NCERT Class 12 Physics Chapter 1 Notes Electric Charges and Fields

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NCERT Notes for Class 12 Chapter 1

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.

Electric Charge:

• Charge of a material body or particle is the property (acquired or natural) due to which it produces and experiences electrical and magnetic effects. Some of naturally occurring charged particles are electrons, protons, $\alpha$-particles etc.
• Charge is a derived physical quantity & is measured in Coulomb in S.I. unit. In practice we use $\mathrm{mC}\left(10^{-3} \mathrm{C}\right), \mu \mathrm{C}\left(10^{-6} \mathrm{C}\right), \mathrm{nC}\left(10^{-9} \mathrm{C}\right)$ etc.
• C.G.S. unit of charge $=$ electrostatic $u n i t=$ esu
• 1 coulomb $=3 \times 10^9 \mathrm{esu}$ of charge
• Dimensional formula of charge $=\left[\mathrm{M}^{\circ} L^{\circ} \mathrm{T}^1 \mathrm{I}^1\right]$
• 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 and 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.

Methods of charging:

• A body can be charged by means of (a) friction, (b) conduction, (c) induction,

(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.

Properties of Charge:

1. Additivity: The total charge (Q) in a system containing multiple charges(q₁,q₂,q₃…………q_n) is the sum of individual charges: q₁+q₂+.....q_n.
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 $e=1.6 \times 10^{-19} \mathrm{C}$ 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:

$\begin{aligned} & F \propto \frac{Q_1 Q_2}{r^2} \\ & F=\frac{K Q_1 Q_2}{r^2}\end{aligned}$

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

$\epsilon_0=8.854\times10^{-12}\frac{C^2}{N-m^2}$, known as absolute permittivity of air or free space

The vector form of Coulomb's Law:

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.

$\begin{aligned} & \vec{F_{12}}=\frac{K q_1 q_2}{r^2} \hat{r}_{12} \\ & \text { here } \hat{r}_{12}=\frac{\vec{r}_2-\vec{r}_1}{\left|r_2-r_1\right|}=\frac{\vec{r}_{12}}{r} \\ & \vec{F_{12}}=\frac{K q_1 q_2}{r^3} \vec{r}_{12}\end{aligned}$

Force when dielectric inserted between the charges:

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.

• 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.

Electrostatic Equilibrium:

The point where the resultant force on a charged particle becomes zero is called equilibrium position.

• Stable Equilibrium: A charge is initially in equilibrium position and is displaced by a small distance. If the charge tries to return back to the same equilibrium position then this equilibrium is called position of stable equilibrium.
• Unstable Equilibrium: If a charge is displaced by a small distance from its equilibrium position and the charge has no tendency to return to the same equilibrium position. Instead it goes away from the equilibrium position.
• Neutral Equilibrium: If charge is displaced by a small distance and it is still in equilibrium condition then it is called neutral equilibrium.

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.

Electric field intensity,

$E=\frac{F}{q_0}$

where,

F is the force experienced by q_0,

• The SI unit of E is, $\frac{\text { Newton }}{\text { Coulomb }}=\frac{\text { Volt }}{\text { meter }}=\frac{\text { Joule }}{\text { Coulomb } \times \text { Meter } \text { }}$ and the dimensional formula is $\left[M L T^{-3} A^{-1}\right]$
• 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.

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.

Electric Field Due To A Point Charge:

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}
$

• If $q$ is positive, $\mathbf{E}$ is directed away from $q$. On the other hand, if $q$ is negative, then $\mathbf{E}$ is directed towards $q$.
• The electric field at a point is a vector quantity. Suppose $\mathbf{E}_1$ is the field at a point due to a charge $q_1$ and $\mathbf{E}_2$ in the field at the same point due to a charge $q_2$. The resultant field when both the charges are present is

$
\mathbf{E}=\mathbf{E}_1+\mathbf{E}_2
$

• If the given charge distribution is continuous, we can use the technique of integration to find the resultant electric field at a point.

Electric field due to a system of charges:

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

$d \phi=\vec{E} \cdot \vec{d} A=E d A \cos \theta$

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).

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.

Electric Dipole Moment:

$(\vec{p})=q(\overrightarrow{2 l})$

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

$
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

$\begin{aligned} & E_{\text {equatorial}}=\frac{k p}{\left(r^2+l^2\right)^{\frac{3}{2}}} \\ & \text { If r>>>l} \\ & E_{\text {equatorial}}=\frac{1}{4 \pi \varepsilon_0} \frac{p}{r^3}\end{aligned}$

Torque Experienced by the Dipole:

$\begin{aligned} & \vec{\tau}=\vec{p} \times \vec{E} \\ & \vec{\tau}=p E \sin \Theta\end{aligned}$

Potential Energy of Dipole

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)
$

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:

$\lambda=\frac{q}{L}=\frac{C}{m}=C m^{-1}$

$(\lambda)-$ 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)

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.

$\phi=\oint \vec{E} \cdot \vec{d} A=\frac{1}{\epsilon_0} q$

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.

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.

Electric Field and Charges Previous year Question and Answer

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

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