Physics [First year Bachelor of Science]

1. PHYSICS PAPER I
COURSE CODE:US-FPH-201
HSNC UNIVERSITY
DEPARTMENT OF PHYSICS
KC COLLEGE

2. SYLLABUS
• UNIT I:
Electric field.
Electric Potential.
• UNIT II:
Magnetic Field
Electromagnetic Induction & Ballistic Galvanometer.
• Unit III:
Electrical Circuits
Network Theorems

3. UNIT 1

4. CHAPTER 1: CHARGE AND MATTER
• Brief History:
• Science of origin of electricity: Thales of Miletus
(600 B.C.) observed rubbed piece of amber
attracted bits of straw.

5. • Magnetism:Roughly 4,000 years ago, a Greek
shepherd named Magnes is said to have been
tending his sheep in a region of northern Greece
called Magnesia. He took a step and suddenly found
that the nails that held his shoe together and the
metal tip of his staff were stuck fast to the rock he
was standing on!

6. • Hans Oersted (1777 – 1851):
• Observed connection between electricity and magnetism.
• Observation: Electric current in a wire can affect a magnetic compass
needle.

7. • Faraday (1791 – 1867):
• Contribution to the study of electromagnetism.
• James Clark Maxwell (1831 – 1879):
• Formulation of classical theory of electromagnetism.
• Unifying electricity, magnetism, light.

8. Electric Charge:
• Existence of two kinds of charge:
• First: Glass Rod rubbed with Silk cloth.
• Two glass rods repel each other when both are rubbed with silk cloth and
brought close to each other.

9. • Second kind:
• Hard rubber rod rubbed with fur.
• Two rubber rods repel each other when rubbed with fur and brought close to
each other.

10. Fig a: Attraction between glass rod Fig b: Repulsion between two
and rubber rod. rubber rods.

11. • Benjamin Franklin (1706 – 1790):
• First American physicist.
• named the kind of electricity that appears on:
• Glass as positive.
• Hard rubber as negative.

12. • Classification of Materials:

13. Figure: Conductivity of different materials

14. Coulomb’s Law:
• Charles Augustin de Coulomb (1736 – 1806):
• First measured electrical attractions and repulsions
quantitatively.
• Deduced the law.
• ‘a’ and ‘b’ are charged.
• Electric force on a will twist the suspension fiber.
• Coulomb cancelled out this effect by turning the suspension
head through small angle 𝜃 needed to keep the two charges at a
particular distance.
• Angle 𝜃 relative measure of electric force acting on charge a.

15. Figure: Coulomb’s experimental set – up

16. • Coulomb’s First Experimental Results can be represented as:
F ∝
1
𝑟2
F magnitude of force acting between the two charges ‘a’ and ‘b’.
𝑟 distance between the two charges.
• Coulomb also studied:
• variation of electrical force with the relative size of the charges on the sphere.

17. • Coulomb law relation between force, charge and the distance between the two
charges:
F ∝
𝑞1𝑞2
𝑟2

18. Unit and Definition of Charge:
• If ends of a long wire are connected to the terminals of a battery:
• Current 𝑖 is set up in the wire.
• Flow of current can be visualized as flow of charge.
• Definition: A coulomb is defined as the amount of charge that flows
through a given cross – section of a wire in 1 second if there is a
steady current of 1 ampere in the wire.
Q = it
• Example: a wire connected to an insulated metal sphere, a charge
of 10−6 can be put on the sphere if a current of 1 amp exists in the
wire for 10−6 sec.

19. Force exerted by a charge on remaining charges: [presence of more
than two charge]
• 𝑞1, 𝑞2, 𝑞3, 𝑞4, ……given distribution of charges.
• Force exerted on any one (say 𝑞1) by all others in vector form is given by:
• where 𝐹12 is force exerted on charge 𝑞1 by 𝑞2.

20. Problem:
• Example: Figure shows three charges 𝑞1, 𝑞2, 𝑞3. What force acts on 𝑞1?
• Assume that 𝑞1 = -1.0 X 10−6 cou, 𝑞2 = +3.0 X 10−6 cou, 𝑞3 = -2.0 X 10−6 cou,
and b= 𝑟12 = 15 cm and a= 𝑟13 = 10 cm and 𝜃 = 300.

