12
CLASS
Electrostatics
ELECTRIC CHARGE
Jayant Nagda
Charges
Charges are everywhere!
Charges
Charges
Charging by friction
Charges
Charging by friction
Charges
Charging by friction
Charges
Charging by friction
Charges
Charging by friction
Charges
Charges are everywhere!
Charges
Charges are everywhere!
Charge is always associated with mass.
Intrinsic property like mass
(always exists along with the particles)
Electric Charge
Charge is the property associated with matter
due to which it produces and experiences
electrical and magnetic effects.
Charge at Rest produces only Electric Effects
Electric Charge
Charge in Motion produces both
Electric and Magnetic Effects
Electric Charge
Accelerated Charge produces both
Electric and Magnetic Effects
& also radiates energy
A stationary electric charges produces -
A. Only electric fields
B. Only magnetic fields
C. Both electric and magnetic fields
D. Neither electric nor magnetic field
The electric charge in uniform motion produces A.
an electric field only
B.
a magnetic field only
C.
both electric and magnetic fields
D.
Neither electric nor magnetic fields
An accelerated or deaccelerated charge produces A.
Electric field only
B.
Magnetic field only
C.
Localised electric and magnetic fields
D. Electric and magnetic fields that are radiated
Electric Charge
Charges produce Electrical and Magnetic effects
v = 0 (i.e. at rest)
Q
v = constant
Q
v ≠ constant (i.e. time varying)
Q
Produces only E
Produces E and B
Produces E, B
(electric field)
(magnetic field)
and radiates energy
But no radiation
Fundamental Charges
S.I. Unit of measurement: Coulomb (C)
Dimensional Formula is
Scalar quantity can be added algebraically.
Examples of Charged particles:
proton,
electron,
α-particles
+ -
Fundamental Charges
Electronic Charge e = 1.6×10-19 Coulomb
Charge on Electron = -e = −1.6×10-19 C
Charge on Proton = +e = +1.6×10-19 C
Charge on Neutron = 0 C
Positive and negative sign were arbitrarily assigned
by Benjamin Franklin
+ -
- +
+
+
+
- +
Types of Charges
Positive
Negative
Deficiency of e-s
Excess of e-s
Positive charge :
It is the deficiency of
electrons as compared
to proton.
Negative charge :
It is the excess of
electrons as compared
to proton.
If an object has a net charge of - 1 coulomb,
the number of excess electrons it possesses is A. 1.6 x 10-19
B. 6.25 x 1018
C. 6.25 x 1020
D. 6.25 x 1017
If a body has a charge of 10-12 coulomb A. the body has 6.25 x 106 excess of electrons
B. the body has 625 x 106 excess of electrons
C. the body has 6.25 x 106 deficiency of electrons
D. the body has 6.25 x 103 deficiency of electrons
Properties of Charges
Attractive/Repulsive
Like charges Repel
Unlike charges Attract
Properties of Charges
Attractive/Repulsive
Like charges Repel
Unlike charges Attract
Properties of Charges
Charge is Conserved
Charge is neither created nor destroyed
In every reaction/transformation
the total charge on an isolated system
remains constant.
Properties of Charges
Relativistic Invariant
Mass of a Particle varies with its speed v
m0: rest mass
m: mass in motion
Unlike mass, the charge on a body
does not vary with speed
Properties of Charges
Charge is Quantized
Charge on a body must always exist as
an integral multiple of fundamental unit of charge e.
Fundamental Unit = 1.6 x 10-19 C
Charge on any Body
Charge vs Mass
Charge
Mass
1.
Electric charge can be positive
or negative
1.
Mass of a body is always positive
2.
Charge is conserved
2.
Mass is not conserved as it can be
converted into energy and vice versa by E
= mc2
3.
Charge is quantized
3.
Quantization of mass is yet to be
established
4. Charge is relativistically
invariant Qrest = Qmotion
4. Mass is relativistically variant.
5.
5.
Force between charges may be
attractive or repulsive.
Force between masses always be
attractive
Which one of the following statement regarding
electrostatics is wrong ?
A. Charge is quantized
B. Charge is conserved
C. There is an electric field near an isolated charge at rest
D. A stationary charge produces both electric and
magnetic fields
Minimum magnitude of charge (quantum of charge)
an object can have if measured in coulomb -
A. 1.6 x 10-17 C
B. 1.6 x 10-19 C
C. 1.6 x 10-10 C
D. 4.8 x 10-10
12
CLASS
Electrostatics
Methods of Charging
Jayant Nagda
Electric Charge
Greek word elektron, meaning Amber
Amber rubbed with wool or silk cloth attracts light objects
Different Methods of Charging
Charging by Friction
requires rubbing of two objects
electrons are transferred from one body to the other
Different Methods of Charging
Charging by Friction
total charge on both bodies remains same
(Conservation of Charge)
Different Methods of Charging
Charging by Friction
total charge on both bodies remains same
(Conservation of Charge)
Different Methods of Charging
Charging by Friction
glass
silk
Silk + Glass Rod
If a glass rod is rubbed with silk, it acquires a positive
charge because A.
Protons are added to it
B. Protons are removed from it
C. Electrons are added to it
D. Electrons are removed from it
Different Methods of Charging
Charging by Friction
Different Methods of Charging
Charging by Friction
Rubber Rod + Fur
Different Methods of Charging
Charging by Friction
Different Methods of Charging
Charging by Friction
Different Methods of Charging
Charging by Friction
Positive charge
Negative charge
Glass Rod
Silk cloth
Fur
Rubber/Ebonite rod
Woolen cloth
Amber, Plastic objects
Dry hair
Comb
Different Methods of Charging
Charging by Conduction
Different Methods of Charging
Charging by Conduction
Requires only contact between two objects and no rubbing.
The conductors will be charged with the same sign.
Rubber Rod
+
Metal
The total charge is distributed between the objects
Different Methods of Charging
Charging by Conduction
Different Methods of Charging
Charging by Conduction
Requires only contact between two objects and no rubbing.
The conductors will be charged with the same sign.
The total charge is distributed between the objects
Different Methods of Charging
Charging by Induction
Different Methods of Charging
Charging by Induction
Electrons inside conductor move to create
Surface charge
Different Methods of Charging
Charging by Induction
Different Methods of Charging
Charging by Induction
Induction is a process in which a charged body
can be used to create other charged bodies
without touching them or losing its own charge.
A neutral metallic sphere is placed near a negative
point charge (outside the sphere). The net charge on
sphere is then A.
negative and distributed uniformly on the surface
B. zero
C. Positive and appears only at point on the sphere
closest to the point charge
D. Positive and distributed non-uniformly on the
surface of sphere
Different Methods of Charging
Different Methods of Charging
Different Methods of Charging
Charging by Induction
Creating charges out of nothing!
Two uncharged metal spheres A and B are in contact as
shown in the diagram. A negatively charged rod is brought
near to A, but not touching it. The two spheres are
separated slightly and the rod is then withdrawn.
As a result of this -
A. Both the spheres acquire +ve charge
B. Both the spheres acquire -ve charge
C. A acquires -ve and B acquires +ve charge
D. A acquires +ve and B acquires -ve charge
Gold Leaf Electroscope
Apparatus to detect the charge on a body.
It consists of a gold leaf
attached to the brass strip
which is suspended with the
help of a brass rod in a glass jar
Brass Cap
Brass Rod
Gold leaves
Gold Leaf Electroscope
Bring the body near the brass cap of
the electroscope.
If the gold leaf diverges outward,
the body is electrically charged.
However if no divergence takes place,
then body has no charge.
Gold Leaf Electroscope
The degree of divergence is an indicator of
the amount of charge.
Gold Leaf Electroscope
Gold Leaf Electroscope
When a charged object touches the metal knob at the top
of the rod, charge flows on to the leaves and they diverge.
12
CLASS
Electrostatics
Coulomb’s Law
Jayant Nagda
Charles Augustin de Coulomb
French Physicist
Charles-Augustin de Coulomb (1736 – 1806)
Began his career as a military engineer
In the West Indies.
In 1776, he returned to Paris and
retired to a small estate
to do his scientific research.
Charles Augustin de Coulomb
He invented a torsion balance to
measure the quantity of a force and
used it for determination of
forces of electric attraction or repulsion
between small charged spheres.
Different Methods of Charging
Charging by Conduction
Charles Augustin de Coulomb
He invented a torsion balance to
measure the quantity of a force and
used it for determination of
forces of electric attraction or repulsion
between small charged spheres.
Coulomb’s Law
In 1785 he arrived at the inverse square law relation,
now known as Coulomb’s law.
The law quantifies the amount of force between
two stationary, electrically charged particles
Coulomb’s Law
According to the law,
magnitude of electrostatic force
acting between two point charges is
directly proportional to product of charges and
inversely proportional to square of the distance
between them (Inverse Square Law).
When the distance between two charged particle is halved,
the force between them becomes-
A. One fourth
B. One half
C. Double
D. Four times
Each of the two point charges are doubled and their
distance is halved. Force of interaction becomes n
times, where n is -
Coulomb’s Law
Two point electric charges q1 and q2 at rest,
separated by a distance r exert a force on each other
Whose magnitude is given by
F= k
F
q 1q2
r2
k is a proportionality constant
k=
Two identical charges repel each other with a force of 2.5 N
when placed 3 m apart. The magnitude of each charge is -
A. 40 μC
B. 50 μC
C. 50 μC
D. None of these
Two similar and equal charges repel each other with
force of 1.6 N, when placed 3m apart. Strength of each
charge is A. 40μC
B. 20μC
C. 4μC
D. 2μC
Ans: A
Fg and Fe represent the gravitational and electrostatic
force respectively between two electrons situated at
some distance. The ratio of Fg to Fe is of the order of
A.
1036
B. 101
C. 100
D. 10-43
Coulomb’s Constant k
k is also known as Coulomb’s constant
F= k
q1 q 2
r2
k can be measured experimentally
k = 8.98755 × 109 Nm2C–2
k = 9 x 109 N-m2/C2
Dimensions of k ?
Permittivity of Free Space 𝜀0
F=k
q1q 2
r2
F=
1
q 1q2
4π𝜀o r2
𝜀0: permittivity of free space or vacuum
𝜀o =
Permittivity of a Medium 𝜀
measure of the electric polarizability of the material of medium
How well the medium allows/permits
electric field to pass through it
Permittivity of a Medium 𝜀
measure of the electric polarizability of the material of medium
How well the medium allows/permits
electric field to pass through it
Depends on atomic/molecular structure of medium,
its dipole characteristics.
𝜀0: permittivity of free space or vacuum
𝜀o =
1
4πk
𝜀o = 8.854 x 10-12 C2 N-1 m-2
The dimensions of permittivity of a medium are -
A. M-1L-3T4A2
B. ML3T-4A-2
C. MLT-3A2
D. None of these
Relative Permittivity of a Medium K or 𝜀r
If
𝜀0: permittivity of free space or vacuum
Permittivity of any other medium
can be expressed as
K: Relative permittivity or Dielectric Constant
of material of the medium
Relative Permittivity of a Medium K or 𝜀r
𝜀 = K 𝜀0
K lies between 1 and ∞
with Kvacuum = 1 and Kconductor= ∞
Now as K ≥ 1
(i.e ε > ε0)
The permittivity of vacuum is 8.86 × 10-12C2/N-m2 and
the dielectric constant of water is 81. The permittivity
of water in C2/N-m2 isA. 81 × 8.86 × 10-12
B. 8.86 × 10-12
C. (8.86 × 10-12)/81
D. 81/(8.86 × 10-12)
The dimensions of dielectric constant or relative
permittivity of a medium are-
Ans: D
A.
M-1L-3T4A2
B.
ML3T-4A-2
C.
MLT-3A2
D.
None of these
Coulomb’s Force in Different Medium
Charges placed in medium other than vacuum
experience a net force that is given by
F=
1
q1 q2
4π𝜀
r2
where 𝜀 is the permittivity of this new medium
&
𝜀 = K 𝜀0
K: Dielectric Constant of material of the medium
Coulomb’s Force in Different Medium
When charges are submerged in water medium
KWater = 81
Two charges are at distance (d) apart in air. Coulomb
force between them is F. If a dielectric material of
dielectric constant (K) is placed between them, the
coulomb force now becomesA. F/K
B. FK
C. F/K2
D. K2F
The force between two point charges placed in vacuum
at distance 1 mm is 18 N. If a glass plate of thickness 1
mn and dielectric constant 6, be kept between the
charges and new force between them would be A.
18 N
B. 108 N
C. 3 N
D. 3 × 10-6 N
Ans: C
Two point charges certain distance apart in air repel
each other with a force F. A glass plate is introduced
between the charges. The force becomes F1 where,
A. F1 < F
B. F1 = F
C. F1 > F
D. data is insufficient
Ans: A
A certain charge ‘Q’ is to be divided into two parts q
and Q - q. What is the relationship of ‘Q’ to ‘q’ if the
two parts, placed at a given distance ‘r’ apart are to
have maximum Coulomb repulsion?
A.
q = Q/2
B. q = Q/3
C. q = 2Q/3
D. q = Q/4
Ans: A
12
CLASS
Electrostatics
Superposition Principle
Jayant Nagda
Coulomb’s Law
Two point electric charges q1 and q2 at rest,
separated by a distance r exert a force on each other
Whose magnitude is given by
F= k
F
q1q2
r2
k is a proportionality constant
k=
1
4π𝜀o
= 9 x 109 N-m2/C2
Coulomb’s Law in Vector Form
r21: position of 2 w.r.t 1
F21: Force on 2 because 1
q1
q2
^
r21: unit vector along r21
Coulomb’s Law in Vector Form
q1q2
^
r21
F21 = k
2
r21
Coulomb’s Law in Vector Form
F21
q1
F21: Force on 2 because 1
^
r21: unit vector along r21
q2
F21 = k
r21: position of 2 w.r.t 1
q1q 2
r21
3
r21
Or
q1q2
^
r21
F21 = k
2
r21
Coulomb’s Law in Vector Form
r12: position of 1 w.r.t 2
F12
F12: Force on 1 because 2
q1
^
r12: unit vector along r12
q2
F12 = k
q1 q2
r123
r12
Or
q 1q2
^
r12
F12 = k
2
r12
Coulomb’s Law in Vector Form
In general,
F
q1
F= k
q2
q1q2
r3
r
Or
q 1 q2
r
F= k 2 ^
r
Coulomb’s Law in Vector Form
F21
q1
i.e. same (like) charges,
q2
F21 = k
q1q2
r213
1. If q1q2 > 0
r21
Coulomb’s Law in Vector Form
2. And q1q2< 0
unlike (opposite) charges
q1
q2
F21 = k
q1 q2
r213
r21
Coulomb’s Law in Vector Form
But for most questions,
For Magnitude we use
Coulomb’s Law in scalar form
and Direction use the fact:
like repel & unlike attract
Three charges +3q, +q and Q are placed on a straight line
with equal separation. In order to make the net force on
q to be zero, the value of Q will b
A. +3q
B. +2q
C. -3q
D. -4q
Three charged particles are placed on a straight lines as
shown in figure. q1 and q2 are fixed but q3 can be moved.
