatively charged making the wool positively charged.
(c) Clouds also get charged by friction. (d) During landing or take-off, the tyres of an aircraft get electrified therefore
special material is used to manufacture them.
(e) A comb moving through dry hair gets electrically charged. It starts attracting small bits of paper.
1.6 Coulomb's Law: - The electrostatic force, exerted on a point charge Q2 by another point charge Q1
separated by a distance r is F.
Statement-“Electrostatic force, is directly proportional to the product of the magnitude of two charges and
inversely proportional to the square of the distance between them and the force act along the line joining them”.
Charge Mass
1. Charge is a physical quantity, which determines the physical
interaction between charged bodies.
1. Mass is the quantity of matter posses by
a body.
2. Charge is quantised i.e. Q = ± n e 2. Mass is not quantised.
3. Charge is conserved. 3. Mass is not conserved
4. Charge posses by a body is positive, negative or zero. 4. Mass is always 0 or +ve.
5. Charge is non-relativistic i.e. not change with speed. 5. Mass is relativistic i.e. with increases in
speed mass increases.
6. Force between the two charges is attractive & repulsive both. 6. Gravitation force is only attractive.
7. Charge is additive 7. Mass is not additive.

So F ∝ Q1 × Q2 - - - - - - - - - -(1) And F ∝ 1/r2
- - - - - - - - - -(2)
Combining Eq 1 & 2 2
21
r
qq
F ∝ 2
21
r
qq
kF = This is Coulomb’s law of electrostatics.
1) +F indicates that Q1 & Q2 are like charges and the force is repulsive, −F indicates that Q1 & Q2 are opposite and
the force is attractive.
2) The proportionality Constant k depends on medium between the charges & system of unit.
(a)For Air (i) in cgs unit k= 1 2
21
r
qq
F = (ii) In mks unit 2
21
04
1
r
qq
F
επ
=
k = 1/ ( 4πε 0 ) = 9 × 109
N m2
/C2
Where ε0 (epsilon zero) is called the permittivity of the vacuum or free space.ε0 = 8.85 × 10−12
C2
/N m2
(The
electrostatic force can act through empty space or a vacuum.)
(b) For a given medium, (i) In cgs unit k = 1/εr . For air, εr = 1.005 ≈ 1, Thus, for air εr ≈ ε0 .
Where K or εr is called dielectric constant or relative permittivity or specific inductive capacity (SIC) of the
medium. εr has no unit.
vacuumoftypermittivi
mediumoftypermittiviAbsolute
r ==
0ε
ε
ε (ii) In SI units, K
k
04
1
επ
=
3. Vector form of Coulomb’s law:−
1.7 Dielectric Constant is define as the ratio of the electrostatic force between two charges separated by some
distance in air to the force experienced by same charges kept at same distance in a medium. It is also called relative
permittivity or specific inductive capacity or dielectric coefficient.
Let Q1 & Q2 are two charges separated by a distance r in air ,then force between them is 2
21
04
1
r
qq
Fa
επ
=
If charges are kept in a medium then force 2
21
04
1
r
qq
K
Fm
επ
= so
med
air
F
F
K =
1.6 Unit of Electric Charge:- S I unit of electric charge is Coulomb. If Q1 = Q2 = Q & r =1 m &force F = 9×10 9
N
F = 9 × 10 9
Q1 × Q2 / r2
So 9 × 10 9
= 9 × 10 9
Q× Q / 12
Hence Q = ± 1
If two equal charges separated by a distance 1 m in air, experience a force of repulsion of 9 × 10 9
N then each
charge is said to be one coulomb. 1 Coulomb = 3 × 10 9
Stat Coulomb
CGS unit of electric charge is state coulomb or e.s.u (electrostatic unit) of charge.
Electrostatic Force Gravitation Force
a. It depends on medium between the charges (it become
1 / K times)
a. Gravitation force does not depend on medium
between the bodies.
b. It is stronger force (1036
times gravitation force) b. It is weak force.
c. It may be attractive or repulsive c. It is always attractive
Similarity:- 1. Both forces obey inverse squire law. 2. Both act in vacuum.
3. Both forces are central forces i.e. act along the line joining the two charges/masses
4. Both are conservative force i.e. work is independent of the path in these field.
1.7 Principle of Superposition: - Statement-“ In the presence of many charges the total force act on a given
charge is equal to vector sum of the forces exerted on it by all other charges”.
This principle gives a method to find out force on a charge 1 when n charges are interacting.
→→→→→
+−−−−−+++= nFFFFF 1141312
Similarly electric field
→→→→→
+−−−−−+++= nEEEEE 1141312
1.8 Continuous Distribution of Charge:-

