NCERT Solutions For Class 12 Physics Chapter 1: Electric Charges and Fields

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NCERT Solutions for Class 12 Physics Chapter 1 Electric charges and fields are provided in this article. Electric charge is the basic property of matter that causes it to experience a force when it is kept in an electric or a magnetic field. The electric field lines start from the positive charge and end on the negative charge.

The chapter is included in the unit Electrostatics which together with Current Electricity has a weightage of 17 marks in the CBSE Class 12 Physics Exam. The NCERT Solutions for Class 12 Physics Chapter 1 covers concepts of electric charge, conductors and insulators, Coulomb’s Law, electric field lines, Gauss’s law, and electric dipole.

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NCERT Solutions for Class 12 Physics Chapter 1

The NCERT Solutions for class 12 physics chapter 1 Electric Charges and fields are as given below –

Check out: Electric Charges and Fields MCQ

Important Topics of Electric Charges and Fields

Electric Charge refers to the property of subatomic particles that leads it to experience a force when it is placed in an electric and magnetic field.

Electric Charges are of two types – like charges and unlike charges

Like Charges repel each other.
Unlike Charges attract each other.

An electric charge has 3 fundamental properties – quantization, additive nature, and conservation of electric charge.

Quantization – The total charge of a body denotes the integral multiple of a basic quantum of charge.
Additive – This property represents the total charge of a body as an algebraic sum of all the singular charges that act on the system.
Conservation – This property expressed that the total charge of a system is not affected with time. Charges can neither be created nor destroyed.

Coulomb’s law states that mutual electrostatic force that exists between two point charges A and B is directly proportional to their product, AB and inversely proportional to the square of the distance between them.

F_e = \({kq_1q_2\over r^2}\)

Electric flux is the total number of electric field lines that pass through a given area in a unit of time.

The electric flux Δθ through an area element of ΔS can be denoted by

Δθ = E.ΔS = EΔScosθ

Conductors are the objects that assist in the movement of electric charge.
Insulators offer resistance to the flow of electricity through them.

Examples of Conductors – Human bodies, Earth, metal, etc.

Examples of Insulators – Nylon, Wood, Porcelain, etc.

Gauss law states that the total amount of electric flux that passes through a closed surface is directly proportional to the enclosed electric charge. The Gauss law formula is expressed by-

Φ = \({Q \over ϵo}\)

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Also Read:

Chapter 1 Related Articles
Applications of Gauss’s Law	Charging by Induction	Dipole in a Uniform External Field
Electric Dipole	Maxwell’s Equations	Cells, EMF and Internal Resistance
Kirchhoffs Laws	Electrostatics	Unit of Electric Field

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CBSE CLASS XII Related Questions

1.
A current of 5 A is passing along +X direction through a wire lying along X-axis. Find the magnetic field \( \vec{B} \) at a point \( \vec{r} = (3 \hat{i} + 4 \hat{j}) \, \text{m} \) due to 1 cm element of the wire, centered at the origin.

View Solution

2.
A ray of light is incident on a refracting face AB of a prism ABC at an angle of \( 45^\circ \). The ray emerges from face AC and the angle of deviation is \( 15^\circ \). The angle of prism is \( 30^\circ \). Show that the emergent ray is normal to the face AC from which it emerges out. Find the refraction index of the material of the prism.

View Solution

3.
The graph shows the variation of current with voltage for a p-n junction diode. Estimate the dynamic resistance of the diode at \( V = -0.6 \) V.

View Solution

4.
Answer the following giving reason:
(a) All the photoelectrons do not eject with the same kinetic energy when monochromatic light is incident on a metal surface.
(b) The saturation current in case (a) is different for different intensity.
(c) If one goes on increasing the wavelength of light incident on a metal sur face, keeping its intensity constant, emission of photoelectrons stops at a certain wavelength for this metal.

View Solution

5.
Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C), and (D) as given below.
Assertion (A): In double slit experiment, if one slit is closed, diffraction pattern due to the other slit will appear on the screen.
Reason (R): For interference, at least two waves are required.

Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
Assertion (A) is true, but Reason (R) is false.
Both Assertion (A) and Reason (R) are false.

View Solution

6.
A point source is placed at the bottom of a tank containing a transparent liquid (refractive index \( n \)) to a depth H. The area of the surface of the liquid through which light from the source can emerge out is:

\( \frac{\pi H^2}{(n-1)} \)
\( \frac{\pi H^2}{(n^2-1)} \)
\( \frac{\pi H^2}{\sqrt{n^2-1}} \)
\( \frac{\pi H^2}{(n^2+1)} \)

View Solution

NCERT Solutions For Class 12 Physics Chapter 1: Electric Charges and Fields

NCERT Solutions for Class 12 Physics Chapter 1

Important Topics of Electric Charges and Fields

Previous Year Questions

CBSE CLASS XII Related Questions

1.
A current of 5 A is passing along +X direction through a wire lying along X-axis. Find the magnetic field \( \vec{B} \) at a point \( \vec{r} = (3 \hat{i} + 4 \hat{j}) \, \text{m} \) due to 1 cm element of the wire, centered at the origin.

3.
The graph shows the variation of current with voltage for a p-n junction diode. Estimate the dynamic resistance of the diode at \( V = -0.6 \) V.

6.
A point source is placed at the bottom of a tank containing a transparent liquid (refractive index \( n \)) to a depth H. The area of the surface of the liquid through which light from the source can emerge out is:

Similar Physics Concepts

Comments

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NCERT Solutions For Class 12 Physics Chapter 1: Electric Charges and Fields

NCERT Solutions for Class 12 Physics Chapter 1

Important Topics of Electric Charges and Fields

Previous Year Questions

CBSE CLASS XII Related Questions

1.A current of 5 A is passing along +X direction through a wire lying along X-axis. Find the magnetic field \( \vec{B} \) at a point \( \vec{r} = (3 \hat{i} + 4 \hat{j}) \, \text{m} \) due to 1 cm element of the wire, centered at the origin.

3.The graph shows the variation of current with voltage for a p-n junction diode. Estimate the dynamic resistance of the diode at \( V = -0.6 \) V.

6.A point source is placed at the bottom of a tank containing a transparent liquid (refractive index \( n \)) to a depth H. The area of the surface of the liquid through which light from the source can emerge out is:

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
A current of 5 A is passing along +X direction through a wire lying along X-axis. Find the magnetic field \( \vec{B} \) at a point \( \vec{r} = (3 \hat{i} + 4 \hat{j}) \, \text{m} \) due to 1 cm element of the wire, centered at the origin.

3.
The graph shows the variation of current with voltage for a p-n junction diode. Estimate the dynamic resistance of the diode at \( V = -0.6 \) V.

6.
A point source is placed at the bottom of a tank containing a transparent liquid (refractive index \( n \)) to a depth H. The area of the surface of the liquid through which light from the source can emerge out is: