Download Coulomb`s law and Electric field

Welcome to my class
I am Mr. Humayun Kabir
Assistant Professor,
Department of Physics,
Jahangirnagar University, Savar,
Dhaka-1342.
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Course No. : PHY-122
 Course Title : Physics-II (Electricity, Magnetism

and Modern Physics)

Number of Credit: 3.0
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BOOKS
RECOMMENDED
BOOK RECOMMENDED:



Resnick, R. and Halliday, D. and Walker, J.;
Fundamental of Physics, part-2; John Wiley and
Sons.
Ahmed G. U.; Physics for Engineers Part-2, Hafiz
Book Center.
Beiser, A.; Concepts of Modern Physics; McGrawHill International.
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Syllabus
Chapter-1: Coulomb’s law and
Electric field
 Concept of charge,
 Coulomb’s law,
 Electric field,
 Electric field strength,
 Field due to a point charge,
 Field due to a circular ring,
 Field due to an electric dipole and
 A dipole in an external electric field.
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Syllabus
Chapter-2: Gauss’s law and its
Applications
 Flux of the electric field,
 Gauss’s law and
 Applications of Gauss’s law.
Deduction of Coulomb’s law, Field due to a
line of charge, Electric field due to a charged
solid sphere, Field due to a uniformly charged
plane (sheet of charge)
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Syllabus
Chapter-3: Electric Potential
 Electric potential,
 Potential due to a point charge,
 Potential due to collection of charge,
 Potential due to a dipole and
 Electric potential energy.
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Syllabus
Chapter-4: Capacitance and Dielectric
 Capacitor, Capacitance,
 Calculation of capacitance for various
capacitors,
Capacitance for parallel plate,
spherical and
cylindrical capacitor.
 Capacitors in series and parallel,
 Energy stored in a charged capacitor,
 Capacitance with dielectric.
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Syllabus
Chapter-5: Current and Resistance
o Concept of electric current,
o Current density,
o Electron drift velocity,
o Ohm’s law,
o Resistivity,
o Conductivity,
o Resistor in series and parallel and
o Electromotive force
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Syllabus
Chapter-6: D.C. Circuits
•
•
•
•
Calculation of current in a single loop,
Multi-loop circuits (Kirchhoff’s laws),
Applications of Kirchhoff’s laws,
RC circuit
 Charging of a capacitor
 Discharging of a capacitor.
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Syllabus
Chapter-7: The Magnetic Field
•
•
•
•
Magnetic field,
Magnetic force on a charge,
Magnetic lines of induction and
Magnetic force on a current carrying
conductor.
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Syllabus
Chapter-8: Electromagnetic
Induction
• Laws of electromagnetic induction (Faraday’s
•
•
•
•
law),
Lenz’s law,
Self induced electromotive force (self
inductance),
Mutual inductance,
Energy stored in a magnetic field and
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Syllabus
Chapter-9: Magnetic Properties
•
•
•
•
•
•
Magnetization,
Magnetic susceptibility
Magnetic permeability,
Diamagnetism,
Paramagnetism and
Ferromagnetism.
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Syllabus
Chapter-10: Modern Physics
• Particle and wave nature of light,
• Atomic models (Thomson model, Rutherford model
and Bohr model),
• Radioactivity and Nuclear Phenomena (Radioactivity,
mean life, half life, disintegration constant, radioactive
decay law, nuclear fission and fusion),
• Basis of Quantum mechanics (Failure of classical
mechanics, origin of quantum mechanics, uncertainty
principle) and
• Relativistic mechanics (Special theory of relativity,
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length contraction, time dilation)
Chapter-1
(Coulomb’s law and Electric
field)
Electric Charge: Electric charge is an intrinsic characteristics
of the fundamental particles whose presence generate static
electricity, electric field, static electric energy in an object
which can attract very tiny objects like paper and due to their
motion electric current, electric and magnetic field are
generated.
