Electricity Full lecture.pptx

Electricity
Electricity is the set of physical phenomena associated with the presence and motion of
matter that has a property of electric charge.
Static Electricity
Current Electricity

Static Electricity
Static electricity is the result of an imbalance between negative and
positive charges in an object. These charges can build up on the
surface of an object until they find a way to be released or
discharged. One way to discharge them is through a circuit.

Electric Charge
Electric charge is the physical property of matter that causes it to
experience a force when placed in an electric or magnetic field.
Benjamin Franklin named two types of charges as ‘positive’ and
‘negative’ (commonly carried by protons and electrons
respectively). Charge is measured in coulombs (C).

Properties of Electric charge
•

Point Charge And Test charge
Point charge: Any charge whether positive or negative, whose electric field is to be found at a
particular distance(point) is called point charge. The charge bodies whose sizes are much
smaller than the distance between them are called point charges.
Test charge: Any charge whose magnitude is very small, in fact negligible, as compared to that
of the point charge, and which does not affect the electric field of the point charge is called test
charge. By test charge, magnitude of point charge is to be found out.

Coulomb’s Law
In 1875, Charles Augustin Coulomb measured electrostatic attraction and repulsion
force quantitatively and deduced the law that governs them.
“The magnitude of electrical force between two point charges is directly
to the product of magnitude of charges and inversely proportional to the square of
the distance between their centers and the force acts along the line connecting the
charges.”

Example 1: What must be the distance between point charge 𝒒𝟏 = 𝟐𝟔. 𝟑 𝝁𝑪 and point charge
𝒒𝟐 = −𝟒𝟕. 𝟏 𝝁𝑪 for the attractive electrical force between them to have a magnitude of 5.66 N.
Solution:
𝑞1= 26.3 𝜇𝐶 = 26.3 × 10−6 𝐶
𝑞2 = −47.1 𝜇𝐶 = −47.1 × 10−6
𝐶
𝐹 = 5.66 𝑁
𝐹 = 𝑘
𝑞1𝑞2
𝑟2
Or,𝑟 =
𝑘𝑞1𝑞2
𝐹
=
9×109 × 26.3×10−6 × 47.1×10−6
5.66
= 1.40 𝑚

Solution:
𝑼𝟐𝟑𝟖 → 𝟐
𝟒
𝑯𝒆 + 236Th
𝑟 = 12 × 10−15 𝑚
𝐹 =?
𝑎 =?
𝐹 = 𝑘
𝑞1𝑞2
𝑟2
= 9 × 109
×
2𝑒 × 90𝑒
12 × 10−15 2
= 288 𝑁
Mass of helium atom 𝑚 = 4 𝑎𝑚𝑢 = 4 × 1.67 × 10−27 𝑘𝑔 = 6.67 × 10−27 𝑘𝑔
From Newton’s second law of motion:
𝐹 = 𝑚𝑎 ⇒ 𝑎 =
𝐹
𝑚
=
288
6.67 × 10−27
= 43.18 × 1023 𝑚𝑠−2
Example 2: In the radioactive decay of 𝑼𝟐𝟑𝟖, the center of the emerging 𝟐
𝟒
𝑯𝒆 particle is at a certain distance 𝟏𝟐 ×
𝟏𝟎−𝟏𝟓
𝒎 from the center of residual 236Th nucleus at that instant. (a) What is the force on helium atom and (b) what is
its acceleration?

Example- 3 The electrostatic force between identical ions that are separated by a distance of 5 ×
10−10 𝑚 is 3.7 × 10−9 𝑁 (a) Find the charge on each ion? (b) How many electrons are missing from
each ion?
3
Solution: 𝐹 = 3.7 × 10−9
𝑁
𝑟 = 5 × 10−10
𝑚
(a) 𝑞1 = 𝑞2 = 𝑞 =?
𝐹 = 𝑘
𝑞1𝑞2
𝑟2
= 𝑘
𝑞2
𝑟2
𝑜𝑟, 𝑞2
=
𝐹𝑟2
𝑘
=
3.7 × 10−9
× 5 × 10−10 2
9 × 109
= 10.27 × 10−38
𝑞 = 3.20 × 10−19
𝐶
(b) 𝑛 = ?
𝑞 = 𝑛𝑒
𝑛 =
𝑞
𝑒
=
3.20 × 10−19
1.6 × 10−19
= 2

The space surrounding an electric charge within which it is capable of exerting a
force on another electric charge is called electric field. An electric field is generated
by electrically charged particles and time-varying magnetic fields.
Figure show an electric field produced by a positive and negative charge.
The Electric Field

Electric Field Lines or Lines of Force
A visual representation of the electric field can be obtained in terms of electric
field lines. Electric field lines provides information about the direction and
strength of the electric field at various places. As electric field line provides the
information about the electric force exerted on a charge, the lines are
commonly called “Lines of Force”.
Properties of Electric Field Lines
i)Electric field lines originate from positive
charges and end on negative charges.
ii)The tangent to a field line at any point gives the
direction of the electric field intensity at that
point.
iii)The lines are closer where the field is strong,
the lines are farther apart where the field is weak.
iv)No two lines cross each other.

