**CBSE NCERT NOTES CLASS 12 PHYSICS CHAPTER 1**

**ELECTROSTATICS**

**Frictional Electricity**

Frictional electricity is the electricity produced by rubbing two suitable bodies and transfer of electrons from one body to other.

For example, electrons in glass are loosely bound than the electrons in silk. So, when glass and silk are rubbed together, the comparatively loosely bound electrons from glass get transferred to silk. As a result, glass becomes positively charged and silk becomes negatively charged.

Similarly, ebonite becomes negatively charged and wool becomes positively charged.

**Properties of Charges**

- There exist only
**two types of charges**, namely positive and negative. - Like charges
**repel**and unlike charges**attract**each other. **Charge is a scalar quantity**and behaves like real numbers, ie. Charge is additive in nature. e.g., +2C + 5C – 3C = +4C**Charge is quantized**Electric charge exists in discrete packets rather than in continuous amount. It can be expressed in integral multiples of a fundamental electronic charge (

*e*= 1.6 × 10^{-19 }C) and any charge q = ±*ne*, where n = 1, 2, 3, ....At a macroscopic or large scale level, the charges used are huge as compared to the magnitude of electronic charge. It is impossible and impractical to tell the exact position of each electron in the bulk. Hence, quantization of electric charge is of no use on macroscopic scale. Therefore, the electric charge is it is considered continuous.

**Charge is conserved**i.e., the algebraic sum of positive and negative charges in an isolated system remains constant.For example when a glass rod is rubbed with silk, negative charge appears on the silk and an equal amount of positive charge appears on the glass rod. The net charge on the glass-silk system remains zero before and after rubbing.

**SI Unit of Charge: **Coulomb, 1C = 3 × 10^{9} statC

**Point Charge**

If the linear size of the charged object is much smaller than the distances being considered, the charge is assumed to be **point charge.**

**Methods of Charging**

**Induction:** When a charged body is brought near an uncharged body without touching it, then an equal and opposite charge appears on the uncharged body. This is called induction. The induced charge is temporary.

**Conduction: **When the bodies are in contact with each other, the charge gets redistributed.

- If two charged objects of same size and material with charge q
_{1}and q_{2}are touched with each other and then separated then the charge on each object after separation will be, $\frac{{\mathrm{q}}_{1}\mathrm{}+\mathrm{}{\mathrm{q}}_{2}}{2}$. - If the charges after separation are q
_{1}' and q_{2}', then$\mathrm{}{\mathrm{q}}_{1}+\mathrm{}{\mathrm{q}}_{2}\mathrm{}=\mathrm{}{\mathrm{q}}_{1}\mathrm{\text{'}}+\mathrm{}{\mathrm{q}}_{2}\mathrm{\text{'}}$

*.*

**Charging by Friction: **As discussed earlier.

**Skin Effect: **Charges reside on the surface of the charged conducting body.

**Electric Force - Coulumb’s Law**

**Force** of interaction between two stationery point charges is,

- directly proportional to the product of the charges,
*F ∝ q*_{1}*q*_{2} - inversely proportional to the square of the distance between them,
$$\mathrm{F}\mathrm{}\propto \mathrm{}\frac{1}{{\mathrm{r}}^{2}}$$

- acts along the straight line joining the two charges.
Or,

$$\mathrm{F}=\frac{1}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}\frac{{\mathrm{q}}_{1}\mathrm{}{\mathrm{q}}_{2}}{{\mathrm{r}}^{2}}$$

Where q

_{1}and q_{2}are charges,ε

_{o}is called the**permittivity of free space***.*The value of ε

_{o}in SI units is*ε*_{o}= 8.854 × 10^{–12}C^{2}N^{–1}m^{–2}And value of $\frac{1}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}$ = 9 × 10

^{9}N m^{2}C^{-2}

**Permittivity** is the measure of a material's ability to resist an electric field and not its ability to ‘permit’ it and may be defined as the amount of charge needed to generate one unit of electric flux in a particular medium. Permittivity of free space is the permittivity when the medium is vaccum.

**Vector form of Coulomb’s Law**

From the diagram above, we can see,

$$\overrightarrow{{\mathrm{r}}_{21}}=\overrightarrow{{\mathrm{r}}_{2}}\mathrm{}\u2013\mathrm{}\overrightarrow{{\mathrm{r}}_{1}}$$

$$\overrightarrow{{\mathrm{r}}_{12}}=\overrightarrow{{\mathrm{r}}_{1}}\mathrm{}\u2013\mathrm{}\overrightarrow{{\mathrm{r}}_{2}}$$

$$\Rightarrow \mathrm{}\overrightarrow{{\mathrm{r}}_{21}}=-\overrightarrow{{\mathrm{r}}_{12}}$$

The vector form of Coulomb’s law can be written as,

$$\overrightarrow{{\mathrm{F}}_{21}}=\frac{1}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}\frac{{\mathrm{q}}_{1}{\mathrm{q}}_{2}}{{\mathrm{r}}_{21}^{2}}\mathrm{}{\widehat{\mathrm{r}}}_{21}$$

$$\mathrm{}\overrightarrow{{\mathrm{F}}_{12}}=\frac{1}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}\frac{{\mathrm{q}}_{1}{\mathrm{q}}_{2}}{{\mathrm{r}}_{21}^{2}}{\widehat{\mathrm{r}}}_{12}$$

We can see that,

$$\mathrm{}\overrightarrow{{\mathrm{F}}_{21}}=-\overrightarrow{{\mathrm{F}}_{12}}$$

Where,

- $\overrightarrow{{\mathrm{F}}_{12}}$
**=**Force on*q*_{1}due to*q*_{2}; - $\overrightarrow{{\mathrm{F}}_{21}}$ = Force on
*q*_{2}due to*q*_{1}

Further,

- If
*q*_{1}*q*_{2}< 0; the force is an attraction force - if
*q*_{1}*q*_{2}> 0; the force is a repulsion force

**Forces between Multiple Charges: **

The exerted by multiple point charges on a point charge obey principle of superposition.