21. • Solution:

22. Quantization of Charge:
• Charge not a “continuous fluid”.
• Experiment shows that the “electric fluid” is not continuous.
• Made up of a certain minimum electric charge. (quantized)
• Fundamental charge: Symbol: e (magnitude 1.602 X 10−19
)
• Any positive or negative charge can be detected as:
q = ne………… n = ±1, ±2, ±3,……

23. Charge and Matter:
• Composition of Matter:
• Made up of elementary particles.
Matter
Proton Neutron Electron

24. Conservation of charge:
• Hypothesis (Conversation of charge)
• Glass rod rubbed with silk cloth develops positive charge.
• Observation from experiment: Measurement shows that a
negative charge of equal magnitude appears on the silk
cloth.
• Conclusion:
• Rubbing does not create charge.
• Disturbs the electrical neutrality of each.
• Conservation of charge is satisfied.

25. • Example 1:
• Pair annihilation of positron and electron.
• Charge on electron = -e and positron = +e
• Two particles disappear converting all mass into energy.
(emission of two oppositely directed gamma rays)
• Net charge before and after the event = 0.

26. • Example 2:
• Radioactive decay of Uranium – 238.
• Atomic number (Z) before the decay: 92 (Uranium)
• After the decay: 90 + 2 (Thorium + Alpha particle)
• Amount of charge before and after the disintegration
= + 92e.

27. CHAPTER 2: THE ELECTRIC FIELD
• Examples of Vector field:
• Gravitational field strength:
• Every point in space near the earth has associated with itself a
gravitational field strength g.
• g gravitational acceleration experienced by a test body at
that point.
• For a body of mass m if F gravitational force acting on it, g is
given by:
g = 𝐹
𝑚

28. • Near the surface of the earth the field lines are almost parallel.
• Field is uniform.

29. Figure: Flow Field for water

30. • Electric Field:
• Space surrounding a charged rod is affected by the presence of the rod.
• We speak of an electric field in this space.

31. • Central concepts in Classical Theory of Electromagnetism:
Classical
Electromagnetism
Electric Field
Magnetic field

32. • How interaction between two charges can be visualized???
• We prefer to think in terms of electric fields as follows:
1) Charge 𝑞1 sets up an electric field in the space around
itself.
2) The field acts on the charge 𝑞2 (shows up in the force
that 𝑞2 experiences)
3) Field plays an intermediary role in forces between the
charges.

33. • Two class of problems exist:
1) Calculating the fields that are set up by given distributions of charge, and
2) Calculating the forces that given fields will exert on charges placed in them.
• We think of:
charge ⇋ field

34. Electric Field Strength:
• Defining electric field operationally:
• We place a test charge 𝑞0 at a point in space to be
examined.
• Electric field strength at the point:
E =
𝐹
𝑞1
• Direction of 𝐸:
• Direction of 𝐸 is the direction of force.
• Direction in which a resting positive charge placed at the
point would tend to move.

35. Lines of Force:
• Lines of force form a convenient way of visualizing
electric – field patterns.
• Relationship between the (imaginary) lines of force
and the electric field strength:
1) The tangent to a line of force at any point gives the
direction of 𝐸 at that point.

36. 2) The lines of force are drawn so that
the number of lines per unit cross –
sectional area is proportional to the
magnitude of 𝐸.
3) 𝐸 is large where the lines of force are
close together and its value is small in
the region of widely spaced lines.

37. • Example 1:
• Lines of force for a section of an infinitely large sheet of positive charge.

38. • Example 2:
• Lines of force for a negatively charged sphere and positively charged sphere.

39. • Example 3:
• Lines of force for two equal positive charges.

40. • Example 4:
• Lines of force for equal and opposite charge.

41. • Figure: patterns of electric lines of force around a) charged plate b) two rods with equal
and opposite charge.
• Patterns made by suspending grass seed in an insulating liquid.

