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At some point have you felt a slight electric shock when you pulled the woollen sweater off or when you touched a metallic object in darkness and you hear a spark? No, nothing magical here, but the phenomenon of static electricity doing all this. In Chapter 1 Electric Charges and Fields in Class 12 Physics, you will be taught the basic concepts of electrostatics wherein you will learn the process by which electric charges may accumulate, interact with one another, and form invisible electric fields that govern many phenomena that we see around us every day. This chapter will cover the science of lightning in a storm all the way to the sudden shock that you get when you touch a doorknob.
The NCERT Class 12 Physics Chapter 1 is a pillar of the Class 12 Physics and lays the foundation of learning electrostatics which is a major part of CBSE board examinations as well as competitive examinations such as JEE and NEET. The concepts presented in this chapter, e.g. Coulomb Law, electric field and field lines, quantization of charge, superposition principle, etc., are all important to have a solid background upon as one progresses towards more advanced problems in physics. The Physics Class 12 Notes of Electric Field and Charges Chapter present an easy-to-read and well-organized overview of everything important such as key facts and definitions in a format that can be understood easily, valuable formulas to refer and solve problems, important diagrams which can help illustrate the visualization of electric field lines and direction of forces and practice questions which will provide assessment of understanding and develop confidence.
Also Read,
NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields
Use the button below to save the PDF version of Electric Charges and Fields NCERT Notes. It can be viewed anytime you desire without internet access. It is most suitable for revision and last-minute study.
The NCERT Notes Class 12 Physics chapter 1 Electric Charges and Fields are a comprehensive and concise representation of the study material in form of concepts, formulas and illustrations. The notes can be used by students to revise the basics of electrostatics in quick time and are therefore, most suitable for CBSE, JEE and NEET exams.
(a) Charging by Friction : When a neutral body is rubbed against other neutral body then some electrons are transferred from one body to other. The body which can hold electrons tightly, draws some electrons and the body which can not hold electrons tightly, loses some electrons. The body which draws electrons becomes negatively charged and the body which loses electrons becomes positively charged.
For example : Suppose a glass rod is rubbed with a silk cloth. As the silk can hold electrons more tightly and a glass rod can hold electrons less tightly (due to their chemical properties), some electrons will leave the glass rod and get transferred to the silk. So, in the glass rod their will be deficiency of electrons, therefore it will become positively charged. And in the silk, there will be some extra electrons, so it will become negatively charged
(b) Charging 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.
(c) Charging 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.
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:
$\begin{aligned} & F \propto \frac{Q_1 Q_2}{r^2} \\ & F=\frac{K Q_1 Q_2}{r^2}\end{aligned}$
Where,
$K=\frac{1}{4 \pi \epsilon_0}$
Where
$\epsilon_0=8.854\times10^{-12}\frac{C^2}{N-m^2}$, known as absolute permittivity of air or free space
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 q1 on q2 is then directed along the unit vector r12, as shown in figure F12.
$\begin{aligned} & \vec{F_{12}}=\frac{K q_1 q_2}{r^2} \hat{r}_{12} \\ & \text { here } \hat{r}_{12}=\frac{\vec{r}_2-\vec{r}_1}{\left|r_2-r_1\right|}=\frac{\vec{r}_{12}}{r} \\ & \vec{F_{12}}=\frac{K q_1 q_2}{r^3} \vec{r}_{12}\end{aligned}$
When a dielectric material with a dielectric constant (k) is completely filled between charges, the force
$F_{m e d}=\frac{q_1 q_2}{4 \pi \varepsilon_0 k r^2}=\frac{q_1 q_2}{4 \pi \varepsilon_0 \epsilon_r r^2}$
$\epsilon_r$ is the relative permittivity / dielectric constant of the medium.
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.
The point where the resultant force on a charged particle becomes zero is called equilibrium position.
The region surrounding a charge, where another charged particle would experience a force, is referred to as the electric field in that space.
