NCERT Class 12 Physics Chapter 1 Notes Electric Charges and Fields - Download PDF

NCERT Class 12 Physics Chapter 1 Notes Electric Charges and Fields - Download PDF

Edited By Vishal kumar | Updated on Jan 23, 2024 11:08 AM IST

Have you ever heard or seen a spark when you take off your synthetic clothes, especially in dry weather? That isn't magic; that is electric discharge! These experiences, like lightning during thunderstorms, occur due to the release of electric charges built up by rubbing insulating surfaces, known as static electricity. In Chapter 1, CBSE electric charges and fields class 12 notes delves into the nitty-gritty of these phenomena. Prepare to investigate electrostatics, where we will investigate the forces, fields, and potentials generated by static charges.

Electric Charges and Fields is the very first chapter of Class 12 NCERT Physics. This chapter is important in CBSE Class 12 exams because it makes up a large portion of the physics paper. The Class 12 Physics Chapter 1 notes provide a concise overview of the topics covered in the NCERT Book, giving you a quick look at the essentials you'll encounter in this foundational chapter.

Also, see,

Electric Charge:

It is observed that glass rods rubbed with wool or silk repel one another and attract lighter objects like paper. Similarly, silk and woollen cloth also repel one another and glass rod and wool attract one another. Similarly, plastic rods rubbed with cat fur repel one another. The plastic rod attracts a glass rod and repels wool and silk. From the above observation, it had been concluded by scientists that by rubbing the glass rod, silk, wool, plastic rods, etc... are electrified.

The charge is the property associated with the matter due to which it produces and experiences the electrical and magnetic effects.

From observations, the scientist has concluded that the electric charge is of two types, namely positive charge and negative charge. A body is positively charged when it loses electrons and negatively charged when it acquires or gains electrons.

  • Electric charge is a scalar quantity

  • SI unit is Coulomb, and it is represented by the symbol "C."

  • Dimension-[AT].

  • Like charges repel each other

  • unlike charges attract each other

Point Charge:

When the size of charged bodies is significantly smaller than the distance between them, their representation is simplified and they are treated as point charges.

Conductors and Insulators:

  • Conductors are materials that allow electric charges to move through them and thus facilitate the flow of electricity.
  • Insulators, on the other hand, are materials that obstruct the passage of electricity, making it difficult for electric charges to pass through them.
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Methods of charging:

  1. By Friction: When two bodies rub together, an equal amount of positive and negative charges appear due to electron transfer.

    For example, When a glass rod is rubbed against silk, electrons are transferred from the glass rod to the silk. As a result, the glass rod gains a positive charge while the silk loses a negative charge.

  1. By induction: When a charged body comes into contact with an uncharged body, one side of the neutral body becomes oppositely charged while the other side remains unchanged.

    For example, if a positively charged glass rod is brought close to a piece of paper, the paper is attracted. This occurs because the rod attracts electrons in the paper, causing the edge of the paper closest to the rod to become negatively charged, while the opposite end becomes positively charged due to an electron deficiency.

3. By conduction: When two conductors come into contact, the charges will be distributed evenly between them.

For example, When a negatively charged plastic rod comes into contact with a neutral pith ball, some electrons from the rod transfer to the pith ball, causing the pith ball to become negatively charged.

Properties of Charge:

  1. Additivity: The total charge (Q) in a system containing multiple charges(q1,q2,q3…………qn,) is the sum of individual charges: q1+q2+.....qn.
  2. Conservation of Charge: It is not possible to create or destroy a charge; it can only be transferred. When rubbing a glass rod with silk, for example, the charge is transferred rather than created.
  3. Independence of Velocity: The velocity of a particle has no effect on its charge.
  4. Quantization: The charge on a body is quantized, meaning it is an integral multiple of the elementary charge (e), where e=1.6\times10^{-19} C is the charge of an electron. Therefore, the charge (q) on a body is q = ne, where n is an integer.