42. Calculation of 𝐸:
• Let a test charge 𝑞0 be placed a distance 𝑟 from a point charge 𝑞.
• Magnitude of force acting on 𝑞0(given by Coulomb’s law:
F =
1
4𝜋𝜀0
𝑞𝑞0
𝑟2
• Electric field strength at the site of the test charge:
E =
𝐹
𝑞0
=
1
4𝜋𝜀0
𝑞
𝑟2
• NOTE:
• Direction of 𝐸 is on a radial line from 𝑞,
• pointing outward if 𝑞 is positive.
• pointing inward if 𝑞 is negative.

43. • To find 𝐸 for a group of point charges:
1) Calculate 𝐸𝑛 due to each charge at the given point
as if it were the only charge present.
2) Add these separately calculated fields vectorially to
find the resultant field 𝐸 at the point.
3) Mathematically:
𝐸 = 𝐸1 + 𝐸2+ 𝐸3 + …… = 𝐸𝑛
𝑛 = 1,2,3, … …

44. • Calculating E for continuous charge distribution:
• Field the charge sets up at any point P can be computed by
dividing the charge into infinitesimal elements 𝑑𝑞.
• Field due to each element at the point in question is
calculated by treating the elements as point charges.
• Magnitude of 𝑑𝐸:
𝑑𝐸 =
1
4𝜋𝜀0
𝑑𝑞
𝑟2
where 𝑟 is the distance from the charge element 𝑑𝑞 to the
point P.

45. • The resultant field at P can be found by adding ( i.e. integrating) the field
contributions due to all the charge elements:
E = 𝑑𝐸

46. Problem: Electric dipole: Calculating electric field
• mmm

47. Electric field due to infinitely line charge:

48. CHAPTER 3: GAUSS’S LAW
• Flux of the Electric Field:
• Flux (Φ):
• Property of any vector field.
• Refers to a hypothetical surface in the field.
• Surface may be closed or open.
• For a flow field: flux is measure of the number of streamlines that cut through
a surface.

49. • Examples Closed Surface and Open Surface:

50. • Defining Φ𝐸 precisely:
• Figure shows an arbitrary closed surface immersed in an
electric field.
• Surface divided into elementary squares Δ𝑆 or ΔA [small
enough to be considered as a plane]
• Such can area element can be represented as a vector Δ𝑺
[magnitude Δ𝑆].
• Direction of Δ𝑆:
• Outward – drawn normal to the surface.
• At every square, we can construct an electric field vector E [ E
is constant if the area element is arbitrarily small]

51. • Vectors E and Δ𝑆 make an angle 𝜃 with each other.
• Semi – quantitative definition of flux:
Φ𝐸 ≅ E . Δ𝑆
• Exact definition of electric flux is in differential limit of above equation:
[Replacing the sum over the surface by integral over the surface]:
Φ𝐸 = E. 𝑑𝑆

52. • Example:
1) If E is everywhere outward; 𝜃 < 900; E.Δ𝑆
is positive (Φ𝐸 is positive for the entire
surface).
2) If E is everywhere inward; 𝜃 > 900
; E.Δ𝑆 is
negative (Φ𝐸 is negative for the entire
surface).

53. Problem: Figure shows a hypothetical cylinder of radius R immersed in an electric
field E, the cylinder axis being parallel to the field. What is Φ𝐸 for this closed
surface??
• Figure:

54. Gauss’s Law:
• Gauss’s Law:
• Refers to any closed hypothetical surface. [Gaussian
Surface]
• Connects Φ𝐸 for the surface and the net charge q
enclosed by the surface.
𝜀0Φ𝐸 = q
𝜀0 𝐸 . 𝑑𝑆 = q
q is the net charge enclosed by the surface. [taking
into account the algebraic sum]

55. • IMPORTANT NOTE:
1) If a surface encloses equal and opposite charges, flux Φ𝐸 is zero.
2) Charge outside the surface makes no contribution to the value of q.
3) Exact location of the inside charges does not effect the value of q.
• APPLICATION:
• Used to evaluate E if charge distribution is symmetric.(by proper choice of the
gaussian surface the integral can be easily evaluated.
• If the electric field E is known for all points on a given closed surface, Gauss’s
law can be used to compute the charge inside.