- Force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges, taken one at a time.
$$\overrightarrow{{\mathrm{F}}_{12}}=\frac{1}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}\frac{{\mathrm{q}}_{1}{\mathrm{q}}_{2}}{{\mathrm{r}}_{12}^{2}}\mathrm{}{\widehat{\mathrm{r}}}_{12}$$

$$\overrightarrow{{\mathrm{F}}_{13}}=\frac{1}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}\frac{{\mathrm{q}}_{1}{\mathrm{q}}_{3}}{{\mathrm{r}}_{13}^{2}}\mathrm{}{\widehat{\mathrm{r}}}_{13}$$

$$\dots \dots \dots \dots \dots .$$

$$\overrightarrow{{\mathrm{F}}_{1\mathrm{n}}}=\frac{1}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}\frac{{\mathrm{q}}_{1}{\mathrm{q}}_{\mathrm{n}}}{{\mathrm{r}}_{1\mathrm{n}}^{2}}\mathrm{}{\widehat{\mathrm{r}}}_{1\mathrm{n}}$$

- The individual forces are unaffected due to the presence of other charges. This is termed as the principle of superposition.

The total force $\overrightarrow{{\mathrm{F}}_{1}}$ on the charge *q*_{1}, due to all other charges is given by the vector sum of the forces $\overrightarrow{{\mathrm{F}}_{12}}$** , **$\overrightarrow{{\mathrm{F}}_{13}}$

**$\overrightarrow{{\mathrm{F}}_{1\mathrm{n}}}$**

*, ...,**is*

$$\overrightarrow{{\mathrm{F}}_{1}}=\overrightarrow{{\mathrm{F}}_{12}}+\mathrm{}\overrightarrow{{\mathrm{F}}_{13}}+\dots .+\mathrm{}\overrightarrow{{\mathrm{F}}_{1\mathrm{n}}}$$

$$\overrightarrow{{\mathrm{F}}_{1}}=\frac{1}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}\frac{{\mathrm{q}}_{1}{\mathrm{q}}_{2}}{{\mathrm{r}}_{12}^{2}}\mathrm{}{\widehat{\mathrm{r}}}_{12}+\frac{1}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}\frac{{\mathrm{q}}_{1}{\mathrm{q}}_{3}}{{\mathrm{r}}_{13}^{2}}\mathrm{}{\widehat{\mathrm{r}}}_{13}+\mathrm{}\dots +\frac{1}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}\frac{{\mathrm{q}}_{1}{\mathrm{q}}_{\mathrm{n}}}{{\mathrm{r}}_{1\mathrm{n}}^{2}}\mathrm{}{\widehat{\mathrm{r}}}_{1\mathrm{n}}$$

$$\overrightarrow{{\mathrm{F}}_{1}}=\frac{{\mathrm{q}}_{1}}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}\left(\frac{{\mathrm{q}}_{2}}{{\mathrm{r}}_{12}^{2}}\mathrm{}{\widehat{\mathrm{r}}}_{12}+\frac{{\mathrm{q}}_{3}}{{\mathrm{r}}_{13}^{2}}\mathrm{}{\widehat{\mathrm{r}}}_{13}+\dots +\frac{{\mathrm{q}}_{\mathrm{n}}}{{\mathrm{r}}_{1\mathrm{n}}^{2}}\mathrm{}{\widehat{\mathrm{r}}}_{1\mathrm{n}}\right)$$

$$\overrightarrow{{\mathrm{F}}_{1}}=\frac{{\mathrm{q}}_{1}}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}\sum _{\mathrm{i}=2}^{\mathrm{n}}\frac{{\mathrm{q}}_{\mathrm{i}}}{{\mathrm{r}}_{1\mathrm{i}}^{2}}{\widehat{\mathrm{r}}}_{1\mathrm{i}}$$

**Electric Field**

Electric field due to a charge q at a point in space may be defined as the force that a unit positive charge would experience if placed at that point.

$$\overrightarrow{\mathrm{E}}\left(\mathrm{r}\right)=\frac{1}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}\frac{\mathrm{Q}}{{\mathrm{r}}^{2}}\mathrm{}\widehat{\mathrm{r}}\mathrm{}$$

Therefore, we can write,

$$\overrightarrow{\mathrm{F}}=\mathrm{q}\mathrm{}\overrightarrow{\mathrm{E}}$$

The charge Q, which is producing the electric field, is called a source charge and the charge q, which tests the effect of a source charge, is called a test charge.

$$\overrightarrow{\mathrm{E}}=\underset{\mathrm{q\; \to}0}{\mathit{lim}}\mathrm{}\mathrm{}\frac{\overrightarrow{\mathrm{F}}}{\mathrm{q}}$$

- SI Unit of E is N C
^{-1}or V m^{-1}.