Electric Field Intensity
The electrostatic force on unit positive charge at a specific field point is called the
electric field intensity. In order to find out electric field intensity, a test charge 𝑞0 is
placed in the electric field of a point charge. The electric field intensity 𝐸 is
expressed as,
𝐸 =
𝐹
𝑞0
Where 𝐹 is the electrostatic force on test charge 𝑞0.
Unit: N/C

Consider a test charge 𝑞0 placed at point P in the electric field of a point charge 𝑞 at a distance 𝑟 apart.
We want to find out electric field intensity at point 𝑃 due to a point charge 𝑞.
The electrostatic force 𝐹 between 𝑞 and 𝑞0 can be find out by using expression,
𝐹 =
1
4𝜋𝜖0
𝑞𝑞0
𝑟2
The electric field intensity 𝐸 due to a point charge 𝑞 can be obtained by putting the value of electrostatic
force in expression of electric field intensity:
𝐸 =
1
4𝜋𝜖0
𝑞𝑞0
𝑟2
𝑞0
𝐸 =
1
4𝜋𝜖0
𝑞
𝑟2
This expression gives the magnitude of electric field intensity due to a point charge 𝑞.
Electric Field Intensity Due To a Point Charge

Problem 4. In an ionized helium atom (a helium atom in which one of the two electrons has
been removed)the electron and nucleus are separated by a distance of 26.5 pm. What electric
field due to the nucleus at the location of the electron.
Total charge of helium nucleus 2𝑒 = 2 × 1.6 × 10−19
= 3.2 × 10−19
𝐶
Distance 𝑟 = 26.5 𝑝𝑚 = 26.5 × 10−12 𝑚
𝐸 = 𝑘
𝑞
𝑟2
= 9 × 109 ×
3.2 × 10−19
26.5 × 10−12 2
= 4.1 × 1012 𝑁𝐶−1

Problem 5. Two equal and opposite charges of magnitude 𝟏. 𝟖𝟖 × 𝟏𝟎−𝟕
𝑪 are held 15.2 cm apart. What is the
direction and magnitude of E at mid-point between the charges? What is the force act on an electron placed
here?
Solution:
𝑞1 = 1.88 × 10−7
𝐶
𝑞2 = 1.88 × 10−7
𝐶
Distance, 𝑟 = 15.2 𝑐𝑚 = 15.2 × 10−2
𝑚
Mid-point, 𝑑 =
𝑟
2
= 7.6 × 10−2
𝑚
Total electric field 𝐸 = 𝐸+ + 𝐸− = ?
Force on an electron placed at the same point 𝐹 = ?
𝐸+ = 𝑘
𝑞1
𝑑2
= 9 × 109
×
1.88 × 10−7
7.6 × 10−2 2
= 1.28 × 106
𝑁
𝐶
𝐸− = 𝑘
𝑞2
𝑑2
= 9 × 109
×
1.88 × 10−7
7.6 × 10−2 2
= 1.28 × 106
𝑁
𝐶
Total electric field 𝐸 = 𝐸+ + 𝐸− = 2.56 × 106 𝑁
𝐶
Now, Force on an electron placed at the same point 𝐹 = 𝑞𝐸 = −1.6 × 10−19
× 2.56 × 106 𝑁
𝐶
= −4.096 × 10−13
𝑁

Electric Flux
The number of electric lines of force passing normally through a certain area is called the electric
flux. It is measured by the product of area and the component of electric field intensity normal to
the area. It is denoted by the symbol 𝜙𝑒. The S.I. unit of electric flux is
𝑁𝑚2
𝐶
Consider a surface placed in a uniform electric field of intensity 𝐸. Let 𝐴 be the area of the surface.
The direction of A taken as the outward drawn normal to the surface. The component of 𝐸 which is
in same direction to the area 𝐴 is 𝐸𝑐𝑜𝑠𝜃 as shown in the figure below. The electric flux through the
surface is given by,

Electric Flux through an Irregular Shaped Object
𝜙𝑒 = 𝑆
𝐸. 𝑑𝐴
By convention, the outward flux is taken as positive and inward flux is taken as negative
When then the sigma is replaced by the surface integral i.e,

Gauss’s Law
“The total electric flux through any close surface is
1
𝜖0
times the total charge enclosed by the surface.”
The Gauss’s law gives the relation between total flux and total charge enclosed by the surface. Consider a collection of positive
positive and negative charges in a certain region of space. According to Gauss’s law,
𝜙𝑒 =
𝑞
𝜖0
−−− − 1
Where 𝑞 is the net charge enclosed by the surface. Also,
𝜙𝑒 = 𝐸. 𝑑𝐴 −− − 2
Comparing (1) and (2) we have,
𝐸. 𝑑𝐴 =
𝑞
𝜖0
Thus we can describe the Gauss’s law as
“The surface normal integral of electric field intensity is equal to
1
𝜖0
times the total charge enclosed by the surface.”

Deduction of Coulomb’s Law from Gauss’s Law
Coulomb’s law can be deduced from Gauss’s law under certain consideration.
Consider positive point charge ‘𝑞’. In order to apply the Gauss’s law, we assume a
spherical Gaussian surface as shown in the figure below.
Considering the integral form of Gauss’s law,
𝐸. 𝑑𝐴 =
𝑞
𝜖0
Because the both vectors 𝐸 and 𝑑𝐴 are directed radially outward, so
𝐸𝑑𝐴𝑐𝑜𝑠0° =
𝑞
𝜖0

As E is constant for all the points on the spherical Gaussian surface,
𝐸 𝑑𝐴 =
𝑞
𝜖0
𝐸 4𝜋𝑟2 =
𝑞
𝜖0
𝐸 =
1
4𝜋𝜖0
𝑞
𝑟2
This equation gives the magnitude of electric field intensity 𝐸 at any point which is at the
distance ‘𝑟’ from an isolated point charge ‘𝑞’. From the definition of electric field intensity,
we know that
𝐹 = 𝑞0𝐸
Where 𝑞0 is the point charge placed at a point at which the value of electric field intensity has to
be determined Therefore
𝐹 =
1
4𝜋𝜖0
𝑞𝑞0
𝑟2
This is the mathematical form of Coulomb’s law.

Show that the uniform spherical shell of charge behaves, for all
external points, as if all its
charges were concentrated at its center.
Consider a thin spherical shell of radius which have the charge with constant surface charge density .
Consider a point ‘𝑃’ outside the shell. We want to find out electric field intensity due to this charge distribution. For this we
consider a spherical Gaussian surface of radius 𝑟 > 𝑅 which passes through point ′𝑃′ as shown in the figure below. According to
Gauss’s law,
𝐸. 𝑑𝐴 =
𝑞
𝜖0
𝐸𝑑𝐴𝑐𝑜𝑠0° =
𝑞
𝜖0
𝐸 ∥ 𝑑𝐴
𝐸 𝑑𝐴 =
𝑞
𝜖0
𝐸 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝐸 4𝜋𝑟2
=
𝑞
𝜖0
𝐸 =
1
4𝜋𝜖0
𝑞
𝑟2
Thus the uniform spherical shell of charge behaves like a point charge for all the points outside the shell.

Show that the uniform spherical shell of charge exerts no
electrostatic force on a charged particle placed inside the shell.
Consider a point inside the shell. We want to fine out electric field intensity at point
due to this symmetrical charge distribution. For this we consider a spherical Gaussian
surface of radius which passes through point as shown in the figure below.
According to Gauss’s law,
Because the Gaussian surface enclose no charge, therefore ‘q = 0’,
𝐸𝑑𝐴𝑐𝑜𝑠0° = 0
𝐸 𝑑𝐴 = 0
As 𝑑𝐴 ≠ 0, therefore
𝐸 = 0
So the electric field does not exist inside a uniform shell of charge. So the test charge
placed inside the charged shell would experience no force.

Electric Potential
The electrical potential is defined as the capability of the charged body to do
work. When the body is charged, either electric electrons are supplied to it, or
they are removed from it. In both the cases, the work is done. This work is
stored in the body in the form of electric potential. The measure of the
electrical potential is the work done to charge a body to one coulomb. It is
denoted by ‘V’. Its unit is ‘volt’ or J/C.

POTENTIAL DIFFERENCE
The electrical potential difference is defined as the amount of work done to carrying a
unit charge from one point to another in an electric field. In other words, the potential
difference is defined as the difference in the electric potential of the two charged
bodies. When a body is charged to a different electric potential as compared to the
other charged body, the two bodies are said to a potential difference.
Potential difference denote by ′Δ𝑉′ .Mathematically
Δ𝑉 =
Δ𝑊
𝑞0

Expression for the Electric Potential Difference due to a
Point Charge
Suppose a unit positive test charge ′𝑞0′ is moved from one point ′𝑎′ to the point
′𝑏′ in the electric field 𝐸 of a large positive charge ′𝑞′ as shown in figure below:
The work done in moving ′𝑞0′ from point ′𝑎′ to the point ′𝑏′ against the electric
field 𝐸 is
𝑊𝑎→𝑏 =
𝑎
𝑏
𝐹. 𝑑𝑟

The electrical force of magnitude 𝐹 = −𝑞0𝐸 must have to supplied in order
to move ′𝑞0′ against the electric field. Therefore
𝑊𝑎→𝑏 =
𝑎
𝑏
−𝑞0𝐸 . 𝑑𝑟
𝑊𝑎→𝑏 = −𝑞0
𝑎
𝑏
𝐸. 𝑑𝑟
𝑊𝑎→𝑏
𝑞0
= −
𝑎
𝑏
𝐸. 𝑑𝑟
Therefore, the electrical potential difference between two points in an
electrical field will be,
𝑉𝑏 − 𝑉
𝑎 = −
𝑎
𝑏
𝐸. 𝑑𝑟