A system of closely spaced charges forms a continuous distribution of charge.
(a) Charge per unit length is define as linear charge density λ
λ = total charge posses by the conductor / total length of the
conductor = q / l
(b) Charge per unit area is define as surface charge density σ
σ = Total charge on the surface / total surface area = q / A
(c) Charge per unit volume is define as volumetric charge
density ρ
ρ= Total charge on the body/Volume of the body = q /V
1.8 Electric Field: - Electric field is the space around a charged body, where electric interaction can be
experience by another charged body.
The Electric field intensity or strength of an electric field
at any point in the field is directly related to the force it
experience by a one coulomb positive (small test charge Qo →0) placed at that point.
The strength can be calculated by using the formula E = F/Q0. OR F = Q0 E.
Hence 02
0
04
1
QE
r
QQ
=
επ
2
04
1
r
Q
E
επ
=
In vector form
In this formula, E stands for the strength of the field, F for the force, and Q0 for the value of the test charge. The
direction of an electric field at any point is the same as the direction of the force on a positive test charge at that
point i. e. outward from positive body & towards negative body.
In SI, the unit of the electric field intensity is Newton per Coulomb ( N/C )
1.9 Electric Lines of Forces:- Like a gravitational field, an electric field is represented by a series of arrow lines
called electric field lines or flux lines this concept of lines of force is given by Michel Faraday.
An electric field line is the path along which a free unit positive charge would move in a free space. It is an
imaginary straight line or curve. OR If a unit positive charge is free to move in an electric field, then
path followed by the unit positive charge is called electric lines of force.
For a positive test charge, field direction radiate outwards. For a negative test charge, field direction radiate inwards.
[1] The electric field lines are always originating normally from a positive charge and terminate normally to the
negative charge. (If not so then electron would flow due to effect of field i.e. electric current would flow).
[2] The lines of force in uniform field are parallel equidistant straight lines.
[3] The electric field intensity is defined as the number (density) of the electric field lines passing through unit area
placed ⊥ to the electric field. More lines are drawn in the area of stronger field.

[4] A tangent drawn at a point on electric line will give direction of electric field at that point.
{5} Two lines of force never intersect each other. If two lines of force intersect each other at a point ‘P’ then there
will be two tangents at point ‘P’ on two curves giving two direction of electric field at same point. But it is
impossible because electric field is a vector quantity which posses both magnitude & definite direction.
[6] Lines of force tend to contract lengthwise (longitudinal tension). This explain unlike charge attract each other.
[7] The lines of force tend
to repel each other (lateral
repulsion). This explains
like charge repels each
other.
[8] Electric lines do not
passes through a close
conductor i.e. electric field
inside a conductor is zero.
Neutral Point: - the point where net electric field intensity is zero called neutral point.
Uniform Electric Field:- The field where magnitude and direction of field intensity are constant and same is called
uniform electric field. 2
04
1
r
dl
dE
λ
επ
=
1.10 Electric Field due to Uniformly Charged Ring: -Consider a positive charged q distributed uniformly
over the surface of a metal ring of radius r having linear charge density λ = dq /dl
Electric field due to the length element ABC is 2
0
2
0 4
1
4
1
a
dq
a
dl
dE
πε
λ
πε
== { λ dl = dq}
From resolution of vector at point P there are two components dE cosθ along the axis and dE sinθ perpendicular to the
axis. Components dE sinθ of segment ABC & DEF are equal & opposite on Y-axis so they cancel out each other. So the
effective value of electric field at P due to element ABC is
a
x
a
dq
dEE
QQ
×== ∫∫ 2
000 4
1
cos
πε
θ
axBOP /cos =∆ θ Hence
∫=
Q
dq
a
x
E
0
3
0
1
4πε
[ ]Q
q
a
x
E 03
0
1
4πε
=
[ ]0
1
4 3
0
−= Q
a
x
E
πε
3
04
1
a
xQ
E
επ
= 22
xraBOPinBut +=∆ 2/322
0 )(4
1
xa
xq
Eso
+
=
επ
Special Case−I At the centre of the ring x = 0 so electric field x = 0
Special Case−II At a point much away from the centre of the cercular coil x >> r so neglecting r2
2/32
0 )(4 x
xq
E
επ
= So 2
04
1
x
q
E
επ
=
Motion of Charge Particle in Electric Field:− (a) Suppose a charge particle having charge Q and mass m is
initially at rest in an electric field of strength E. The particle will experience an electric force which causes it's motion.
(i) Force and acceleration : The force experienced by the charged particle is QEF = .

Acceleration produced by this force is
m
QE
m
F
a ==
(ii) Velocity : Suppose at point A particle is at rest and in time t, it reaches the point B where it's velocity becomes
v. Also if ∆V = Potential difference between A and B, S = Separation between A and B
⇒
m
tEQ
v =
m
VQ∆
=
2
(2) When a charged particle enters with an initial velocity at right angle to the uniform field
When charged particle enters perpendicularly in an electric field, it describe a parabolic path as shown
(i) Equation of trajectory : Throughout the motion particle has uniform velocity along x-axis and horizontal
displacement (x) is given by the equation x = ut
Since the motion of the particle is accelerated along y–axis
So
2
2
1












=
u
x
m
QE
y ; this is the equation of parabola which shows 2
xy ∝
(ii) Velocity at any instant : At any instant t, uvx = and
m
QEt
vy = so
2
222
222
||
m
tEQ
uvvvv yx +=+== &
mu
QEt
v
v
x
y
==βtan
If β is the angle made by v with x-axis .
Suspended charge System of three collinear charge
Freely suspended charge
In equilibrium
mgQE =
Q
mg
E =⇒
Suspension of charge from string
In equilibrium
QET =θsin ….(i)
mgT =θcos ….(ii)
From equations (i) and (ii)
( ) ( )22
mgQET += and mg
QE
=θtan
In the following figure three charges Q1, Q and
Q2 are kept along a straight line, charge Q will be
in equilibrium if and only if
|Force applied by charge Q1|
= |Force applied by charge Q2 |
i.e. 2
2
2
2
1
1
x
QQ
x
QQ
= ⇒
2
2
1
2
1








=
x
x
Q
Q
This is the necessary condition for Q to be in
equilibrium.
If all the three charges (Q1, Q and Q2) are similar,
Q will be in stable equilibrium.
If extreme charges are similar while charge Q is
of different nature so Q will be in unstable
equilibrium.
1.11Electric Dipole:-A system of two equal but opposite charges separated by a small distance is called electric dipole.
Consider a dipole of charge +q & −q separated by distance 2l.The strength of the dipole is measured by a physical
quantity called dipole moment. It is a vector quantity; its direction is from negative to positive charge.
The product of magnitude of one of the charge and distance between the charges is called dipole moment “p”.
p = q × 2l Its SI unit is Coulomb × meter.
1.12 Electric field intensity at a point on
axial line: - Consider a dipole of charge +q & −q
separated by a distance 2l so dipole moment p = 2ql
We have to find field intensity at point P, at distance r
from centre of the dipole. So electric field due to +q
charge is
( )
( )2
4
1
4
1
2
0
12
0
1 −−−−−−
−
==
lr
q
ESo
BP
q
E
πεπε {along OP}
Intensity at P due to –q is
( )
( )3
4
1
4
1
2
0
22
0
2 −−−−
+
==
lr
q
ESo
AP
q
E
πεπε {along PO}
So resultant intensity at P is E = E1 + E2 =
( ) ( )2
0
2
0 4
1
4
1
lr
q
lr
q
+
−
− πεπε
E

A B
S
Y
X
E
u
P(x, y)
vy
β
v
+Q
F = QE
mg
E

QQ1
Q2
x
x1
x2
QE
E

θ
θ
l
mg
T sin θ
T cos θ
T

( ) ( )






+
−
−
= 22
0
11
4 lrlr
q
E
πε
So
( ) ( )
( ) ( ) 







+−
−−+
= 22
22
04 lrlr
lrlrq
E
πε
( ) ( )
( )( ) 2
2222
0 }{
22
4 lrlr
rllrrllrq
E
+−
−+−++
=
πε
So
( )222
0
4
4 lr
lrq
E
−
=
πε
( )222
0
22
4
1
lr
lqr
E
−
××
=
πε Therefore ( )222
0
2
4
1
lr
pr
E
−
=
πε Since p = q×2l , is electric dipole moment.
Special case:− If dipole is very small, then r > l so r2
>>l2
, therefore neglecting r2
.
( )22
0
2
4
1
r
pr
E
πε
= Therefore 3
0
2
4
1
r
p
E
πε
= In vector form 3
0
2
4
1
r
p
E
→
→
=
πε
.
1.12Electric Field at a point on equatorial line of a dipole:- Consider a dipole with charges + q & − q
separated by a distance of 2 l. So dipole moment p = 2 q l - - - -(1)
We have to find intensity of electric field at point A on the equatorial line
of the dipole at distance ‘r’ from the centre of the dipole. So
222
lra +=
Electric field intensity at A due to +q charge is
2
0
2
0
1
4
1
4
1
a
q
AP
q
E
επεπ
== along BA
22
0
1
4
1
lr
q
E
+
=
επ - -- - - - (2) Electric field at A due to –q charge is
2
0
2
0
2
4
1
4
1
a
q
BP
q
E
∈
==
πεπ 22
0
2
4
1
lr
q
E
+
=
επ - - -
- (3) {Along AC} Since ( ) .021 LetEEE ==
θθ SinESinESo 21 = . Since the vertical components are equal but opposite so the cancels out each other but
the horizontal component are along same direction so resultant field intensity at A due to dipole is
θθθθ coscoscoscos 0021 EEEEE +=+= Therefore θcos2 0EE =
θ
επ
Cos
lr
q
E
o
222
4
1
2
+
×= 2222
0
2
4
1
lr
l
lr
q
+
×
+
=
επ Therefore E = 2/322
0 )(4
1
lr
p
+επ
Where P= 2ql is dipole moment. Its direction is parallel to the dipole axis and from +ve to –ve.
Special case:- For a small dipole 2l is very small so l < r hence l2
<< r2
. Now we can neglect l2.
E = 2/32
0 )(4
1
r
p
επ E = 3
04
1
r
p
επ
1.13a. Torque and Force acting on a dipole placed in uniform
electric field:- Consider a dipole with charges + q & - q separated by a
distance of 2 l, angle between the dipole moment & electric field intensity is θ.
So dipole moment p = 2 q l
Force on each charge is F = q E -- - --- (1)
Force on + q is along the direction of field & on – q opposite to the direction of
the field so net force on the dipole is zero. The system of two equal but
opposite charges constitute a couple. The torque of the couple
τ = Force × perpendicular distance τ = F × AC - - - - - - - - (2)
In ∆ABC sin θ = BC/AB ⇒ BC = AB sin θ = 2l sin θ - - - - -- -(3)
Putting values from eq1 & eq3 in eq 2 τ = qE × 2l Sin θ
But the dipole moment p = 2ql so τ = p E sin θ - - - - - - - - - -(4)
In vector form
→→→
×= Epτ
The direction of the torque is perpendicular to both E and p & given by Fleming’s right hand rule.
Case-I θ = 0 Then the torque τ = pE sin 0 = 0
Hence if dipole is placed parallel to the field torque on it is minimum.
Case-II θ = 90 Then the torque τ = pE sin 90 = p E