Under the action of the forces from q1 and q2, q3 is in
equilibrium. What is the relation between q1 and q2 ?
A. q1 = 4q2
B. q1 = -q2
C. q1 = -4q2
D. q1 = q2
Three charges +4q, Q and q are placed in a straight line
of length l at point distance 0, l/2 and l respectively.
What should be the value of Q in order to make the net
force on q to be zero?
A. +q
B. -2 q
C. -q/2
D. -q
Equal charges of each 2μC are placed at a point x = 0,
2, 4, and 8 cm on the x-axis. The force experienced by
the charge at x = 2 cm is equal toA. 5 Newton
B. 10 Newton
C. 0 Newton
D. 15 Newton
Ans: B
Principle of Superposition in Coulomb’s Law
q1
q2
q0
q3
Principle of Superposition in Coulomb’s Law
Net force on a charge can be found by
vector addition of the forces due to
each of these charges as if they were acting alone.
q1
F3
q2
q0
q3
Fnet
F1
F2
Net force on any one charge is
unaffected by the presence of other charges.
Principle of Superposition in Coulomb’s Law
All Forces do not necessarily follow the Principle of Superposition.
10 kg
10 kg
Normal Reaction here is affected
by the presence of other forces.
A point charge q₁ exerts a force F upon another charge q₂.
If one other charge q3 be placed quite near to charge q2,
then the force that charge q₁ exerts on the charge q2 will beA. F
B. >F
C. <F
D. Zero
Principle of Superposition in Coulomb’s Law
q1
Net force on a charge can be found by
Fi
q2
Fnet
q0
qi
vector addition of the forces due to
each of these charges
as if they were acting alone.
F1
F2
Principle of Superposition in Coulomb’s Law
The fact that we must understand is,
in Superposition Principle each force is individual
and not affected by other forces on particle.
Three equal charges (q) are placed at corners of an
equilateral triangle. The force on a fourth charge q
kept at the centroid of this triangle is
A.
B.
C.
D.
zero
A point charge + Q is placed at the centroid of an equilateral
triangle. When a second charge +Q is placed at a vertex of
the triangle, the magnitude of the electrostatic force on the
central charge is 4N. What is the magnitude of the net force
on the central charge when a third charge +Q is placed at
another vertex of the triangle?
A. Zero
B. 4N
C. 4√2N
D. 8N
Ans: B
Six charges +Q each are placed at the corners of a
regular hexagon of side a, force on a charge +Q kept at
the centre of hexagon isA. zero
B.
C.
D.
Five point charges, each of value +q coulomb, are placed
on five vertices of a regular hexagon of side L metre.
The magnitude of the force on a point charge of value -q
coulomb placed at the centre of the hexagon is -
A.
B. Zero
C.
D.
Three identical charges are placed at the vertices of an
equilateral triangle. The force experienced by each change
(if k = 1/4πε0) is -
Ans: C
A.
B.
C.
D.
Three charges each of +q, are placed at the vertices of an
equilateral triangle. The charge needed at the centre of
the triangle for the charges to be in equilibrium is
A. -q/√3
B. -√3q
C. √3q
D. q/√3
Ans: A
Three equal charges are placed on the three corners of
a square. If the force between q₁ and q2 is F12 and that
between q₁ and q3 is F13, then the ratio of magnitudes
(F12/F13) is
A. 1/2
B. 2
C. 1√2
D. √2
Ans: B
Four charges equal to +q are placed at the four corners
of a square a. Then the coulomb force experienced by one
charge due to the rest of three is -(k = 1/πε0)
A. (2√2 + 1)kq2/2a2
B. 3kq2/a2
C. 2√2 kq2/a2
D. Zero
Ans: A
Four charges equal to -Q are placed at the four corners
of a square and a charge q is at its centre. If the system
is in equilibrium the value of q is-
Ans: B
A.
B.
C.
D.
Four charges are arranged at the corners of a square
ABCD as shown in figure. The force on a charge kept at
the centre O is A. Zero
B. Along diagonal AC
C. Along diagonal BD
D. Perpendicular to the side AB
Ans: A
Vertices of a regular hexagon of sides 3 cm have three
positive and negative charges each of magnitude θ = 1.5 nC
as shown in diagram. A point charge of Q 5nC is placed at
centre of hexagon. The net force on 5 nC isA. 9 X 109 N
B. Zero
C. 90 X 10-9 N
D. 90 N
Six charges are placed at the corners of a regular
hexagon as shown. If an electron is placed at its
centre, net force on it will be -
A. Zero
B. Along OF
C. Along OC
D. None of these
12
CLASS
Electrostatics
Electric Field
Jayant Nagda
Electric Field
Region or Space around charged particle
into which when another charge is brought in,
it experiences electrostatic force.
Electric Field
Electrostatic interaction
between charged particles
is a two step process:
1. Charge creates Electric Field around them
Charge
Field
Electric Field
Electrostatic interaction
between charged particles
is a two step process:
1. Charge creates Electric Field around them
1. Another charge feels the force
when it is brought into this Electric Field.
Charge
Field
Force
Electric Field Intensity
Intensity of Electric Field at a point is equal to
electrostatic force experienced by a unit positive charged
Electric Field Intensity
Intensity of Electric Field at a point is equal to
electrostatic force experienced by a unit positive charged
E=
F
q0
q0 is a test charge.
Vector quantity: directed along force on positive charge.
S.I units is N/C (Newton/Coulomb).
If Q = 2 coulomb and force on it is F = 100 Newton,
then the value of field intensity will be
A. 200 N/C
B. 100 N/C
C. 50 N/C
D. 10 N/C
The dimensions of Electric Field Intensity are
A. M-1L-1T3A1
B. M1L1T-2A-2
C. M1L1T-3A-1
D. None of these
General Method of determining Electric Field Intensity
Electric Field Intensity E
1. Bring in a positive test charge qo
2. Determine force F on it
due to a Point Charge
Electric field intensity due to
a point charge q
at a distance r from it
q
Electric Field Intensity E
1. Bring in a positive test charge qo
2. Determine force F on it
due to a Point Charge
F=k
E
qo
q
r
P
F
E=
q0
E=
E=
Directed away from q
Its dimensional formula is
[M1L1T–3A–1]
r2
r3
r2
Its unit is
Newton/Coulomb
kq
kq
qq0
r
Electric Field Intensity E
due to a Point Charge
kq ^
r
E=
2
r
Electric Field Intensity E
due to a Point Charge
kq ^
r
E=
2
r
Electric Field Intensity E
due to a Point Charge
kq ^
r
E=
2
r
Electric Field Intensity E
due to a Point Charge
kq ^
r
E=
2
r
Electric Field Intensity E
due to a Point Charge
kq ^
r
E=
2
r
E=
kq
r3
r
Three identical point charges, as shown are placed at the
vertices of an isosceles right angled triangle. Which of
the numbered vectors coincides in direction with the
electric field at the midpoint M of the hypotenuse
A. 1
B. 2
C. 3
D. 4
Properties of Electric Field intensity
It is a vector quantity. Its direction is the same as
the force experienced by positive charge.