The electric field intensity at any point is defined as the force experienced by a unit positive charge placed at that point.
Electric field intensity,
$E=\frac{F}{q_0}$
where,
F is the force experienced by q0,
Properties of electric field intensity $\vec{E}$ :
(i) It is a vector quantity. Its direction is the same as the force experienced by positive charge.
(ii) Direction of electric field due to positive charge is always away from it while due to negative charge, always towards it.
(iii) Its S.I. unit is Newton/Coulomb.
(iv) Its dimensional formula is $\left[\mathrm{MLT}^{-3} \mathrm{~A}^{-1}\right]$
(v) Electric force on a charge $q$ placed in a region of electric field at a point where the electric field intensity is $\vec{E}$ is given by $\vec{F}=q \vec{E}$. Electric force on point charge is in the same direction of electric field on positive charge and in opposite direction on a negative charge.
(vi) It obeys the superposition principle, that is, the field intensity at a point due to a system of charges is vector sum of the field intensities due to individual point charges,
$
\text { i.e. } \overrightarrow{\mathrm{E}}=\overrightarrow{\mathrm{E}}_1+\overrightarrow{\mathrm{E}}_2+\overrightarrow{\mathrm{E}}_3+\ldots .
$
(vii) It is produced by source charges. The electric field will be a fixed value at a point unless we change the distribution of source charges.
The electric field produced by a point charge $q$ can be obtained in general terms from Coulomb's law. First note that the magnitude of the force exerted by the charge $q$ on a test charge $q_0$ is
$
F_e=\frac{1}{4 \pi \varepsilon_o} \cdot \frac{q q_o}{r^2}
$
then divide this value by $q_o$ to obtain the magnitude of the field.
$
E=\frac{1}{4 \pi \varepsilon_o} \cdot \frac{q}{r^2}
$
$
\mathbf{E}=\mathbf{E}_1+\mathbf{E}_2
$
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.
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.
The Electric flux through an area is the number of electric field lines passing normally through that area.
$d \phi=\vec{E} \cdot \vec{d} A=E d A \cos \theta$
here $\theta$ is the angle between the area vector and the electric field.
Total flux through area A is
$\phi=\int \vec{E} \cdot \vec{d} A$
Flux is a scalar quantity and the SI unit of flux is volt-meter or weber (Wb).
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.
$(\vec{p})=q(\overrightarrow{2 l})$
Here, 2l is the dipole length.
i) At the axial point
$
E_{\text {arial }}=\frac{2 k p r}{\left(r^2-l^2\right)^2}
$
if $r>>>l$
$
E_{axial}=\frac{1}{4 \pi \epsilon_0} \frac{2 p}{r^3}
$
ii) At Equatorial Point
$\begin{aligned} & E_{\text {equatorial}}=\frac{k p}{\left(r^2+l^2\right)^{\frac{3}{2}}} \\ & \text { If r>>>l} \\ & E_{\text {equatorial}}=\frac{1}{4 \pi \varepsilon_0} \frac{p}{r^3}\end{aligned}$
$\begin{aligned} & \vec{\tau}=\vec{p} \times \vec{E} \\ & \vec{\tau}=p E \sin \Theta\end{aligned}$
When an electric dipole is placed in an electric field $\mathbf{E}$, a torque $\vec{\tau}=\mathbf{p} \times \mathbf{E}$ acts on it. If we rotate the dipole through a small angle $d \theta$, the work done by the torque is
$
\begin{aligned}
& d W=\tau d \theta \\
& d W=-p E \sin \theta d \theta
\end{aligned}
$
The work is negative as the rotation $d \theta$ is opposite to the torque. The change in electric potential energy of the dipole is therefore
$
d U=-d W=p E \sin \theta d \theta
$
Now, at angle $\theta=90^{\circ}$, the electric potential energy of the dipole may be assumed to be zero as net work done by the electric forces in bringing the dipole from infinity to this position will be zero.
