JEE-Physics
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ELECTROSTATICS
ELECTRIC CHARGE
Charge is the property associated with matter due to which it produces and experiences electrical and magnetic
effects. The excess or deficiency of electrons in a body gives the concept of charge.
Types of charge :
(i) Positive charge : It is the deficiency of electrons as compared to proton.
(i) Negative charge : It is the excess of electrons as compared to proton.
SI unit of charge : ampere × second i.e. Coulomb
Dimension : [A T]
eri
ng
.in
Practical units of charge are ampere × hour (=3600 C) and faraday (= 96500 C)
•
Millikan calculated quanta of charge by 'Highest common factor' (H.C.F.) method and it is equal to charge of
electron.
•
1 C = 3  109 stat coulomb, 1 absolute - coulomb = 10 C, 1 Faraday = 96500 C.
SPECIFIC PROPERTIES OF CHARGE
Charge is a scalar quantity : It represents excess or deficiency of electrons.
•
Charge is transferable : If a charged body is put in contact with an another body, then charge can be
transferred to another body.
•
Charge is always associated with mass
gin
e
•
Charge cannot exist without mass though mass can exist without charge.
•
In charging, the mass of a body changes.
•
When body is given positive charge, its mass decreases.
•
When body is given negative charge, its mass increases.
Charge is quantised
En
So the presence of charge itself is a convincing proof of existence of mass.
w.
Le
arn
•
•
The quantization of electric charge is the property by virtue of which all free charges are integral multiple of a
basic unit of charge represented by e. Thus charge q of a body is always given by
q = ne
n = positive integer or negative integer
The quantum of charge is the charge that an electron or proton carries.
Note : Charge on a proton = (–) charge on an electron = 1.6 × 10–19 C
•
Charge is conserved
E
•
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NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
In an isolated system, total charge does not change with time, though individual charge may change i.e. charge
can neither be created nor destroyed. Conservation of charge is also found to hold good in all types of reactions
either chemical (atomic) or nuclear. No exceptions to the rule have ever been found.
Charge is invariant
Charge is independent of frame of reference. i.e. charge on a body does not change whatever be its speed.
Accelerated charge radiates energy
v = 0 (i.e. at rest)
+
Q
produces only E
(electric field)
•
v = constant
v constant (i.e. time varying)
+
Q
+
Q
produces both E and B
(magnetic field)
but no radiation
produces E, B
and radiates energy
Attraction – Repulsion
Similar charges repel each other while dissimilar attract
1
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METHODS OF CHARGING
•
F ri ct i on : If we rub one body with other body, electrons are transferred from one body to the other.
Transfer of electrons takes places from lower work function body to higher work function body.
Negative charge
Glass rod
Silk cloth
Woollen cloth
Rubber shoes, Amber, Plastic objects
Dry hair
Comb
Flannel or cat skin
Ebonite rod
Note : Clouds become charged by friction
•
Electrostatic
induction
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Positive charge
If a charged body is brought near a metallic neutral body, the charged body will attract opposite charge and
repel similar charge present in the neutral body. As a result of this one side of the neutral body becomes
negative while the other positive, this process is called 'electrostatic induction'.
q'=-ve
charging
body
uncharged body is
connected to
earth
step-2
w.
Le
arn
charged body is
brought near
uncharged body
step-1
charging
body
En
q'=0
charging
body
gin
e
• Charging a body by induction (in four successive steps)
q'=-ve
uncharged body is
disconnected
from the earth
step-3
q'=-ve
charging
body
is removed
step-4
Some important facts associated with inductionInducing body neither gains nor loses charge
(ii)
The nature of induced charge is always opposite to that of inducing charge
(iii)
Induction takes place only in bodies (either conducting or non conducting) and not in particles.
The process of transfer of charge by contact of two bodies is known as conduction. If a charged body is put in
contact with uncharged body, the uncharged body becomes charged due to transfer of electrons from one body
to the other.
• The charged body loses some of its charge (which is equal to the charge gained by the uncharged body)
• The charge gained by the uncharged body is always lesser than initial charge present on the charged body.
• Flow of charge depends upon the potential difference of both bodies. [No potential difference  No conduction].
• Positive charge flows from higher potential to lower potential, while negative charge flows from lower to
higher potential.
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Conduction
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•
(i)
E
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GOLDEN KEY POINTS
•
Charge differs from mass in the following sense.
(i)
In SI units, charge is a derived physical quantity while mass is fundamental quantity.
(ii) Charge is always conserved but mass is not (Note : Mass can be converted into energy E=mc2
(iii) The quanta of charge is electronic charge while that of mass it is yet not clear.
(iv) For a moving charged body mass increases while charge remains constant.
•
True test of electrification is repulsion and not attraction as attraction may also take place between a charged
and an uncharged body and also between two similarly charged bodies.
•
For a non relativistic (i.e. v << c) charged particle, specific charge
•
For a relativistic charged particle
m
eri
ng
.in
q
q
m
=constant
decreases as v increases, where v is speed of charged body.
Example
When a piece of polythene is rubbed with wool, a charge of – 2 × 10 –7 C is developed on polythene. What
is the amount of mass, which is transferred to polythene.
gin
e
Solution
From Q = ne, So, the number of electrons transferred n 
Q
e
=
2  10 7
1.6  10 19
= 1.25 × 1012
En
Now mass of transferred electrons = n × mass of one electron = 1.25 × 1012 × 9.1 × 10–31 = 11.38 × 10–19 kg
Example
w.
Le
arn
10 12  – particles (Nuclei of helium) per second falls on a neutral sphere, calculate time in which sphere gets
charged by 2C.
Solution
Number of  – particles falling in t second = 1012t
Charge on  – particle = +2e , So charge incident in time t =
(1012t).(2e)

 t
Given charge is 2 C

2 ×
10–6 = (1012t).(2e)
10 18
1.6  10
19
 6.25s
E
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NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
COULOMB'S LAW
The electrostatic force of interaction between two static point electric charges is directly proportional to the
product of the charges, inversely proportional to the square of the distance between them and acts along the
straight line joining the two charges.
If two points charges q1 and q2 separated by a distance r. Let F be the electrostatic force between these two
charges. According to Coulomb's law.
F  q1 q2 and F 
Fe =
kq 1 q 2
r2
1
+
2
F2 on1 q1
r
+
q2 F1 on 2
–
q1
F2on1
+
F1 on 2 q2
2

1
9 Nm 
 k = coulomb's constant or electrostatic force constant
where k  4    9  10
C2 
0

3
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Coulomb's law in vector form

F12 = force on q1 due to q2 =

F21 =
F12
kq 1 q 2
r̂21
r2
F21
r
q1
q2
r12
kq 1 q 2
r̂12 (here r̂12 is unit vector from q1 to q2 )
r2
y
Coulomb's law in terms of position vector
q1

kq 1 q 2  
F12 =   3 (r1  r2 )
r1  r2
r1
r21
q2
r2
x
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.in
O
Principle of superposition
The force is a two body interaction, i.e., electrical force between two point charges is independent of presence
or absence of other charges and so the principle of superposition is valid, i.e., force on a charged particle due



to number of point charges is the resultant of forces due to individual point charges, i.e., F1 = F12 + F13 +...
Note : Nuclear force is many body interaction, so principle of superposition is not valid in case of nuclear force.
F
gin
e
When a number of charges are interacting, the total force on a given charge is vector sum of the forces exerted
on it by all other charges individually
n

kq 0 q 1 kq 0 q 2 ... kq 0 q i
kq 0 q n
qi
ˆr




...
F

kq
in vector form
0
2
2
2
2
2 i
r1
r2
ri
rn
i 1 ri

The law is based on physical observations and is not logically derivable from any other concept. Experiments till
today reveal its universal nature.
•
The law is analogous to Newton’s law of gravitation : F = G
w.
Le
arn
•
m 1m 2
with the difference that :
r2
(a)
Electric force between charged particles is much stronger than gravitational force, i.e., FE >>FG. This is
why when both FE and FG are present, we neglect FG.
(b)
Electric force can be attractive or repulsive while gravitational force is always attractive.
(c)
Electric force depends on the nature of medium between the charges while gravitational force does not.
(d)
The force is an action–reaction pair, i.e., the force which one charge exerts on the other is equal and
opposite to the force which the other charge exerts on the first.
The force is conservative, i.e., work done in moving a point charge once round a closed path under the action
of Coulomb’s force is zero.
•
The net Coulomb’s force on two charged particles in free space and in a medium filled upto infinity are
ww
•
1
F = 4

q1 q 2
1
q1 q 2
F
and F' =
. So
=  = K,
2
2
r
4 r
F'
0
0

Dielectric constant (K) of a medium is numerically equal to the ratio of the force on two point charges in free
space to that in the medium filled upto infinity.
•
The law expresses the force between two point charges at rest. In applying it to the case of extended bodies of
finite size care should be taken in assuming the whole charge of a body to be concentrated at its ‘centre’ as this
is true only for spherical charged body, that too for external point.
Although net electric force on both particles change in the presence of dielectric but force due to
one charge particle on another charge particle does not depend on the medium between them.
•
Electric force between two charges does not depend on neighbouring charges.
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En
SOME IMPORTANT POINTS REGARDING COULOMB’S L AW AND ELECTRIC FORCE
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Example
If the distance between two equal point charges is doubled and their individual charges are also doubled,
what would happen to the force between them?
Solution
1
F = 4
qq
....(i)
r2
0
1
Again, F' = 4
0
(2q) (2q)
1
or F' =
(2r) 2
4
1
4q 2
=
2
4
4r
0
0
q2
= F
r2
So, the force will remain the same.
Example
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ng
.in
A particle of mass m carrying charge '+q 1' is revolving around a fixed charge '–q 2' in a circular path of
radius r. Calculate the period of revolution.
Solution
1 q1q2
4 2mr
2
=
mr
=
4 0 r 2
T2
( 4 0 )r 2 ( 4 2mr )
T =
q1q2
2
or
T = 4r
 0mr
q1q2
gin
e

where r is the vector drawn from source charge is test charge.
Example
En
The force of repulsion between two point charges is F, when these are at a distance of 1 m. Now the point
charges are replaced by spheres of radii 25 cm having the charge same as that of point charges. The distance
between their centres is 1 m, then compare the force of repulsion in two cases.
Solution
w.
Le
arn
In 2nd case due to mutual repulsion, the effective distance between their centre of charges will be increased
(d' > d) so force of repulsion decreases
as F 
1
d2
d
d'
E
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NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
EQUILIBRIUM OF CHARGED PARTICLES
In equilibrium net electric force on every charged particle is zero. The equilibrium of a charged particle, under
the action of Colombian forces alone can never be stable.