Properties of Electric Field intensity
Electric field due to positive charge is always away from it
while due to negative charge always towards it.
Properties of Electric Field intensity
Force on a point charge is in the same direction of electric field
on positive charge and in opposite direction on a negative charge.
+
-
Properties of Electric Field intensity
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
q1
q2
q0
q3
Three identical charges each of 1 μC are kept on the
circumference of a circle of radius 1 metre forming
equilateral triangle. The electric intensity at the center of
the circle in N/C is
A. 9 x 103
B. 13.5 x 103
C. 27 x 103
D. Zero
Electric Field Intensity
Electric Field Intensity at O in Each Case Shown Below is zero
Two unlike charges of the same magnitude Q are placed
at a distance d. The intensity of the electric field at the
middle point in the line joining the two charge is -
Ans: B
A.
B.
C.
D.
Find electric field at O -
A.
B.
C.
D. zero
Charges Q1 and Q2 are at points A and B of a right angle
triangle OAB (see figure). The resultant electric field at point
0 is perpendicular to the hypotenuse, then Q1/Q2 is
proportional to
[JEE Main - 2020]
A.
B.
C.
Ans: C. For detailed soln check Abhyas link in description
D.
Two charges Q1 = 18 μC and Q2 = - 2 μC are separated by a
distance R and Q1 is to the left of Q2. The distance of the
point where the net electric field is zero is A. between Q1 and Q2
B. left of Q1 at R/2
C. right of Q2 at R
D. right of Q2 at R/2
For detailed soln check Abhyas link in description
A charge Q is placed at each of two opposite corners
of a square. A charge q is placed at each of the two
opposite corners of the square. If the resultant electric
field on Q is zero then-
A. Q = -q/2√2
B. Q = -2√2q
C. Q = -2q
D. Q = 2√2 q
Two particles of masses in the ratio 1 : 2 with charges in
the ratio 1 : 1, are laced at rest in a uniform electric field.
They are released and allowed to move for the same
time. The ratio of their kinetic energies will be finally
A. 2 : 1
B. 8 : 1
C. 4 : 1
D. 1 : 4
Two point charges q1 (√10 μC) and q2 (- 25 μC) are placed on
the x-axis at x = 1 m and x = 4 m respectively. The electric
field (in V/m) at a point y = 3 m in y-axis is,
[JEE Main - 2019]
A. (63 î - 27 ĵ) x 102
B. (81 î - 81 ĵ) x 102
C. (- 81 î + 81 ĵ) x 102
D. (- 63 î + 27 ĵ) x 102
Find the electric field at point P (as shown in figure) on
the perpendicular bisector of a uniformly charged thin
wire of length L carrying a charge Q. The distance of the
point P from the centre of the rod is a = √3/2L
A.
B.
C.
D.
12
CLASS
Electrostatics
E due to Ring
Jayant Nagda
Types of Charge Distributions
Linear Charge λ
+
+
+
+
+
+
+
Charge spread along length
Charge per unit length of the object
is called Linear Charge Density λ
Types of Charge Distributions
Surface Charge σ
𝛔
Charge spread on the surface,
in an area
+ + + + + + +
+ + + + + + +
+ + + + + + +
The charge per unit area of the body
is called Surface Charge Density σ
+ + + + + + +
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Types of Charge Distributions
Volume Charge ρ
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Charge spread in a volume
+
+
+
+
The charge per unit volume of the body
is called Volume Charge Density ρ
Types of Charge Distributions
Point Charge
When linear size of charged body is much smaller than
the distance under consideration, the size may be ignored
and the charge body is called point charge.
Electric Field due to a Continuous Charge Distributions
At any point P, field due to ‘dq’ is
Electric Field due to a Continuous Charge Distributions
At any point P, field due to ‘dq’ is
dE
dq
r
dE =
k dq
r2
Total field due to all such elements
E=k
∫
dq
r2
Net field is vector addition of
tiny field vectors
E=
dE
∫
A circular ring has a uniform charge Q distributed over it.
What kind of charge distribution is it?
A. Point charge
B. Linear charge
C. Surface charge
D. Volume charge
Electric Field Intensity E
due to a uniform circular ring
at its centre
A Uniform Ring of charge Q and radius R.
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Electric Field Intensity E
due to a uniform circular ring
at a point on its axis
Ring of charge Q and radius R.
P is the point at a distance x from the centre of the ring on its axis
+ +
+
R
+
+
+
+
+
+
+
+
+
+
+
Electric Field Intensity E
due to a uniform circular ring at a point on its axis
Component of dE perpendicular to line OP dEY
will be cancelled due to circular symmetry
+ +
+
R
+
+
+
+
+
+
+
+
+
+
+
dEY
dEX
Electric Field Intensity E
due to a uniform circular ring at a point on its axis
+ +
+
R
+
+
+
+
+
+
+
+
+
+
+
A small circular ring has a uniform charge distribution.
On a far-off axial point distance x from the centre of the
ring, the electric field is proportional to-
A. x-1
B. x-3/2
C. x-2
D. x5/4
A circular ring carries a uniformly distributed positive charge
and lies in y-z plane with centre at origin of coordinate system.
If at a point (x,0,0) electric field is E, which of the following
graphs is correct -
A.
C.
x
x
B.
D.
x
x
The electric field strength due to a ring of radius R at a
distance x from its centre on the axis of the ring carrying
charge Q is given by:
At what distance from the centre will the electric field
be maximum?
A. X = R
B. X = R/2
C. X = R/√2
Detailed solution in Abhyas session
1st link in description
D. X = √2 R
Electric Field Intensity E
𝜆 : charge per unit length
R : Radius of arc
due to a uniformly charged arc
⍺ : angle of arc
+
+
+
+
+
+
+
⍺
+
R
+
+
+
+
+
+
Electric Field Intensity E
𝜆 : charge per unit length
R : Radius of arc
due to a uniformly charged arc
⍺ : angle of arc
+
+
+
+
+
+
+
⍺
+
R
+
+
+
+
+
+
E=
2kλ
sin (α/2)
R
Determine electric field at P due to Uniformly charged
Half Ring (𝜆, R).
+
+
+
+
+
+
A.
+
+
+
B.
C.
D.
Determine electric field at P due to Uniformly charged
quarter Ring (𝜆, R)
A.
+
+
B.
+
+
+
+
C.
D.
A thin conducting ring of radius r has an electric charge
+ Q. If a point charge q is placed at the center of the
ring, then tension of the wire of ring will be -
A.