But the electric field intensity due to point charge: 𝐸 =
1
4𝜋𝜖0
𝑞
𝑟2 𝑟
So,
𝑉𝑏 − 𝑉
𝑎 = −
𝑟𝑎
𝑟𝑏
1
4𝜋𝜖0
𝑞
𝑟2
𝑟. 𝑑𝑟
𝑉𝑏 − 𝑉
𝑎 = −
𝑞
4𝜋𝜖0
𝑟𝑎
𝑟𝑏
1
𝑟2
𝑟. 𝑑𝑟
So, 𝑉𝑏 − 𝑉
𝑎 = −
𝑞
4𝜋𝜖0 𝑟𝑎
𝑟𝑏 𝑑𝑟
𝑟2
𝑉𝑏 − 𝑉
𝑎 = −
𝑞
4𝜋𝜖0
−
1
𝑟 𝑟𝑎
𝑟𝑏
=
𝑞
4𝜋𝜖0
1
𝑟 𝑟𝑎
𝑟𝑏
𝑉𝑏 − 𝑉
𝑎 =
𝑞
4𝜋𝜖0
1
𝑟𝑏
−
1
𝑟𝑎
This is the expression for the potential difference between two points ′𝑎′𝑎𝑛𝑑 ′𝑏

Expression for the Absolute Electric
Potential due to a Point Charge
The electric potential at any point is the amount of work done per unit charge in moving a unit
positive charge (test charge) from infinity to that point, against the electric field. If the point ′𝑎′ is at
infinity then
𝑉
𝑎 = 𝑉 ∞ = 0, 𝑟𝑎 = ∞
Putting this value in equation in the expression of electric potential difference due to point charge, we get:
𝑉𝑏 − 0 =
𝑞
4𝜋𝜖0
1
𝑟𝑏
−
1
∞
𝑉𝑏 =
𝑞
4𝜋𝜖0
1
𝑟𝑏
In general, the electric potential at point due to a point charge ′𝑞′ is
𝑉 =
1
4𝜋𝜖0
𝑞
𝑟

Problem 9. Two protons in the nucleus of 𝑼𝟐𝟑𝟖 are 𝟔 𝒇𝒎 apart.
What potential energy associated with the electric force that acts
between them?
Problem 10. What is the electric potential at the surface of the gold
nucleus. The radius of the gold nucleus is 7 × 10−15
𝑚 and atomic
number is 79.

Example-. Two protons in the nucleus of 𝑈238
are 6 𝑓𝑚 apart. What potential
energy associated with the electric force that acts between them?
Solution:
𝑟 = 6 𝑓𝑚 = 6 × 10−15 𝑚
𝛥𝑈 = ?
𝑞1 = 𝑞2 = 1𝑒 = 1.6 × 10−19 𝐶
As,
𝛥𝑉 = 𝑘
𝑞
𝑟
= 9 × 109 ×
1.6 × 10−19
6 × 10−15 = 2.34 × 105 𝑉
Now,
𝛥𝑈 = 𝑞1. 𝛥𝑉 = 1.6 × 10−19
× 2.34 × 105
= 3.744 × 10−14
𝐽
𝛥𝑈 =
3.744 × 10−14
1.6 × 10−19
= 2.34 × 105 𝑒𝑉

Example-. What is the electric potential at the surface of the gold nucleus. The
radius of the gold nucleus is 7 × 10−15 𝑚 and atomic number is 79.
Solution: 𝑟 = 7 × 10−15 𝑚
𝑍 = 79
𝑞 = 79𝑒 = 79 × 1.6 × 10−19 𝐶
𝑉 = ?
As, 𝑉 = 𝑘
𝑞
𝑟
= 9 × 109 ×
79×1.6×10−19
7×10−15
= 1.6 × 107 𝑉

Expression for the Electric Potential
Difference due to a Point Charge
Suppose a unit positive test charge ′𝑞0′ is moved from one point ′𝑎′ to the point
′𝑏′ in the electric field 𝐸 of a large positive charge ′𝑞′ as shown in figure below:
The work done in moving ′𝑞0′ from point ′𝑎′ to the point ′𝑏′ against the electric
field 𝐸 is
𝑊𝑎→𝑏 =
𝑎
𝑏
𝐹. 𝑑𝑟

Example-. Two protons in the nucleus of 𝑈238
are 6 𝑓𝑚 apart. What potential
energy associated with the electric force that acts between them?
Solution:
𝑟 = 6 𝑓𝑚 = 6 × 10−15 𝑚
𝛥𝑈 = ?
𝑞1 = 𝑞2 = 1𝑒 = 1.6 × 10−19 𝐶
As,
𝛥𝑉 = 𝑘
𝑞
𝑟
= 9 × 109 ×
1.6 × 10−19
6 × 10−15 = 2.34 × 105 𝑉
Now,
𝛥𝑈 = 𝑞1. 𝛥𝑉 = 1.6 × 10−19
× 2.34 × 105
= 3.744 × 10−14 𝐽
𝛥𝑈 =
3.744 × 10−14
1.6 × 10−19
= 2.34 × 105 𝑒𝑉

Capacitor
Capacitor is a device which is used to store charge. A simple capacitor consists
of two conductors which are separated a small distance. There may be vacuum
or some dielectric medium between the conductors of a capacitor.