Equilibrium of three point charges
(i)
Two charges must be of like nature as Fq 
KQ 1 q
KQ 2 q

0
x2
 r  x 2
(ii) Third charge should be of unlike nature as FQ1 
Therefore x 
Q1
Q1  Q 2
r
and
q=
x
Q1
q
r
KQ 1 Q 2 KQ 1 q

0
r2
x2
Q 1 Q 2
( Q 1  Q 2 )2
5
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Q2
JEE-Physics

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Equilibrium of symmetric geometrical point charged system
Value of Q at centre for which system to be in state of equilibrium
q
a
a
q
a
a
q
a
Q
Q
q
q
a
q
q

(ii) For square Q =
3
Equilibrium of suspended point charge system
For equilibrium position
Tcos  = mg
x2
Fe
kQ 2
 tan  =
= 2
mg
x mg
If  is small then tan
kQ 2
x
x

= 2
2
2 x mg
 x3 =

w.
Le
arn
Example
Fe

Q
  
T
Tsin
Q
x
mg
 Q2
2 kQ 2 
 x  

mg
2  0 mg
If whole set up is taken into an artificial satellite (geff 
~ 0) then T = Fe=
•

Tcos
1
3
En
 ~
~ sin  =
4
gin
e
•
and T sin  = Fe =
kQ 2
 q(2 2  1)
eri
ng
.in
(i) For equilateral triangle Q =
q
a

2
kq 2
42
q
180°
q
a
Q
(a,a)
For the system shown in figure find Q for which resultant force on q is zero.
Solution
q
(0,a)
For force on q to be zero, charges q and Q must be of opposite of nature.
Net attraction force on q due to charges Q = Repulsion force on q due to q
2
kq
kQq
2 a 2 = ( 2a )2  q = 2 2 Q Hence
q=–2 2Q
FR
ww
2 F A = FR 
(0,0)
Q a FA
FA
q
Example
Two identically charged spheres are suspended by strings of equal length. The strings make an angle of 300
with each other. When suspended in a liquid of density 0.8 g/cc the angle remains same. What is the dielectric
constant of liquid. Density of sphere = 1.6 g/cc.
300
0
15
F
mg
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NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
a
E
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Solution
When set up shown in figure is in air, we have tan 150 =
F
When set up is immersed in the medium as shown
n
mg
F
in figure, the electric force experienced by the ball will reduce and will be equal to 
r
F
F
1

 r 
=2


  mg
1 
mg r  1  
s
s 

eri
ng
.in

 
gravitational force will become mg  1   Thus we have tan 150 =
s 

and the effective
0
30
15°
F
r

mgeff=mg 1 
gin
e

Example
Given a cube with point charges q on each of its vertices. Calculate the force exerted on any of the charges due
to rest of the 7 charges.
En
Solution
The net force on particle A can be given by vector sum of force experienced by this particle due to all the other
w.
Le
arn

 
charges on vertices of the cube. For this we use vector form of coulomb's law F  kq 1 q 2  r1  r2 
3
r1  r2
 

kq 2  akˆ
From the figure the different forces acting on A are given as FA 1 =
a3



 
Z

E
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NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
kq 2  aˆi  aˆj  akˆ 

k q 2 ajˆ  akˆ

kq 2 aiˆ  akˆ
F

FA 2 
A3
3
3
,
; FA 4 
3
3a
2a
2a


 
 


(a,0,a)
(a,a,a)
 
(0,a,a)
2
3

kq 2  aˆi  ajˆ

kq 2 aiˆ

kq 2  aˆj
FA 6 
3
FA 5 
,
, FA 7 
2a
a3
a3
 
1
4
 
X
A (0,0,0)
5
(a,0,0)
(0,a,0)
6
(a,a,0)
7
y
The net force experienced by A can be given as










1
 kq 2  1

 1 ˆi  ˆj  kˆ 
Fnet  FA1  FA 2  FA 3  FA 4  FA 5  FA 6  FA 7 

2


a
2
 3 3


7

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Example


f is the force due to sixth charge and F due to remaining five charges.

 

F  f  0  F  f  F  f 
1 qq
q2

2
4  0 L
4  0 L2
gin
e
Now if
eri
ng
.in
Five point charges, each of value +q are placed on five vertices of a regular hexagon of side Lm. What
is the magnitude of the force on a point charge of value –q coulomb placed at the centre of the hexagon?
Solution
If there had been a sixth charge +q at the remaining vertex of hexagon force due to all the six charges
on –q at O will be zero (as the forces due to individual charges will balance each other).
w.
Le
arn
En
ELECTRIC FIELD
In order to explain ‘action at a distance’, i.e., ‘force without contact’ between charges it is assumed that a
charge or charge distribution produces a field in space surrounding it. So the region surrounding a charge (or
charge distribution) in which its electrical effects are perceptible is called the electric field of the given charge.

Electric field at a point is characterized either by a vector function of position E called ‘electric intensity’ or by
a scalar function of position V called ‘electric potential’. The electric field in a certain space is also visualized
graphically in terms of ‘lines of force.’ So electric intensity, potential and lines of force are different ways of
describing the same field.
Intensity of electric field due to point charge
p
r
Electric field intensity is defined as force on unit test charge.

E  Lim
q 0 0

F
kq
kq 
= 2 r̂ = 3 r
q0
r
r
q
Pr ope rti e s o f e lec tri c fi eld i n tens i ty :
(i)
It is a vector quantity. Its direction is the same as the force experienced by positive charge.
(ii)
Electric field due to positive charge is always away from it while due to negative charge always
towards it.
(iii)
Its unit is Newton/coulomb
(iv)
Its dimensional formula is [MLT –3A –1]
(v)
Force on a point charge is in the same direction of electric field on positive charge and in
opposite direction on a negative charge.


E  qE
(vi)
It obeys the superposition principle that is the field intensity point due to charge distribution is
vector sum of the field intensities due to individual charge
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NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
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Note : Test charge (q0) is a fictitious charge that exerts no force on nearby charges but experiences forces due to them.
E
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GOLDEN KEY POINTS
•
Charged particle in an electric field always experiences a force either it is at rest or in motion.
•
In presence of a dielectric , electric field decreases and becomes
1

E
r
times of its value in free space.

Ftest
•
Test charge is always a unit (+ ve) charge.
•

If identical charges are placed on each vertices of a regular polygon, then E at centre = zero.
eri
ng
.in
test charge
E LE CT R IC F IE L D IN T EN SI T IE S D UE TO VAR IO U S CH A RG E DIS T R IB U T I ON S
p
Due to discrete distribution of charge
Field produced by a charge distribution for discrete distribution:–
th
By principle of superposition intensity of electric field due to i charge

ri
q4
q1
q3
q2
qi

Ei
gin
e
 Net electric field due to whole distribution of charge E p 

kq 
E ip  3 ri
ri
i 1
Continuous distribution of charge
1
dq 
r
4  0  r 3
En

Treating a small element as particle E 
Due to linear charge distribution E = k
r
dq
d 
r
2
[ = charge per unit length]

[ = charge per unit area]
dv
Due to volume charge distribution E = k  r 2
[ = charge per unit volume]
w.
Le
arn
ds
Due to surface charge distribution E = k  r 2
s
E
P
Electric field strength at a general point due to a uniformly charged rod
As shown in figure, if P is any general point in the surrounding of rod,
to find electric field strength at P, we consider an element on rod of
length dx at a distance x from point O as shown in figure. Now if dE be
the electric field at P due to the element, then
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NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
v
kdq
dE = x 2  r 2

1 
2
r
Q
O
L
 Here dq = L dx
dEcos
dE
Electric field strength in x–direction due to dq at P is
 kdq 
kQ sin 
dx
dEx = dE sin  =  x 2  r 2  sin  =
L
x2  r2





P
dEsin


r
Here we have x = r tan  and dx = r sec2  d 
Thus dEx 
kQ
L
kQ
r sec2  d
sin  , Strength =
sin  d
2
2
Lr
r sec 
9
dx
x
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P
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Net electric field strength due to dq at point P in x–direction is
E x   dE x 
kQ
Lr

1
2
sin  d  
kQ
Lr
  cos 
1
2
kQ
 cos 2  cos 1 
Lr

Similarly, electric field strength at point P due to dq in y–direction is dE y  dE cos  =
Again we have x = r tan and dx = r sec2 d. Thus we have dEy=
kQ dx
L
r  x 
2
2
× cos 
r sec 2 
kQ
kQ
d =
cos  × 2
cos d
L
Lr
r sec 2 
Ey =  dE y
eri
ng
.in
Net electric field strength at P due to dq in y–direction is
1
kQ
=
Lr
kQ
1
 cos  d = kLQr   sin 
2 = L r  sin 1  sin 2 
2
Thus electric field at a general point in the surrounding of a uniformly charged rod which subtend angles 1 and
2 at the two corners of rod can be given as
in x–direction : Ex =
kQ
Lr
kQ
 cos 2  cos 1  and in y–direction E y  L r  sin 1  sin 2 
Case – I : At its centre
gin
e
Electric field due to a uniformly charged ring
B
+
A+
+
+
+
+
+
+
En
+
+
+
+
+
O
+
+
w.
Le
arn
+
+
R
+
+
D
+
+
C
+
+
+
+
Here by symmetry we can say that electric field strength at centre due to every small segment on ring is
cancelled by the electric field at centre due to the segment exactly opposite to it. The electric field strength at
centre due to segment AB is cancelled by that due to segment CD. This net electric field strength at the centre
of a uniformly charged ring is E0=0
+
+
+
+ +
C
+
+
+
R

+
+
dE cos
P

dE sin
dE
+
+
+
+
+
Q
10
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
d
+
+
ww
Case II : At a point on the axis of Ring
E
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Here we'll find the electric field strength at point P due to the ring which is situated at a distance x from the ring
centre. For this we consider a small section of length d on ring as shown. The charge on this elemental section
is
dq =
Q
d  [Q= total charge of ring]
2 R
Kdq
R  x 
Due to the element d, electric field strength dE at point P can be given as dE=
2
2
2 R
E p   dE cos  =

O
=
kdq
R  x 
2
2

2 3/2

kQx
 2 R R 2  x2
=
R 2  x2
kQ x
2 R R 2  x
2 R
x
×
eri
ng
.in
The component of this field strength dE sin  which is normal to the axis of ring will be cancelled out due to the
ring section opposite to d . The component of electric field strength along the axis of ring dE cos  due to all
the sections will be added up. Hence total electric field strength at point P due to the ring is

O
 2 R  

d =
kQx

2 R R
2
2 R
 d
3/2
O
 x2

kQ x
R  x 
2
2
3/2
gin
e
Electric field strength due to a charged circular arc at its centre :
Figure shows a circular arc of radius R which subtend an angle  at its centre. To find electric field strength at
C, we consider a polar segment on arc of angular width d at an angle  from the angular bisector XY as
shown.
x
Rd
+++++++++
+
+
++
++
En
++
d
++
+
+++
++
++
w.
Le
arn
++
R


C
dEsin


dE
dEcos
Y
Q
E
ww
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
The length of elemental segment is Rd, the charge on this element d is dq =    .d 
Due to this dq, electric field at centre of arc C is given as dE 
kdq
R2
Now electric field component due to this segment dEsin which is perpendicular to the angular bisector gets
cancelled out in integration and net electric field at centre will be along angular bisector which can be calculated
by integrating dEcos within limits from –
 / 2
=

kQ
 / 2 R
2
cos d  =
kQ


to . Hence net electric field strength at centre C is E C   dE cos 
2
2
 / 2

R 2  / 2
cos d  =
 
kQ
R
 / 2
sin  / 2 =
2
11
kQ  
  2kQ sin  2 
sin

sin

2  =
R 2  2
R 2
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Electric field strength due to a uniformly surface charged disc :
If there is a disc of radius R, charged on its surface with surface charge density  we wish to find electric field
strength due to this disc at a distance x from the centre of disc on its axis at point P shown in figure.

dy
y
x
dE
P
eri
ng
.in
To find electric field at point P due to this disc, we consider an elemental ring of radius y and width dy in the disc
dq = .2 ydy [Area of elemental ring ds = 2y dy]
as shown in figure. The charge on this elemental ring
Now we know that electric field strength due to a ring of radius R, charge Q, at a distance x from its centre on
its axis can be given as
E
kQx
2 3/2
x  R 
2
Here due to the elemental ring electric field strength dE at point P can be given as
kdqx
2 3 / 2
x  y 
2
=
k 2 y dy x
2 3/2
gin
e
dE=
x  y 
2
Net electric field at point P due to this disc is given by integrating above expression from 0 to R as
0
R
k 2 xydy
x  y 
2
2
3 / 2 = k  x 
R
 
1


1
= 2k x  
=
2 0 
2
2 
2
2 3/2
x  y 0

x y
2 y dy
En
R
E   dE  
0

w.
Le
arn
Example

x


x2  R 2 
Calculate the electric field at origin due to infinite number of charges as shown in figures below.
O
Solution
q
1
2
4
fig (b)
x(m)
O
q
–q
q
1
2
4
1 1 1

kq.1
4kq
E 0  kq   
     =
=
(1  1 / 4)
3
 1 4 16

x(m)
[ S =
1
a
, a = 1 and r =
]
1r
4
kq.1
1 1 1

4kq
E 0  kq   
     =
=
1


1
/
4




1
4
16
5


ww
(a)
q
(b)
Example
A charged particle is kept in equilibrium in the electric field between the plates of millikan oil drop experiment.
If the direction of the electric field between the plates is reversed, then calculate acceleration of the charged
particle.
Solution
Let mass of the particle = m, Charge on particle = q
Intensity of electric field in between plates = E, Initially mg = qE
After reversing the field ma = mg + qE ma = 2mg
 Acceleration of particle  a = 2g
12
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
fig (a)
q
E
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JEE-Physics
Example
Calculate the electric field intensity E which would be just sufficient to balance the weight of an electron.
If this electric field is produced by a second electron located below the first one what would be the
distance between them? [Given : e = 1.6 × 10-19 C, m = 9.1 × 10 –31 kg and g = 9.8 m/s 2]
Solution
As force on a charge e in an electric field E
F e = eE
So according to given problem
 E 
eri
ng
.in
Fe = W  eE = mg
mg 9.1  10 31  9.8
V

= 5.57 × 10 –11
19
e
1.6  10
m
As this intensity E is produced by another electron B, located at a distance r below A.
1 e
r 
4 0 r 2
 9  10 9  1.6  10 19 
e
So, r  
5 m
4 0 E
5.57  10 11


gin
e
E
Example
w.
Le
arn
Solution
En
A block having mass m = 4 kg and charge q = 50 C is connected to a spring having a force constant
k = 100 N/ m. T he b lock li es on a fri ctionless hori zontal track and a uni form electri c fi eld
E = 5 × 105 V/m acts on the system as shwon in figure. The block is released from rest when the spring
is unstretched (at x = 0)
(a) By what maximum amount does the spring expand?
(b) What is the equilibrium position of the block?
(c) Show that the block's motion is simple harmonic and determine the amplitude and time period of the motion.
(a) As x increases, electric force qE will accelerate the block while elastic force in the spring kx will
oppose the motion. The block will move away from its initial position x = 0 till it comes to rest, i.e., work
done by the electric force is equal to the energy stored in the spring.
So if x max is maximum stretch of the spring.
1 2
2qE
2  (50  10 6 )  (5  10 5 )
kx max  (qE )x max xmax =
 x max =
= 0.5 m
2
k
100
E
ww
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
(b) In equilibrium position F R = 0, so if x 0 is the stretch of the spring in equilibrium position
kx 0 = qE  x 0 = (qE/k) =
1
x
= 0.25 m
2 max
(c) If the displacement of the block from equilibrium position (x 0) is x, restoring force will be
F = k(x ± x0)  qE = kx
[ax kx0 = qE]
and as the restoring force is linear the motion will be simple harmonic with time period
T  2
and
m
4
 2
 0.4  s
k
100
amplitude = x max – x 0 = 0.5 – 0.25 = 0.25 m
13
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ELECTRIC LINES OF FORCE
Electric lines of electrostatic field have following properties
+
A
B–
(i)
Imaginary
(ii)
Can never cross each other
(iii)
Can never be closed loops
(iv)
The number of lines originating or terminating on a charge is proportional to the magnitude of charge. In
rationalised MKS system (1/o) electric lines are associated with unit charge, so if a body encloses a charge q ,
total lines of force associated with it (called flux) will be q/o.
(v)
Lines of force ends or starts normally at the surface of a conductor.
(vi)
If there is no electric field there will be no lines of force.
(vii)
Lines of force per unit area normal to the area at a point represents magnitude of intensity , crowded lines
represent strong field while distant lines weak field.
(viii)
Tangent to the line of force at a point in an electric field gives the direction of intensity. So a positive charge free
to move follow the line of force.
gin
e
eri
ng
.in
qA>qB
GOLDEN KEY POINTS
Lines of force starts from (+ve) charge and ends on (–ve) charge.
•
Lines of force start and end normally on the surface of a conductor
w.
Le
arn
+
–– – – – –
+s +
+
E=0 +
+
+
fixed point charge near
infinite metal plate
.
– –s
– E=0
–
–
–
edge effect
The lines of force never intersect each other due to superposition principle.
•
The property that electric lines of force contract longitudinally leads to explain attraction between opposite
charges.
•
The property that electric lines of force exert lateral pressure on each other leads to explain repulsion
between like charges.
ww
•
Electric flux ()
The word "flux" comes from a Latin word meaning "to flow" and you can consider the flux of a vector field to be
a measure of the flow through an imaginary fixed element of surface in the field.
Electric flux is defined as
 
E   E  dA

This surface integral indicates that the surface in question is to be divided into infinitesimal elements of area dA
 
and the scalar quantity E  dA is to be evaluated for each element and summed over the entire surface.
Important points about electric flux :
(i) It is a scalar quantity
(ii) Units (V–m) and N – m2/C Dimensions : [ML 3T –3 A –1 ]
(iii) The value of  does not depend upon the distribution of charges and the distance between them inside the
closed surface.
14
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
En
•
E
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Electric Flux through a circular Disc :
Figure shows a point charge q placed at a distance  from a disc of radius R. Here we wish to find the electric
flux through the disc surface due to the point charge q. We know a point charge q originates electric flux in
radially outward direction. The flux is originated in cone shown in figure passes through the disc surface.

eri
ng
.in
q
To calculate this flux, we consider on elemental ring an disc surface of radius x and width dx as shown. Area of
dS  2 x dx . The electric field due to q at this elemental ring is given as E=
gin
e
If d is the flux passing through this elemental ring, then
kq
x   
2
2


this ring (strip) is
dS
x


w.
Le
arn
En
q
d = E d S c o s  =
R
   d = 
0
kq
x   
q
2 0
2
2
× 2 x dx ×
x dx
2 3/2
  x 
2

2
=
2
x 
0
  x 
2
2
3/2
R
R
q
=
2 0 
2 kq  x dx


1
1
q 
q  1
 

 =
= 2   2
2 0  
0 
2  x2 
  x 2 0
x dx
2 3/2
  x 
2
E
ww
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
The above result can be obtained in a much simpler way by using the concept of solid angle and Gauss's law.
Electric flux through the lateral surface of a cylinder due to a point charge :
Figure shows a cylindrical surface of length L and radius R. On its axis at its centre a point charge q is placed. Here
we wish to find the flux coming out from the lateral surface of this cylinder due to the point charge q. For this we
consider an elemental strip of width dx on the surface of cylinder as shown. The area of this strip is dS = 2 R.dx
dS

E

x
C
q
R
L
15
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The electric field due to the point charge on the strip can be given as E 
d   EdS cos  =
through the strip, then
. If d is the electric flux
x  R 
2
R
Kq
x  R 
2
kq
2
2
× 2  Rdx ×
   d 
Total flux through the lateral surface of cylinder
2
x R
qR 2
2 0
= 2  KqR 2 ×
2
L / 2

L / 2
dx
2 3/2
x  R 
2
dx
2 3/2
x  R 
=
2
q 0 
2  4R 2
eri
ng
.in
This situation can also be easily handled by using the concepts of Gauss's law.
GAUSS'S LAW
It relates with the total flux of an electric field through a closed surface to the net charge enclosed by that surface
and according to it, the total flux linked with a closed surface is 1 / 0  times the charge enclosed by the closed
  q
surface i.e., S E.ds = 
gin
e
0
E

n
ds
O
+q

r
En
r
Flux through gaussian suface is independent of its shape.
(ii)
Flux through gaussian suface depends only on charges present inside gaussian surface.
(iii)
Flux through gaussian suface is independent of position of charges inside gaussian surface.
(iv)
Electric field intensity at the gaussian surface is due to all the charges present (inside as well as out side)
(v)
In a close surface incoming flux is taken negative while outgoing flux is taken positive.
(vi)
In a gaussian surface  = 0 does not employ E = 0
(vii)
ww
(i)
but E = 0 employs  = 0.
Gauss's law and Coulomb's law are equivalent, i.e., if we assume Coulomb's law we can prove Gauss's law and
vice–versa. To prove Gauss's law from Coulomb's law consider a hypothetical spherical surface [called Gaussian–
surface] of radius r with point charge q at its centre as shown in figure. By Coulomb's law intensity at a point
P on the surface will be,

E
1
4 0 r
3

 

r And hence electric flux linked with area ds  E.ds =
1
 


Here direction of r and ds are same, i.e., 
 E.ds = 4 
S
q
0 r
2
1
1
1 q  
r.ds
4  0 r 3
 
q
 S ds  4 0 r 2  4 r 2   S E.ds = 
0
Which is the required result. Though here in proving it we have assumed the surface to be spherical, it is true
for any arbitrary surface provided the surface is closed.
16
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
Note :
w.
Le
arn
REGARDING GAUSS'S LAW IT IS WORTH NOTING THAT :
E
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(viii)
(a) If a closed body (not enclosing any charge) is placed in an electric field (either uniform or non–uniform) total
flux linked with it will be zero.
sphere
 E=0
(A)
(b) If a closed body encloses a charge q, then total flux
q
(A)
q
(B)
 E=(q/ 0)
eri
ng
.in
 E=0
(B)
(C)
(D)
q
q
 E=(q/ 0)
 E=(q/ 0)
 
linked with the body will be
gin
e
 E=(q/ 0)
q
 S E.ds = 
0
From this expression it is clear that the flux linked with a closed body is independent of the shape and size of the
body and position of charge inside it.[figure]
En
Note : So in case of closed symmetrical body with charge at its centre, flux linked with each half will be
w.
Le
arn
 q 
1
E  = 

and the symmetrical closed body has n identical faces with point charge at its centre, flux linked
 2 0 
2
 q 
 E 
with each face will be   =  n  
 0
n
E
Gauss's law is a powerful tool for calculating electric intensity in case of symmetrical charge distribution by

choosing a Gaussian– surface in such a way that E is either parallel or perpendicular to its various faces. As an
example, consider the case of a plane sheet of charge having charge density . To calculate E at a point P close
to it consider a Gaussian surface in the form of a 'pill box' of cross–section S as shown in figure.
ww
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
(ix)
E
n
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ + +
+ + + + +
+ + + + +
+ + + + +
+ + + + +
+ + + + +
+ + + + +
+ + + + +
+ + + + +
+ + + + +
+ + + + +
+ + + + +
+ + + + +
+ + + + +
+ + + + +
+ + + +
+
E
n
n
n
E
n
Ein=0


Sheet of charge
(A)
Conductor
(B)
The charge enclosed by the Gaussian–surface = S and the flux linked with the pill box = ES + 0 + ES = 2ES
(as E is parallel to curved surface and perpendicular to plane faces)
1
1
So from Gauss's law,  E =  (q), 2ES = 
0
0
 S 
17

 E = 2
0
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 
 


E.ds
E.ds
If E =0,   
=0,
so
q
=
0
but
if
q=0,
=0
So
may or may not be zero.
E


 

If a dipole is enclosed by a closed surface then, q =0, so 
 E.ds =0, but E  0
Note : If instead of plane sheet of charge, we have a charged conductor, then as shown in figure (B) Ein=0.