**Properties of Electric field**

- Electrical field exists at all points in the space, but the magnitude may vary.
- $\overrightarrow{\mathrm{E}}$
- For a positive charge, the electric field will be
**directed radially outwards from the charge**. On the other hand, if the source charge is negative, the electric field vector, at each point,**points radially inwards.** - Since the magnitude of the force $\overrightarrow{\mathrm{F}}$ on charge q due to charge Q depends only on the distance r of the charge q from charge Q, the magnitude of the electric field $\overrightarrow{\mathrm{E}}$ will also depend only on the distance r. Thus at equal distances from the charge Q, the magnitude of its electric field $\overrightarrow{\mathrm{E}}$ is same.

**Physical Significance of Electric Field**

Time dependent electromagnetic phenomenon can be explained elegantly by electric field.

The accelerated motion of q_{1} produces EM waves, which propagate with the speed of light **c** and reach q_{2}, causing the force on q_{2}. There is a delay between the effect (the force on q_{2}) and the cause (accelerated motion of q_{1}). Since the electrical and magnetic fields can be experienced only by their effect, they are considered **real physical entities** rather than **mere mathematical constructs**.

They have their own **independent dynamics**. They can **transport energy**. If a time dependent E-F is turned on briefly and switched off, leaving behind a propagating EM field, transporting energy, even when the cause has been turned off.

**Electric field due to a system of Charges**

At any point P, the electric fields due to various charges follow laws of superposition.

Let the displacement vectors from *q*_{1}*, q*_{2}*, …, q*_{n} to P be $\overrightarrow{\mathrm{r}}}_{1\mathrm{P}$*, *$\overrightarrow{\mathrm{r}}}_{2\mathrm{P}$*, …, *$\overrightarrow{\mathrm{r}}}_{\mathrm{n}\mathrm{P}$ respectively.

Let us represent the electric fields at P due to *q*_{1}*, q*_{2}*,…,q*_{n} be $\overrightarrow{\mathrm{E}}}_{1$*, *$\overrightarrow{\mathrm{E}}}_{2$*, …, *$\overrightarrow{\mathrm{E}}}_{\mathrm{n}$ respectively, then,

We can write,

$$\overrightarrow{{\mathrm{E}}_{1}}=\frac{1}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}\frac{{\mathrm{q}}_{1}}{{\mathrm{r}}_{1\mathrm{P}}^{2}}\mathrm{}{\widehat{\mathrm{r}}}_{1\mathrm{P}}$$

$$\overrightarrow{{\mathrm{E}}_{2}}=\frac{1}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}\frac{{\mathrm{q}}_{2}}{{\mathrm{r}}_{2\mathrm{P}}^{2}}\mathrm{}{\widehat{\mathrm{r}}}_{2\mathrm{P}}\mathrm{}$$

$$\dots \dots \dots \dots \dots .$$

$$\overrightarrow{{\mathrm{E}}_{\mathrm{n}}}=\frac{1}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}\frac{{\mathrm{q}}_{\mathrm{n}}}{{\mathrm{r}}_{\mathrm{n}\mathrm{P}}^{2}}\mathrm{}{\widehat{\mathrm{r}}}_{\mathrm{n}\mathrm{P}}$$

By the superposition principle, the electric field $\overrightarrow{\mathrm{E}}$ at P due to the system of charges *q*_{1}*, q*_{2}*, …., q*_{n}* *is

$$\overrightarrow{\mathrm{E}}\left(\mathrm{r}\right)=\overrightarrow{{\mathrm{E}}_{1}}+\mathrm{}\overrightarrow{{\mathrm{E}}_{2}}+\dots .+\mathrm{}\overrightarrow{{\mathrm{E}}_{\mathrm{n}}}$$

$$\overrightarrow{\mathrm{E}}\left(\mathrm{r}\right)=\frac{1}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}\frac{{\mathrm{q}}_{1}}{{\mathrm{r}}_{1\mathrm{P}}^{2}}\mathrm{}{\widehat{\mathrm{r}}}_{1\mathrm{P}}+\frac{1}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}\frac{{\mathrm{q}}_{2}}{{\mathrm{r}}_{2\mathrm{P}}^{2}}\mathrm{}{\widehat{\mathrm{r}}}_{2\mathrm{P}}+\dots +\frac{1}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}\frac{{\mathrm{q}}_{2}}{{\mathrm{r}}_{\mathrm{n}\mathrm{P}}^{2}}\mathrm{}{\widehat{\mathrm{r}}}_{\mathrm{n}\mathrm{P}}$$

$$\overrightarrow{\mathrm{E}}\left(\mathrm{r}\right)=\frac{1}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}\mathrm{}\sum _{\mathrm{i}=1}^{\mathrm{n}}\frac{{\mathrm{q}}_{1}}{{\mathrm{r}}_{\mathrm{i}\mathrm{P}}^{2}}{\widehat{\mathrm{r}}}_{\mathrm{i}\mathrm{P}}$$

**Electric Field Lines**

An electric field line is a curve drawn in such a way that the tangent to it at each point is in the direction of the net field at that point. The direction of field is indicated by an arrow on the curve. A field line is a space curve, i.e., a curve in three dimensions.