B.
C.
D.
Ans: A: Detailed solution in Abhyas session
The electric field at the centre of a uniformly charged
ring is zero. What is the electric field at the centre of a
half ring if the charge on it be Q and its radius be R?
A.
B.
C.
D.
Ans: A
A ring of radius R is marked in six equal parts and these
parts are charged uniformly with a charge of magnitude
Q but positive and negative alternately as shown. Then
the electric field at centre of ring will be A.
B.
C.
D.
Ans: D
12
CLASS
Electrostatics
Electric Field
due to
Line of Charge
& Sphere
Jayant Nagda
Electric Field Intensity E
due to a line of charge
: angles subtended by line at P.
+
+
+
+
+
+
+
+
+
P
r
Electric Field Intensity E
due to a line of charge
: angles subtended by line at P.
+
+
+
+
+
+
+
+
+
P
r
Electric Field Intensity E
due to a line of charge
: angles subtended by line at P.
+
+
+
+
+
+
+
+
+
P
r
Electric Field Intensity E
due to a line of charge
: angles subtended by line at P.
+
+
+
+
+
+
+
+
+
P
r
Electric Field Intensity E
due to a line of charge
: angles subtended by line at P.
+
+
+
+
+
+
+
+
+
P
r
Electric Field Intensity E
due to a line of charge
: angles subtended by line at P.
+
+
+
+
+
+
+
+
+
P
r
Electric Field Intensity E
due to a line of charge
: angles subtended by line at P.
+
+
+
+
+
+
+
+
+
kλ (sin α + sin β)
E⟂ =
r
P
r
E∥ =
kλ
(cos β - cos α)
r
Electric Field Intensity E
due to an infinite line of charge
+
+
+
+
+
+
+
+
+
kλ (sin α + sin β)
E⟂ =
r
P
r
E∥ =
kλ
(cos β - cos α)
r
Electric Field Intensity E
due to a semi-infinite line of charge
E⟂ =
+
+
+
+
+
E∥ =
P
r
kλ
(cos β - cos α)
r
kλ (sin α + sin β)
r
and
Determine electric field at centre O
+
+
+
+
+
A.
+
+
+
+
+
+
+
+
B.
C. Zero
D.
+
+
+
+
Electric Field Intensity E
𝜆 : charge per unit length
R : Radius of arc
due to a uniformly charged arc
+
⍺ : angle of arc
+
+
+
+
+
+
+
+
+
+
⍺
O
+
+
+
+
+
+
E=
2kλ
sin (α/2)
R
+
+
+
+
+
+
+
R
+
+
+ +
Determine electric field at centre O
+
+
+
+
A.
B.
C.
D.
Zero
O
+
+
+
+
+
+
+
+
+
Electric Field Intensity E
due to a uniform Spherical Shell
+
+Q
+
+
+
+
+
+
+
+
+
+
+
+
+
R: Radius
r: distance of P
1.
Outside the shell (r > R)
Electric Field Intensity E
due to a uniform Spherical Shell
1.
Outside the shell (r > R)
Electric Field Intensity E
due to a uniform Spherical Shell
1.
Cross check your intergration from Abhyas link in description
Outside the shell (r > R)
Electric Field Intensity E
due to a uniform Spherical Shell
+
+Q
+
+
1.
+
Outside the shell (r > R)
+
+
+
+
E=k
+
+
+
+
+
+
R: Radius
r: distance of P
2.
Q
r2
Inside the shell (r < R)
E=0
A thin spherical shell of radius R has charge Q spread
uniformly over its surface. Which of the following
graphs most closely represents the electric field E (r)
produced by the shell in the range 0 ≤ r < ∞, where r is
AIEEE 2008
the distance from the centre of the shell?
A.
C.
B.
D.
Electric Field Intensity E
due to a uniform Spherical Shell
E
1.
Outside the shell (r > R)
E=k
r=R
+
+
+
+
+Q
+
r
2.
Q
r2
Inside the shell (r < R)
E=0
+
+
+
+
Electric Field Intensity E
due to a uniform Solid Sphere
+Q
+
+
+
+
+
+
+
+
+
+
+
+
Outside the sphere (r > R
+
+
+
+
+
1.
E=k
+
Q
r2
+
+
R: Radius
r: distance of P
2.
Inside the sphere (r < R)
E=k
Qr
R3
There is a solid sphere of radius R having uniformly
distributed charge throughout it. What is the relation
between electric field E and distance r from the centre
(r<R)?
A. E
∝ r-2
B. E
∝ r-1
C. E
∝r
D. E
∝ r2
The electric field due to a uniformly charged sphere of
radius R as a function of the distance from its centre is
represented graphically by
A.
C.
B.
D.
A thin spherical shell of radius R has charge spread
uniformly over its surface. Which of the following
graphs most closely represents the electric field E(r)
produced by the shell in the range 0 ≤ r < ∞, where r is
the distance from the centre of the shell?
A.
C.
Ans: A
B.
D.
Two concentric conducting thin spherical shells A, and B
having radii rA and rB ((rB > rA) are charged to QA and -QB
(|QB| > |QA|). The electric field along a line (passing
through the centre) is
A.
C.
Ans: A
B.
D.
If an insulated non-conducting sphere of radius R has
uniform volume charge density ρ. The electric field at a
distance r from the centre of sphere (r>R) will be -
A.
B.
C.
D.
Ans: C
Two definitely long charged wires with linear densities λ
and 3λ are placed along x and y axis respectively
determined the slope of electric field at any point on
the line y = √3 x A. 3√3
B.
C.
D. √3
Ans: C
12
CLASS
Electrostatics
Electric Field Lines
Jayant Nagda
Electric Field Lines
Electric field line is a concept, given by Faraday
to visualise electric field
Lines of force
+
-
starts from (+ve) charge
ends on (–ve) charge
Electric Field Lines
They are imaginary lines, pictorially mapping the electric field.
Electric Field Lines
They start out of a positive charge or infinity
+
-
and terminate at negative charge or at infinity.
Electric Field Lines
Two equal positive charges
+
+
Electric Field Lines
Two equal positive charges
+
+
Electric Field Lines
Dipole
(Equal magnitude Positive & Negative charges)
+
-
Electric Field Lines
Dipole
(Equal magnitude Positive & Negative charges)
+
-
Electric Field Lines
Two equal positive charges
+
+
Dipole
(Equal magnitude Positive & Negative charges)
+
-
In the adjoining figure, the electric field lines for
charges q1 and q2 are shown. Identify the sign of
the charges A. both negative
B. upper charge is negative and lower is
positive
C. both positive
D. upper charge is positive and lower is
negative
Which one of the following diagrams correct lines of force?
A.
C.
B.
D.
Electric Field Lines
Tangent to the line of force at a point in an electric field
gives the direction of intensity.
+
Electric Field Lines
Positive charge kept at a point feels a force
in direction along tangent to Field Line
+
+
Electric Field Lines
Positive charge kept at a point feels a force
in direction along tangent to Field Line
-
+
Electric Field Lines
However it is not necessary that a charge will follow
one particular line of field if released from rest.
+
-
Electric Field Lines
Can never cross each other
Can never be closed loops
Electric Field Lines
Crowded lines represent strong field while distant lines weak field.