So E  ES and hence in this case E=  . This result can be verified from the fact that intensity at the surface
0
q
0
eri
ng
.in
1
of a charged spherical conductor of radius R is, E = 4
1
So for a point close to the surface of conductor, E =
4 0 R
Example
2

2
× 4 R 

=
0
gin
e
If a point charge q is placed at the centre of a cube.
with q = 4 R 2 
R2
What is the flux linked (a) with the cube? (b) with each face of the cube?
Solution
According to Gauss's law flux linked with a closed body is (1/0) times the charge enclosed and here the
En
(a)
1
closed body cube is enclosing a charge q so,  T   (q)
w.
Le
arn
Now as cube is a symmetrical body with 6–faces and the point charge is at its centre, so electric flux
linked with each face will be F 
N ot e : (i)
Here flux linked with cube or one of its faces is independent of the side of cube.
If charge is not at the centre of cube (but anywhere inside it), total flux will not change, but the flux linked
with different faces will be different.
ww
(ii)
1
q
T  

6
6 0
Example
If a point charge q is placed at one corner of a cube, what is the flux linked with the cube?
Solution
In this case by placing three cubes at three sides of given cube and four cubes above, the charge will be in the centre.
q
q
So, the flux linked with each cube will be one–eight of the flux  .  Flux associated with given cube = 8
0
18
0
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
(b)
0
E
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FLUX CALCULATION USING GAUSS LAW
ds E
E
R
ds
ds
E
in= –R 2 E and  out= R2 E total = 0
E
R
in= circular = –R2 E and out=   curved = R2 E  total = 0
y
ds
eri
ng
.in
E
E
a
ds
in= –a2 E and out= a2 E  total = 0
ds
x
a
a
E
E 
R q
kq
R
2
2
R 2 E and out 
2
   2 R 
w.
Le
arn
Note : here electric field is radial
1
gin
e
E
in  –
q
4  0 R
q
ww
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
R 2 E  total = 0
q
2 0
q
q
 cylinder = 2 
0
R
R
E

2
q
 hemisphere= 2 
0
q
q
q
2
1
En
z
q

q
cube = 2 
0
q
 =
q
q
8 0
q
q
q
q
q
 =
19
4 0
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Example
As shown in figure a closed surface intersects a spherical conductor. If a negative charge is placed at point P.
What is the nature of the electric flux coming out of the closed surface ?
conductor
++++
close surface
++
+ ++ +++
P
–Q
+
eri
ng
.in
++
Solution
Point charge Q induces charge on conductor as shown in figure.
Net charge enclosed by closed surface is negative so flux is negative.
Example
gin
e

Consider E = 3 × 103 î (N/C) then what is the flux through the square of 10 cm side, if the normal of its plane
makes 60° angle with the X axis.
Solution
 = EScos = 3 × 10 3 × [10 × 10 –2] 2 × cos60° = 3 × 10 3 × 10 –2 × ½
En
Example
= 15 Nm 2/C
Find the electric field due to an infinitely long cylindrical charge distribution of radius R and having linear charge
density  at a distance half of the radius from its axis.
2k r
R
2
=
FG R IJ = 
4  R
R H 2K
2k 
2
0
E
ELECTRIC FIELD DUE TO SOLID CONDUCTING OR HOLLOW SPHERE
+
+
q
0
ww



; E.ds = E ds cos 0° = E ds
 At every point on the Gaussian surface E  d s
q

O R
+
 E.d s  
0
+
q
q
0
+
Gaussian surface
 E.ds   [E is constant over the gaussian surface]  E  4 r   E = 4 r
2
P dA
+
 
Using Gauss's theorem
r
+
For outside point (r > R)
+
•
+
p
2
0
q
•
For surface point r = R :
ES 
•
For Inside point (r < R) :
Because charge inside the conducting sphere or hollow is zero.
4  0 R2
(i.e. q = 0) So
 
q
E
 .ds  0 = 0  Ein = 0
20
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
E=
+
R
point will be inside so
2
+
w.
Le
arn
r =
+
Solution
E
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ELECTRIC FIELD DUE TO SOLID NON CONDUCTING SPHERE
•
 
q
q
2
P
0
s
0
2
0
At surface (r = R)
Gaussian surface
q
ES 
40 R2
Put r = R
Inside (r < R) :
E
 
From Gauss's theorem
eri
ng
.in
•
r
O R
q
 E.ds    E  4 r    E  4 r
•
E
P ds
Outside (r > R)
From Gauss's theorem
q
 E.ds  
0
s
ds
P
r
O R
Where q charge contained within Gaussian surface of radius r
E(4 r 2 ) 
q
q
E
...(i)
0 
4 r 2 0
Gaussian surface
As the sphere is uniformly charged, the volume charge density (charge/volume)  is constant throughout the
 q   4 3
qr 3
 4 3
 charge enclosed in gaussian surface q    r  =  (4 3)R3   3 r    q  3
3
R
q
4
R 3
3
gin
e
sphere  
1 qr
put this value in equation (i) E in  4   3
0 R
En
ELECTRIC FIELD DUE TO AN INFINITE LINE DISTRIBUTION OF CHARGE

s1
w.
Le
arn
Let a wire of infinite length is uniformly charged having a constant linear charge density .
P is the point where electric field is to be calculated.
Let us draw a coaxial Gaussian cylindrical surfaces of length  .
dS1 
From Gauss's theorem
 
E.dS 1 

 
E.dS 2 
s2
 
q
E.dS 3 
0
s
r

E
3



 

 
E  dS 1 so E.dS 1  0 and E  dS 2 so E.dS 2  0
E  2 r  
q
0
ww
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
E
(f)
E  2 r  

0

 E 

or
2  0 r
E
dS2
q = .
E
dS3
E


[ E  dS 3 ]
Charge enclosed in the Gaussian surface
So
Gaussian
surface
2k
where k =
r
1
4 0
Charged cylindrical nonconductor of infinite length
Electric field at outside point
Electric field at inside point

2k
ˆr
EA 
r


r
EB 
2 0 R2
r > R
Gaussian
Surface
r < R
21
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DIELECTRIC IN ELECTRIC FIELD


Let E 0 be the applied field, Due to polarisation, electric field is E p .

The resultant field is E . For homogeneous and isotropic dielectric,


the direction of E p is opposite to the direction of E 0 .
E0
–
–
–
–
–
E0
Ep
E
+ E
0
+
+
+
+

So, Resultant field is E=E0–E P
GOLDEN KEY POINTS
Electric field inside a solid conductor is always zero.
•
Electric field inside a hollow conductor may or may not be zero (E  0 if non zero charge is inside the sphere).
•
The electric field due to a circular loop of charge and a point charge are identical provided the distance of the
observation point from the circular loop is quite large as compared to its radius
i.e. x >>> R.
eri
ng
.in
•
Example
For infinite line distribution of charge draw the curve between log E and log r.
 E 

2  0 r

A
r
where A 

2  0
gin
e
Solution
log E
 constant
log r
En
take log on both side log E = log A – log r
log A
Example
w.
Le
arn
A point charge of 0.009 C is placed at origin.
Calculate intensity of electric field due to this point charge at point  2, 7,0  .
Solution

E

qr
4  0 r
3
 9  10 5  9  10 9 ( 2 ˆi  7 ˆj)

; where r  x i  yj  2 i  7 j , E 
=  3 2 ˆi  3 7 ˆj  NC 1
(3) 3
ww
Potential energy of a system of particles is defined only in conservative fields. As electric field is also
conservative, we define potential energy in it. Potential energy of a system of particles we define as the work
done in assembling the system in a given configuration against the interaction forces of particles. Electrostatic
potential energy is defined in two ways.
(i) Interaction energy of charged particles of a system
(ii) Self energy of a charged object
•
Electrostatic Interacti on Energy
Electrostatic interaction energy of a system of charged particles is defined as the external work required to
assemble the particles from infinity to the given configuration. When some charged particles are at infinite
separation, their potential energy is taken zero as no interaction is there between them. When these charges
are brought close to a given configuration, external work is required if the force between these particles is
repulsive and energy is supplied to the system, hence final potential energy of system will be positive. If the
force between the particle is attractive, work will be done by the system and final potential energy of system
will be negative.
22
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
ELECTROSTAT IC POTENTIAL ENERGY
E
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•
Interaction Energy of a system of two charged particles
Figure shows two + ve charges q1 and q2 separated by a distance r. The electrostatic interaction energy of this
system can be given as work done in bringing q2 from infinity to the given separation from q1.
x
r
q1
dx
F
r  
r
kq 1 q 2
dx [ – ve sign shows that x is decreasing]
W =  F.dx = – 
x2


W =
kq 1 q 2
= U [ interaction energy]
r
eri
ng
.in
It can be calculated as
q2
If the two charges here are of opposite sign, the potential energy will be negative as U = –
•
kq 1 q 2
r
Interaction Energy for a system of charged particles
gin
e
When more than two charged particles are there in a system, the interaction energy can be given by sum of
interaction energies of all the pairs of particles. For example if a system of three particles having charges q 1,
q2 and q3 is given as shown in figure.
q1
r3
En
r2
r1
w.
Le
arn
q2
The total interaction energy of this system can be given as U =
q3
kq 1 q 2
kq 1 q 3
kq 2 q 3
+
+
r3
r2
r1
ELECTRIC POTENTIAL
Electric potential is a scalar property of every point in the region of electric field. At a point in electric field
potential is defined as the interaction energy of a unit positive charge. If at a point in electric field a charge q0
E
U
V = q joule/coulomb
0
ww
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
has potential energy U, then electric potential at that point can be given as
Potential energy of a charge in electric field is defined as work done in bringing the charge from infinity to the
given point in electric field. Similarly we can define electric potential as "work done in bringing a unit positive
charge from infinity to the given point against the electric forces.
Example
A charge 2C is taken from infinity to a point in an electric field, without changing its velocity, if work
done against electrostatic forces is –40J then potential at that point is ?
Solution
V
Wext 40 J

