CBSE Class 12 Physics Notes Chapter 1 Electric Charges and Fields

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Have you ever felt a zap from static electricity on a dry winter day, or watched lightning flash across the sky, lighting it up for a moment? Have you ever wondered what makes these things happen? It's all because of a hidden force in the universe called electric charge.

Inside every atom, there are tiny particles called electrons.

When some atoms have more electrons, they get a negative charge.
Others, with fewer electrons, get a positive charge.
This difference creates an electric field, an invisible force around anything charged.

Imagine holding two magnets, one north and one south, far apart. They pull toward each other with an unseen force. Electric fields work the same way, influencing and even attracting charged things nearby. CBSE Class 12 Physics Notes for Chapter 1 Electric Charges and Fields are given in the article below for easy preparation and understanding of the concepts involved.

Read More:

Additional Resources for Preparation
Electric Charges and Fields	NCERT Solutions for Class 12 Physics Chapter 1
Electric Charges and Fields MCQs	Electric Charges and Fields Questions

Class 12 Physics Chapter 1 Notes - Electric Charges and Fields

Electric Charge

Electric charge is a fundamental property of matter.
It is carried by subatomic particles like electrons and protons.

Basic Properties of Electric Charge:

Additivity: Charges add up algebraically; they can be positive or negative.
Charge is Conserved: The total electric charge in an isolated system remains constant over time.
Quantization of Charge: Electric charge exists in discrete, quantized units, multiples of the elementary charge.

Coulomb’s Law:

Coulomb’s Law describes the electrostatic force between two charged objects.
Formula:

F = kq₁q₂/r²

Where

F is the force,
q₁ and q₂ are charges,
r is the distance, and
k is Coulomb's constant.

The unit of electric charge is the coulomb (C).

Coulomb’s Law

Coulomb’s Law

Conductors and Insulators:

Conductors: Materials that allow the easy flow of electric charge or current through them due to the presence of free electrons.
Insulators: Materials that resist the flow of electric charge, preventing the movement of electrons and maintaining high electrical resistance.

Forces Between Multiple Charges:

Superposition Principle: Net force on a charge is the vector sum of forces from all other charges.
In simpler terms, the total force on a charge in the presence of several other charges is the combined effect of the forces each charge would exert individually.

Forces Between Multiple Charges

Forces Between Multiple Charges

Electric Field:

Electric field is a region around a charged object where a force would be exerted on another charge.

Mathematical Expression:

E = F/q

where

E is the electric field,
F is the force, and
q is the test charge.

Electric Field

Electric Field

Physical Significance of Electric Field:

A test charge in an electric field experiences a force, indicating the presence of electric influence.

Electric Field Lines:

Imaginary lines that represent the direction and strength of the electric field.

Electric Flux:

Electric flux measures the flow of an electric field through a surface.

Electric Dipole:

Electric dipole is a pair of equal and opposite charges separated by a distance.
Electric Dipole Moment: Measured by the product of charge and separation

p = q * 2l.

Dipole in a Uniform External Field:

An electric dipole experiences a torque in an external electric field.
Torque in a uniform electric field, τ = pE sin θ.
Stable at θ = 0°, unstable at θ = 180°.
Work Done in Rotating Dipole: Work done is W = pE (cos θ1 – cos θ2).
Potential Energy of Electric Dipole: U = – pE cos θ.
Net Force and Torque on Dipole exist in a non-uniform electric field.

Work Done in Various Rotations:

From stable to unstable: W = 2pE.
From stable to max torque: W = pE.

Continuous Charge Distribution:

When the charge is spread continuously along a line, surface, or volume.
Electrostatic Force due to Continuous Charge Distribution describes the force between point charges and continuous charge distributions.
They are often represented by linear, surface, or volume charge densities.
In the case of continuous charge distribution, the total force on a point charge is determined by integrating the contributions of infinitesimally small charge elements along the distribution.

Continuous Charge Distribution

Continuous Charge Distribution

Gauss’s Law:

Gauss Law Principle states the total electric flux through a closed surface is proportional to the enclosed charge.

Gauss’s Law

Gauss’s Law

Applications of Gauss’s Law:

The applications of Gauss’s law include -

Gauss’s Law is applied to simplify the calculation of electric fields for symmetric charge distributions.
Useful in analyzing electric fields inside conductors in electrostatic equilibrium.
Essential for calculating electric flux through closed surfaces, aiding in determining the enclosed charge.

Electric Field Intensity:

Force per unit positive test charge. It is expressed as EFI.

Electric Field due to a Dipole:

Space around a dipole where electric effects are experienced.