Point Charge: When the size of the charged body is
extremely small, then that charged body is called point
charge. Those charged bodies are so small compared to the
distances between them that they may be considered as
mathematical points.
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Chapter-1
(Coulomb’s law and Electric
field)
Coulomb’s law: This law states that “At a particular medium,
the magnitude of the electric force (attractive or repulsive)
between two point charges is directly proportional to the
product of their charges and inversely proportional to the
square of the distance between the charges. The force acts
along the straight line joining the two charges”.
F
A
q1
r
B
q2
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Chapter-1
(Coulomb’s law and Electric
field)
Explanation: If two point charges q1 and q2 are separated
by distance r and the force between them is F, then according
to Coulomb’s law, F ∞ q1q2, when r is constant,
and F ∞ 1/r2, when q1 and q2 are constant.
When q1 and q2 and r all vary, then F ∞ q1q2/ r2,
or, F= Kq1q2/ r2,
where K is the proportionality constant and is equal to 1/4πε
in S.I. or M.K.S. unit. here, ε (epsilon) is the permittivity of
the medium in which the charges are located.
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Chapter-1
(Coulomb’s law and Electric
field)
Therefore, Coulomb’s law becomes, F= q1q2/4πε r2.
In air or vacuum, Coulomb’s law is F0 = q1q2/4πε0 r2.
In S.I. unit F and F0 is expressed in Newton (N),
charge in Coulomb (C) and distance in meter (m),
so the experimental value of εo is 8.85X10-12 C2N-1m-2
and 1/4πε0 =1/(4π×8.854×10-12) = 9 × 109 Nm2C-2.
Thus, F0 = 9 × 109 (q1q2/ r2)
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Chapter-1
(Coulomb’s law and Electric
field)
Example-1: Two charge having magnitude 3 ×
10-6 C and 8 × 10-6 C are separated at a distance
of 2 × 10-3 m. Calculate the force exerted by one
another.
Solution: Given, q1=3X10-6 C, q2= 8X10-6 C,
r= 2 X10-3 m, εo =8.85X10-12 C2/(Nm2).
We know, F= q1q2/4πεo r2,
or, F=(3X10-6 X 8X10-6 /(4X3.14X8.85X10-12X2 X10-3)
= 107.95 N
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Chapter-1
(Coulomb’s law and Electric
field)
Example-2: Determine the force between two
free electrons spaced 1 A0 (10-10 m) apart.
Solution: Given, q1= q2= -1.6 × 10-19,
r= 1 A0 = 1 ×10-10 m, εo =8.85X10-12 C2/(Nm2).
We know, F = 9 × 109 (q1q2/ r2)
= 9 × 109 × (-1.6 × 10-19)2/(1 ×10-10)2
= 2.3 × 10-8 N.
Self Assesent-1: The distance between the
electron and the proton in the hydrogen atom is
about 5.3 × 10-11 meter. What is the magnitude
of the electric force? [ Ans: 8.1 × 10-8 N].
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Chapter-1
(Coulomb’s law and Electric
field)
Electric field: The space around a charged body where its
influence is experienced is called the electric field of that
charged body.
Electric field strength / Electric field intensity/ Electric
intensity: It is defined as the amount of electric force acting on
a unit charge at a point in the electric field. It is denoted by E. It
is a vector quantity. Mathematically, it can be written as,
E = F/q0 or, F = q0E. Its S.I. unit is N/C.
Example-3: What is the magnitude of the electric field
strength E such that an electron placed in the field,
would experience an electrical force equal to its weight?
Solution: We know, E = F/q0 = mg/ q0
= (9.1 × 10-31 × 9.8) / 1.6 × 10-19 = 5.6 × 10-11 N/C.
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Chapter-1
(Coulomb’s law and Electric
field)
Example-4: A pith ball of mass 0.002 kg is charged with
10-4 C. What is the magnitude of electric field needed to
keep the ball at rest in gravitational field?