56. APPLICATION OF GAUSS LAW TO CHARGE DISTRIBUTIONS WITH
SPHERICAL, CYLINDRICAL AND PLANAR SYMMETRIES:

57. Problem: Computation of electric field E at a distance r from the line of an
infinite rod of charge[linear charge density 𝜆 = q/h, charge per unit length]
• From Gauss Law:
𝜀0 𝐸 . 𝑑𝑆 = q
𝑑S = 2𝜋𝑟ℎ , q = 𝜆ℎ, 𝜃 = 00
E =
𝜆
2𝜋𝜀0𝑟

58. Problem: Computation of E at a distance r in front of the plane.[A sheet of charge
having surface charge density 𝜎 = q/A, charge per unit area]
• From Gauss Law:
𝜀0 𝐸 . 𝑑𝑆 = q
E =
𝜎
2𝜀0

59. Problem: Computation of electric field E at a distance r from a point charge.

60. Problem: Find whether the flux is positive, negative or zero for the different
areas shown in the figure.
Figure:

61. Gauss Law in differential form:

62. Interpretation of Differential Form of Gauss Law:

63. CHAPTER 4: LAPLACE and POISSON’S
EQUATION, FIRST UNIQUENESS THEOREM.
• NEED FOR Poisson’s and Laplace Equation:
• Primary task in electrostatics:
• To compute the electric field of a given stationary charge distribution given by
the equation:
𝐸(𝑟) =
1
4𝜋𝜀0
𝑟
𝑟2 𝜌 𝑟′ 𝑑𝜏′
• Integrals of the type given above are difficult to calculate.
• Gauss Law can be applied only for charge distributions having symmetry.

64. • A simplified approach to find electric field:
• Calculating the potential:
V(𝑟) =
1
4𝜋𝜀0
1
𝑟
𝜌 𝑟′ 𝑑𝜏′
• Evaluating the integral is again a cumbersome task!!!!!!
• In problems involving conductors 𝜌 is not known in advance.
• Charges in a conductor are free to move, we can control the total charge of the
conductor.

65. • Poisson’s Equation:
• Electric field can be written as a gradient of a scalar function:
E = - ∇𝑉
• Taking divergence of above equation:
∇. E = ∇2
V = -
𝜌
𝜀0
∇2
V = -
𝜌
𝜀0
-------------------[1]
• Equation [1] is known as Poisson’s Equation.

66. • Laplace Equation:
• In regions where there is no charge, 𝜌 = 0 (Poisson’s equation becomes Laplace
Equation given by:
∇2
V = 0 ------[Laplace Equation]----[2]

67. Laplacian operator in Cartesian, Cylindrical and Spherical co-ordinates:

68. Poisson’s Equation:[∇2
V = -
𝜌
𝜀0
]
• Example: Uniformly charged sphere.
• We choose spherical co-ordinates system.
• Since distribution of charge is uniform, Potential (V):
Independent of 𝜃 and 𝜑.
Dependent only on the radial distance 𝑟.
• To find the potential within the sphere. [𝑟 = constant]

69. • Solution:
• To find potential within the sphere at a certain distance 𝑟 we use:
• Poisson’s Equation:
∇2V = -
𝜌
𝜀0
• Laplacian operator in spherical co-ordinates:

70. • Since potential V is a function of 𝑟. [i.e.𝑉 = 𝑉(𝑟)]:
• Differentiation of 𝜃 and 𝜑 term in Laplacian operator vanishes.
1
𝑟2
𝜕
𝜕𝑟
𝑟2 𝜕𝑉
𝜕𝑟
= -
𝜌
𝜀0
1
𝑟2
𝑑
𝑑𝑟
𝑟2 𝑑𝑉
𝑑𝑟
= -
𝜌
𝜀0
• Rearranging and integrating:
𝑑 𝑟2 𝑑𝑉
𝑑𝑟
= −
𝜌
𝜀0
𝑟2
𝑑𝑟
𝑟2 𝑑𝑉
𝑑𝑟
= -
𝜌𝑟3
3𝜀0
+ 𝐶1