+
Lines of force per unit area normal to the area
at a point represents magnitude of intensity
In fig. shown the electric lines of emerging from a
charged body. If the electric fields at A and B are EA
and EB are respectively, If the distance between A and
B is r then -
A. EA > EB
B. EA < EB
C. EA = EB
D. EA = (EB)/r2
Figure shows electric field lines. If EA and EB are
electric fields at A and B then A. Field is uniform and EA > EB
B. Field is non uniform and EA > EB
C. Field is uniform and EB > EA
D. Field is non uniform and EB > EA
Electric Field Lines
The number of lines originating or terminating on a charge is
proportional to the magnitude of charge.
The spatial distribution of the electric field due to two
charges (A and B) is shown in figure. Which one of the
following statements is correct?
A. A is +ve and B -ve and |A| > |B|
B. A is -ve and B +ve and |A| = |B|
C. Both are +ve but A > B
D. Both are -ve but A > B
A few electric field lines for a system of two charges Q₁ and
Q₂ fixed at two different points on the x-axis are shown in
the figure. These lines suggest that
IIT JEE - 2010
A. |Q1| > |Q2|
B. |Q1| < |Q2|
C. At a finite distance to the left of Q1
the electric field is zero
D. At a finite distance to the right of Q2
the electric field zero
Relation between Plane Angle and Solid Angle
Ω = 2π ( 1 - cos α )
α
R
Two charges +q1 and -q2 are placed at A and B respectively.
A line of force emanates from q1 at an angle α with the line
AB. At what angle β will it terminate at -q2.
A.
α
A +
q1
β
- B
q2
B.
C.
D.
Check Abhyas link in description for detailed solution
Electric Field Lines
Lines of force end or start normally
at the surface of a conductor
+
+
+
+
+
Electric Field Lines
Lines of force end or start normally
at the surface of a conductor
If there is no electric field there will be no lines of force
A metallic solid sphere is sphere is placed in a uniform
electric field. The lines of force follow the path(s)
shown in figure as
IIT - 1996
A. 1
B. 2
C. 3
D. 4
Which of the following statements concerning the
electrostatics is correctA. electric line of force never intersect each
other
B. electric lines of force start from positive
charge and end at the negative charge
C. electric lines of force start or ends
perpendicular to the surface of a charged
metal.
D. all of the above
Ans: D
The tangent drawn at a point on a line of electric force
shows theA. intensity of gravity field
B. intensity of magnetic field
C. intensity of electric field
D. Direction of electric field
Ans: D
Electric lines of forces A. Exist everywhere
B. Are imaginary
C. Exist only in the immediate vicinity of
electric charges
D. None of the above
Ans: B
In the electric field is uniform, then the electric lines of
Forces areA. Divergent
B. Convergent
C. Circular
D. Parallel
Ans: D
Three positive charges of equal value q are placed at
the vertices of an equilateral triangle. The resulting lines
of force should be sketched as inIIT-JEE 2001
A.
Check link in description for detailed solution
C.
B.
D.
12
CLASS
Electrostatics
Electric
Potential Energy
+
+
+
+
Jayant Nagda
Potential Energy
Gravitational Potential Energy
Increase in P.E.
Working against gravity
increases potential energy
of system.
Potential Energy
Increase in P.E.
Gravitational
Potential Energy
Electrostatic
Potential Energy
Working against gravity or electrostatic force
increases potential energy of system.
Potential Energy
Change in Potential energy of a system of particles is defined as
negative of the work done by conservative force
in assembling the system from initial to final configuration.
Electrostatic Potential Energy
Work done against the Electrostatic Force between two charged particles
in bringing them together from infinity to a particular separation.
Electrostatic Potential Energy
Work done against the Electrostatic Force between two charged particles
in bringing them together from infinity to a particular separation.
r
q2
q1
Electrostatic Potential Energy
r
q2
q1
Electrostatic Potential Energy
ΔU = Uf - Ui = kq1q2
1
1
rf ri
Electrostatic Potential Energy
r
q2
U=
kq1q2
r
q1
● Scalar quantity
● S.I. unit Joule (J)
● Reference at infinity
Electrostatic Potential Energy
q1 is kept fixed, ensuring q2 moves slowly
r
q1
q2
Electrostatic Potential Energy
If particle is moved slowly then,
r
q2
q1
Electrostatic Potential Energy
Work done by an external force in slowly bringing together
two charged particles from infinity to a particular separation.
r
q2
q1
Definitions of Electric Potential Energy
Negative of the work done by an electrostatic force in
bringing together a system of charged particles from
infinity to a particular configuration.
Or
Work done against the electrostatic force in bringing
together a system of charged particles from infinity to a
particular configuration.
Or
Work done by an external force in slowly bringing
together a system of charged particles from infinity to a
particular configuration.
Electrostatic Potential Energy
r
q1
q2
U=
kq1q2
r
Works for all combinations of the + ve and -ve charges:
1. Like Charges
2. Unlike Charges
Electrostatic Potential Energy
Work done by an external force get stored as Potential Energy
1. Like Repel
Positive Work needs to be
done
Electrostatic Potential Energy
Work done by an external force get stored as Potential Energy
2. Unlike Attract
Negative Work needs to be
done
A.
C.
B.
D.
A.
B.
C.
D.
Ans: B; check out Abhyas link (Electric Potential Energy) for detailed solution
A.
B.
C.
D.
12
CLASS
Electrostatics
Electric Potential
Jayant Nagda
Electrostatic Potential V
If U is the electrostatic potential energy of interaction
of a system of charge with ‘q0’ at a particular point P,
then, Electric potential V at P is given by:
Electric potential is a A. Vector quantity
B. Scalar quantity
C. Neither vector Nor scalar
D. Fictitious quantity
Electrostatic Potential V
Electric Potential V due to configuration of charge Q
at a point is electric potential energy of interaction
of Q with the test charge q0, divided by the test charge.
System of charge
Q
P
q0
U
V=
q0
Scalar Quantity
Units: J/C or Volt (V)
Dimension M1L2T-3A-1
Electrostatic Potential V
due to a Point Charge
Electric potential due to a point charge q at a distance r
q
r
At a certain distance from a point charge the electric field
is 500 V/m and the potential is 3000V. What is the
distance?
A. 6 m
B. 12 m
C. 36 m
D. 144 m
Three equal charges are placed at the three corners of
an equilateral triangle as shown in the figure.
The statement which is true for electric potential V
and the field intensity E at the centre of the triangle is-
A. V = 0, E = 0
B. V = 0, E ≠ 0
C. V ≠ 0, E = 0
D. V ≠ 0, E ≠ 0
Electrostatic Potential V
due to a uniform circular ring at its centre
+
+
+
A Uniform Ring of charge Q and radius R.
+
+
+
+
+
+
+
+
+
+
+
+
Electrostatic Potential V
due to a uniform circular ring at a point on its axis
P is the point at a distance x from the centre of the ring on its axis
+ +
+
R
+
+
+
+
+
+
+
+
+
+
+
Electrostatic Potential V
due to a uniform circular ring at a point on its axis
kQ
V=
√R2 + x2
O
A circular ring carries a uniformly distributed positive
charge and lies in x-y plane with centre at origin of
coordinate system. If at a point (0, 0, z) electric potential
is V, which of the following is correct?
