1.0 Preview:

This chapter revolves around the intriguing nature of electric charges and the fields that they produce. We shall begin by defining electric charge as the property that causes the protons and electrons in an atom to experience a force when placed in an electromagnetic field. There are two types of electric charges: positive and negative. This property of electric charges to exhibit two types of charges is what we call, polarity. You will see how the Gold-leaf electroscope, a simple apparatus can be used to detect charge on a body and the polarity of the charge. In the next section, we shall study about conductors and insulators. Conductors are substances through which electric charges can flow easily while Insulators are substances through which electric charges cannot flow easily.

There are two ways by which bodies are charged: Conduction – The process by which an uncharged body is charged when it brought in contact with a charged body and Induction – The process by which an uncharged body is charged when it is brought near (but not in contact) with a charged body.

The properties of electrical charges shall be discussed briefly. Followed by this, we will state and apply Coulomb’s law. The law states that, “The electrostatic force between two stationary point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.” The SI unit of electric charge was thus coined, Coulomb and is defined as that charge which when placed at rest in air at a distance of 1m from an identical charge that repels it and is repelled by a force of 9 x $10^9$ N. Consequently, we will obtain an expression for dielectric constant from the Coulomb’s law.

Dielectric constant or the relative permittivity of a medium is the ratio of the electrostatic force between two stationary point charges separated by a certain distance in free space to the electrostatic force between the same two point charges separated by the same distance in the medium. We shall also obtain an expression for force due to multiple charges in vacuum from the Coulomb’s law.

We shall then learn about electric field. The electric field due to a charge at a point in space is defined as the force that a unit positive charge would experience if placed at that point. Then, we define another important quantity—the electric intensity.

The force experienced by a unit positive charge placed at any point in an electric field is known Electrical Intensity.

Then we will move to an interesting way of representing the electric field using lines and curves. The representation is called, Electric field lines.

Electric Flux will then be defined, as the total number of electric field lines passing through the surface along the normal.

Further, we will study an electric dipole as a pair of equal and opposite point charges q and –q, separated by a distance 2a.

Next, you will learn to express some important charge densities, such as, surface density which is the charge per unit area around that point; linear density which is the charge per unit length; similarly, volume density, which is charge per unit volume.

Gauss’s theorem and its applications shall then be dealt with. We will be using it to find the flux inside a sphere.

1.1 Introduction

We have all seen a spark or heard a crackle when we take off our synthetic clothes or sweater, especially in dry weather. It is most evident in materials like polyester.
Other such examples of Electric discharge include:

1. Lightning in the sky during a thunderstorm.

2. A sensation of an electric shock while opening the door of a car.

3. Holding an iron bar in a bus for support suddenly after sliding from the seat.

The reason for the experiences in points 2 and 3 is the discharge of electric charges through our body, which were accumulated due to rubbing of insulating surfaces (that is, due to the generation of static electricity).

Electrostatics deals with the study of forces, fields and potentials arising from static charges.

Questions for the section:

1. What is an electric charge?

1.2 Electric charge

• The name electricity is coined from the Greek word elektron meaning amber. Electrons, along with protons and neutrons are the constituent elementary particles of an atom. Atoms, in turn, are the building blocks of all matter. A proton and an electron attract each other while two electrons or two protons repel each other. It should be noted that the repulsive force between two protons or two electrons is very much greater than the attractive force between a proton and an electron.
• Electric charge: The physical property that causes the protons and electrons in an atom to experience a force when placed in an electromagnetic field. This is known as electric charge.
• Some combinations of materials display the properties of exchange of charges in a more pronounced way than others.
• If a glass rod is rubbed with a silk cloth, some electrons pass from the glass rod to the silk cloth. (To understand why this happens in detail, check Concept Box)
• The glass rod, after having lost electrons becomes positively charged; the silk cloth attains an equal and opposite negative charge. It has been shown experimentally that like charges repel and unlike charges attract.
• The property which differentiates the two kinds of charges is called the polarity of charge.

Points to Note:

1. Charges are not created by the rubbing action. There is only a transfer of charges from one body to another. This is in line with the Law of conservation (See Concept Box).