Hence if dipole is placed perpendicular to the field torque on it is maximum.
1.13b. Work done & Potential Energy in deflecting a Dipole in Uniform Electric Field:-If the dipole
rotate through an angle dθ against the torque τ = pE sin θ
Then work done dW = τ dθ = pE sin θ × dθ
So Total work done in deflecting the dipole from angle θ12 to θ2 is ∫=
2
1
θ
θ
τW dθ = ∫
2
1
sin
θ
θ
θEp dθ
W = θ
θ
θ
θ dEp ∫
2
1
sin ⇒ ( ) ⇒= − θ θ
θ
cos 2
1
EpW ( )12 coscos θθ −−= EpW
Work done : -In equilibrium dipole is parallel to the field so θ1 = 0 . Hence work done in deflecting the dipole
through an angle θ1 = θ. ( ) ⇒−−= 0coscosθEpW ( )θcos1 −= EpW
Potential Energy: -Work done in deflecting the dipole perpendicular to the field so θ1 = 90 . Hence work done in
deflecting the dipole through an angle θ2 =θ. W = U = − p E ( cos θ − cos 90)
Hence potential energy U = − p E cos θ or
→→
−= EpU .
θ = 0o
, τ = 0 & Umin = – pE (minimum) θ = 90O
, τmax = pE, U = 0, θ = 180o
, τ = 0, Umax = pE(maximum)
Stable equilibrium Not in equilibrium Unstable equilibrium
1.15 Electric Flux:-Total number of electric lines passing through any surface area in the electric field is called
electric flux φ. It is a scalar quantity.
dφ = dS . E Cos θ So dφ
= E dS Cos θ
Hence total flux through the
hole surface
φ = ∫
S
dφ
= ∫ θCosdSE
Hence ∫
→→
=
S
SdE .φ
Area Vector: - area is considered as vector quantity having direction
perpendicular to the surface area.
Solid Angle:-In two dimension, the angle subtended by an arc of length dl of a circle of radius r is,
Angle θ = arc / radius = dl/r
Due to identical properties concept of solid angle is given. Consider a sphere of radius ‘r’ & small surface area
element dS on it. The solid angle subtended by the surface element is 2
/ rdSd =Ω
If area vector makes an angle θ ,with the line joining point O & surface then 2
r
CosdS
d
θ
=Ω
SI unit of solid angle is Steradian. The total solid angle over the whole surface is 2
2
2
4
r
r
r
S π
==Ω OR Ω = 4π
1.16 Gauss`s Theorem of Electrostatics: -Karl Fiedric Gauss gave a theorem which gives the value of net
outward electric flux through any closed hypothetical surface called Gaussian surface.
Statement (SI unit): -“The surface integral of the normal component of the electric field over any closed
hypothetical surface is equal to 1/ε0 total charge enclosed by the surface”.
OR “The net outward electric flux through any closed hypothetical surface is equal to 1/ε0 total charge
enclosed by the surface”. ∫
→→
=
s
Q
dSE
0
.
ε OR 0ε
φ
Q
=
Proof:- Consider a closed surface S with charge + q lies inside it at O. dS is a small surface area element
surrounding point P, which is at distance ‘r’ from point O. Electric field intensity at point P due to charge + q is
2
04
1
r
q
E
επ
= - - - - - - - - - - - - - - - - (1)
p
→
E
→ p
→
E
→
p
→
E
→