Integrating,
From $90^{\circ}$ to $\theta$, we have or
$
U(\theta)-U\left(90^{\circ}\right)=p E[-\cos \theta]_{90^{\circ}}^\theta
$
$
\therefore U(\theta)=-p E \cos \theta=-\mathbf{p} \cdot \mathbf{E}
$
If the dipole is rotated from an angle $\theta_1$ to $\theta_2$, then
$
\begin{aligned}
& \text { Work done by external forces }=U\left(\theta_2\right)-U\left(\theta_1\right) \\
& \qquad \begin{array}{l}
W_{\text {ext forces }}=-p E \cos \theta_2-\left(-p E \cos \theta_1\right) \\
W_{\text {ext. foreses }}=p E\left(\cos \theta_1-\cos \theta_2\right)
\end{array}
\end{aligned}
$
or
and work done by electric forces,
$
W_{\text {electric force }}=-W_{\text {ext. forsec }}=p E\left(\cos \theta_2-\cos \theta_1\right)
$
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:
$\lambda=\frac{q}{L}=\frac{C}{m}=C m^{-1}$
$(\lambda)-$ charge per unit length.
Example: wire, circular ring
ii) Surface charge distribution:
$\sigma=\frac{Q}{A}=\frac{C}{m^2}=C m^{-2}$
$(\sigma)-$ charge per unit Area
Example: plane sheet
iii) Volume Charge distribution
$\rho=\frac{Q}{V}=\frac{C}{m^3}=C m^{-3}$
$(\rho)-$ charge per unit volume.
(charge on a dielectric sphere etc)
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.
$\phi=\oint \vec{E} \cdot \vec{d} A=\frac{1}{\epsilon_0} q$
Notes:
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.
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
Let the Gaussian surface be a cylinder of cross-sectional area A
$\phi=E \times 2 A$
( the flux passes only through 2 circular cross-sections of the cylinder)
According to Gauss's law
$\begin{aligned} & \phi=\frac{q}{\varepsilon_o} \\ & E \times 2 A=\frac{\sigma A}{\varepsilon_0} \\ & E=\frac{\sigma}{2 \varepsilon_0}\end{aligned}$
3. Field due to a uniformly charged thin spherical shell
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
4. Electric field due to a point charge
The electric field due to a point charge is everywhere radial. We wish to find the electric field at a distance $r$ from the charge $q$. We select Gaussian surface, a sphere at distance $r$ from the charge. At every point of this sphere the electric field has the same magnitude $E$ and it is perpendicular to the surface itself. Hence, we can apply the simplified form of Gauss's law,
$
E S=\frac{q_{\text {in }}}{\varepsilon_0}
$
Here, $S=$ area of sphere $=4 \pi r^2$ and
$\begin{aligned} & q_{\text {in }}=\text { net charge enclosing the Gaussian surface }=q \\ & E\left(4 \pi r^2\right)=\frac{q}{\varepsilon_0}\end{aligned}$
$
E=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2}
$
It is nothing but Coulomb's law.
Q1: Consider three metal spherical shells A, B and C, each of radius R. Each shell has a concentric metal ball of radius $\mathrm{R} / 10$. The spherical shells A, B and C are given charges $+6 q,-4 q$, and $+14q$ respectively. Their inner metal balls are also given charges $-2 q,+8 q$ and $-10 q$ respectively. Compare the magnitude of the electric fields due to shells A, B and C at a distance 3 R from their centres.
Answer:
The electric field at a distance $3 R$ from the center of a spherical shell depends only on the net charge enclosed and is given by Gauss's law:
$
E=\frac{1}{4 \pi \epsilon_0} \frac{Q_{\mathrm{net}}}{r^2}
$
where $Q_{\text {net }}$ is the total charge enclosed by each shell.