 20V
q
2 C
Note : Always remember to put sign of W and q.
23
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Electric Potential due to a point charge in its surrounding :
p
r
q
U
The potential at a point P at a distance r from the charge q VP = q . Where U is the potential energy of
0
kqq 0
. Thus potential at point P is
r
Electric Potential due to a charge Rod :
VP =
kq
r
eri
ng
.in
charge q0 at point p, U =
Figure shows a rod of length L, uniformly charged with a charge Q. Due to this we'll find electric potential at a
point P at a distance r from one end of the rod as shown in figure.
x
Q
dx
r
L
P
Charge on this element is
dQ =
gin
e
For this we consider an element of width dx at a distance x from the point P.
Q
dx
L
The potential dV due to this element at point P can be given by using the result of a point charge as
kdq
kQ
=
dx
x
Lx
En
dV =
r L
w.
Le
arn
Net electric potential at point P : V =  dV =

r
kQ
r L
kQ
dx =
ln 
Lx
r 
L
Electric potential due to a charged ring
Case – I : At its centre
To find potential at the centre C of the ring, we first find potential dV at centre due to an elemental charge
kdq
kdq
kQ
Total potential at C is V=  dV = 
=
.
R
R
R
+
Q
+ +
+
+
+
+ +
C
+
R
dq
+
+
+
+
+
As all dq's of the ring are situated at same distance R from the ring centre C, simply the potential due to all dq's
kQ
.
R
Here we can also state that even if charge Q is non–uniformly distributed on ring, the electric potential C will
remain same.
is added as being a scalar quantity, we can directly say that the total electric potential at ring centre is
24
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
+
+
ww
+
dq on ring which is given as dV =
E
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Case II : At a point on axis of ring
We find the electric potential at a point P on the axis of ring as shown, we can directly state the result as
here also all points of ring are at same distance
x 2  R 2 from the point P, thus the potential at P can be given
kQ
R 2  x2
+
x 2
+R 2
R
x
+
Electric potential due to a uniformly charged disc :
P
eri
ng
.in
+ + + + +
+ + +
+
+ +
as VP =
Figure shows a uniformly disc of radius R with surface charge density  coul/m2. To find electric potential at
point P we consider an elemental ring of radius y and width dy, charge on this elemental ring is dq=2y dy.
kdq
gin
e
Due to this ring, the electric potential at point P can be given as dV =
2
x y
k..2 y dy
2
=
x2  y2

dy
w.
Le
arn
En
y
x
P
Net electric potential at Point P due to whole disc can be given as
R 
V =  dV  0 2  ·
0
y dy
R


2
2
= 2   x  y  = 2 
x y
0
0
0
2
2
 x2  R 2  x 


ELECTRIC POTENTIAL DUE TO HOLLOW OR CONDUCTING SPHERE
E
At outside sphere
ww
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
•
R
r>R
P
According to definition of electric potential, electric potential at point P
r
r
r


q
q
1
q
q 1 

E


dr
V


dr



out
2
2
2 ;


4  0  r
4  0 r
4  0 r 
4  0  r  40 r



V    E .dr  

r

q

25
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At surface
R
R
R

q
q 
1 
q
q 1 

E

V
dr


V


dr

out
 2


2
2 ;


4  0  r 
40 R
4  0 r
4 0 r 

4  0  r 


R
 
V    E .dr  


•
Inside the surface
q
:
 Inside the surface E  
dV
dr
 0  V = constant so V 
q
4  0 R
•
At outside sphereSame as conducting sphere.
•
At Surface
•
Inside the sphere
eri
ng
.in
ELECTRIC POTENTIAL DUE TO SOLID NON CONDUCTING SPHERE
Same as conducting sphere.
r
 R  kq 
 kqr  
V       dr   
dr 
2
3 
   r 
R  R  

r 

R
kq  r 2  
 1

V   kq    
 
3 
 
  r
 R  2 R 


 1
r2
R2 
V  kq   


 R 2R 3 2R 3 


V    E .dr

gin
e

r
R

V     E 1 dr   E 2 dr 
 

R
r
V 
kq
3R 2  r 2 

2R
3 
En
Potential Difference Between Two points in electric field
Potential difference between two points in electric field can be defined as work done in displacing a unit
positive charge from one point to another against the electric forces.
w.
Le
arn
A
B
B  
VA
VB
If a unit +ve charge is displaced from a point A to B as shown work required
can be given as V B – V A = –  E.dx
A
If a charge q is shifted from point A to B, work done against electric forces can be given as W = q (VB – VA)
If in a situation work done by electric forces is asked, we use W = q (VA – VB)
ww
Example
1C charge is shifted from A to B and it is found that work done by external force is 80J against
electrostic forces, find V A – V B
Solution
WAB = q(VB – VA)
80 J = 1C (VB – VA)  VA – VB = – 80 V
Equipotential
surfaces
For a given charge distribution , locus of all points having same potential is called 'equipotential surface'.
•
Equipotential surfaces can never cross each other (otherwise potential at a point will have two values which
is absurd)
26
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
If V B < V A, then charges must have tendency to move toward B (low potential point) it implies that electric
forces carry the charge from high potential to low potential points. Hence we can say that in the direction of
electric field always electric potential decreases.
E
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•
Equipotential surfaces are always perpendicular to direction of electric field.
•
If a charge is moved from one point to the other over an equipotential surface then work done
WAB = – UAB = q (VB–VA) = 0
•
[ VB = VA]
Shapes of equipotential surfaces
V=V1
V=V2
V=V1
V=V2
V=V2
•
for a point charge, spherical
conductor equipotential
surfaces are spherical
for a line distribution of
charge equipotential
surfaces are cylinderical
eri
ng
.in
for uniform electric field
equipotential surfaces
are parallel plane
V=V1
The intensity of electric field along an equipotential surface is always zero.
Electric Potential Gradient
gin
e
The maximum rate of change of potential at right angles to an equipotential surface in an electric field is defined


as potential gradient. E  V = – grad V
Note : Potential is a scalar quantity but the gradient of potential is a vector quantity

Example
15 z then find magnitude of electric field at point (x,y,z).
w.
Le
arn
If V = –5x + 3y +
Solution
LM V i  V j  V k OP
N x y z Q
En
In cartesian co–ordinates  V =

 V ˆ V ˆ V ˆ 
E  –
i
j
k  = –(–5 i + 3 j +
y
z 
 x
Example


15 k )  |E|  25  9  15  49  7 unit
E
ww
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
The four charges q each are placed at the corners of a square of side a. Find the potential energy of one
of the charges
The electric potential of a point A due to charges B, C and D is
1
V = 4

0
1
q
+ 4
a
0
q
2a
1
+ 4
0
1
q
= 4
a
0

1  q
 2 

2 a

1
Potential energy of the charge at A is PE = qV = 4
0
27

1  q2
 2 

.
2 a

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Example
A proton moves with a speed of 7.45 × 105 m/s directly towards a free proton originally at rest. Find the
distance of closest approach for the two protons.
1
Given : 4 
= 9 × 10 9
0
N  m2
; m p = 1.67 × 10 –27 kg and e = 1.6 × 10 –19 C
C2
Solution
1
u
2
mu = mv + mv v =
And by conservation of energy
1
1
1
1
mu2 =
mv2 +
+ 4
2
2
2
0
4e 2
u
[as v =
]
4  0 mu 2
2
And hence substituting the given data,
ELECTRIC DIPOLE
4  (1.6  10 19 )2
= 10 –12 m
1.67  10 27  (7.45  10 3 ) 2
En
r = 9 × 10 9 ×
e2
r
gin
e
So, r=
eri
ng
.in
As here the particle at rest is free to move, when one particle approaches the other, due to electrostatic
repulsion other will also start moving and so the velocity of first particle will decrease while of other will
increase and at closect approach both will move with same velocity. So if v is the common velocity of
each particle at closest approach, by 'conservation of momentum'.
w.
Le
arn
A system of two equal and opposite charges separated by a certain distance is called electric dipole, shown in
figure. Every dipole has a characteristic property called dipole moment. It is defined as the product of magnitude
of either charge and the separation between the charges, given as
d

p


p = qd
+q
+q
p
q
(i)
Vector quantity , directed from negative to positive charge
(ii)
Dimension : [LTA], Units : coulomb × metre (or C-m)
(iii)
Practical unit is "debye"  Two equal and opposite point charges each having charge 10–10 frankline (  e)

and separation of 1Å then the value of dipole moment (p) is 1 debye.
1 Debye = 10–10 × 10–10 Fr × m
= 10–20 ×
28
C m
 3.3 × 10–30 C × m
3  10 9
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
In some molecules, the centres of positive and negative charges do not coincide. This results in the formation of
electric dipole. Atom is non – polar because in it the centres of positive and negative charges coincide. Polarity
can be induced in an atom by the application of electric field. Hence it can be called as induced dipole.


Dipole Moment : Dipole moment p = q d
ww
•
-q
E
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Example
A system has two charges qA = 2.5 × 10–7 C and qB = – 2.5 × 10–7 C located at points A: (0, 0, – 0.15 m)
and B ; (0, 0, + 0.15 m) respectively. What is the total charge and electric dipole moment of the system?
Solution
Total charge = 2.5 × 10 –7 – 2.5 × 10 –7 = 0
Electric diople moment,
p = Mangitude of either charge × separation between charges
= 2.5 × 10 –7 [0.15 + 0.15] C m = 7.5 × 10 –8 C m. The direction of dipole moment is from B to A.
Dipole Placed in uniform Electric Field
eri
ng
.in
•

Figure shows a dipole of dipole moment p placed at an angle  to the direction of electric field. Here the




charges of dipole experience forces qE in opposite direction as shown. Fnet =  qE  ( q) E   0
qE
+
p 
E
gin
e
–
– qE
En
Thus we can state that when a dipole is placed in a uniform electric field, net force on the dipole is zero. But as
equal and opposite forces act with a separation in their line of action, they produce a couple which tend to align
the dipole along the direction of electric field. The torque due to this couple can be given as
w.
Le
arn
 = Force × separation between lines of actions of forces = qE × d sin  = pE sin 
   

   
  r  F  d  qE  qd  E  p  E
Work done i n Rotation of a Dipole in Electric field
When a dipole is placed in an electric field at an angle  , the torque on it due to electric field is   pE sin 
Work done in rotating an electric dipole from 1 to 2 [ uniform field]
dW =  d  so W =
ww
W0 180 = pE [1– (–1)] = 2 pE
1
2
1
2
1
W0  90
= pE (1–0)
=
pE
If a dipole is rotated from field direction ( = 0° ) to  then W = pE ( 1– cos)
–
E
p
+
p
E
+
+
E
 dW =   d and W  = W =  pE sin d = pE (cos – cos )
–
E
p
–
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
e.g.
2
=0
 = 90°
 = 180°
 = minimum = 0
 = maximum = pE
 = minimum = 0
W = minimum = 0
W = pE
W = maximum = 2pE
29
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Electrostatic potential energy :
Electrostatic potential energy of a dipole placed in a uniform field is defined as work done in rotating a dipole from


a direction perpendicular to the field to the given direction i.e., W90   pE sin  d  = – pE cos = – p.E
90 