If θ is the angle between E & dS then net outward electric flux ∫
→→
=
s
dSE .φ
φ = ∫
S
E dS cos θ =φ⇒ ∫
S
2
04
1
r
q
επ
dS cos θ
=φ
04 επ
q
∫
S r
dS
2
cos θ
=φ⇒
04 επ
q
∫
S
Ωd =⇒ φ
04 επ
q
× 4π
∫
S
Ωd = 4π is the total solid angle subtended by the whole surface area.
Hence net outward flux =φ
0ε
q
For large number of charges the total outward flux is −−−−+φ+φ+φ=φ 321 OR =φ
o
q
ε
∑
Hence
0ε
φ
Q
= Where ∑ q = Q is total charge enclosed by the surface. ∫
→→
==
s
Q
dSE
0
.
ε
φ
1.17 Application of Gauss’s Theorem :-(a)[I]Electric Field Intensity due to uniformly Charged Hollow
Sphere:-Consider a charged sphere of radius ‘R’ having charge + q. P is a point outside the sphere at distance ‘r’
from the centre of the sphere where we want to find electric field intensity.
We will draw an imaginary sphere of radius ‘r’ with ‘O’ as a centre; this surface is Gaussian surface & point ‘P’ lies
on it. Electric field at each point on the Gaussian surface is same & directed perpendicularly outward so angle
between E & dS is θ = 0
Therefore net outward flux through the Gaussian surface is
∫∫ ×==
s
1dSE
s
0cosdSEφ
− − − − − (1)
From Gauss’s theorem, net outward flux 0/ ∈=qφ − − (2) Comparing eq1 &
eq2
0
2 q
rπ4E
ε
=× Hence 2
0 r
q
π4
1
E
ε
=
Hence electric field out side the sphere will be same as if whole charge is
concentrate at the centre of the sphere.
(II)At a Point lies on the surface of the Conductor: - In this case the radius of the Gaussian surface is equal
to radius of the conductor i.e. r = R Hence electric field intensity
2
0 R
q
π4
1
E
ε
=
(III)At a Point lies inside the surface of the Conductor: - Consider a
charged sphere of radius ‘R’ having charge + q.P is a point in side the
sphere at distance ‘r’ from the centre of the sphere where we want to find
electric field intensity. We will draw an imaginary sphere of radius ‘r’ with
‘O’ as a centre; this surface is Gaussian surface & point ‘P’ lies on it.
Electric field at each point on the Gaussian surface is same & directed
perpendicularly outward so angle between E & dS is θ = 0.
Therefore net outward flux through the Gaussian surface is
∫∫ ×==
s
1dSE
s
0cosdSEφ
− − − − − (1)
From Gauss’s theorem, net outward flux
0
q
ε
φ = OR 0
0
0
==
ε
φ − − − (2) Comparing eq1 and eq2. So
∫ =×
s
0
1dSE
0ε

(Since there is no charge lies inside the Gaussian surface so q = 0.)
02rπ4E =× Hence E = 0 Hence electric field inside the hollow conductor is zero.
All three cases are represented graphically below. It shows the variation of electric field intensity E with distance
from the centre of a uniformly charged spherical shell.
(b)Electric Field Intensity due to uniformly Charged Plane Metal Sheet:-Consider a thin metallic
plate of uniformly charged density σ = q / dS - - - - - - - - -(1)
We have to find field intensity at a point P at distance r from
plate. Consider an imaginary cylinder across the plate, it act as
Gaussian surface & point P lies on it.
(I ) Electric field is parallel to the plane surface ‘A’ i.e. area
vector is parallel to the electric field θ = 0,so flux is maximum&
maximum electric line passes through the surface.
∫=
s
0CosdSE1φ ∫=
s
dSE1φ
= SE ×
( II )Electric field is parallel to the plane surface ‘B’ i.e. area
vector is parallel to the electric field θ = 0,so flux is maximum&
maximum electric line passes through the surface.
SE
s s
dSE0CosdSE 222 ×=⇒∫ ∫=⇒= φφφ
(III) Electric field is perpendicular to the curved surface i.e. area vector is perpendicular to the electric field θ = 90,
so flux is minimum& no electric line passes through the surface. ∫ ==
s
090cosdSE3φ
The total electric flux through the Gaussian surface is
221 φ+φ+φ=φ
ES20SESE =++=φ − − − − −(1)
From Gaussian theorem )2(
0
−−−−−=
ε
φ
q
Comparing eq1 & eq2
⇒=⇒=
0
0
2
/qSΕ2
ε
ε
S
q
E
02ε
σ
=E
So electric field is independent of the distance of the point i.e. constant.
(c)Electric Field Intensity due to two uniformly Charged Plane Metal Sheet:-Consider two parallel long
metallic plate having + ve charge with surface charge density σ1 and σ2. Electric field intensity near plate A & B is E1 & E2.
(a)Intensity in Ist
region:-In this region electric field due to both plates is from right to left. So intensity in the first
region is E = −E1 − E2
From Application of Gaussian theorem. )1(
2
)σσ(
2
σ
2
σ
0
21
0
2
0
1
−−−−−
+−
=−−=
εεε
E
(b)Intensity in II nd
region:-In this region electric field due to plate B is from right to left & due to plate A is from
left to right. So intensity in the second region is, E = E1 − E2 From Application of Gaussian theorem.
)2(........
2
)σσ(
2
σ
2
σ
0
21
0
2
0
1
εεε
−
=−=E
Spacial Case:-If second plate is – ve and having same charge density σ then E = E1 + E2
So intensity in the second region is
000 2
σ2
2
σ
2
σ
εεε
=+=E )3(
σ
0
−−−−=
ε
E
(c)Intensity in III rd
region:-In this region electric field due to both plates is from left to right. So intensity in first
region is E = E1 + E2
From Application of Gaussian theorem. )4(
2
)σσ(
2
σ
2
σ
0
21
0
2
0
1
−−−−−
+
=+=
εεε
E
( d ) Electric Field Intensity due to two uniformly Charged infinite long Charged Conductor:-
Consider a thin infinite long positively charged straight long conductor of linear charge density
λ = q / l - - - -(1).
We have to find electric field intensity at point P, which is at distance ‘r’ from the plate.