Coulomb's Law:

Coulomb's Law states that the force of attraction or repulsion between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Mathematically, it is expressed as:

\\F\propto \frac{Q_{1}Q_{2}}{r^{2}}\\F=\frac{KQ_{1}Q_{2}}{r^{2}}

Where,

  • F is the electrostatic force between the charges,
  • K is Coulomb's constant
  • Q1 and Q2 are the magnitudes of the charges
  • r is the separation between the charges.

K=\frac{1}{4\pi\epsilon_0}

Where

\epsilon_0=8085\times10^{-12}\frac{C^2}{N-m^2}, known as absolute permittivity of air or free space

The vector form of Coulomb's Law:

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Consider two charges, q1 and q2, separated by r. Let the position vectors of q1 be r1 and that of q2 be r2. The force due to q_2on q2 is then directed along the unit vectorr_{12}, as shown in figureF_{12}.

\\ F_{12}=\frac{Kq_1q_2}{r^2}\hat r_{12}\\here\ \hat r_{12}=\frac{\vec r_2-\vec r_1}{|r_2-r_1|}=\frac{\vec r_{12}}{r}\\ F_{12}=\frac{Kq_1q_2}{r^3}\vec r_{12}

Force when dielectric inserted between the charges:

When a dielectric material with a dielectric constant (k) is completely filled between charges, the force

F_{med}=\frac{q_{1}q_{2}}{4\pi \varepsilon _{0}kr^{2}}=\frac{q_{1}q_{2}}{4\pi \varepsilon _{0}\epsilon_rr^{2}}

\epsilon_r is the relative permittivity / dielectric constant of the medium. The dielectric constant is the ratio of the permittivity of a substance to the permittivity of free space. (dielectric will be explained later in detail in this chapter)

Principle of Superposition:

It states that the total force acting on a given charge as a result of a group of other charges is the vector sum of the individual forces acting on that charge as a result of each of the other charges.

Electric Field :

The region surrounding a charge, where another charged particle would experience a force, is referred to as the electric field in that space.

Electric Field Intensity:

The electric field intensity at any point is defined as the force experienced by a unit positive charge placed at that point.

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Electric field intensity,

E=\frac{F}{q_0}

where,

F is the force experienced by q0,

The SI unit of E is,\frac{Newton}{Columb}=\frac{Volt}{meter}=\frac{Joule}{Coulomb\times Meter}and the dimensional formula is [MLT^{-3}A^{-1}]

The electric field is a vector quantity and due to the positive charge is away from the charge and for the negative charge, it is towards the charge.

Electric Field Due To A Point Charge:

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Consider a point charge placed at the origin O. Let a test charge q0 is placed at P which is at a distance r from O. Force F on test charge q0 is

F=\frac{1}{4\pi \epsilon_0}\frac{qq_0}{r^2}\hat r

The electric field at point P due to q is

E=\lim_{q_0\rightarrow 0}\frac{F}{q_0}=\lim_{q_0\rightarrow 0}\frac{1}{q_0}\frac{1}{4\pi \epsilon_0}\frac{qq_0}{r^2}\hat r=\frac{1}{4\pi \epsilon_0}\frac{q}{r^2}\hat r

(As q0 tends to zero the electric field produced by q is not affected by q0.)

The magnitude of the electric field

E=\frac{1}{4\pi \epsilon_0}\frac{q}{r^2}

Electric field due to a system of charge:

The electric field obeys the superposition principle. That is the electric field due to a system of charge at a point is equal to the vector sum of all the electric fields.

Electric Lines of Force:

An electric field line is an abstract line or curve that is traced through a region of space so that its tangent at any point indicates the direction of the electric field at that point.