Concept box Why do some pairs of materials lose electrons and why do others gain electrons when they are rubbed against one another? This can be answered by looking at the Triboelectric effect. Once two materials are rubbed, i.e, come in contact with each other, a chemical bond is formed between parts of the two surfaces which is called adhesion. Charges move to equalize their electrochemical potential exactly like water moves to attain the same height from one place to another. Hence, now there is a flow of charges to equalize the electrochemical potential.  Now, when the two materials are separated, some of the bonded atoms have a tendency to keep extra electrons, and some a tendency to give them away. The material that loses electrons becomes positively charged. The material that gains electrons becomes negatively charged.   You can remember it this way: Electrons are negatively charged. So, when a material loses electrons, it becomes positively charged.

Concept Box 1.2 Law of conservation According to the Law of Conservation, the magnitude of properties within a chemical system, such as mass, energy, or charge, remain unchanged during a chemical reaction. These properties may be exchanged between components of the system; however, the total amount in the system does not increase or decrease

2. Electrons have mass, as do protons and neutrons. Now, when a silk cloth receives electrons (and becomes negatively charged), its mass increases due to the addition of new electrons.

3. When the glass rod and the silk cloth lose or acquire electrons, they are said to be electrified.

Concept Box All matter is made up of atoms and/or molecules. Normally materials are electrically neutral but they always contain charges; their charges, however, are exactly balanced and hence they remain electrically neutral. a. Forces that hold the molecules together, b. Forces that hold atoms together in a solid, c. The adhesive forces, d. Forces associated with surface tension, are all basically electrical in nature, arising from the forces between charged particles

Now, if the electrified glass rod and the electrified silk cloth are brought in touch with each other, they no longer attract each other. They also do not attract or repel other light objects as they did on being electrified.

Thus, the charges acquired after rubbing are lost when the charged bodies are brought in contact with each other. This happens because the two electrified bodies have unlike charges. Unlike charges tend to nullify or neutralize each other.

The charges are called positive and negative. By convention, the charge on the proton is considered to be positive and that on an electron is said to be negative.

The Gold Leaf electroscope

The Gold-leaf electroscope is a simple apparatus to detect charge on a body and the polarity of the charge.

It consists of a vertical metal rod housed in a box, with two thin gold leaves attached to its bottom end. When a charged object touches the metal knob at the top of the rod, charge flows on to the leaves and they diverge. The degree of divergence is an indicator of the amount of charge.

Questions from section 1.2:

1. What were the observations made and inferences drawn from the glass rod and silk cloth experiment?
2. What is a gold leaf electroscope?

1.3 Conductors and Insulators

There are 3 categories that substances are divided into:

1. Conductors

2. Semiconductors

3. Insulators

We will concentrate only on Conductors and Insulators.

Conductors are substances through which electric charges can flow easily i.e transferring charges through them is easy.

Ex: Silver, Copper, Iron, Aluminium, Mercury etc.

Insulators are substances through which electric charges cannot flow easily i.e transferring charges through them is relatively more difficult.

Ex: Glass, Rubber, Mica, Plastic, Wood etc.

When some charge is transferred to a conductor, it readily gets distributed over the entire surface of the conductor. In contrast, if some charge is put on an insulator, it stays at the same place or charges are said to remain localized.

Concept box Earthing When we bring a charged body in contact with the earth, all the excess charge on the body disappears. The charges now flow to the earth. This process of sharing the charges with the earth is called grounding or earthing. Earthing is a safety measure for electrical circuits and appliances. A thick metal plate is buried deep into the earth and the earth wire s connected to it. The electric wiring in our houses has three wires: live, neutral and earth. The first two carry electric current from the power station and the third is earthed by connecting it to the buried metal plate. Metallic bodies of the electric appliances such as electric iron, refrigerator, TV are connected to the earth wire. When any fault occurs or live wire touches the metallic body, the charge flows to the earth without damaging the appliance. Earthing also protects humans. In case of a fault in a metal appliance, a great majority of the charges flows through the earth wire which is a conductor (and hence has less resistance). The human body does not experience a shock unless the magnitude of the fault is pretty high as the human body is not a good conductor of electricity.

Questions from section 1.2:

1. What are insulators? Give two examples.
2. What are conductors? Give two examples.
3. What is earthing? Why is it done?

1.4 Charging by Induction

As we have previously seen, a body is charged either by adding electrons (body gets negatively charged) or by removing electrons (body gets positively charged). This process of a body acquiring electric charge is known as electrification.

An uncharged body can be charged in two ways:

1. Conduction – An uncharged body is charged when it brought in contact with a charged body.

2. Induction – An uncharged body is charged when it is brought near (but not in contact) with a charged body.

Example 1.1 How can you charge a metal sphere positively without touching it?

Solution: Figure 1.2(a) shows an uncharged metallic sphere on an insulating metal stand. Bring a negatively charged rod close to the metallic sphere, as shown in Fig. 1.2(b). As the rod is brought close to the sphere, the free electrons in the sphere move away due to repulsion and start piling up at the farther end. The near end becomes positively charged due to deficit of electrons. This process of charge distribution stops when the net force on the free electrons inside the metal is zero. Connect the sphere to the ground by a conducting wire. The electrons will flow to the ground while the positive charges at the near end will remain held there due to the attractive force of the negative charges on the rod, as shown in Fig. 1.2(c). Disconnect the sphere from the ground. The positive charge continues to be held at the near end [Fig. 1.2(d)]. Remove the electrified rod. The
positive charge will spread uniformly over the sphere as shown in Fig. 1.2(e).

In this experiment, the metal sphere gets charged by the process of induction and the rod does not lose any of its charge. Similar steps are involved in charging a metal sphere negatively by induction, by bringing a positively charged rod near it. In this case the electrons will flow from the ground to the sphere when the sphere is connected to the ground with a wire. Can you explain why? A conductor can be charged by induction as follows:

1. Two metal spheres A and B, on insulating stands, are brought in contact with each other. (Fig 1.2 (a)).

2. A positively charged rod (Fig 1.2 (b)) is brought near Sphere A. The rod should not touch the sphere.

The free electrons (i.e the negative charges) in the Sphere is now attracted to the left of the sphere. The positive charges accumulate on the opposite side of the Sphere B.

The accumulated charges continue to hold their position till the positively charged glass rod is held in the same position. Once the glass rod is removed, there is no external force acting on the charges in the sphere. Hence, the charges redistribute inside the body and the spheres return to their original neutral state.

3. Now, the two spheres are separated by a small distance (Fig 1.2 (c)) while holding the glass rod intact near the Sphere A. The two spheres are found to be oppositely charged and attract each other.

4. Remove the rod. The charges on spheres rearrange themselves as shown in Fig. 1.2(d).

5. Now, separate the spheres quite apart. The charges on them get uniformly distributed over them, as shown in Fig. 1.2(e).

Questions from section 1.4:

1. What is electrification?
2. What are the ways by which a body can be charged? Explain each of them.

1.5 Basic Properties of Electric charge

If the sizes of charged bodies are very small as compared to the distances between them, we treat them as point charges. Importantly, all the charge content of the body is assumed to be concentrated at one point in space.

1.5.1 Additivity of charges:

Charges can be added algebraically. Charges add up like real numbers. Proper signs have to be used while adding the charges in a system. If a system contains n charges $q_1, q_2, q_3$….$q_n$, then the total charge in the system is said to be:

$$ Q_1 = q_1 + q_2 + q_3 + ……..+ q_n $$

Ex: If $q_1$ = 0.5C and $q_2$ = 0.3C and $q_3$ = -0.4C, then the total charge in the system is:
$Q_1$ = 0.5 + 0.3 – 0.4
$Q_1$ = 0.4C

1.5.2 Charge is conserved:

Within an isolated system consisting of many charged bodies, due to interactions among the bodies, charges may get redistributed but the total charge of the isolated system is always conserved. Conservation of charge has been established experimentally.

It is not possible to create or destroy net charge carried by any isolated system although the charge carrying particles may be created or destroyed in a process.

Sometimes nature creates charged particles: a neutron turns into a proton and an electron. The proton and electron thus created have equal and opposite charges and the total charge is zero before and after the creation.