Capacitance
The capacitance of a capacitor is its ability to store electrical charge.
When the plates of a capacitor are connected with the terminals of the battery of
emf V, then the charge q is stored in the capacitor. This charge stored is directly
proportional to the potential difference applied between the plates.
𝑞 ∝ 𝑉
𝑞 = 𝐶𝑉
Here C is constant of proportionality, called the capacitance of a capacitor.
Unit : Farad.

Problem 11. A storage capacitor on a random access memory (VRAM)
chip has a capacitance of 55 fF. If it is charged to 5.3 V, how many excess
electrons are there on its negative plate?
Solution:
𝐶 = 55𝑓𝐹 = 55 × 10−15𝐹,
𝑉 = 5.3 𝑉
𝑞 =?
For case of a capacitor,
𝑞 = 𝐶𝑉 = 55 × 10−15 × 5.3 = 291 × 10−15𝐶
Number of electrons 𝑛 =?
As 𝑞 = 𝑛𝑒
𝑛 =
𝑞
𝑒
=
291 × 10−15𝐶
1.6 × 10−19𝐶
= 181.8 × 104 = 1.818 × 106 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠

Capacitance of Parallel Plate Capacitor
Let the two plates of a parallel plate capacitor has area 𝐴, and charge 𝑞 on them.
The plates are separated by a distance 𝑑. Suppose that the length of plate is very
large as compared to the distance between the plates. So, 𝐸 inside the plates of the
capacitor is uniform. We want to find out the expression of electric field between
the plates of capacitor.

For this we consider a box shaped Gaussian surface which encloses positive charge +𝑞 on the top plate. We
can write
𝜀𝑜 𝐸. 𝑑𝐴 = 𝑞
∴ 𝜀𝑜 𝐸𝑑𝐴 𝐶𝑜𝑠𝜃 = 𝑞
∴ 𝜀𝑜 𝐸𝑑𝐴 𝐶𝑜𝑠0° = 𝑞
𝑆𝑖𝑛𝑐𝑒 𝐸 𝑎𝑛𝑑 𝑑𝐴 𝑎𝑟𝑒 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙, 𝑖. 𝑒. 𝜃 = 0°
∴ 𝜀𝑜𝐸 𝑑𝐴 = 𝑞
∴ 𝜀𝑜𝐸𝐴 = 𝑞
∴ 𝐸 =
𝑞
𝜀𝑜𝐴
The equation is expression of electric field intensity inside the plates of capacitor.

Now potential difference between the plates
𝑉 = 𝐸𝑑 =
𝑞𝑑
𝜀𝑜𝐴
Hence the capacitance between the plates,
𝐶 =
𝑞
𝑉
=
𝑞
𝑞𝑑
𝜀𝑜𝐴
=
𝑞𝜀𝑜𝐴
𝑞𝑑
=
𝜀𝑜𝐴
𝑑
This is the expression for capacitance of a parallel plate capacitor with the free
space between the plates.

Dielectrics
Dielectrics are the insulating materials through which the electric current
cannot pass easily, because these materials have very high value of electrical
resistance. For example, paper, pyrex, polystyrence, transformer oil, pure
water, silicon etc.
Polar Dielectrics
Nonpolar Dielectrics

Capacitance with Dielectrics
Consider a parallel plate capacitor which is connected with a battery of emf ′𝑉′. Let 𝐴 is the area
of each plate and ′𝑑′ is the separation between the plates.
If 𝑞 charge is stored in the capacitor when there is vacuum or air as medium between the plates, then
𝑞 = 𝐶𝑉
Where 𝐶 is the capacitance of the capacitor, which, for the case of parallel plate capacitor is
expressed as
𝐶 =
𝜖0𝐴
𝑑
Micheal Faraday, in 1937, investigated that if the space between the plates of a parallel plate
capacitor is filled with some dielectric medium, then the charge stored in the capacitor
increased to 𝑞′. And, hence, the capacitance of the capacitor also increases to 𝐶′.