Properties of Electric Lines of Force (Electric Field Lines):

i) Electric forces diverge from positive charges and converge on negative charges.

ii) The number of field lines is proportional to the charge magnitude.

iii) Force lines never intersect because this would imply a contradiction in the direction of the electric field at the intersection point.

iv) Electric field lines do not form closed loops, as they cannot start and end on the same charge.

v) The direction of the force on a charged particle placed at any point on an electric field line is given by the tangent.

vi) Electric field lines do not represent a particle's actual path; they can only indicate the path if the electric field lines are straight.

vii) Electric field lines are normal to conductors if they originate or terminate on a conductor.

viii) Straight, parallel, and equidistant lines represent a uniform electric field.

ix) An electric field does not exist within a conductor.

Electric Flux (\phi):

The Electric flux through an area is the number of electric field lines passing normally through that area.

  • flux through an area dA is given by

d\phi=\vec E.\vec dA=EdAcos\theta

here \theta is the angle between the area vector and the electric field.

  • Total flux through area A is

\phi=\int\vec E.\vec dA

  • Flux is a scalar quantity and the SI unit of flux is volt-meter or weber (Wb).

Electric Dipole:

An electric dipole is made up of two equal and opposing charges separated by a small vector distance. The direction of the dipole moment is from the negative to the positive charge.

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Electric Dipole Moment:

\left ( \vec{P} \right )=q\left ( \vec{2l} \right )

Here, 2l is the dipole length.

The electric dipole moment(p) is a vector quantity from negative to positive charge and Its SI unit is Coulomb-meter (C·m).

Electric Field Due To An Electric Dipole At Axial and Equatorial Point:

i) at the axial point

1624881871762

E_{axial}=\frac{2kpr}{\left ( r^{2}-l^{2} \right )^{2}}

if r>> l

E_{axi}=\frac{1}{4\pi \epsilon _{0}}\frac{2P}{r^{3}}

ii) at Equatorial Point

1624881871895

E_{equi}= \frac{kP}{\left ( r^{2}+l^{2} \right )^{\frac{3}{2}}}

If r >> l

E_{equi}= \frac{1}{4\pi \varepsilon _{0}}\frac{P}{r^{3}}

Torque Experienced by the Dipole:

1624881871616

\vec{\tau }=\vec{P}\times \vec{E}

\vec{\tau }=PE\sin \Theta

Continuous Charge Distribution:

A continuous charge distribution refers to the distribution of an amount of charge either uniformly or non-uniformly on a body. There are three main types of continuous charge distribution:

i) Linear charge distribution:

\dpi{100} \lambda=\frac{q}{L}=\frac{C}{m}=Cm^{-1}

\left ( \lambda \right ) - charge per unit length.

Example: wire, circular ring

ii) Surface charge distribution:

\dpi{100} \sigma=\frac{Q}{A}=\frac{C}{m^{2}}=Cm^{-2}

\left ( \sigma \right )- charge per unit Area

Example: plane sheet

iii) Volume Charge distribution

\dpi{100} \rho=\frac{Q}{V}=\frac{C}{m^{3}}=Cm^{-3}

\left ( \rho \right )- charge per unit volume.

(charge on a dielectric sphere etc)

Gauss's Law:

The total electric flux through a charged closed surface is equal to \frac{1}{\epsilon_0} times the total charge q enclosed by the surface.

\phi=\oint \vec E.\vec dA=\frac{1}{\epsilon_0}q

Notes:

  • Gauss's law can be applied to any closed surface. The surface's size and shape do not matter.

  • Gauss' law considers the electric field (E) as the result of all charges, whether inside or outside the Gaussian surface.

  • The electric field outside the Gaussian surface has no effect on the net flux through it.

  • The Gaussian surface is the surface chosen for the application of Gauss' law. Any Gaussian surface can be chosen, but be careful not to choose one that passes through any discrete charge. This is due to the fact that the electric field generated by a system of discrete charges is undefined at the location of any charge.

  • Continuous charge distributions are allowed to pass through the Gaussian surface.

  • Gauss' law is especially useful for simplifying the calculation of the electrostatic field in systems with some symmetry.