**Properties of Electric Field Lines**

- Field lines start from positive charges and end at negative charges. If there is a single charge, they may start or end at infinity.
- In a charge-free region, electric field lines can be taken to be continuous curves without any breaks.
- Two field lines can never cross each other. (If they did, the field at the point of intersection will not have a unique direction, which is absurd.)
- Electrostatic field lines do not form any closed loops. This follows from the conservative nature of electric field.
- The electric field lines are always normal to the surface.

**Magnitude of electric field is represented by the density of field lines.**

Let us consider a point charge q at origin O as shown in figure.

The solid angle subtended by a small perpendicular plane area ΔS, at a distance *r*, can be written as,

*ΔΩ = *$\frac{\mathrm{\Delta}\mathrm{S}}{{\mathrm{r}}^{2}}$*. *

The area at P_{1},

*ΔS*_{1 }*=*_{ }*r*_{1}^{2 }*ΔΩ *

and the area at P_{2, }

*ΔS*_{ 2 }*=*_{ }*r*_{2}^{2}* ΔΩ.*

The number of lines (say *n*) cutting these area elements are the same. Density of field lines is,

At P_{1}

$${\mathrm{\rho}}_{1}=\frac{\mathrm{n}}{\mathrm{\Delta}{\mathrm{S}}_{1}}=\frac{\mathrm{n}}{\mathrm{}{{\mathrm{r}}_{1}}^{2}\mathrm{}\mathrm{\Delta}}$$

At P_{2}

$${\mathrm{\rho}}_{2}=\frac{\mathrm{n}}{\mathrm{\Delta}{\mathrm{S}}_{2}}=\frac{\mathrm{n}}{\mathrm{}{{\mathrm{r}}_{2}}^{2}\mathrm{}\mathrm{\Delta}}$$

From the above two equations, we can see, that,

$$\mathrm{\rho \; \propto}\mathrm{}\mathrm{}\frac{1}{{\mathrm{r}}^{2}}$$

We observe that both E and ρ are proportional to $\frac{1}{{\mathrm{r}}^{2}}\mathrm{}$,^{ }that is,

*E ∝ ρ*

**Electric Dipole**

**An electric dipole** is a pair of equal and opposite point charges *q *and –*q,* separated by a distance 2*a*. The line from –*q *to *q *is said to be the direction of dipole space and is called the **axis of the dipole**. The mid-point of locations of –*q *and *q *is called the** centre of the dipole.**

The total charge of the electric dipole is zero. However the field of the electric dipole is not zero. The electric field due to a dipole falls off at large distance, faster than that due to a single charge *q*).

The strength of dipole or dipole moment is defined as *p = q* (2*a*).

**Point Diople:**

The dipole with finite value of dipole moment (*q × 2a*) is considered a **point dipole**, if the distances considered are much larger than the size of the dipole (i.e., *r >>* 2*a*)

**Electric Field Due to a Dipole**

**For points on the axis**

Let the point P be at distance r from the centre of the dipole on the side of the charge q, then,

$${\overrightarrow{\mathrm{E}}}_{-\mathrm{q}}=-\frac{1}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}\frac{\mathrm{q}}{{\left(\mathrm{r}+\mathrm{a}\right)}^{2}}\mathrm{}\widehat{\mathrm{p}};\mathrm{}\mathrm{}$$

$${\overrightarrow{\mathrm{E}}}_{+\mathrm{q}}=\frac{1}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}\frac{\mathrm{q}}{{(\mathrm{r}-\mathrm{a})}^{2}}\mathrm{}\widehat{\mathrm{p}}$$

Total field at P,

$$\overrightarrow{\mathrm{E}}={\overrightarrow{\mathrm{E}}}_{+\mathrm{q}}+{\overrightarrow{\mathrm{E}}}_{-\mathrm{q}}$$

$$=\frac{\mathrm{q}}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}\left[\frac{1}{{(\mathrm{r}+\mathrm{a})}^{2}}-\mathrm{}\frac{\mathrm{q}}{{(\mathrm{r}-\mathrm{a})}^{2}}\right]\mathrm{}\widehat{\mathrm{p}}$$

$$=\frac{\mathrm{q}}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}\frac{4\mathrm{a}\mathrm{r}}{{({\mathrm{r}}^{2}-{\mathrm{a}}^{2})}^{2}}\mathrm{}\widehat{\mathrm{p}}$$

For *r** >> **a**,*

$$\overrightarrow{\mathrm{E}}=\frac{4\mathrm{q}\mathrm{a}}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}{\mathrm{r}}^{3}}\mathrm{}\widehat{\mathrm{p}}$$

Now, since, 2 *q a = p,* we can write,

$$\overrightarrow{\mathrm{E}}=\mathrm{}\frac{2\mathrm{}\overrightarrow{\mathrm{p}}}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}{\mathrm{r}}^{3}}\mathrm{}$$

**For points on the equatorial plane**

The magnitudes of the electric fields due to the two charges +*q *and –*q *are given by

$$\left|{\overrightarrow{\mathrm{E}}}_{+\mathrm{q}}\right|=\left|{\overrightarrow{\mathrm{E}}}_{-\mathrm{q}}\right|=\frac{1}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}\frac{\mathrm{q}}{{\mathrm{r}}^{2}+{\mathrm{a}}^{2}}\mathrm{}$$