A.
z
B.
z
C.
z
D.
z
Electrostatic Potential V
due to a uniform Spherical Shell
+
+
Q, R
+
+
+
+
+
+
+
+
+
+
+
+
+
P1
r
1.
Outside the shell (at P1)
Electrostatic Potential V
1.
due to a uniform Spherical Shell
+
+
Q, R
+
+
+
+
+
+
+
+
+
+
+
+
+
P1
r
Outside the shell (at P1)
Electrostatic Potential V
due to a uniform Spherical Shell
+
+
Q, R
+
+
+
+
+
+
+
+
+
+
+
+
+
P1
r
1.
Outside the shell (at P1)
Electrostatic Potential V
due to a uniform Spherical Shell
Q, R
+
+
+
+
+
+
+
+
P2
r
+
+
+
+
+
+
+
2.
Inside the shell (at P2)
Electrostatic Potential V
due to a uniform Spherical Shell
1.
Outside the shell (at P1)
V=
2.
kQ
r
Inside the shell (at P2)
V=
kQ
R
A hollow spherical shell of radius R is charged uniformly.
The electrostatic potential V is plotted as a function of
distance r from the centre of the sphere. Which of the
following best represents the resulting curve?
A.
B.
C.
D.
Electrostatic Potential V
due to a uniform Solid Sphere
Q, R
+
+
+
+
+
+
+
+
+
+
+
+
+
P1
+
+
+
+
+
+
+
1.
r
Outside the sphere (at P1)
Electrostatic Potential V
due to a uniform Solid Sphere
Q, R
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Outside the sphere (at P1)
P1
+
+
1.
r
Electrostatic Potential V
due to a uniform Solid Sphere
Q, R
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
r+ P +
2
+
2.
Inside the sphere (at P2)
Electrostatic Potential V
due to a uniform Solid Sphere
Q, R
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
r+ P +
2
+
+
2.
Inside the sphere (at P2)
Electrostatic Potential V
1.
due to a uniform Solid Sphere
Q, R
+
+
+
+
+
+
+
+
+
+
+
+
+
+
r+ P +
2
+
+
V=
P1
+
+
Outside the sphere (at P1)
r
2.
kQ
r
Inside the sphere (at P2)
V=
kQ
[ 3R2 - r2 ]
2R3
A solid sphere of radius R is charged uniformly.
The electrostatic potential V is plotted as a function of
distance r from the centre of the sphere. Which of the
following best represents the resulting curve?
A.
B.
C.
D.
Electrostatic Potential V
1.
due to a uniform Solid Sphere
Outside the sphere (at P1)
V
V=
2.
r
r=R
+
+
+
+
+
+
+
+
+
+
+
+
kQ
r
Inside the sphere (at P2)
V=
kQ
[ 3R2 - r2 ]
2R3
A solid sphere of radius R is charged uniformly.
At what distance from its surface is the electrostatic
potential half of the potential at the centre
A. R
B. R/2
C. R/3
D. 2R
Ans: C; detailed solution in Abhyas link in description
A hollow charged metal sphere has radius r. If the potential
difference between its surface and a point at a distance 3r
from the centre is V, then electric field intensity at
distance 3 r from the centre is A. V/3r
B. V/4r
C. V/6r
D. V/2r
Ans: C; detailed solution in Abhyas link in description
A half ring of radius R has a charge of λ per unit length.
The potential at the centre of the half ring is -
A. K λ/R
B. K λ/πR
C. K λ/R2
D. Kπλ
Ans: D; detailed solution in Abhyas link in description
A charge + q is fixed at each of the point x = xo, x = 3x0,
x = 5 x0, .......ad inf. on the x-axis, and charges –q is fixed at
each of the point x = 2x0, x = 4x0, x = 6 xo..........ad inf. Here xo
is a positive constant. Take the electric potential at a
point due to a charge Q at a distance r from it to be Q /
(4πε0r). Then the potential at the origin due to the above
system of charges is
A.
B.
C.
D.
Ans: D; detailed solution in Abhyas link in description
12
CLASS
Electrostatics
Equipotential
Surfaces
Jayant Nagda
Relation between Electric Field E and Potential V
E
Relation between Electric Field E and Potential V
dr
E
Perpendicular to Electric Field Lines,
Potential does not change,
i.e. Potential remains Constant
dV = - E.dr
dV = - E dr Cos90o
dV = 0
V = constant
Equipotential Surfaces
dr
Locus of all points on which Potential is constant.
E
V = constant
Electric Potential at all points on the
same Equipotential Surface is equal.
The points resembling equal potentials are A. P and Q
B. S and Q
C. S and R
D. P and R
At a point in space, the electric field points towards
north. In the region surrounding this points, electric
potential will be constant along A. North
B. South
C. North-South
D. East-West
Equipotential Surfaces
Uniform Electric Field
For uniform electric field, shape of
the equipotential surfaces are ?
Equipotential Surfaces
V1
Uniform Electric Field
V
V
2
3
For uniform electric field, shape of
the equipotential surfaces are
a family of planes
perpendicular to Field Lines
Direction of electric field is perpendicular
to the equipotential surfaces.
As perpendicular to the Electric Field E,
Electric Potential V remains constant.
Equipotential Surfaces
Infinite Plane
Equipotential surfaces are
planes parallel to each other
perpendicular to field.
Equipotential Surfaces
Point Charge
Direction of electric field is
perpendicular to the
equipotential surfaces.
+
Equipotential Surfaces
Point Charge
The equipotential surfaces produced by a point charge
are a family of concentric circles/spheres
+
Equipotential Surfaces
Spherical Charge
Equipotential surfaces shape ?
Equipotential Surfaces
Spherical Charge
The equipotential surfaces are a
family of concentric spheres (3D).
Equipotential Surfaces
The net work done in
moving a charge on
same equipotential surface is zero.
Equipotential Surfaces
Infinite Line of Charge
Equipotential surfaces shape ?
Equipotential Surfaces
Infinite Line of Charge
Equipotential surfaces are co-
axial cylinders having their
common axes at the line charge.
Equipotential surfaces can
never intersect each other
What is not true for equipotential surface for uniform
electric field?
A. Equipotential surface is flat
B. Equipotential surface is spherical
C. Electric lines are perpendicular to
equipotential surface
D. Work done is zero
Relation between Electric Field E and Potential V
While moving at angle 𝜃 to the Electric Field
E
dV = - E.dr
Relation between Electric Field E and Potential V
dr
θ
E
1. Moving along the Field θ < 90o
dV = - E dr Cosθ
dV = -ve
Potential Decreases along the Field
Relation between Electric Field E and Potential V
dr
2. Moving opposite to the Field θ > 90o
θ
E
dV = - E dr Cosθ
dV = +ve
Potential Increases opposite to the Field
Figure shows three points A, B and C in a region of
uniform electric field E. The line AB is perpendicular
and BC is parallel to the field lines. And which of the
following holds good?