E is a conservative field so what ever work is done in rotating a dipole from 1 to 2 is just equal to change in
electrostatic potential energy W1 2  U 2  U 1 = pE (cos 1– cos 2)
Work done in rotati ng an electric dipole i n an electric field
2
eri
ng
.in
Suppose at any instant, the dipole makes an angle  with the electric field. The torque acting on dipole.  = qEd
= (q 2 sin)E = pE sin The work done in rotating dipole from 1 to 2
2
W   d    pE sin  d 
1
1
B
2

W = pE (cos1 – cos2) = U2 – U1
qE
(  U = – pE cos)
qE
d

q A

+q
C

E
gin
e
Force on an electric dipole in Non–uniform electric field :
If in a non–uniform electric field dipole is placed at a point where electric field is E, the interaction energy of
dipole at this point
U
 
U   p.E . Now the force on dipole due to electric field F = –
r
w.
Le
arn
Example
En
If dipole is placed in the direction of electric field then F= – p
dE
dx
-q
p
+q
r

Solution
In the situation shown in figure, the electric field strength due to the wire, at the position
of dipole as E =
2k 
r
Thus force on dipole is F = – p.
 2k  
dE
2kp 
= – p  2  =
r
dr
r2


Here –ve charge of dipole is close to wire hence net force an dipole due to wire will
be attractive.
30
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
+++++++++++
ww
Calculate force on a dipole in the surrounding of a long charged wire as shown in the figure.
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ELECTRIC POTENTIAL DUE TO DIPOLE
At axial point
–q
kq
V1 
(r  )
Electric potential due to +q charge
kq
(r  )
Electric potential due to –q charge
V2 
Net electric potential
V = V1 + V2 =
kq  2 
kp
kq
kq
 2

 2
2
(r  ) (r  ) (r   )
r  2
At equitorial point
kq
V1 =
x
V2 = 
Electric potential of P due to –q charge
A t ge n e r a l po i n t
V =
 
p cos 
p.r

4 0 r 2
4 0 r 3
kq
x
kq kq

= 0  V = 0
x
x
–q
ww



p  q d electric dipole moment
y
Enet
E
Figure shows an electric dipole placed on x–axis at origin. Here we wish to
find the electric field and potential at a point O having coordinates (r, ). Due
to the positive charge of dipole electric field at O is in radially outward direction
and due to the negative charge it is radially inward as shown in figure.
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
q
O

Electric field due to an electric dipole
E
x
r
x
w.
Le
arn
•
V = V1 + V2 =
P
En
Net potential
kp
r2
gin
e
Electric potential of P due to +q charge
P
r
If r > > >    V 
•
q
O
eri
ng
.in
•
E 2r  E 2 =
kp
r3
O
r

2kp cos 
V
1 V kp sin 


Er =–
and E   
3
r
r
r 
r3
Thus net electric field at point O, Enet =
Er

-q
+q
1  3 cos 2 
E  1

If the direction of Enet is at an angle  from radial direction, then  = tan–1 E =  tan 
2
r
Thus the inclination of net electric field at point O is (
31
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x
JEE-Physics
At a point on the axis of a dipole :

Electric field due to +q charge E1 =
Net electric field
E2

E1
r
kq
(r  )2
kq
(r  )2
Electric field due to –q charge E2 =
kq
kq
kq  4r 

2
2 =
(r  )
(r  )
(r 2  2 )2
E = E1 – E2 =
2kpr
2kp
2 2 If r >>> then E =
(r   )
r3
E=
•
q
O
[ p = q × 2 = Dipole moment]
2
At a point on equitorial line of dipole :
kq
kq
Electric field due to +q charge E 1  2 ; Electric field due to –q charge E 2  2
x
x
 P

E
Vertical component of E1 and E2 will cancel each other and horizontal
E = E1cos + E2cos [ E1 = E2]
E = 2E1 cos =
E2
2
gin
e
components will be added So net electric field at P
x
x
r
2kq

cos   cos  
and x  r 2  2
2
x
x

–q


En
 kp
2kq 
2kq 
kp
kp
E 3  2
E
=
If r > > >  then E  3 or  3
x
(r  2 ) 3 2 (r 2  2 ) 3 2
r
r
w.
Le
arn
E1 sin
E1
E sin
–q
eri
ng
.in
•
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O
q

GOLDEN KEY POINTS
•
For a dipole, potential is zero at equatorial position, while at any finite point E  0
•
In a uniform E , dipole may feel a torque but not a force.
•
If a dipole placed in a field E (Non-Uniform) generated by a point charge, then torque on dipole may be zero,
but F  0

Point charge
Dipole
Potential proportional to
r–1
r–2
E proportional to
r
–2
r
–3
Force between
Point charge
Proportional to
r
–2
Dipole and
Dipole-dipole
point charge
–3
r
–4
r
Example
A short electric dipole is situated at the origin of coordinate axis with its axis along x–axis and equator along
y–axis. It is found that the magnitudes of the electric intensity and electric potential due to the dipole are equal
at a point distant r =
5 m from origin. Find the position vector of the point in first quadrant.
Solution
 E P  VP 
kp
r3
1  3 cos2  =

Position vector r of point P is r =
1
kp cos 
 1+3 cos2  = 5 cos2   cos  =
  = 450
2
2
r
5
2
ˆi  ˆj 
32
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
Distribution
ww
•

E
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Example
Prove that the frequency of oscillation of an electric dipole of moment p and rotational inertia I for small
amplitudes about its equilibrium position in a uniform electric field strength E is
 pE 
 I 
1
2
Solution
Let an electric dipole (charge q and –q at a distance 2a apart) placed in a uniform external electric field of
strength E.
2a


E

P

F
-q
  pE sin  = –pE  (as  is small)
gin
e
Restoring torque on dipole

F
eri
ng
.in
+q
Here – ve sign shows the restoring tendency of torque.    I  angular acceleration =  
pE
I
En
For SHM   2  comparing we get  =
ELECTROSTATIC
 pE 
 I 

1
=
2
2
w.
Le
arn
Thus frequency of oscillations of dipole n =
PRESSURE
Force due to electrostatic pressure is directed normally outwards to the surface .
+ + + + ++
Force on small element ds of charged conductor
+
+
+
++
2
E1
Inside E1–E2 = 0  E1 = E2
ww
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65


ds
dF = (Charge on ds ) x Electric field = ( ds) 2 dF 
2 0
0
E
 PE
=

I
I
Just outside E = E1 + E2 = 2E2  E 2 
+
ds
+
+
++
E2
E1
+

2 0
(E1 is field due to point charge on the surface and E2 is field due to rest of the sphere).
The electric force acting per unit area of charged surface is defined as electrostatic pressure.
Peleectrostatic 
dF
dS

2
2 0
33
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Equilibrium of liquid charged surfaces (Soap bubble)
Pressures (forces) act on a charged soap bubble , due to
(i)
Surface tension PT (inward)
(ii)
Air outside the bubble Po (inward)
(iii)
Electrostatic pressure Pe (outward)
(iv)
Air inside the bubble Pi (outward)
in state of equilibrium inward pressure = outward pressure PT + Po = Pi +Pe
+
+ + +++ +
+
+
R+
+
+
+
++ + +
+
but
PT =
eri
ng
.in
Excess pressure of air inside the bubble (Pex) = Pi – Po = PT – Pe
2
4T
2
4T
2
4T


and Pe =
 Pex =
if Pi = Po then
2 0
r
2 0
r
2 0
r
Example
Brass has a tensile strength 3.5 × 108 N/m2. What charge density on this material will be enough to break it by
electrostatic force of repulsion? How many excess electrons per square Å will there then be? What is the value
of intensity just out side the surface?
Solution
So the conductor will break by this force if,
gin
e
dF
2
We know that electrostatic force on a charged conductor is given by ds  2 
0
2
> Breaking strength i.e.,  2  2  9  10 12  3.5  10 8
2 0
En
i.e.  min   3 7   10 2  7.94  10 2  C / m 2 
Now as the charge on an electron is 1.6 × 10–19 C, the excess electrons per m2
 7.94  10 2

 8.8  10 9 V/m
0
9  10 12
(vii)
ww
CONDUCTOR AND IT 'S PROPERT IES [FOR ELECT ROSTATIC CONDITION]
(i)
Conductors are materials which contains large number of free electrons which can move freely inside the
conductor.
(ii)
n electrostatics, conductors are always equipotential surfaces.
(iii)
Charge always resides on outer surface of conductor.
(iv)
f there is a cavity inside the conductor having no charge then charge will always reside only on outer
sur face of conductor.
(v)
Electric field is always perpendicular to conducting surface.
(vi)
Electric lines of force never enter into conductors.
Electric field intensity near the conducting surface is given by formula E =

n̂
0






E A  A nˆ ; E B  B nˆ and E C  C nˆ
0
0
0
(viii)
When a conductor is grounded its potential becomes zero.
34
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
w.
Le
arn
Further as in case of a conductor near its surface E 
E
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(ix)
(x)
When an isolated conductor is grounded then its charge becomes zero.
When two conductors are connected there will be charge flow till their potential becomes equal.
(xi)
Electric pressure at the surface of a conductor is givey by formula P =
2
2 0
where  is the local surface charge
density.
Solution
eri
ng
.in
Example
Prove that if an isolated (isolated means no charges are near the sheet) large conducting sheet is given a
charge then the charge distributes equally on its two surfaces.
Let there is x charge on left side of sheet and Q–x charge on right side of sheet.
Since point P lies inside the conductor so EP = O
2x
x
Q x
–
= 0 
2 A O
2 A O
2 A O
Q–x=
=
Q
Q
x=
2
2 A O
Q
2
Q-x
2A 0
gin
e
So charge is equally distributed on both sides
Example
Q-x
x
P
x
2A 0
If an isolated infinite sheet contains charge Q1 on its one surface and charge Q2 on its other surface then prove
En
that electric field intensity at a point in front of sheet will be
Solution
Q
2AO
, where Q = Q1 + Q2
w.
Le
arn



Q1
Q2
Q1  Q 2
Q
Electric field at point P : E = E Q1  E Q 2 = 2A n̂ + 2A n̂ = 2 A n̂ = 2A n̂
0
0
0
0
E
ww
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
[This shows that the resultant field due to a sheet depends only on the total charge of the sheet and not on the
distribution of charge on individual surfaces].
Example
Three large conducting sheets placed parallel to each other at finite distance contains charges Q, –2Q and
3Q respectively. Find electric field at points A, B , C, and D
Q
·A
Sol.
EA = EQ + E–2Q + E3Q.
EA =
-2Q
·B
3Q
·C
(i) Here EQ means electric field due to ‘Q’.
(Q  2Q  3Q )
2Q
Q
= 2A = A  , towards left
2A  0
0
0
35
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EB =
Q  ( 2Q  3Q )
, towards right = 0
2A  0
(iii)
EC =
(Q  2Q)  (3Q )
4Q
2Q
2Q