APPLICATIONS OF GAUSS’S LAW:

1. Electric field due to an infinitely long straight uniformly charged wire:

The electric field produced by an infinitely long straight uniformly charged wire at a distance x from the wire must be determined.

1624881872033

We have a cylindrical Gaussian surface of radius r=x and length l.
charge enclosed by the surface
E= \frac{\lambda }{2\pi x\varepsilon _{0}}

2. Field due to a uniformly charged infinite plane sheet:

Let the plane sheet be positively charged

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Let the Gaussian surface be a cylinder of cross-sectional area A

\phi = E\times 2A

( the flux passes only through 2 circular cross-sections of the cylinder)

According to Gauss's law

\phi =\frac{q}{\varepsilon _{o}}

E\times 2A=\frac{\sigma A}{\varepsilon _{0}}

E=\frac{\sigma }{2\varepsilon _{0}}

3. Field due to a uniformly charged thin spherical shell

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The radius of the shell is R and the radius of the Gaussian surface (sphere) is r.

(i) Outside shell

E = \frac{q}{4\pi \varepsilon _{o}r^{2}}

(ii) Inside the shell

Since inside a spherical shell charge =0, E=0

Importance of NCERT Class 12 Physics Chapter 1 Notes-

Electric charge and field notes class 12 are useful for a quick review of the chapters and can be used for both state and CBSE board exams as well as entrance exam like JEE or NEET. According to an analysis of CBSE Previous Year Physics Papers, this chapter typically accounts for 4 to 6 marks of the total paper. Furthermore, the availability of physics class 12 chapter 1 notes pdf format allows for convenient offline use, which aids in preparation even further.

Key features of CBSE Class 12 Physics Notes Chapter 1 Electric Charges and Fields

  1. Comprehensive Coverage: The electric charge and field notes class 12 provide thorough coverage of essential topics, encompassing electric charges, fields, Gauss's law, and more.

  2. Clarity in Concepts: CBSE class 12 physics ch 1 notes Presented in an easy-to-understand language for clarity in conceptual understanding.

  3. Exam-Centric Approach: Ch 1 physics class 12 notes aligned with CBSE exam patterns, addressing key concepts and potential exam questions.

  4. Chapter-Wise Organization: Physics class 12 chapter 1 notes pdf organized systematically, facilitating a step-by-step understanding of the chapter.

  5. Useful for Quick Revision: Class 12 physics ch1 notes suitable for quick reviews and last-minute revisions before exams.

  6. CBSE Exam Focus: Emphasizes topics often examined in CBSE Class 12 Physics papers.

  7. PDF Format for Offline Use: Class 12 chapter 1 physics notes available for free PDF download, enabling offline access for convenience.

  8. Clear Diagrams and Formulas: CBSE class 12 physics ch 1 notes Includes diagrams and formulas for visual clarity and quick reference.

  9. Resource for State Board Exams: ch 1 physics class 12 notes Useful not only for CBSE exams but also applicable to various state board examinations.

NCERT Class 12 Notes Chapterwise

Subject Wise NCERT Exemplar Solutions

Subject Wise NCERT Solutions

NCERT Books and Syllabus

Frequently Asked Questions (FAQs)

1. What are the important topics covered in the notes for Class 12 Physics chapter 1?

The main topics covered here are coulombs law, the concept of the electric field and electric dipole, Gauss’s law and its applications.

2. Whether only the Class 12 Physics chapter 1 notes will be sufficient for the CBSE board exam?

No, this is a short note which can be used to revise the chapter. The necessary derivations are not mentioned here in the NCERT Class 12 Physics chapter 1 notes. Along with this, you can also go through the NCERT Solutions and also CBSE previous year papers and derivations.

3. What is the meaning of electrostatics?

The study of charge under rest is known as electrostatics

4. Iron is a conductor of electricity. Why we do not get an electric shock when we touch an isolated iron?

When iron is isolated, then the net charge of the iron is zero. There is no drift of charge to start the conduction. 

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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