Therefore, the relation between the stored charges in the capacitor to the capacitance will become
𝑞′ = 𝐶′𝑉
The factor by which the capacitance of capacitor increases as compared to the capacitance with
air as the medium is called the dielectric constant 𝜅𝑒 . The dielectric constant 𝜅𝑒 is a
dimensionless quantity. It is also called relative permittivity of the medium. The dielectric
constant is described mathematically as:
𝜅𝑒 =
𝐶′
𝐶
𝐶′ = 𝜅𝑒𝐶
𝐶′ = 𝜅𝑒
𝜖0𝐴
𝑑
So, 𝐶′ > 𝐶 by a factor of ′𝜅𝑒′

Current Electricity
Current or dynamic electricity is defined as an electrical charge in
motion. It consists of a flow of negatively charged electrons from
atom to atom through a conductor in an electrical circuit. The
external force that causes the electron current flow of electric
charge is called the electromotive force (emf) or voltage.

Electric Current
The time rate of flow of charge through a conductor is called current. If a
charge 𝑑𝑞 flows through any cross-section of a conductor in time 𝑑𝑡, then the
current I is given by
𝐼 =
𝑑𝑞
𝑑𝑡
The SI unit of current is Ampere, which can be defined as, “when one coulomb
charge flows through a cross-section in one second, then the current flowing is
one ampere”.

Problem 12. A current of 4.82A exist in a resistor for 4.6 minutes. (a)
Find out charge, (b) How many electrons pass through resistor in this
time?

Ohm’s Law
Statement: “The current flowing through a conductor is directly
proportional to the applied potential difference if all physical states
remain same.”
If 𝑉 is the potential difference between the ends of conductor and 𝐼 is
the current flowing through it, then the Ohm’s law is described
mathematically as:
𝐼 ∝ 𝑉
𝐼 = 𝐺𝑉
Where, G is the constant of proportionality, sometimes called
‘conductance’.
It is related to the resistance by, 𝐺 =
1
𝑅
So, 𝐼 =
1
𝑅
𝑉
𝑜𝑟, 𝑉 = 𝐼𝑅
Where, 𝑅 is the resistance of the conductor.

• Junction: A point at which three or more elements are joints
together is called junction. In the figure we can say that point A & B
are called junction.
• Branch: A continuous path in the circuit from one junction to the
only next junction called branch. In one branch value of current
don’t change. In fig there are three branches AE1r1B, ARB, AE2r2B.
• Loop: Loop is combination of different branch. A continuous path
which is starting and ending at the same point. In figure AE1r1BRA,
ARBE2r2A, AE1r1Br2E2 (not necessary to start with point A, you
can choose other point also.)
Kirchhoff's law terminology

‘The algebraic sum of the currents meeting at any junction in a circuit is zero, i.e. 𝑰 = 𝟎
The convention is that, the current flowing towards a junction is positive and the current
flowing away from the junction is negative.
Let 𝐼1
, 𝐼2
, 𝐼3
, 𝐼4
𝑎𝑛𝑑 𝐼5
be the currents passing through the conductors respectively. According
to Kirchhoff’s first law.
𝐼1
+ 𝐼2
+ −𝐼3
+ − 𝐼4
+ −𝐼5
= 0
𝑜𝑟 𝐼1
+ 𝐼2
= 𝐼3
+ 𝐼4 + 𝐼5
i.e., the sum of the currents entering the junction = the sum of the currents leaving the
junction.
Kirchhoff’s First Law/Current law

“The algebraic sum of all differences in potential around a complete circuit loop is
zero, i.e. 𝑽 = 𝟎”
Let’s apply the Kirchhoff’s voltage law for the single circuit loop shown in figure below:
Since the two resistors, R1 and R2 are wired together in a series connection, they are both part
of the same loop so the same current must flow through each resistor.
Thus, the voltage drop across resistor, 𝑅1
= 𝐼𝑅1
and the voltage drop across resistor, 𝑅2
=
𝐼𝑅2
. Now, by KVL:
𝑉
𝑠 + −𝐼𝑅1 + −𝐼𝑅2 = 0
∴ 𝑉
𝑠 = 𝐼𝑅1 + 𝐼𝑅2
∴ 𝑉
𝑠 = 𝐼 𝑅1 + 𝑅2
∴ 𝐼 =
𝑉
𝑠
𝑅1 + 𝑅2
This expression gives the current flow through a circuit containing single loop.
Kirchhoff’s Second Law/Voltage law/ Loop law

RC CIRCUIT (CHARGING OF A CAPACITOR)
A circuit containing a series combination of a resistor and a
capacitor is called an 𝑅𝐶 circuit. Consider an 𝑅𝐶 circuit in series
with the battery of emf 𝜀 as shown in figure.
When the switch 𝑆 is closed , the capacitor starts charging. By
using Kirchhoff’s 2nd rule, we get:
ℰ − 𝑉𝑅 − 𝑉𝐶 = 0
Where, 𝑉𝑅 → 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑎𝑐𝑟𝑜𝑠𝑠 𝑡ℎ𝑒 𝑟𝑒𝑠𝑖𝑠𝑡𝑜𝑟 𝑎𝑛𝑑 𝑉𝐶 →
𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑎𝑐𝑟𝑜𝑠𝑠 𝑡ℎ𝑒 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑜𝑟.
⇒ ℰ = 𝑖𝑅 +
𝑞
𝐶
−−−− − 1