2A  0
2A 0 = A 0 , towards right  A  0 towards left
(iv)
ED =
(Q  2Q  3Q )
2Q
Q
=
=
2A 0
2A0
A0 , towards right
eri
ng
.in
(ii)
Example
Two conducting plates A and B are placed parallel to each other. A is given a charge Q1 and B a charge Q2.
Prove that the charges on the inner facing surfaces are of equal magnitude and opposite sign. Also find the
charges on inner & outer surfaces.
Solution
gin
e
Consider a Gaussian surface as shown in figure. Two faces of this closed surface lie completely inside the
conductor where the electric field is zero. The flux through these faces is, therefore, zero. The other parts of
the closed surface which are outside the conductor are parallel to the electric field and hence the flux on these
parts is also zero. The total flux of the electric field through the closed surface is, therefore zero. From Gauss’s
law, the total charge inside this closed surface should be zero. The charge on the inner surface of A should be
equal and opposite to that on the inner surface of B.
Q1–q
E=0
Q1
En
A
E
B
Q2+q
w.
Le
arn
E=0
A
–q
B
Q2
P
•
+q
The distribution should be like the one shown in figure. To find the value of q, consider the field at a point P
inside the plate A. Suppose, the surface area of the plate (one side) is A. Using the equation E =  / (20), the
electric field at P
Q1  q
q
due to the charge Q1 – q = 2A  (downward); due to the charge + q = 2A (upward),
0
0
0
Q2  q
(downward), and due to the charge Q2 + q = 2A  (upward).
0
The net electric field at P due to all the four charged surfaces is (in the downward direction)
Ep =
Q1  q
Q q
q
q


 2
2A  0
2 A 0 2A  0
2A  0
As the point P is inside the conductor, this field should be zero.
Hence, Q1 – q – q + q – Q2 – q = 0  q 
Q1  Q 2
2
This result is a special case of the following result. When charged conducting plates are placed parallel to each
other, the two outermost, surfaces get equal charges and the facing surfaces get equal and opposite charges.
36
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
ww
q
due to the charge – q = 2A
E
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JEE-Physics
Example
Figure shows three large metallic plates with charges – Q, 3Q and Q respectively. Determine the final charges
on all the surfaces.
3Q
Solution
Q
eri
ng
.in
–Q
We assume that charge on surface 2 is x. Following conservation of charge, we see that surfaces 1 has charge
(– Q – x). The electric field inside the metal plate is zero so fields at P is zero.
1
2
-Q-x
x
3Q
4
3
Q
6
gin
e
P
5
Q  x
x  3Q  Q
5Q
Resultant field at P : EP = 0  2A
=
 –Q – x = x + 4Q  x =
2
A

2
0
0
En
Note : We see that charges on the facing surfaces of the plates are of equal magnitude and opposite sign. This
can be in general proved by gauss theorem also. Remember this it is important result.
w.
Le
arn
Thus the final charge distribution on all the surfaces is :
3Q
+
2
Example
- 5Q
2
5Q
2
+Q
2
-Q
2
+3 Q
2
E
ww
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
An isolated conducting sheet of area A and carrying a charge Q is placed in a uniform electric field E, such
that electric field is perpendicular to sheet and covers all the sheet. Find out charges appearing on its two
surfaces. Also
Q
E
37
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Solution
Let there is x charge on left side of plate and Q – x charge on right side of plate
x
Q-x
2A 0
Q–x
x
+E
2A 0
eri
ng
.in
P
x
Qx
x
Q
Q
EP = 0  2A   E  2A   A   2A   E  x  2  EA  0
0
0
0
0
Q
Q
– EA  o and other side
+ EA  o
2
2
The resultant electric field on the left and right side of the plate.
So charge on one side is
0
Q
+ E towards right and on left side 2 A 
gin
e
Q
On right side E = 2 A 
– E towards left.
0
(ii)
w.
Le
arn
En
SOME OTHER IMPORTANT RESULTS FOR A CLOSED CONDUCTOR.
(i)
f a charge q is kept in the cavity then –q will be induced on the inner surface and +q will be induced on the
outer surface of the conductor (it can be proved using gauss theorem)
If a charge q is kept inside the cavity of a conductor and conductor is given a charge Q then –q charge will be
induced on inner surface and total charge on the outer surface will be q + Q. (it can be proved using gauss
theorem)
+q+Q
Resultant field, due to q (which is inside the cavity) and induced charge on S1, at any point outside S1 (like B,C)
is zero. Resultant field due to q + Q on S2 and any other charge outside S2 , at any point inside of surface
S2 (like A, B) is zero
S2
.B
.C
S1
.q
q+Q
.A
–q
38
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
(iii)
ww
–q
q
E
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(iv)
Resultant field in a charge free cavity in a closed conductor is zero. There can be charges outside the conductor and on the surface also. Then also this result is true. No charge will be induced on the inner most surface
of the conductor.
No
charge
Charge distribution for different types of cavities in conductors
eri
ng
.in
(v).
S2
S2
•q
S1 •C
S1
q
•
C
(A)
(B)
charge is at the common centre
(S1, S2  spherical)
gin
e
S2
q
•
S1
charge is not at the common centre
(S1, S2  spherical)
(C)
w.
Le
arn
charge is at the centre of S2
(S2  spherical)
En
C
S2
S1
(D)
charge is not at the centre of S2
(S2   spherical)
S2
S2
S1
·q
C
E
C
S1 ••q
(F)
ww
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
(E)
•
C •q
charge is at the centre of
S1 (Spherical)
charge not at the centre
of S1(Spherical)

Using the result that E res in the conducting material should be zero and using result (iii) We can show that
Case
S1
A
Uniform
S2
Uniform
B
C
D
Nonuniform Nonuniform Nonuniform
Uniform
Uniform
Uniform
E
Uniform
F
Nonuniform
Nonuniform Nonuniform
Note : In all cases charge on inner surface S1 = – q and on outer surface S2 = q. The distribution of charge on S1 will
not change even if some charges are kept outside the conductor (i.e. outside the surface S2). But the charge
distribution on S2 may change if some charges(s) is/are kept outside the conductor.
39
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Example
An uncharged conductor of inner radius R1 and outer radius R2 contains a point charge q at the centre as
shown in figure
S2
S1 q R1
·C
O
R2
·A
(i)
(ii)

Find E and V at points A,B and C
eri
ng
.in
·B
If a point charge Q is kept out side the sphere at a distance ‘r’ (>>R 2) from centre then find out
resultant force on charge Q and charge q.
+q
Solution
–q
·q
gin
e

Kq Kq K (  q )
Kq 
OA
At point A : VA  OA  R  R
, EA 
OA 3
2
1
Kq K (  q) Kq Kq
Kq 
Kq 
OC
At point B : VB  OB  OB  R  R , EB = 0; At point C : VC =
, EC 
OC 3
OC
2
2
En
Example

KqQ
Force on point charge Q : FQ  2 ˆr (r = distance of ‘Q’ from centre ‘O’)
r

Force on point charge q: Fq  0 (using result (iii) & charge on S1 uniform)
w.
Le
arn
(ii)
An uncharged conductor of inner radius R1 and outer radius R2 contains a point charge q placed at point P
(not at the centre) as shown in figure? Find out the following :
B
A
1
D
ww
S2
(i) VC
(ii) VA
(iii) VB
R2
(iv) EA
(v) EB
(vi) force on charge Q if it is placed at B
Solution
Kq K (  q) Kq
(i) VC  CP  R  R
Kq
(ii) VA = R
(iii) VB =
Kq
CB
(iv) EA = O (point is inside metallic conductor)
(v) EB =
Kq ^
CB
CB 2
1
2
2
40
(vi) FQ =
KQq ^
CB
CB 2
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
R1 P
q
S C
E
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(vi)
JEE-Physics
S h ar i n g o f ch a r g es :
Two conducting hollow spherical shells of radii R 1 and R 2 having charges Q 1 and Q 2 respectively and
seperated by large distance, are joined by a conducting wire. Let final charges on spheres are q 1 and q 2
respectively.
q1
q2
R1
R2
Potential on both spherical shell become equal after joining, therefore
Kq 1 Kq 2 q 1 R 1


R1
R 2 ; q 2 R 2 ...(i) and
(Q 1  Q 2 )R 1
(Q  Q 2 )R 2
q  1
R1  R 2 ; 2
R1  R 2
q 1 R 1 1 4 R 12 R 1

;

ratio of charges
q 2 R 2 2 4 R 22 R 2
gin
e
1 R 2

2 R 1
ratio of surface charge densities
q1 R 1

q2 R 2
En
Ratio of final charges
w.
Le
arn
Ratio of final surface charge densities.
Example
......(ii)
eri
ng
.in
from (i) and (ii) q 1 
q1 + q 2 = Q1 + Q2
1 R 2

2 R 1
The two conducting spherical shells are joined by a conducting wire and cut after some time when
charge stops flowing.Find out the charge on each sphere after that.
Q
-3Q
R
2R
E
ww
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
Solution
After cutting the wire, the potential of both the shells is equal
Thus, potential of inner shell V in =
Kx
K( 2Q  x ) k x  2Q 
+
=
R
2R
2R
and potential of outer shell V out =
Kx
K( 2Q  x )
KQ
+
=
2R
2R
R
As
Vout
= Vin  – KR =
R
x
–2Q – x
K x – 2Q 
 –2Q = x – 2Q  x = 0
2R
So charge on inner spherical shell = 0 and outer spherical shell = – 2Q.
41
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5Q
Example
3
-2Q
Q
2
Find charge on each spherical shell after joining the inner most shell and
1
R
2R
outer most shell by a conducting wire. Also find charges on each surface.
3R
Solution
6Q – x
3
Let the charge on the innermost sphere be x.
-2Q
x
2
1
Finally potential of shell 1 = Potential of shell 3
eri
ng
.in
2R
Kx K ( 2Q) K (6Q  x) KQ k  –2q  k  5Q 





R
2R
3R
3R
3R
3R
3x –3Q + 6Q – x = 4Q ; 2x = Q ;
Charge on innermost shell =
R
x =
Q
2
3R
Q
+3Q/2
–3Q/2
–Q/2
Q/2
Q
5Q
, charge on outermost shell =
2
2
gin
e
middle shell = –2Q
Final charge distribution is as shown in figure.
Example
3Q
-Q
En
Two conducting hollow spherical shells of radii R and 2R carry charges –
R
Q and 3Q respectively. How much charge will flow into the earth if inner
2R
w.
Le
arn
shell is grounded ?
Solution
When inner shell is grounded to the Earth then the potential of inner shell will bcome zero because
potential of the Earth is taken to be zero.
3Q
Kx
K3Q
+
= 0
R
2R
R
the charge that has increased
3Q
Q
– (–Q)=
2
2
hence charge flows into the Earth =
Q
2
2R
Example
An isolated conducting sphere of charge Q and radius R is connected to a similar uncharged sphere
(kept at a large distance) by using a high resistance wire. After a long time what is the amount of heat
loss ?
Solution
When two conducting spheres of equal radius are connected charge is equally distributed on them.
So we can say that heat loss of system
H = Ui – Uf
 Q2
  Q2 / 4 Q2 / 4 
Q2