1.5.3 Quantisation of charge

All free charges are integral multiples of a basic unit of charge. The basic unit of charge is denoted by e. Thus charge q on a body is always given by:

q = ne

Where n is any integer, positive or negative.

This basic unit of charge is the charge that an electron (-e) or proton (+e) carries.

The fact that electric charge is always an integral multiple of e is termed as quantisation of charge.

e = 1.602192 × $10^{–19}$ C

C – Coulomb – SI unit of charge.

There are about 6 × $10^{18}$ electrons in a charge of –1C.

A charge of 1 coulomb is very large. In practice, we deal with charges in terms of µC. Hence, the step size i.e the magnitude of the basic unit of charge (e) is very small.

Quantisation of charge at the macroscopic level is not meaningful.

Since e = 1.6 × $10^{–19}$ C, a charge of magnitude, say 1 µC, contains something like $10^{13}$ times the electronic charge. At this scale, the fact that charge can increase or decrease only in units of e is not very different from saying that charge can take continuous values. Thus, at the macroscopic level, the quantisation of charge has no practical consequence and can be ignored.

Quantisation of charge at the microscopic level is meaningful

At the microscopic level, where the charges involved are of the order of a few tens or hundreds of e, i.e., they can be counted. Such charges appear in discrete lumps and quantisation of charge cannot be ignored.

Mini Summary Note:

1. Like charges repel each other while unlike charges attract each other. This is referred to as Law of Charges and Law of Electrical Charges.

2. A charged body attracts uncharged bodies

3. Charge always resides on the outer surface of a charged conductor.

4. Charge is a scalar quantity and it also has polarity.

5. Charge is independent of the velocity of the body. It is unaffected by motion.

6. Accelerated charge radiates energy in the form of Electromagnetic waves.

7. Charge cannot exist without mass; however, mass can exist without charge.

Questions from section 1.5:

1. State the basic properties of electric charge.

2. Explain the following properties:

a) Additivity of charges

b) Conservation of charges

c) Quantisation of charge (Also explain its significance at macroscopic and microscopic levels)

Example 1.2 If $10^9$ electrons move out of a body to another body every second, how much time is required to get a total charge of 1 C on the other body?

Solution In one second $10^9$ electrons move out of the body. Therefore the charge given out in one second is $1.6 \times10^{–19} \times 10^9 C = 1.6 \times 10^{–10} C$.

The time required to accumulate a charge of 1 C can then be estimated to be $ 1C \div (1.6 \times 10^{–10} C/s) = 6.25 \times 10^{9} s = 6.25 × 10^{9} \div (365 \times 24 \times 3600)$ years = 198 years. Thus, to collect a charge of one coulomb, from a body from which $10^9$ electrons move out every second, we will need approximately 200 years. One coulomb is, therefore, a very large unit for many practical purposes. It is, however, also important to know what is roughly the number of electrons contained in a piece of one cubic centimetre of a material. A cubic piece of copper of side 1 cm contains about $2.5 \times 10^{24}$ electrons.

Example 1.3 How much positive and negative charge is there in a cup of water?

Solution Let us assume th at the mass of one cup of water is250 g. The molecular mass of water is 18g. Thus, one mole (= 6.02 × 10$^{23}$ molecules) of water is 18 g. Therefore, the number of molecules in one cup of water is $(250/18) \times 6.02 \times 10^{23}$.Each molecule of water contains two hydrogen atoms and one oxygen atom, i.e., 10 electrons and 10 protons. Hence the total positive and total negative charge has the same magnitude. It is equal to $(250/18) \times 6.02 \times 10$^{23}$ \times 10 \times 1.6 \times 10^{–19}4 C = 1.34 \times 10^7 C$.

1.6 Coulomb’s law

Coulomb’s law, or Coulomb’s inverse-square law, is a quantitative statement about the force between two point charges.

Recap: When the linear size of charged bodies is much smaller than the distance separating them, the size may be ignored and the charged bodies are treated as point charges.

Definition box: The Coulomb’s law “The electrostatic force between two stationary point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.”