71. 𝑑𝑉
𝑑𝑟
= -
𝜌𝑟
3𝜀0
+
𝐶1
𝑟2
• Integrating:
𝑉 = -
𝜌𝑟2
6𝜀0
-
𝐶1
𝑟
+ 𝐶2
• Verifying the constants 𝐶1 and 𝐶2:
• 𝐶1= 0 [since the point 𝑟 = 0 is included in the given region], potential will be
infinite. (𝑉 = ∞)
• At the surface 𝑟 = R, the potential 𝑉 = 𝑉0
𝑉0 = -
𝜌𝑅2
6𝜀0
+ 𝐶2

72. 𝑉 - 𝑉0 =
𝜌𝑅2
6𝜀0
-
𝜌𝑟2
6𝜀0
𝑽 = 𝑽𝟎 +
𝝆
𝟔𝜺𝟎
(𝑹𝟐 - 𝒓𝟐)

73. Problems (Electric Potential):
• Problem 1:
• What is the potential at the centre of the square in the figure given below?
Assume that 𝑞1 = +1.0 X 10−8𝑐𝑜𝑢𝑙, 𝑞2 = -2.0 X 10−8𝑐𝑜𝑢𝑙, 𝑞3 = +3.0 X 10−8𝑐𝑜𝑢𝑙 𝑞4
= +2.0 X 10−8𝑐𝑜𝑢𝑙 and 𝑎 = 1.0 meter.

74. • Solution:
V =
1
4𝜋𝜀0
𝑞1+𝑞2+𝑞3+𝑞4
𝑟
r = 0.71 meter.
V = 500 volts.

75. • Problem 2:
• What is the electric potential at the surface of a gold nucleus? The radius is 6.6
X 10−15 meter and atomic number Z = 79.
• Solution:
V =
1
4𝜋𝜀0
𝑞
𝑟
Answer: 1.7 X 107 volts.

76. Potential due to a charged disk:
• Problem 3:
• A charged disk. Find the electric potential for points on the axis of a uniformly
charged circular disk whose surface charge density is 𝜎.

77. • Solution:

78. Laplace Equation:[∇2V = 0]
• Laplace Equation holds good in a charge free region.
• Example:
• V = 𝑉0 at z = L
• Z - axis
• V = 0 at z = 0

79. • Infinite plane.
• Problem: To find the potential in volume between the two planes????
• Solution:
• Potential varies along z – direction.
• Potential remains constant in the xy plane.
V(x,y,z) = V(z)

80. •
𝜕2𝑉
𝜕𝑥2 +
𝜕2𝑉
𝜕𝑦2 +
𝜕2𝑉
𝜕𝑧2 = 0
•
𝜕2𝑉
𝜕𝑧2 = 0
•
𝑑2𝑉
𝑑𝑧2 = 0

81. • Integrating:
𝑑𝑉
𝑑𝑧
= 𝐶1
• Integrating again:
V = 𝐶1𝑧 + 𝐶2 -------------[1]
• Using the boundary condition in equation [1]: (V = 0 at z = 0)
𝐶2 = 0 V = 𝐶1𝑧 --------[2]
• From the second boundary condition:(V = 𝑉0 at z = L)
𝐶1 =
𝑉0
𝐿

82. V =
𝑉0
𝐿
𝑧 ------------[3] (Equation of straight line passing through origin)
𝑉0
L

83. IMPORTANT CONCLUSIONS:
Solutions of the Laplace equation doesn’t give maxima or minima in the interior
of the region as seen from the graph (i.e. within the region between the
boundary)

84. Extreme values of V occur at the boundary.
Maximum value of V = = 𝑉0 at z = L
Minimum value of V = 0 at z = 0

85. Average of the function at the boundaries is equal to the value at the centre.
V (at L/2)

86. Previous rule can be generalized:
V(x) is the average of V(x+a) and V(x-a), for any a:
V(x) =
1
2
[V(x+a) + V(x-a)]

87. Laplace’s Equation in 3 – Dimension:
• The value of V at point r is the average value V over a spherical surface of
radius R centered at r:
𝑉 𝑟 =
1
4𝜋𝑅2 𝑉𝑑𝑎
• V can have no maxima or minima
• Extreme values of V must occur at the boundaries.