The directions are as shown in the diagram. As the vertical components of these cancel each other and the horizontal components add up. The resultant field will be

$$\overrightarrow{\mathrm{E}}=-\left({\mathrm{E}}_{+\mathrm{q}}+{\mathrm{E}}_{-\mathrm{q}}\right)\mathit{cos\; \theta}\mathrm{}\widehat{\mathrm{p}}$$

$$=-\frac{2\mathrm{q}\mathrm{a}}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}{{(\mathrm{r}}^{2}+{\mathrm{a}}^{2})}^{3/2}}\widehat{\mathrm{p}}$$

For *r** >> **a***,**

$$=-\frac{2\mathrm{q}\mathrm{a}}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}{\mathrm{r}}^{3}}\widehat{\mathrm{p}}$$

Now, since, 2 *q a = p,* we can write,

$$\overrightarrow{\mathrm{E}}=\mathrm{}\frac{-\mathrm{}\overrightarrow{\mathrm{p}}}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}{\mathrm{r}}^{3}}$$

**Physical significance of dipole**

In most molecules, the centre’s of positive charges and of negative charges** **lie at the same place. Therefore, their dipole moment is zero. CO_{2} and CH_{4} are of this type of molecules. However, they develop a dipole moment when an electric field is applied. This dipole is **induced dipole**.

But in some molecules, the centres of negative charges and of positive charges do not coincide. Therefore they have a **permanent dipole** moment, even in the absence of an electric field. Such molecules are **called polar molecules**. Examples HCl, H_{2}O.

**Dipole in a Uniform External Field **

Consider a permanent dipole of dipole moment **p **in a uniform external field **E**, as shown in **Fig. **There is a force *q***E **on *q *and a force –*q***E **on –*q*. The net force on the dipole is zero, since **E **is uniform. However, the charges are separated, so the forces act at different points, resulting in a torque on the dipole.

Magnitude of torque = qE × 2a sin θ

= 2qa E sin θ.

Its direction is normal to the plane of the paper, coming out of it. Or we can write,

$$\overrightarrow{\mathrm{\tau}}=\overrightarrow{\mathrm{p}}\times \overrightarrow{\mathrm{E}}$$

In case $\overrightarrow{\mathrm{E}}$ and $\overrightarrow{\mathrm{p}}$ are perpendicular to each other, , i.e.,

τ = p E [as sin θ = 1]

In case $\overrightarrow{\mathrm{E}}$ and $\overrightarrow{\mathrm{p}}$ are **parallel **or **anti-parallel** to each other,

τ = 0 [as sin θ = 0]

**Dipole in a Non-uniform External Field **

In this case, the net force be non-zero apart from torque on the system.

Let us consider simple situations when $\overrightarrow{\mathrm{p}}$** **is parallel to $\overrightarrow{\mathrm{E}}$** **or antiparallel to $\overrightarrow{\mathrm{E}}$. In either case, the net torque is zero, but there is a net force on the dipole if $\overrightarrow{\mathrm{E}}$** **is not uniform.

- When $\overrightarrow{\mathrm{p}}$
- When $\overrightarrow{\mathrm{p}}$

Q: How can you explain the fact that a comb run through dry hair attracts pieces of paper?

The comb, acquires charge through friction. But the paper is not charged. The charged comb ‘polarizes’ the piece of paper, i.e., induces a net dipole moment in the direction of field. Further, the electric field due to the comb is not uniform. As a result the paper moves in the direction of the comb.

**Continuous Charge Distribution**

It is impractical to specify the charge distribution in terms of the locations of the microscopic charged constituents. It is more feasible to consider a length element, area element, or volume element of conductor (which is very small on the macroscopic scale but big enough to include a very large number of electrons) and specify the charge on that element.

**Linear Charge Distribution**

The **linear charge density **of a wire is defined by *l = *$\frac{\mathrm{\Delta}\mathrm{Q}}{\mathrm{\Delta}\mathrm{l}}$, where *Dl *is a small line element of wire and *DQ *is the charge contained in that line element.

The unit of linear charge density (l) is C m^{-1}.

**Surface Charge Distribution**

The **surface charge density σ** at the area element ΔS

**,**

*σ =*$\frac{\mathrm{\Delta}\mathrm{Q}}{\mathrm{\Delta}\mathrm{S}}$

*.**DQ*is the charge contained in that surface element.

The units for *s* are C m^{-2}.

**Volume Charge Distrbution**

The **volume charge density** (sometimes simply called charge density) is defined as *ρ = *$\frac{\mathrm{\Delta}\mathrm{Q}}{\mathrm{\Delta}\mathrm{V}}$**, **where *DQ* is the charge included in the macroscopically small volume element *DV*. The unit for volume charge density *ρ* is C m^{-3}.

**The field due to a continuous charge distribution**

Suppose a continuous charge distribution (volume) in space has a charge density *ρ*. Let the position vector of any point in the charge distribution be $\overrightarrow{\mathrm{r}}}_{1$. The charge density *ρ* may vary from point to point, i.e., it is a function of $\overrightarrow{\mathrm{r}}}_{1$. Divide the charge distribution into small volume elements of size *ΔV*. The charge in a volume element *ΔV *is *ΔQ = ρ ΔV*.