Electric Field due to a Dipole

Electric Field due to a Dipole

There are Some important List Of Top Physics Questions On Electric Charges And Coulomb's Law Asked In CBSE CLASS XII

A thorough understanding of Electric Charges and Fields will help in the CBSE Class 12 Physics Exam 2024, as the topic carries a significant 5-mark weightage.

CBSE CLASS XII Related Questions

1.
Capacitors are manufactured with certain standard capacitances and working voltages. However, these standard values may not be the ones that are actually needed in a particular application. Two or more capacitors can be grouped in series or in parallel to achieve desired capacitance and voltage. When connected in series, the total capacitance decreases while the voltage rating increases, whereas in parallel connections, the total capacitance increases and maintains the same voltage rating. A capacitor stores energy in the electric field between its plates and stored energy is proportional to the square of the voltage and capacitance $U = \frac{1}{2}CV^2$, where symbols have their usual meanings.
Two capacitors, one of $3 \ \mu$F and the other of $6 \ \mu$F, are connected in series in the circuit as shown in the figure, for a long time. }

2.
A student sets up the circuit as shown in the figure to find the value of unknown resistance X and records a set of readings of the voltmeter and the ammeter by using the rheostat.

3.
Consider a cylindrical conductor of length $ l $ and area of cross-section $ A $. Current $ I $ is maintained in the conductor and electrons drift with velocity $ \vec{v}_d \, (|\vec{v}_d| = \frac{eE}{m} \tau) $, where symbols have their usual meanings. Show that the conductivity of the material of the conductor is given by \[ \sigma = \frac{n e^2 \tau}{m}. \]

View Solution

4.
A charged particle $+q$ in an electric field $\vec{E}$ experiences a force in the direction of the electric field. As a result, its kinetic energy changes. Similarly, the charged particle also experiences a force when it moves in a magnetic field $\vec{B}$. But this magnetic force is perpendicular to both velocity $\vec{v}$ of the charged particle and the magnetic field $\vec{B}$, so it cannot change the kinetic energy of the charged particle. Consider two charged particles 1 and 2 of masses $m$ and $\frac{m}{2}$ having charges $-q$ and $+2q$ respectively. They are accelerated from rest through the same potential difference $V$ and acquire kinetic energy $K_1$ and $K_2$. Then they enter in a region of uniform magnetic field $\vec{B}$ perpendicular to their velocities.

5.
Read the following paragraph and answer the questions that follow.
A p-type or n-type semiconductor can be converted into a p-n junction by doping it with suitable impurity. The motion of majority charge carriers causes diffusion current across the junction while the barrier electric field causes motion of minority carriers for drift current. In case of unbiased diode, the diffusion and drift currents are equal. This equilibrium is disturbed by the biasing batteries. Diodes, therefore, allow currents in one direction. This property of diode is used in making rectifiers.

6.
The resistance of a metal wire at $ 20^\circ \text{C} $ is $ 1.05 \, \Omega $ and at $ 100^\circ \text{C} $ is $ 1.38 \, \Omega $. Determine the temperature coefficient of resistivity of this metal.

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CBSE Class 12 Physics Notes Chapter 1 Electric Charges and Fields

Class 12 Physics Chapter 1 Notes - Electric Charges and Fields

CBSE CLASS XII Related Questions

2.
A student sets up the circuit as shown in the figure to find the value of unknown resistance X and records a set of readings of the voltmeter and the ammeter by using the rheostat.

6.
The resistance of a metal wire at \( 20^\circ \text{C} \) is \( 1.05 \, \Omega \) and at \( 100^\circ \text{C} \) is \( 1.38 \, \Omega \). Determine the temperature coefficient of resistivity of this metal.

Similar Physics Concepts

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CBSE Class 12 Physics Notes Chapter 1 Electric Charges and Fields

Class 12 Physics Chapter 1 Notes - Electric Charges and Fields

CBSE CLASS XII Related Questions

2.A student sets up the circuit as shown in the figure to find the value of unknown resistance X and records a set of readings of the voltmeter and the ammeter by using the rheostat.

6. The resistance of a metal wire at \( 20^\circ \text{C} \) is \( 1.05 \, \Omega \) and at \( 100^\circ \text{C} \) is \( 1.38 \, \Omega \). Determine the temperature coefficient of resistivity of this metal.

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

2.
A student sets up the circuit as shown in the figure to find the value of unknown resistance X and records a set of readings of the voltmeter and the ammeter by using the rheostat.

6.
The resistance of a metal wire at \( 20^\circ \text{C} \) is \( 1.05 \, \Omega \) and at \( 100^\circ \text{C} \) is \( 1.38 \, \Omega \). Determine the temperature coefficient of resistivity of this metal.