Now, as 𝑖 =
𝑑𝑞
𝑑𝑡
, therefore, equation (1) becomes,
ℰ =
𝑑𝑞
𝑑𝑡
𝑅 +
𝑞
𝑐
⇒ ℰ𝐶 =
𝑑𝑞
𝑑𝑡
𝑅𝐶 + 𝑞
⇒
𝑑𝑞
𝑑𝑡
𝑅𝐶 = ℰ𝐶 − 𝑞
⇒
𝑑𝑞
ℰ𝐶 − 𝑞
=
1
𝑅𝐶
𝑑𝑡
Integrating, we get;
𝑑𝑞
ℰ𝐶 − 𝑞
=
𝑑𝑡
𝑅𝐶
⇒ − ln ℰ𝐶 − 𝑞 =
𝑡
𝑅𝐶
+ 𝐴 −− − 2
Where 𝐴 is the constant of integration. To find 𝐴, we make use of initial conditions.
At 𝑡 = 0; 𝑞 = 0, we have from (2),
− ln ℰ𝐶 = 𝐴
RC CIRCUIT (CHARGING OF A CAPACITOR)……..

Now the equation (2) will become,
− ln ℰ𝐶 − 𝑞 =
𝑡
𝑅𝐶
− ln ℰ𝐶
⇒ ln ℰ𝐶 − 𝑞 − ln ℰ𝐶 = −
𝑡
𝑅𝐶
⇒ ln
ℰ𝐶 − 𝑞
ℰ𝐶
= −
𝑡
𝑅𝐶
⇒
ℰ𝐶 − 𝑞
ℰ𝐶
= 𝑒−
𝑡
𝑅𝐶
⇒ ℰ𝐶 − 𝑞 = ℰ𝐶𝑒−
𝑡
𝑅𝐶
⇒ 𝑞 = ℰ𝐶 − ℰ𝐶𝑒−
𝑡
𝑅𝐶
⇒ 𝑞 = ℰ𝐶 1 − 𝑒−
𝑡
𝑅𝐶
This equation shows that at 𝑡 = ∞; 𝑞 = ℰ𝐶 = 𝑞0, where 𝑞0is the maximum value of charge on the capacitor. Therefore,
𝑞 = 𝑞0 1 − 𝑒−
𝑡
𝑅𝐶 −−− − 3
This equation gives the growth of charge in an 𝑅𝐶 circuit. The equation shows that the charge 𝑞 goes on increasing and ultimately attains the
maximum value 𝑞0 after a long time.
RC CIRCUIT (CHARGING OF A CAPACITOR)…….

Capacitive Time Constant (Charging)
In the equation (3), the factor 𝑅𝐶 has the dimensions of time and is called capacitive time constant. It is denoted by 𝜏𝑐.
Hence, the equation (3) is written as:
𝑞 = 𝑞0 1 − 𝑒
−
𝑡
𝜏𝑐
Special Cases:
At 𝑡 = 0; 𝑞 = 0
At 𝑡 = ∞; 𝑞 = 𝑞0 = 𝐸𝐶
At 𝑡 = 𝜏𝑐; 𝑞 = 𝑞0 1 − 𝑒
−
𝜏𝑐
𝜏𝑐
. ⇒ 𝑞 = 𝑞0 1 − 𝑒−1
= 𝑞0 1 −
1
𝑒
= 𝑞0 1 − 0.37
. ⇒ 𝑞 = 0.63 𝑞0
So, the capacitive time constant is the time after which the charge on the capacitor grows to 63% of its maximum value.

DECAY OF CHARGE IN AN RC CIRCUIT
(DISCHARGING)
Consider the circuit which consists of a capacitor carrying an initial charge 𝑞, a resistor 𝑅, and a switch 𝑆 as shown in figure below
When the switch is open, a potential difference 𝑉
𝑐 =
𝑞
𝐶
exists across the capacitor and 𝑉𝑅 = 0 because 𝑖 = 0. If the switch is closed at
𝑡 = 0, the capacitor begins to discharge through the resistor. Applying Kirchhoff’s rule:
−𝑉𝑅 − 𝑉𝐶 = 0
⇒ 𝑖𝑅 = −
𝑞
𝐶
⇒
𝑑𝑞
𝑑𝑡
= −
𝑞
𝑅𝐶
⇒
𝑑𝑞
𝑞
= −
1
𝑅𝐶
𝑑𝑡

Integrating both sides, we obtain
𝑑𝑞
𝑞
= −
𝑑𝑡
𝑅𝐶
⇒ 𝑙𝑛 𝑞 = −
𝑡
𝑅𝐶
+ 𝐴 −−− − 1
By applying the initial conditions, 𝑎𝑡 𝑡 = 0; 𝑞 = 𝑞0, 𝑤𝑒 𝑔𝑒𝑡 𝑓𝑟𝑜𝑚 1
𝐴 = 𝑙𝑛𝑙𝑛 𝑞0
The equation (1) implies:
𝑙𝑛𝑙𝑛 𝑞 = −
𝑡
𝑅𝐶
+ 𝑙𝑛𝑙𝑛 𝑞0
⇒ 𝑙𝑛𝑙𝑛
𝑞
𝑞0
= −
𝑡
𝑅𝐶
⇒
𝑞
𝑞0
= 𝑒−
𝑡
𝑅𝐶
∴ 𝑞 = 𝑞0𝑒−
𝑡
𝑅𝐶 −− − 2
This is the expression for the decay of charge of capacitor in an 𝑅𝐶 circuit.
DECAY OF CHARGE IN AN RC CIRCUIT
(DISCHARGING)…………………….

Capacitive Time Constant(Discharging)
The factor 𝑅𝐶 = 𝜏𝑐 is called the capacitive time constant. The equation (2) will become:
𝑞 = 𝑞0𝑒
−
𝑡
𝜏𝑐
Special Cases:
At 𝑡 = 0; 𝑞 = 𝑞0
At 𝑡 = ∞; 𝑞 = 0
At 𝑡 = 𝜏𝑐; 𝑞 = 𝑞0𝑒
−
𝜏𝑐
𝜏𝑐 = 𝑞0𝑒−1
= 0.37 𝑞0
So, after time 𝜏𝑐, the charge of capacitor reduces to 37% of the initial value.

Problem 14. A 150𝐹 capacitor is connected through a 500 resistor to 40𝑉 battery. (a) What is the time constant of the
circuit? (b) What is the final charge 𝑞0 on a capacitor plate? (c) How long does it take for the charge on a capacitor plate to
reach 0.8𝑞0?

Problem 15. A resistor 𝑅 = 6.2𝑀𝛺 and a capacitor 𝐶 = 2.4𝜇𝐹 are connected in series and a 12V battery of negligible
internal resistance is connected across their combination. (a) What is capacitive time constant of this circuit. (b) At what
time after the battery is connected does the potential difference across the capacitor equal to 5.6 V.

Example- A capacitor 𝐶 discharge through a resistor 𝑅. (a) After how many time does
its charge fall to one half of its initial value?
Solution: As we know,
𝑞 = 𝑞0𝑒−
𝑡
𝜏
As we want to find out the time in which 𝑞 =
𝑞0
2
. Therefore,
𝑞0
2
= 𝑞0𝑒−
𝑡
𝜏
⇒
1
2
= 𝑒−
𝑡
𝜏
⇒ −
𝑡
𝜏
= 𝑙𝑛
1
2
⇒
𝑡
𝜏
= 𝑙𝑛 2
⇒ 𝑡 = 𝜏 𝑙𝑛 2 = 0.693 𝜏

Consider a capacitor with the capacitance C, which is connected to the battery of emf V. If dq charge
is transferred from one plate to other, then the work done dW will be,
𝑑𝑊 = 𝑉𝑑𝑞
This work done is stored in the form of electric potential energy dU
𝑑𝑈 = 𝑉𝑑𝑞
When the capacitor is fully charged then the total energy stored is,
𝑈 = 𝑑𝑈 =
𝑞
0
𝑉𝑑𝑞
As, 𝑞 = 𝐶𝑉; ⇒ 𝑉 =
𝑞
𝐶
𝑈 =
𝑞
0
𝑞
𝐶
𝑑𝑞
𝑈 =
1
𝐶
𝑞
0
𝑞𝑑𝑞
𝑈 =
1
𝐶
𝑞2
2 0
𝑞
∴ 𝑈 =
1
2
𝑞2
𝐶
− − − − − (1)
Energy Stored in an Electric Field

This is the expression of energy stored in the electric filed between the plates of capacitor.
Again as, 𝑞 = 𝐶𝑉, we can write,
𝑈 =
1
2
𝐶2
𝑉2
𝐶
𝑈 =
1
2
𝐶𝑉2
The energy stored in the capacitor is the energy store in the electric field between its plates. So, the
energy stored can be expressed in terms of electric filed strength 𝐸.
As, 𝐸 = 𝑉𝑑 and 𝐶 =
𝜖0𝐴
𝑑
, therefore,
𝑈 =
1
2
𝜖0𝐴
𝑑
𝐸2
𝑑2
𝑈 =
1
2
𝜖0𝐸2
𝐴𝑑

Energy Density:
The energy density ′𝑢′ is described as the energy stored ′𝑈′ per unit volume ′𝑉′.
Mathematically,
𝑢 =
𝑈
𝑉
=
𝑈
𝐴𝑑
=
1
2
𝜖0𝐸2
𝐴𝑑
𝐴𝑑
∴ 𝑢 =
1
2
𝜖0𝐸2

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