Electric field due to this charge at a point whose position vector is $\overrightarrow{\mathrm{r}}}_{2$_{ }will be

$$\mathrm{\Delta}\overrightarrow{\mathrm{E}}=\frac{1}{4\mathrm{\pi}\mathrm{\epsilon}}\frac{\mathrm{\rho}\mathrm{}\mathrm{\Delta}\mathrm{V}}{{\mathrm{r}}^{2}}\widehat{\mathrm{r}}$$

Where $\overrightarrow{\mathrm{r}}$** = **$\overrightarrow{\mathrm{r}}}_{1$

**$\overrightarrow{\mathrm{r}}}_{2$**

*-*_{ }

Total field at P due to the charge distribution will be

$$\overrightarrow{\mathrm{E}}=\frac{1}{4\mathrm{\pi}\mathrm{\epsilon}}\sum _{\mathrm{over\; all\; \Delta V}}\frac{\mathrm{\rho}\mathrm{}\mathrm{\Delta}\mathrm{V}}{{\mathrm{r}}^{2}}\widehat{\mathrm{r}}\mathrm{}\mathrm{}$$

If we let $\mathrm{\Delta}\mathrm{V\; \to}\mathrm{}\mathrm{}0$ then this sum becomes an integral.

**Electric Flux**

Represented by *Φ*, is the electric field coming out of the surface and perpendicular to it.

Surface area vector $\overrightarrow{\mathrm{S}}$** **defined as a vector whose magnitude is equal to the area of the surface (*S*) and whose direction is in the direction of the vector perpendicular to the surface ($\mathrm{}\widehat{\mathrm{n}}\mathrm{}$). This direction may be different at different points.

Let us consider a surface $\overrightarrow{\mathrm{S}}$,** **in an external electric field $\overrightarrow{\mathrm{E}}$, making an angle θ with it.

Let us further consider an area element $\mathrm{\Delta}\overrightarrow{\mathrm{S}}$, so that,

$$\mathrm{\Delta}\overrightarrow{\mathrm{S}}=\mathrm{}\mathrm{\Delta}\mathrm{S}\mathrm{}\widehat{\mathrm{n}}$$

The flux through$,\mathrm{\Delta}\overrightarrow{\mathrm{S}},$* *

*ΔΦ = *$\overrightarrow{\mathrm{E}}$** .**$\mathrm{}\mathrm{\Delta}\overrightarrow{\mathrm{S}}$

*= **E ΔS cos θ. *

Total flux will be, *Φ = *** ∑ **$\overrightarrow{\mathrm{E}}$

**$\mathrm{}\mathrm{\Delta}\overrightarrow{\mathrm{S}}$**

*.*

*.*

**Gauss’s Law **

Electric flux through a closed surface S is given by,

$$\mathrm{\Phi}=\mathrm{}\frac{\mathrm{q}}{{\mathrm{\epsilon}}_{\mathrm{o}}}$$

where *q =* total* *charge enclosed by *S*.

**Proof of Gauss’s Law**

Let us consider the **total flux through a sphere** of radius r, which encloses a point charge q at its centre. Divide the sphere into small area elements.

The flux through an area element, *ΔS* is

$${\mathrm{\Delta \Phi}=\overrightarrow{\mathrm{E}}.\mathrm{}\mathrm{\Delta}\overrightarrow{\mathrm{S}}=\frac{1}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}\frac{\mathrm{q}}{{\mathrm{r}}^{2}}\mathrm{}\widehat{\mathrm{r}}.\mathrm{\Delta}\overrightarrow{\mathrm{S}}}_{.}$$

Since$\mathrm{}\overrightarrow{\mathrm{E}}$** **and** **$\mathrm{\Delta}\overrightarrow{\mathrm{S}}$** **have the same direction (radially outwards) and |$\mathrm{}\widehat{\mathrm{r}}\mathbf{}$| = 1,

$${\mathrm{}\mathrm{\Delta \Phi}\mathrm{}=\frac{1}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}\frac{\mathrm{q}\mathrm{\Delta}\mathrm{S}}{{\mathrm{r}}^{2}}\mathrm{}}_{.}$$

Total flux through the surface of the sphere

$$\mathrm{\Phi}=\mathrm{}\mathrm{}\mathrm{\sum \Delta \Phi}=\frac{1}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}\frac{\mathrm{q}}{{\mathrm{r}}^{2}}\mathrm{\sum \Delta}\mathrm{S}$$

$$=\mathrm{}\frac{1}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}}\frac{\mathrm{q}}{{\mathrm{r}}^{2}}\mathrm{}\times \mathrm{}4\mathrm{\pi}\mathrm{}{\mathrm{r}}^{2}$$

$$=\mathrm{}\frac{\mathrm{q}}{{\mathrm{\epsilon}}_{\mathrm{o}}}$$

- The law implies that the total electric flux through a closed surface is zero if no charge is enclosed by the surface.

**Total flux through surface of cylinder**

Consider a closed cylindrical surface, with its axis parallel to the uniform field, $\overrightarrow{\mathrm{E}}$.

The total flux *Φ* through the surface is *Φ = Φ*_{1}* + Φ*_{2}* + Φ*_{3}, where *Φ*_{1} and *Φ*_{2} represent the flux through the surfaces 1 and 2 (of circular cross-section) of the cylinder and *Φ*_{3} is the flux through the curved cylindrical part of the closed surface.