Solution: When the weight of the ball is equal to the electric
force, the ball will remain at rest. Now,
We know, W = mg and F = q0E so, q0E = mg or,
E = mg/ q0 = 0.002 × 9.8 / 10-4 = 196 N/C.
Self Assesent-2: A plastic ball of charge 3.23 × 10-19 C is
kept hanging in an uniform electric field of 2.6 × 104
V/m. If the acceleration due to gravity at that place is
9.8 ms-2, find the mass of the ball. [ Ans: 8.4 × 10-16 kg]
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Chapter-1
(Coulomb’s law and Electric
field)
Calculation of electric field strength (E): a) For a point charge:
Let a test charge (a charge of exceedingly small magnitude whose
presence does not affect other charges) q0 be placed at a distance r
from a point charge q. The magnitude of the force acting on q0 is
given by Coulomb’s law is
The electric field strength at the site of the test charge is given
by,
The direction of E is on a radial line from q, pointing outward if q is
positive and inward if q is negative.
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Chapter-1
(Coulomb’s law and Electric
field)
b) For a group of point charge: If there are more
than one charge in the field then (i) the electric field
strength at the given point should be calculated for
each charge as if it were the only charge present and
(ii) these separately calculated fields should be added
vectorially to find the resultant field strength E at the
point. In equation form E = E1 + E2 +E3 + ……=
where n = 1, 2, 3, ….
The sum is a vector sum, taken over all the charges.
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Chapter-1
(Coulomb’s law and Electric
field)
If the charge distribution is not discrete but continuous, then the charge
must be divided into infinitesimal elements of charge dq. The field dE due to
each element of charge at the point in question is then calculated. The
magnitude of dE is given by,
where r is the distance of the given point from the charge element dq. The
resultant field at the point is then obtained by adding (that is by integrating)
the field contributions due to all charge element. Thus,
In carrying out the integration, the only suitable method is to resolve the field
contributions from the charge elements into components, add the
components by integration giving say Ex, Ey and Ez from which E can be
obtained.
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Chapter-1
(Coulomb’s law and Electric
field)
c) For at a point on the axis of a charged circular ring:
Let us consider a circular ring of radius ‘a’ carrying a charge ‘q’. We would like
to calculate E at a point P on the axis of the ring at a distance x from its
center as shown in fig-1.
Fig-1: Field at appoint P due to a charged circular ring.
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Chapter-1
(Coulomb’s law and Electric
field)
We consider a differential element of the ring of length dl at the
top of the ring. Then the charge contained in this element is
given by dq= (q/2πa) × dl, where 2πa is the circumference of
the ring. This element produces a field dE at the point P. The
resultant field E at P is found by integrating the effects of all the
elements that make up the ring. Thus,
The component of dE perpendicular to the axis is cancelled out
by an equal opposite component established by the charge
element on the opposite side of the ring. Thus only the
component of dE parallel to the axis of the ring contributes to
the resultant field. Thus,
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Chapter-1
(Coulomb’s law and Electric
field)
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Chapter-1
(Coulomb’s law and Electric
field)
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Chapter-1
(Coulomb’s law and Electric
field)
d) For at an electric dipole: An electric dipole consists of two
equal but opposite charges separated by a small distance. Fig-2
shows an electric dipole. The charges are +q and –q and they
are separated by a distance 2a.
We would like to calculate E at a point P due to the dipole. P is
located at a distance x along the perpendicular bisector of the
line joining the charges. We assume that x>> a.
Let the distance of the point P be y from both the charges and
E1 and E2 are the electric field at the point P due to the charge
+q and –q respectively. The total field at P due to the dipole is
obtained by vector addition of E1 and E2, that is, E = E1 + E2.
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Chapter-1
(Coulomb’s law and Electric
field)
Since the charges have the same magnitude and the distance of
P from the charges is also same, the magnitude of E1 and E2
are equal, i.e.,
The direction of E1 and E2 are indicated by arrows in fig-2. The
horizontal components of E1 and E2 cancel each other. Hence
the vector sum of E1 and E2 points vertically downwards and
has the magnitude, E = 2E1cosθ………..(1)
From fig-2,
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Chapter-1
(Coulomb’s law and Electric
field)
Putting the values of E1 and cosθ in equ. (1), we get,
Since x >> a, a2 can be neglected in the denominator.