(where VA, VB and VC represent the electric potential at
the points A, B and C respectively)
A. VA = VB = VC
B. VA = VB > VC
C. VA = VB < VC
D. VA, VB and VC
Relation between Electric Field E and Potential V
E
While moving at 𝜃 with the Electric Field
Relation between Electric Field E and Potential V
dr
E
Along the Electric Field,
Potential changes at the maximum rate,
and Potential decreases
High V
Low V
dV
=-E
dr
Relation between Electric Field E and Potential V
Along the Electric Field,
Potential decreases
Work needs to be done to move a positive charge
opposite to the Field
If a positive charge is shifted from a low-potential
region to a high-potential region, the electric potential
energy
A. increases
B. decreases
C. Remains the same
D. May increase or decrease
The diagram shows a uniform electric field in which the
lines of equal potential are spaced 2.0 cm apart. What
is the value of the electric force which is exerted on a
charge of +5.0 μC when placed in the field?
A. 6.0 x 10-6 N
B. 1.5 x 10-2 N
C. 3.0 x 103 N
D. 6.0 x 108 N
12
CLASS
Electrostatics
Electric Dipole
Jayant Nagda
Electric Dipole
Arrangement of two equal and opposite charges separated by a distance.
Electric Dipole
Arrangement of two equal and opposite charges separated by a distance.
-q
2a
Dipole Moment
+q
➝
➝
p=qx2a
product of magnitude of either charge (q)
and distance between them (2a)
directed from negative to positive.
Electric Field of a Dipole
1. On the Axis of Dipole (end-on position)
P
2a
r
Electric Field of a Dipole
1. On the Axis of Dipole (end-on position)
P
p
r
Electric Field of a Dipole
2. On the Equatorial line/ Perpendicular bisector (broadside-on position)
P
r
a
Electric Field of a Dipole
2. On the Equatorial line/ Perpendicular bisector (broadside-on position)
P
r
a
Electric Field of a Dipole
On the Equatorial line/ Perpendicular bisector (broadside-on position)
➝
Eeq=
➝
- kp
r3
p
➝
Eaxis
➝
2kp
=
r3
On the Axis of Dipole (end-on position)
Electric Field of a Dipole
On the Equatorial line/ Perpendicular bisector (broadside-on position)
➝
Eeq=
➝
- kp
r3
p
➝
Eaxis
➝
2kp
=
r3
Potential due to a Dipole
P
r
Ө
a
a
‘V’ at point P
Potential due to a Dipole
Potential due to a Dipole
a
a
EquiPotential Surface of a Dipole
Field due to Dipole
P
kp cos θ
V=
r2
r
Ө
a
‘E’ at point P
a
Field due to Dipole
P
r
Ө
a
a
Field due to Dipole
EӨ = kp sin θ
r3
P
r
Ө
a
a
Er = 2kp cos θ
r3
Electric Field and Potential due to Dipole
Enet =
P
1 + 3 cos2 θ
tan α =
tan θ
2
α: angle of Enet with r
r
Ө
a
kp
r3
V=
a
kp cos θ
r2
Electric Field and Potential due to Dipole
Enet =
kp
r3
1+3
cos2
θ
At θ = 0° ; point on axis
a
tan α =
tan θ
2
EAxis =
2kp
r3
a
At θ = 90° ; point on perpendicular bisector Eeq =
kp
r3
A short dipole is placed along x-axis with centre at origin.
The electric field at a point P, which is at a distance r
from origin such that OP = r makes an angle of 45° with xaxis, is directed along a direction making -
A. tan-¹(0.5) with x-axis
B. π/4 + tan-¹(0.5) with x-axis
C. π/4 + tan-¹(0.5) with y-axis
D. tan-¹(0.5) with y-axis
An electric dipole is placed is placed at the origin and is
directed along the x-axis. At a point P, far away from the
dipole, the electric field is parallel to the y-axis. OP makes
an angle θ with the x-axis, thenp-
A. tan θ = √3
B. tan θ = √2
C. θ = 450
D. tan θ = (1/√2)
Ans: B
12
CLASS
Electrostatics
Dipole in Uniform E
F = qE
p
F = qE
θ
Jayant Nagda
Electric Dipole
Arrangement of two equal and opposite charges separated by a distance.
-q
2a
Dipole Moment
+q
➝
➝
p=qx2a
product of magnitude of either charge (q)
and distance between them (2a)
directed from negative to positive.
Torque on Dipole in a Uniform Electric Field
E
p
Torque on Dipole in a Uniform Electric Field
E
p
θ
Torque on Dipole in a Uniform Electric Field
E
p
θ
Torque on Dipole in a Uniform Electric Field
E
p
τ=pxE
τ = p E sinθ
θ
{ θ : angle b/w p & E }
Direction Right Hand Thumb Rule
When an electric dipole P is placed in a uniform
electric field E then at what angle between P and E
the value of torque will be maximumA. 90°
B. 0°
C. 180°
D. 45°
An electric dipole consists of two opposite charges each
of magnitude 1 × 10-6 C separated by a distance 2 cm.
The dipole is placed in an external field of 1 × 105N/C.
The maximum torque on the dipole is -
A. 0.2 × 10-3 N-m
B. 1.0 × 10-3 N-m
C. 2 × 10-3 N-m
D. 4 × 10-3 N-m
An electric dipole is placed at an angle of 30° with an
electric field of intensity 2 × 105 N/C. It experiences a
torque equal to 4 N-m. Calculate the charge on the
dipole if the dipole length is 2 cm -
A. 8 mC
B. 4 mC
C. 9 μC
D. 2 mC
Potential Energy of Dipole in Electric Field
τ=pxE
E
p
Potential Energy of Dipole in Electric Field
E
dθ
p
θ
Potential Energy of Dipole in Electric Field
E
θ2
p
θ1
Potential Energy of Dipole in Electric Field
E
θ2
p
θ1
ΔU = U2 - U1 = pE (cos θ1 - cos θ2)
An electric dipole of moment p placed in a uniform
electric field E has minimum potential energy when
the angle between p and E is-
A. Zero
B. π/2
C. π
D. 3π/2
Potential Energy of Dipole in Electric Field
E
p
θ
U = - pE cosθ
U = - p.E
U
+ PE
O
- PE
θ
An electric dipole of length 2 cm is placed with its axis
making an angle of 30° to a uniform electric field 105 N/C.
If it experiences a torque of 10 √3 Nm, then potential
energy of dipole –
A. -10J
B. -20J
C. -30J
D. -40J
Work done in Rotating Dipole in Electric Field
E
θ2
p
θ1
ΔU = pE (cos θ1 - cos θ2)
ΔU = - WElectric = Wext
Work done in Rotating Dipole in Electric Field
Electric dipole is kept in an uniform electric field by making an angle θ1 with the field,
is turned very slowly (without increasing K.E.) so that it makes angle θ2 with the field,
work done by external force in this process is given by the formula
Wext = ΔU = pE (cos θ1 - cos θ2)
θ2
p
θ1
If θ1 = 0o and θ2 = θ
i.e. initially dipole is kept along the field
then it turn through θ so work done
The work done in deflecting a dipole through 180o
from field direction is
A. pE
B. 2pE
C. 1/2 pE
D. Zero
An electric dipole of moment is placed normal to the
lines of force of electric intensity, then the work
done in deflecting it through an angle of 180o is
A. pE
B. +2pE
C. -2pE
D. Zero
Ans: C
Ans: B