Now the normal to the surface 3 at every point is perpendicular to $\overrightarrow{\mathrm{E}}$, so by definition of flux, *Φ*_{3}* =* 0.

The outward normal to 2 is along E while the outward normal to 1 is opposite to E. Therefore *Φ*_{1}* = - *$\overrightarrow{\mathrm{E}}.\mathrm{}\mathrm{\Delta}{\overrightarrow{\mathrm{S}}}_{1}$ and *Φ*_{2}* = *$+\mathrm{}\overrightarrow{\mathrm{E}}.\mathrm{}\mathrm{\Delta}{\overrightarrow{\mathrm{S}}}_{2}$_{.}

Since *S*_{1 }*= S*_{2}, total flux *Φ* *=* 0.

**Important Points Regarding Gauss’s Law:**

- Gauss’s law is true for any closed surface, no matter what its shape or size.
- The term
*q*on the right side of Gauss’s law includes the sum of all charges enclosed by the surface. The charges may be located anywhere inside the surface. - If there are some charges inside and some outside, the electric field [whose flux appears on the left side of the eq. is due to all the charges, both inside and outside
*S*. The term*q*on the right side of Gauss’s law, however, represents only the total charge inside*S*. - The surface chosen for Gauss’s law is called the Gaussian surface. We should not let the Gaussian surface to pass through any discrete charge. This is because electric field due to a system of discrete charges is not well defined at the location of any charge. However, the Gaussian surface can pass through a continuous charge distribution.
- Gauss’s law is often useful towards a much easier calculation of the electrostatic field when the system has some symmetry. This is facilitated by the choice of a suitable Gaussian surface.
- Gauss’s law is based on the inverse square dependence on distance contained in the Coulomb’s law. Any violation of Gauss’s law will indicate departure from the inverse square law.

**APPLICATIONS OF GAUSS’S LAW**

**Field due to an infinitely long straight uniformly charged wire**

Consider an infinitely long thin straight wire with uniform linear charge density λ. The direction of electric field at every point must be radial (outward if λ > 0, inward if λ < 0).

Consider a pair of line elements P_{1} and P_{2} of the wire.

The electric fields produced by the two elements of the pair when summed give a resultant electric field which is radial (the components normal to the radial vector cancel). This is true for any such pair and hence the total field at any point P is radial.

Also since the wire is infinite, electric field does not depend on the position of P along the length of the wire. That is, the electric field is everywhere radial in the plane cutting the wire normally, and its magnitude depends only on the radial distance *r.*

To calculate the field, imagine a cylindrical Gaussian surface. Since the field is everywhere radial, flux through the two ends of the cylindrical Gaussian surface is zero.

At the cylindrical part of the surface, E is normal to the surface at every point, and its magnitude is constant, since it depends only on r. The surface area of the curved part is 2πr*l*, where *l* is the length of the cylinder.

Flux through the Gaussian surface = flux through the curved cylindrical part of the surface

*= E × 2πrl*

For other two surfaces (top and bottom circular surfaces), it is 0.

The surface includes charge equal to *λl*.

Gauss’s law then gives

$$\mathrm{E}\mathrm{}\times \mathrm{}2\mathrm{\pi}\mathrm{r}\mathrm{l}\mathrm{}=\frac{\mathrm{\lambda}\mathrm{l}}{{\mathrm{\epsilon}}_{\mathrm{o}}}$$

$$\Rightarrow \mathrm{}\mathrm{E}=\frac{\mathrm{\lambda}}{2\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}\mathrm{r}}$$

Vectorially,

$$\overrightarrow{\mathrm{E}}=\frac{\mathrm{\lambda}}{2\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}\mathrm{r}}\widehat{\mathrm{n}}\mathrm{}$$

**Field due to a uniformly charged infinite plane sheet**

Let σ be the uniform surface charge density of an infinite plane sheet. We take the x-axis normal to the given plane. By symmetry, the electric field will not depend on y and z coordinates.

Let us take Gaussian surface to be a rectangular parallelepiped of cross sectional area A (A cylindrical surface will also do.) Only the two faces 1 and 2 will contribute to the flux; electric field lines are parallel to the other faces and they, therefore, do not contribute to the total flux.

The unit vector normal to surface 1 is in –x direction while the unit vector normal to surface 2 is in the +x direction. Therefore, flux, $\overrightarrow{\mathrm{E}}.\mathrm{}\mathrm{\Delta}\overrightarrow{\mathrm{S}}$, through both the surfaces are equal and add up. Therefore the net flux through the Gaussian surface of area A is *2 E A*. The charge enclosed by the closed surface is *σ A*. Therefore by Gauss’s theorem,

$$2\mathrm{E}\mathrm{A}\mathrm{}=\frac{\mathrm{\sigma}\mathrm{A}}{{\mathrm{\epsilon}}_{\mathrm{o}}}\mathrm{}\mathrm{o}\mathrm{r},\mathrm{}\mathrm{E}\mathrm{}=\frac{\mathrm{\sigma}}{2{\mathrm{\epsilon}}_{\mathrm{o}}}\mathrm{}$$

Vectorially,

$$\mathrm{}\overrightarrow{\mathrm{E}}\mathrm{}=\frac{\mathrm{\sigma}}{2{\mathrm{\epsilon}}_{\mathrm{o}}}\mathrm{}\widehat{\mathrm{n}}$$

$\overrightarrow{\mathrm{E}}$ is directed away from the plate if σ is positive and toward the plate if σ is negative.