With conductor as an axis we will draw an imaginary cylinder of radius ‘r’ & length ‘l’ it is called Gaussian surface
& point ‘P’ lies on it. Electric field at each point on the conductor is same & directed perpendicularly outward.
From Gauss’s theorem the outward flux through the Gaussian surface is
(i)Flux through the plane surfaces A & B is minimum because area vector is perpendicular to the field i.e. θ = 90o
hence minimum lines pass through it ∫ ===
s
dSE 090cos21 φφ --------( 1 )
Due to same reason flux through plane surface B is also zero.
( ii )Flux through the curved surface is maximum because area vector is parallel to the field i.e. θ = 0 hence
maximum lines pass through it.
)2(20cos 33 −−−×=⇒== ∫ ∫ lrEdSEdSE
s s
πφφ
Hence net outward flux through the Gaussian surface is φ = φ1 + φ2 + φ3
From eq 1 & 2,
)3(2200 −−−=⇒×++= ElrlrE πφπφ
From Gauss’s theorem )4(
0
−−−−=
ε
φ
q
Comparing eq3 & 4
r
EFromeq
rl
q
E
q
Elr
000 2
1
2
2
πε
λ
πεε
π ==⇒=
Hence electric field at a point near a straight long conductor is inversely
proportional to distance of the point.
( e )To Derive Coulomb’s Law:-From application (a) I , derive the
formula of field intensity at a point on the conductor 2
04
1
r
q
E
επ
= If we keep a point charge q0 at point P
,then the force on the test charge is F = q0 E = 2
0
0
4
1
r
q
q
επ
× 2
0
04
1
r
qq
F
επ
= This is Coulomb’s
Law.
Principle of superposition
1. Name the law which is a mathematical equivalent
of Coulomb's law and superposition principle.(92) 1
Coulomb force
1. How does the Coulomb force between two point
charges depend upon the dielectric constant of the
intervening medium? [2005](1
2. A charge ‘q’ is placed at the center of the line joining
two equal charges Q. Show that the system of three
charges will be in equilibrium if small q = −Q/4 [05](2
3. Two fixed point +4e and +e units are separated by a
distance ‘a’. Where should the third point charge be
placed for it to be in equilibrium? [2005](2
Electric field, strength/intensity
1. Define electric field at a point. (1990, 91, 92) 1
2. If an oil drop of weight 3.3×10−13
N is balanced in an
electric field of 5×105
V/m, find charge on the drop.
3. Is electric field-intensity a scalar or a 'vector’
quantity? (1999, 2000) 1
4. Electric field inside a conductor is zero. Explain.
5. Define the term 'electric field intensity'.(97,2000)1
6. Name the physical quantity which has joule/coulomb as
its unit. Is it a scalar or a vector quantity? (2003)1
SI unit of Electric Field
Write the S.I, unit of electric field intensity.(1999, 2000,
2003) 1
Types of field
1. What is a conservative field? (1990) 1
Electric field due to one or more point charges
1. Two point charges q1=+0.2 C and q2 = + 0.4C are
placed 0.1 m apart. Calculate the electric field at (a) the
mid-point between the charges. (b) a point on the line
joining q1and q2 such that it is 0.05 m away from q2
& 0.15 m away from q1. (1993) 3
2. Derive an expression for the electric field intensity
at a distance V from a point charge q. (1994) 2
3.Two point electric charges of unknown magnitude
and sign are placed at a distance‘d’ apart. The electric
field intensity is zero at a point, not between the charges
but on the line joining them. Write two essential
conditions for this to happen. (1997) 2
4.Two point charges of 5×10−19
C & 20×10−19
C are
separated by a distance of 2 m. Find a point on the line
joining them at which electric field intensity is 0. (01) 2
5. Two point charges qA= +3μC & qB= −3 μC are
located 20 cm apart in vacuum. (i) Find die electric field
at the midpoint of the line AB joining the two charges.
(ii)If a negative test charge of magnitude 1.5×10−1
C
is placed at the centre, find the force experienced by the
test charge. (2003)2
6. Two similarly and equally charged identical metal
spheres A and B repel each other with a force of
2 ×10−5
N. A third identical uncharged sphere C is touched
with A and then placed at mid point between A and B.
Calculate the net electric force on C.(2003)2
7. A spherical conducting shell of inner radius r, and
outer radius r, has a charge 'Q'. A charge 'q' is placed at
the centre of the shell. (a) What is the surface charge