 0  


 8 0 R
  8  0 R 8 0 R  16  0 R
42
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
3Q
x =
,
2
ww
=
x
E
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SOME WORKED OUT EXAMPLES
Example#1
By using Gauss law
(Note : Check dimensionally that   r6)
  4 r 2 dr 
0
 k 0 r 9   r 2 dr
sia
us
O
c
dr
gin
e
 kr 7   4 r 2   
n Surfa
Ga
  4 r 2 dr 
 
q

2


 E  dS  0   E  4 r  0
e
eri
ng
.in

For a spherically symmetrical charge distribution, electric field at a distance r from the centre of sphere is E  kr 7 ˆr ,
where k is a constant. What will be the volume charge density at a distance r from the centre of sphere ?
(B)  = 5k0r3
(C)  = 3k0r4
(D) =9k 0r 0
(A)  = 9k0r6
Solution
Ans. (A)
En
Example#2
Two positrons (e+) and two protons (p) are kept on four corners of a square of side a as shown in figure. The mass
of proton is much larger than the mass of positron. Let q denotes the charge on the proton as well as the
positron then the kinetic energies of one of the positrons and one of the protons respectively after a very long
time will be–
e+
w.
Le
arn
p
e+
p
q2 
1 
q2 
1 
(A) 4   a  1 
 , 4   a  1 

2 2
2 2
0
0
q2
(B) 2   a ,
q2
q2
(C) 4   a , 4   a
q2 
1 
q2
(D) 2   a  1 
 ,
4 2 8 2  0 a
0
E
ww
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
0
0
0
q2
4 2  0 a
Solution
Ans. (D)
As mass of proton >>> mass of positron so initial acceleration of positron is much larger than proton. Therefore
positron reach far away in very short time as compare to proton.
e+
a
a
p
a
a
p
a
a
e+
p
p
a
a
kq 2
q2
 4kq 2 2kq 2  kq 2
q2 
1 
2K


0

K

2K e   



K

1


p
p
and


e
 a
2  0 a 
a 2
8 2  0 a
a 2 a 2
4 2
43
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Example#3
Four charges are placed at the circumference of a dial clock as shown in figure. If the clock has only hour
hand, then the resultant force on a charge q0 placed at the centre, points in the direction which shows the time
as :–
+q
12
3
q 9
6
q
(A) 1:30
Solution
(B) 7:30
(C) 4:30
+q
+q
Ans. (B)
Fnet
6
q
En
7.3
0
q0
(D) 10:30
gin
e
q 9
+q
eri
ng
.in
q0
w.
Le
arn
Example#4
A small electric dipole is placed at origin with its dipole moment directed along positive x-axis. The direction of
electric field at point (2, 22,0) is
(A) along z-axis
(B) along y-axis
(C) along negative y-axis
(D) along negative z-axis
Solution
Ans. (B)
tan  
q
(2,2 2)

x
+q

y
1
tan 
1
 2; cot  

 cot    +  = 90° i.e., E is along positive y-axis.
Also tan  
x
2
2
2
Example#5
Uniform electric field of magnitude 100 V/m in space is directed along the line y= 3 + x. Find the potential
difference between point A (3, 1) & B (1,3).
(A) 100 V
(B) 2002V
(C) 200 V
(D) zero
Solution
Ans. (D)
Slope of line AB =
3 1
 1 which is perpendicular to direction of electric field.
13
44
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
ww
y
E
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Example#6
The diagram shows a uniformly charged hemisphere of radius R. It has volume charge density . If the electric
field at a point 2R distance above its centre is E then what is the electric field at the point which is 2R below its
centre?
A
R
(A) 6   E
R
(B) 12   E
0
eri
ng
.in
B
R
(C) 6   E
0
0
Solution
0
Ans. (B)

+
+
R
(D) 24   E
gin
e
Apply principle of superposition
R
R
Electric field due to a uniformly charged sphere = 12  ; E resultant = 12   E
0
0
m  2 2
2e 
Solution
(B)
m  2 2
e
(C)
w.
Le
arn
(A)
En
Example#7
A metallic rod of length l rotates at angular velocity  about an axis passing through one end and perpendiuclar
to the rod. If mass of electron is m and its charge is –e then the magnitude of potential difference between
its two ends is
m 2 
e
(D) None of these
Ans. (A)
When rod rotates the centripetal acceleration of electron comes from electric field E 

e
eE
E
ww
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65


mr 2
m  2 2
dr 
Thus, V    E.dr   
e
2e
0
E
mr  2
e
Example#8
Consider a finite charged rod. Electric field at Point P (shown) makes an angle with horizontal dotted line then
angle  is :-
320
0
88
(A) 60°
(B) 28°
(C) 44°
45
(D) information insufficient
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Solution.
Ans. B
Required angle =
2  1
88 0  32 0 56 0
0
=
=
= 28
2
2
2
Example#9
The electric potential in a region is given by the relation V(x) = 4 + 5x2. If a dipole is placed at position (–1,0)

with dipole moment p pointing along positive Y-direction, then
eri
ng
.in
(A) Net force on the dipole is zero.
(B) Net torque on the dipole is zero
(C) Net torque on the dipole is not zero and it is in clockwise direction
(D) Net torque on the dipole is not zero and it is in anticlockwise direction
Solution
Ans. (AC)
F

V(x) = 4 + 5x 2  E  10xiˆ
p
( 1,0)
 Net force will be zero and torque not zero
(0,0)
F
and rotation will be along clockwise direction
gin
e
Example#10 to 12
A thin homogeneous rod of mass m and length  is free to rotate in vertical plane about a horizontal axle pivoted
at one end of the rod. A small ball of mass m and charge q is attached to the opposite end of this rod. The whole
system is positioned in a constant horizontal electric field of magnitude E 
En
position from rest.
mg
. The rod is released from shown
2q

m
What is the angular acceleration of the rod at the instant of releasing the rod?
(A)
11.
8g
9
(B)
3g
2
(C)
9g
8
(D)
2g
9
8g
7
What is the acceleration of the small ball at the instant of releasing the rod?
8g
9
(B)
ww
(A)
12.
E
9g
8
(C)
7g
8
(D)
(C)
3g 
2
(D)
What is the speed of ball when rod becomes vertical?
(A)
2g 
3
(B)
3g 
4
4g 
3
Solution
1 0 . Ans. (C)
/2
/2
mg
mg

m 2
9g
By taking torque about hinge I  mg    mg   when I 
 m 2   
 2
3
8
46
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
10.
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arn
m
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11.
JEE-Physics
Ans. (B)
Acceleration of ball =    9g    9 g
 8 
8
12.
Ans. (C)
From work energy theorem
1 2
 
I  mg    mg   qE 
 2
2
14
3
mg 
2
2
2
 m 2  2  mg    
 m     mg  
23
2
2
3
3g 
2
eri
ng
.in
Speed of ball =  
3g
2
T  2

g eff
geff = g –
qE
1  5  10 6
 5  15 
M
1  10 6
2
E
5m/s2
En
g eff = 10 = 
T = 2 sec
gin
e
Example#13
A simple pendulum is suspended in a lift which is going up with an acceleration of 5 m/s 2 . An electric field of
magnitude 5 N/C and directed vertically upward is also present in the lift . The charge of the bob is 1 C and
mass is 1 mg . Taki ng g =  2 and length of t he simple pendulum 1m, fi nd t he t ime period of t he simple
pendulum (in sec).
Solution
Ans. 2
w.
Le
arn
Example#14
The variation of potential with distance x from a fixed point is shown in figure. Find the magnitude of the electric
field (in V/m) at x =13m.
V(volt)
60
45
30
20
10
E
2
4
6
8
10 12 14 16
x(m)
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NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
0
Solution
E = –
Ans. 5
dV 20

= 5 volt/meter
dx
4
Example#15
The energy density u is plotted against the distance r from the centre of a spherical charge distribution on a
log-log scale. Find the magnitude of slope of obtianed straight line.
Solution.
Ans. 2
2
 q 


1
q2
q2
1
2


 log u  log 
 log k  2 log r
u =
 0E =
0 
2
2

2
2
2
32  0 r
 4  0 r 
2
 32  0 r 
47
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Example#16
The figure shows four situations in which charges as indicated (q>0) are fixed on an axis. How many
situations is there a point to the left of the charges where an electron would be in equilibrium?
+q
(1)
+4q
(3)
–4q
–q
–q
–4q
(2)
+4q
(4)
q
Solution
For (1)
Let the electron be held at a distance x from +q charge.
Ans. 2
x
+q
eri
ng
.in
–e
d
q( e)
(  e)( 4q )
For equilibrium 4   x 2  4   (x  d)2
0
0
–4q
We can find value of x for which F net = 0 which means that electron will be in equilibrium.
For (2)
:
( e)(  q)
( e)4 q
For equilibrium 4   x 2  4   (x  d)2
0
0
e
x
d
–q
+4q
(x+d)
gin
e
We can find value of x for which F net = 0 which means that electron will be in equilibrium.
In case (3) and (4) the electron will not remain at rest, since it experiences a net non–zero force.
OR
Equilibrium is always found near the smaller charge
Example#17
Solution
w.
Le
arn
En

N

An electric field is given by E   yˆi  xjˆ . Find the work done (in J) in moving a 1C charge from rA   2ˆi  2ˆj 
C

m to rB   4ˆi  ˆj  m .
Ans. 0
B
B 

A= (2,2) and B = (4, 1); W A  B  q  VB  VA   q  dV    qE.dr
B
A
B
 4,1
A
2,2 
A
 4,1
 q   yiˆ  xjˆ.  dxiˆ  dyjˆ  q   ydx  xdy    q  d  xy    q  xy 2,2    q  4  4   0
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Example#18
The arrangement shown consists of three elements.
(i) A thin rod of charge –3.0 C that forms a full circle of radius 6.0 cm.
(ii) A second thin rod of charge 2.0 C that forms a circular arc of radius 4.0 cm and
concentric with the full circle, subtending an angle of 90° at the centre of the full
circle.
(iii) An electric dipole with a dipole moment that is perpendicular to a radial line and
has magnitude 1.28 × 10 –21C-m.
Find the net electric potential in volts at the centre.
Solution
Potential due to dipole at the centre of the circle is zero.
Potentials due to charge on circle = V1 =
Potential due to arc V2 =
3
1
2
Ans. (0)
K.( 3  10 6 )
6  10 2
K.(2  10 6 )
Net potential = V1 + V 2= 0
4  10 2
48
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Example#19
Six charges are kept at the vertices of a regular hexagon as shown in the figure. If
magnitude of force applied by +Q on +q charge is F, then net electric force on the +Q
is nF. Find the value of n.
Solution
+q
+q
+q
+3q
-2q
+3q
+Q Ans. 9
-3q
-4q
5F
-2q
4F
-3q
5F
+q
F net = 9F
eri
ng
.in
-4q
Example#20

Electric field in a region is given by E  4xiˆ  6 yjˆ . The charge enclosed in the cube of side 1m oriented as
shown in the diagram is given by 0. Find the value of .
gin
e
z
y
x
Solution
w.
Le
arn
En
z
Ans. 2
q
2
2
y  = (6y) Area – (4x) Area = 6 × 1 × (1) – 4 × 1 × (1) = 2 therefore 
x
Example#21
0
 2  q = 2
0
C
is tilted at a 37° angle to the vertical direction as shown below.
m2
Find the potential difference, VA–VB in volts, between points A and B at 5 m distance apart. (where B is vertically
above A).
E
B
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NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Electrostatics\English\Theory.p65
An infinite plane of charge with   2 0
A
Solution
Ans. 3
B
53°
A
E
    

5 cos 53   3V
 1 N/C  VB  VA    E.d   
2 0
 2 0 
37°
49
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