The equation above then reduces to
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Chapter-1
(Coulomb’s law and Electric
field)
The product 2aq is called the electric dipole moment and is
denoted by ‘p’. It is a vector having the direction along the axis
of the dipole from negative to the positive charge. Thus, equ
(2) becomes,
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Chapter-1
(Coulomb’s law and Electric
field)
The product 2aq is called the electric dipole moment and is
denoted by ‘p’. It is a vector having the direction along the axis
of the dipole from negative to the positive charge. Thus, equ
(2) becomes,
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Chapter-1
(Coulomb’s law and Electric
field)
A dipole in an external electric field: Let an electric dipole
is placed in a uniform external electric field E; its dipole
moment p making an angle θ with this field as shown in fig-3.
The two forces (F and -F) acting on the charges are equal and
opposite where
F = qE.
The net force on the dipole is clearly zero. But since the forces
do not act along the same line, there is a net torque on the
dipole about an axis passing through the center of the dipoles
and is given by, τ = magnitude of the force × perpendicular
distance between the forces.
= F × d sinθ = qE × d sinθ = pE sinθ = p × E. [As, p = d sinθ]
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Chapter-1
(Coulomb’s law and Electric
field)
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Chapter-1
(Coulomb’s law and Electric
field)
This torque tends to align the dipole with the electric field. Hence,
work (positive or negative) must be done by an external agent to
change the orientation of the dipole. This work is stored as
potential energy U in the system. The work done to change the
orientation of the dipole through a small angle dθ is given by,
dW = τdq
where is the torque exerted by the agent and does
the work.
Then the work done to turn the dipole from an initial orientation
θ0 to a final orientation θ is given by,
where θ0 and θ is the initial and final values of the angle between the
dipole axis and external field.
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Chapter-1
(Coulomb’s law and Electric
field)
This work is stored as potential energy U. Thus,
Since we are interested only the change in potential energy, the
reference orientation θ0 can be chosen to have any convenient
value. This gives, U = -pE cosθ = - p . E .
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Chapter-1
(Coulomb’s law and Electric
field)
Example-5: A molecule of water vapour (H2O) has an electric
dipole moment of magnitude
6.2 ×10-30 C.m. The dipole moment arises because the effective
center of positive charge does not collide with the effective center
of negative charge.
(i) How far apart are the effective centers of positive and negative
charge in a molecule of H2O?
(ii) What is the maximum torque on a molecule of H2O in an
electric field of magnitude 1.5 × 104 N/C?
(iii) Suppose the dipole moment of a molecule of H2O is initially
pointing in a direction opposite to the field. How much work is
done by the electric field in rotating the molecule into alignment
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with the field?
Chapter-1
(Coulomb’s law and Electric
field)
Solution: (i) There are 10 electrons and correspondingly, 10
positive charges in this molecule. Thus, we have,
p = iqd = 10e ×d or, d = p/10e = 6.2 ×10-30/(10 × 1.6 × 10-19)
= 3.9 × 10-12m = 3.9 pm.
(ii) Maximum torque, τ = pE sin 900 = 6.2 ×10-30 × 1.5 × 104 × 1
= 9.3 10-26 Nm.
(iii) Work done,
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Chapter-1
(Coulomb’s law and Electric
field)
Self Assesent-3: An electric dipole consists of two
opposite charges of magnitude 2.0 × 10-6 C separated
by a distane 1.0 cm. It is placed in an external electric
field of 2.0 × 105 N/C.
What maximum torque does the field exert on the
dipole? [4 × 10-3 Nm]
How much work must an external agent do to turn the
dipole from its initial alignment 00 to 900?
[Ans: 4 × 10-3 Joules]
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Chapter-1
(Mechanics)
Thank you all
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