88. • Example:
• Problem:
• Calculating the average potential over a spherical surface of radius R due to a single point
charge q located outside the sphere??
• Solution:
• Centre the sphere at the origin.
• Choose coordinates such that q lies on the z – axis.

89. Spherical Polar coordinates:

90. V =
1
4𝜋𝜀0
𝑞
𝑟
where 𝑟2= 𝑧2 + 𝑅2 − 2𝑅𝑧 cos 𝜃
𝑉
𝑎𝑣𝑔 =
𝑞
4𝜋𝜀0
1
4𝜋𝑅2 [𝑧2
+𝑅2
− 2𝑅𝑧 cos 𝜃]−1 2
𝑅2
sin 𝜃 𝑑𝜃 𝑑𝜑
• On integrating:
𝑉
𝑎𝑣𝑔 =
1
4𝜋𝜀0
𝑞
𝑧
• Above equation is precisely potential due to 𝑞 at the centre of the sphere.
• Conclusion: The average potential over the sphere is equal to the net potential
at the centre.

91. • Summary:
• To calculate the potential due to a charge 𝑞 at a point, we have:
𝑉
𝑎𝑣𝑔 =
𝑉𝑑𝑎
𝑑𝑎
= 𝑉𝑐𝑒𝑛𝑡𝑟𝑒
point charge spherical surface (charge free region)

92. • V cannot be maximum or minimum at any point in the interior region.
• Maxima or minima will occur only at the surface or boundary. (Contradicting in the example
below)
• If Maximum value (V must be less on all points on the surface of the enclosed region)

93. First Uniqueness Theorem:
• Statement: The solution to Laplace’s equation in some volume 𝒱 is uniquely
determined if V is specified on the boundary surface 𝒮.
1) If the potential is defined throughout the surface - and also,
2) If the region satisfies the Laplace equation [∇2
V = 0], then –
3) there can be only one unique potential (V)

94. • Different possible paths between two points:
1) Straight Line.
2) Semi – circle.
3) Random Path.
4) Parabola.

95. • If
𝑑2𝑓
𝑑𝑥2 = 0 condition is imposed, then the only allowed solution is straight line.
• Proof: [First Uniqueness Theorem]
• Let 𝑉1(𝑟) and 𝑉2(𝑟) be the solution of the Laplace equation.
• Implies Laplace equation is valid for the above two potential.
• ∇2𝑉1(𝑟) = 0 and ∇2𝑉2(𝑟) = 0 ------------------ [1]
• 𝑉1𝑠(𝑟) = 𝑉
𝑠(𝑟) (Value of potential at the surface) --------------[2]
• 𝑉2𝑠(𝑟) = 𝑉
𝑠(𝑟) (Value of potential at the surface) --------------[3]

96. • Considering the difference of 𝑉1(𝑟) and 𝑉2(𝑟):
𝑉3(𝑟) = 𝑉1(𝑟) - 𝑉2(𝑟)
𝑉3𝑠(𝑟) = 𝑉1𝑠(𝑟) - 𝑉2𝑠(𝑟)
∇2
𝑉3 = ∇2
𝑉1 - ∇2
𝑉2
• From equation [1]:
∇2𝑉3 = 0 Solution to the Laplace’s Equation
• 𝑉3𝑠(𝑟) = 𝑉1𝑠(𝑟) - 𝑉2𝑠(𝑟) = 0 ------------[from 2]

97. Potential 𝑉3 is zero on the surface. (𝑉3𝑠(𝑟) = 0)
𝑉3 cannot be maximum or minimum inside the region.
𝑉3(r) = 0 everywhere.
𝑉3(𝑟) = 𝑉1(𝑟) - 𝑉2(𝑟) = 0
𝑉1(𝑟) = 𝑉2(𝑟) 𝑉3(r) = 0 𝑉3𝑠(𝑟) = 0