density on the (i) inner surface, (ii) outer surface of the
shell ? (b) Write the expression for the electric field at.a
point x > r2 from the centre of the shell. (2) 2010
Electric dipole, Electric dipole moment
1. Define electric dipole moment. (92,93,94,95,99,02)1
2. Two charges, + 5μC and −5μC are placed 1mm, apart.
Calculate dipole moment. (1994) 2
3. Draw electric lines of force for a dipole. (93, 95) 2
5. Write the S.I unit of electric dipole moment. (1992,
93, 95, 02, 03,08) 1
6. Define the term electric dipole moment. Is it a scalar or
vector quantity? [2006, 08](1
Electric field of a dipole at a point on its axis
1. An electron is separated from a proton through a
distance of 0.53A. Calculate the electric field at the
location of electron. (1992) 2
2. Calculate electric intensity due to an electric dipole at
a point on the axial line. (1992) 2
Ele. Field of dipole at a point on its equator
1. Two-point charges + q and − q are placed at a
distance 2a apart. Calculate the electric field at a point P
situated at a distance ‘r’ along the perpendicular
bisector of she line joining the charges. What is the
field when r >> a?(1990) 5
2. An election moves a distance of 6 cm when
accelerated from rest by an electric field of strength
2×104
N/C. Calculate the time of travel. (91)5
3. Calculate the electric intensity due to an electric dipole at
a point on its equatorial line.(92,93,94)2
4. At what points of the electric dipole is electric field
intensity parallel to the line joining the charges? (97) 2
5. Show mathematically that the electric field intensity
due to a short dipole at a distance ‘d’ along its axis is
twice the intensity at the same distance along the
equatorial axis. (1999) 5
Net force on an electric dipole placed in a field
1. What is the net force on an electric dipole placed in
a uniform electric field? (1992) 2
Net torque on an electric dipole placed in a field
1. Show that in a uniform electric field a dipole
experience only a torque but no net force. Derive an
expression for the torque.(92, 93, 94, 97, 98. 99, 02)
2. An electric dipole is held at an angle 'θ' in uniform
external electric field 'E'. Will there be any (i) net force
(ii) torque acting on it? Explain what happens to the
dipole on being released. (93, 94)5
3. An electric dipole, when held at 300
with respect to a
uniform electric field of 104N/C, experiences a torque of 9 ×
10−26
Nm. Calculate dipole moment of the dipole. (96) 2
4. An electric dipole is held in a uniform electric field.
Show that no translatory force acts on it (1997, 98) 3
5. Which orientation of an electric dipole in a uniform
electric field would corresponds
to stable equilibrium? (1 (08)
6. (a) Derive an expression for
the torque experienced by an
electric dipole kept in a uniform
electric field. (b) Calculate the
work done to dissociate the system of three charges placed
on the vertices of a triangle as shown. Here q = 1.6 × 10−10
C 2008 (5
Work in rotating elec. dipole in an electric field
1. Derive an expression for the total work done in
rotating an electric dipole through an angle θ in a
uniform electric field of 105
N/C, if it experiences
a torque of 8√3 Nm, calculate the
(i) magnitude of the charge on the dipole and
(ii) potential energy of the dipole. (2000) 2
2. What orientation of an electric dipole in a uniform
electric field corresponds to its stable equilibrium? (01)1
3. An electric dipole of length 4 cm, when placed with
its axis making an angle of 60° with a uniform electric
field experiences a torque of 4√3 Nm. Calculate (i)
magnitude of the electric field, (ii) potential energy of
the dipole, if the dipole has charge of ± 9 nC. (04)2
Electric lines of force & their properties
1. Why two electric line. Of force do not intersect?
(1992, 2001, 03) 1
2. Draw one Equipotential surface (I) in a uniform
electric field and (ii) for a point charge (Q <0).(01)2
3. Why electric field lies never cross each other?[05](1
3. A positive point charge (+q) is kept in the vicinity of an
uncharged conducting plate. Sketch electric field line
originating from the point on to the surface of the plate.
Derive the expression for the electric field at the surface of
a charged conductor. 2009(3
Electric lines of force of isolated charges
1. Give two properties of electric lines of force. Sketch
them for an isolated positive point charge. (92, 2003) 2
2. Sketch the electric lines of force due to point charges
(i) q < 0 and (ii) q > 0. (1995) 1
Electric lines of multiple charges
1. Draw a diagram to show lines of force in a place
containing two equal point charges of opposite sign
separated by a small distance. Giving reason, indicate on
the diagram a point where a small positive charge would
experience a force parallel to the line joining the two
charges. (1993) 2
2. What is an electric line of
force? Sketch lines of uniform
electric field. (92, 95) 2
3. The dipole is aligned parallel to the field. Calculate
work done in rotating it through 180°.(97) 5
P.E. of an electric dipole in an electric field
1. An electric dipole of length 2 cm is placed with its
axis making an angle of 60° with respect to a force due
to two equal positive charges placed at a small
distance apart in air. (1997, 2003)
3. What is an electric line of force? (2003) 1
Electric lines of uniform electric field
1. Draw lines of force to represent a uniform electric field.
(1992, 95) 1
2. Explain the physical significance of electric
potential and field.
3. Show that the integral of electric field intensity
between any two points depends only on the position