Net Charge on Each Shell -
For Shell A:
$
Q_A=6 q+(-2 q)=4 q
$
For Shell B:
$
Q_B=-4 q+8 q=4 q
$
For Shell C:
$
Q_C=14 q+(-10 q)=4 q
$
Since the total charge enclosed for all three shells is the same $(4 q)$, the magnitude of the electric field at a distance $3 R$ is identical for all:
$
E_A=E_B=E_C=\frac{1}{4 \pi \epsilon_0} \frac{4 q}{(3 R)^2}
$
Hence, the electric fields due to shells $\mathrm{A}, \mathrm{B}$, and $\mathrm{C}$ at a distance $3 R$ are equal.
Q2: A charge $Q$ in fixed in position. Another charge q is brought near charge $Q$ and released from rest. Which of the following graphs is the correct representation of the acceleration of the charge q as a function of its distance r from charge $Q$?
A.
B.
C.
D.
Answer:
When a charge $q$ is brought close to a stationary charge $Q$, the force on $q$ due to $Q$ is provided by Coulomb's Law:
$
F=k \frac{|Q q|}{r^2}
$
where $k$ is Coulomb's constant and $r$ is the distance between the charges. The acceleration $a$ of charge $q$ can be expressed using Newton's second law:
$
a=\frac{F}{m}=\frac{k|Q q|}{m r^2}
$
where $m$ is the mass of charge $q$.
From this formula, we can observe that as distance $r$ gets smaller, acceleration $a$ gets much greater, tending towards infinity as $r$ tends towards zero. Therefore, the graph that best illustrates this relationship is graph (1), which illustrates that acceleration gets smaller as distance gets greater, in a hyperbolic manner.
Hence, the answer is the option (A).
Q3: State Gauss's law on electrostatics and derive an expression for the electric field due to a long straight thin uniformly charged wire (linear charge density $\lambda$ ) at a point lying at a distance r from the wire.
Answer:
Gauss' Law
The total electric flux over a closed surface is equal to $\ \frac {1}{\varepsilon_0}$ times the net charge enclosed within the surface.
$
\phi_E=\oint E \cdot d s=\frac{1}{\varepsilon_0} q
$
Electric field due to an infinitely long straight charged wire
Consider infinitely long straight charged wire of linear charge density $ \lambda C / m$
For calculating electric field consider an imaginary cylindrical Gaussian Surface of radius r and length l.
Here the field is radial everywhere, so flux through the two ends of the cylinder is Zero.
At the Gaussian cylindrical surface, the electric field E is normal to the surface at every point. The magnitude of E depends only on radius 'r', so it is constant.
Therefore flux through the Gaussian surface
$
=E \times 2 \pi r l---(1)
$
According to Gauss's law, flux
$
\phi=\frac{\text { charge enclosed }}{\varepsilon_0}
$
Here, total charge enclosed $=$ linear charge density $\times$ length$=\lambda l$
Therefore flux $\phi=\frac{\lambda l}{\varepsilon_0}---(2)$
Using equation (1) and (2)
$
E \times 2 \pi r l=\frac{\lambda l}{\varepsilon_0}
$
That is,
$
E=\frac{\lambda}{2 \pi \varepsilon_0 r}
$
The vector notation is
$
\vec{E}=\frac{\lambda}{2 \pi \varepsilon_0 r} \hat{n}
$
where $\hat{n}$ is the radial unit vector normal to the line charge
NCERT Class 12 Physics Chapter 1 Notes |
The main topics covered here are coulombs law, the concept of the electric field and electric dipole, Gauss’s law and its applications.
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.
The study of charge under rest is known as electrostatics
When iron is isolated, then the net charge of the iron is zero. There is no drift of charge to start the conduction.
Yes, Electric Charges and Fields Class 12 notes are important for JEE preparation, covering fundamental concepts frequently tested in both JEE Main and JEE Advanced exams.
Exam Date:22 July,2025 - 29 July,2025
Exam Date:22 July,2025 - 28 July,2025
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