**Field due to a uniformly charged thin spherical shell**

Let σ be the uniform surface charge density of a thin spherical shell of radius R. The field at any point P, outside or inside, can depend only on r (the radial distance from the centre of the shell to the point) and must be radial (i.e., along the radius vector).

**Field outside the spherical shell**

Consider a point P outside the shell with radius vector r. To calculate E at P, we take the Gaussian surface to be a sphere of radius r and with centre O, passing through P. All points on this sphere are equivalent relative to the given charged configuration. The electric field at each point of the Gaussian surface, therefore, has the same magnitude E and is along the radius vector at each point.

Thus, E and ΔS at every point are parallel and the flux through each element is EΔS.

Summing over all ΔS, the flux through the Gaussian surface is E × 4πr^{2}.

The charge enclosed is σ × 4πR^{2}.

By Gauss’s law,

$$\mathrm{E}\times 4\mathrm{\pi}{\mathrm{r}}^{2}=\frac{\mathrm{\sigma}}{{\mathrm{\epsilon}}_{\mathrm{o}}}\times 4\mathrm{\pi}{\mathrm{R}}^{2}\mathrm{}$$

$$\Rightarrow \mathrm{}\mathrm{E}=\frac{\mathrm{\sigma}{4\mathrm{\pi}\mathrm{R}}^{2}}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}{\mathrm{r}}^{2}}$$

$$=\mathrm{}\frac{\mathrm{\sigma}{\mathrm{R}}^{2}}{{\mathrm{\epsilon}}_{\mathrm{o}}{\mathrm{r}}^{2}}$$

$$=\frac{\mathrm{q}}{{4\mathrm{\pi}\mathrm{\epsilon}}_{\mathrm{o}}{\mathrm{r}}^{2}}$$

Vectorially,

$$\overrightarrow{\mathrm{E}}=\frac{\mathrm{q}}{{4\mathrm{\pi}\mathrm{\epsilon}}_{\mathrm{o}}{\mathrm{r}}^{2}}\mathrm{}\widehat{\mathrm{n}}$$

The electric field is directed outwards if q > 0 and inwards if q < 0.

- This is exactly the field produced by a charge q placed at the centre O. Thus for points outside the shell, the field due to a uniformly charged shell is as if the entire charge of the shell is concentrated at its centre.

**Field inside the shell**

Let us take the point P inside the shell. The Gaussian surface is again a sphere through P centred at O. The flux through the Gaussian surface, calculated as before, is E × 4 π r^{2}.

However, in this case, the Gaussian surface encloses no charge. Gauss’s law then gives,

*E × 4πr*^{2}* = 0 *

⇒ E = 0 (r < R).

**Field Due to Non-conducting Solid Sphere**

The charge distribution is spherically symmetric with the charge density given by

$$\mathrm{\rho}=\frac{\mathrm{q}}{\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}}=\frac{\mathrm{q}}{\frac{4}{3}\mathrm{\pi}{\mathrm{a}}^{3}}$$

Where, a = radius of the sphere. In this case, the electric field **E** is radially symmetric and directed outward. The magnitude of the electric field is constant on spherical surfaces of radius r.

**Case I, when r ≤ a**

*Φ = EA = E (4π r*^{2}*)*

The charge enclosed in the Gaussian sphere of radius r is

$$\mathrm{q}\mathrm{}=\mathrm{}\mathrm{\rho}\left(\frac{4}{3}\mathrm{\pi}{\mathrm{r}}^{3}\right)$$

As per Gauss’s law ,

$$\mathrm{E}.\mathrm{S}\mathrm{}=\mathrm{}\mathrm{}=\frac{\mathrm{q}}{{\mathrm{\epsilon}}_{\mathrm{o}}}$$

$$\Rightarrow \mathrm{}\mathrm{}\mathrm{E}.\mathrm{}\mathrm{}\left(4\mathrm{\pi}{\mathrm{r}}^{2}\right)=\frac{\mathrm{\rho}}{{\mathrm{\epsilon}}_{\mathrm{o}}}\left(\frac{4}{3}\mathrm{\pi}{\mathrm{r}}^{3}\right)$$

$$\Rightarrow \mathrm{}\mathrm{}\mathrm{E}=\frac{\mathrm{\rho}\mathrm{r}}{3{\mathrm{\epsilon}}_{\mathrm{o}}}=\mathrm{}\frac{\mathrm{q}\mathrm{r}}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}{\mathrm{a}}^{3}}$$

**Case II, when r ≥ a**

*Φ = EA = E (4π r*^{2}*)*

$$\Rightarrow \mathrm{}\mathrm{E}\mathrm{}\left(4\mathrm{\pi}\mathrm{}{\mathrm{r}}^{2}\right)=\frac{\mathrm{q}}{{\mathrm{\epsilon}}_{\mathrm{o}}}\mathrm{}\mathrm{}$$

$$\Rightarrow \mathrm{}\mathrm{}\mathrm{E}=\frac{\mathrm{q}}{4\mathrm{\pi}{\mathrm{\epsilon}}_{\mathrm{o}}{\mathrm{r}}^{2}}$$