of these points and is independent of the path
followed between these points.
4. Using Gauss's theorem, derive an expression
for electric field intensity at a point (a) outside
(b) inside due to (i) a line of charge
(ii) a solid sphere of charge
4. Explain the concept of electric potential energy.
Derive an expression for potential energy of a
system of two point charges. Generalize the
expression for N discrete charges. State Gauss's
theorem in electrostatics. How will you prove it for
spherically symmetric surfaces?
6. Deduce Coulomb's law from Gauss's theorem.
7. Derive a relation for work done in moving a
charge in an electric field
8. An electrostatic field lines cannot be discontinuous
why? [2005] (1
Gauss's theorem
1. Give mathematical expression for Gauss's theorem.
(92, 95, 2001, 03) 1
2. State Gauss's theorem. (92,95,96,00,01,03,04) 1
3. Apply Gauss' theorem to obtain an expression for
electric field due to an infinite plane sheet of charge.
(2003, 04) 3
Electric flux
1. A spherical Gaussian surface encloses a charge of
8.85 × 10−8
C. (i) Calculate the electric flux passing
through the surface; (ii) If the radius of the Gaussian
surface is doubled, how would the flux change?(01)2
3. Define electric flux. Write its SI unit. A charge q is
enclosed by a spherical surface of radius R. If the radius is
reduced to half, how would the electric flux through the
surface change? 2009(2
Meaning of electric flux linked with a surface
1. An electric flux of−6 × 103
Nm2
/C passes normally
through a spherical Gaussian surface of radius 10 cm,
due to a point charge placed at the center. (i)What is
the charge enclosed by the Gaussian surface? (ii)If
the radius of the Gaussian surface is doubled, how
much flux would pass through the surface? (99)
2. Figure shows electric lines of force due to two point
charges q1 and q2 placed at points A and B respectively.
Write the nature of charge on them.(03)1
Misc Qs on Gauss' theorem
1. If Coulomb's law involved l/r dependence
(instead of 1/r2
), would Gauss's law still hold.(91) 1
Electric field & potential due to a point charge
1. The electric field at a point due to a point charge is
30 N/C and the electric potential at that point is 15 J/C.
Calculate the distance of the point from the charge and
the magnitude of the charge. (1996) 2
Electric field & potential near a line charge
1. Use it to derive an expression for the electric
field of a thin infinitely long straight line of charge,
with a uniform charge density of λ C /m. (91,04)3
2. State Guass theorem. Apply this theorem to obtain the
expression for the electric field intensity at a point due to
infinitely long, thin, uniformly charged straight wires.
[2005] (3
3. State Guass’s theorem in electrostatics. Apply this
theorem to derive an expression for electric field intensity at
a point near an infinitely long straight charged wire. [07](3
Electric field of a thin sheet of charge
1. Calculate electric field of an infinite plane sheet
of charge. (1992, 95, 96, 2000, 01) 5
2. State Guass theorem. Apply this theorem to obtain
expression for the electric field intensity at a point due to
an infinitely large thin plane sheet of charge. [05](3
3. Show that the electric freld at the surface of a
charged conductor is given by
∧→
= nE
0
σ
ε
, where
∧
n
is the surface charge density and f, is a unit vector
normal to the surface in the outward direction.(2) 2010
Electric field due to a thin spherical shell of charge
1. Derive expressions for the electric field due to
a uniformly charged spherical shell at a point
(i) inside and (ii) outside the shell.(91, 92, 01) 2
2. Derive expressions for the electric field flue to a
uniformly charged spherical shell at a point inside the
shell. (91, 92, 01, 04) 2
3. Apply this to show that for a spherical shell, the electric
field inside the shell vanishes, whereas outside it , field is as if
all the charge had been concentrated at the centre. (92)5
4. Apply this theorem to calculate electric field
inside a hollow conducting sphere.(95, 97, 01) 3
5. A charge of 12 μ C is given to a hollow metallic
sphere of radius 0.1m. Find potential at. (97, 2000)1
(1) the surface of the sphere , and (2) the centre of
the sphere. (1995) 2
6. A charge of+l0μC is given to a hollow metallic
sphere of radius 0.l m. Find the potential at the (i) outer
surface and (ii) centre of the sphere.(1999) 2
7. Prove that no electric field exists inside a hollow
charged sphere. (2002) 2
8. Using the Gauss theorem show mathematically that for
any point outside the shell, the field due to a uniformly
charge thin spherical shell is the same as if the entire
charge of the shell is concentrated at the center. Why do
you expect the electric filed inside the shell to be zero
according to this theorem? [2006](3
9. State Guass’s theorem in electrostatics. Apply this
theorem to derive an expression for electric field intensity at
a point outside a uniform charge thin spherical shell. [07](5
9. (a) Using Gauss’ law, derive an expression for the
electric field intensity at any point outside a uniformly
charge thin spherical shell of radius R and charge
densityσC /m2 Draw the field lines when the charge
density of the sphere is (i) positive, (ii) negative. 08 (5
Elec. field due to a circular ring of charge
1. A metal wire is bent in a circle of radius 10 cm. It is
given charge of 200μC which spreads on it uniformly.
Calculate electric potential at its centre. (95) 5
Misc. Qs on applications of Gauss's theorem
1. An infinitely long cylinder of radius R m carries a
uniform volume charge of density ρ Cm−3
Obtain

expressions for the electric field at a point: (i) Inside,
(ii) Outside the cylinder. (1991) 2
8. TWO point charges 10μC and 20μC are separated by a
distance 'r' in air. If an additional charge of − 8μC is given
to each by what factor does the force between the
charges change. (1993) 1
9. How does the force between two, point charges change
if the dielectric constant of the medium in which they
are kept, increases (1999) 1
10. In an electric field an electron is kept freely. If electron
is replaced by a proton, what will be the relationship
between forces experienced by them? (2000) 1
11. Force of attraction between two point charges
placed at distance ‘d’ apart in a medium is 'F'. What
should be the distance apart in the same medium so that
the force of attraction between them becomes 9 F? (01)2
12. Two point charges q1 and q2. are placed close to
each Other. What is the nature of force between the
charge when q1q2 < 0? (2003) 1

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