Mathematically, this can be written as:
$$ F∝ \frac {|q_1.q_2 |}{r^2} $$
$$ F = k. \frac {|q_1.q_2 |}{r^2} $$

Where:
K is called the electrostatic force constant or Coulomb constant. The value of K depends upon:

1. The nature of the medium separating the charges

2. The system of units chosen.

$$ K= \frac {1}{4πε_0}$$

Where:

$ε_0$ is called the permittivity of free space

The experimental value of $ε_0 = 8.854 \times 10^{-12} \ C^2N^{-1}m^{-2}$ or $Fm^{-1}$

$ε_0 = 8.854 \times 10^{-12} \ C^2N^{-1}m^{-2}$ or $Fm^{-1}$

$K = \frac{1}{4\pi ε_0} = 8.9875 \times 10^9 = 9 \times 10^9 \ Nm^2C^{-2}$

Now substituting this in the equation for force:

$$F = \frac{1}{4\pi ε_0} \cdot \frac{|q_1 q_2|}{r^2}$$

In case there is a material medium, then

$$ ε= ε_o ε_r $$

Hence, force in a material medium is now given by the equation:

$$ F= \frac {1}{4πε_0 ε_r }.   \frac {|q_1.q_2 |}{r^2}  $$

Direction of the Electric force:

The electrostatic force between two stationary point charges is always acting along the straight line joining the two charges.

1. Two positive charges repel
2. Two negative charges repel
3. One positive and one negative charge attract

Fig 1.3 Repulsive and attractive forces between like and unlike charges alongwith the resultant force in vector notation (Vector form will be explained in the next segment)

The SI Unit of Charge

Recap: The SI unit of charge is Coulomb (C)

The definition of Coulomb (a measure of electric charge) can be obtained from Coulomb’s law:

Consider,

$$ F= \frac {1}{4πε_0 ε_r }. \frac {|q_1.q_2 |}{r^2} $$

Let us take unit charges and unit distance between them. Hence:

$q_1 = q_2$ = 1C; r=1m

Then,

$$ F= \frac {1}{4πε_0 ε_r}. \frac {1 \times 1}{1^2} $$

F = 9 x $10^9$ N

In words,

Thus, One Coulomb of charge is that charge which when placed at rest in air at a distance of 1m from an identical charge that repels it and is repelled by a force of 9 x $10^9$ N.

Limitations of Coulomb’s Law

1) Coulomb’s law does not hold good for moving charges. It applies only to stationary charges.
2) Coulomb’s law is dependent on the intervening medium. This intervening medium has to be taken into consideration for all calculations.

Relative permittivity or Dielectric Constant

The electrostatic force between two point charges $q_1$ and $q_2$ separated by a distance, d in free space is given by:

$$ F_a=\frac {1}{4πε_0 ε_r}. \frac { |q_1.q_2 |}{d^2} $$

The force between the same two charges in a material medium other than air is given by:

$$ F= \frac {1}{4πε_0 ε_r}. \frac {|q_1.q_2|}{r^2} $$

From the above two equations,

$$ \frac {F_a}{F_m} =ε_r=K $$

In words,

“The dielectric constant or the relative permittivity of a medium can be defined as the ratio of the electrostatic force between two stationary point charges separated by a certain distance in free space to the electrostatic force between the same two point charges separated by the same distance in the medium.”

Note box: K=1, for vacuum K=1.00059 for air medium 2) For all other dielectric media, K>1

Coulomb’s law in vector form

Consider two point charges $q_1$ and $q_2$ with position vectors $ \vec r_1$ and $ \vec r_2$respectively as shown in Fig 1.4a. The force on $q_2$ due to $q_1$ is denoted by $ \vec {F_{21}}$.

The vector leading from q1 $q_1$ to q2 $q_2$ is $ \vec {r_{21}}$ and from $q_2$ to $q_1$ as $ \vec {r_{12}}$.The magnitude of $ \vec {r_{21}}$ is $r_{21}$ and $ \vec {r_{12}}$ is $r_{12}$

As the direction of a vector is specified by a unit vector along the vector, we define:
$$ r_{21} = r_2-r_1$$

In the same way, the vector leading from 2 to 1 is defined as:
$$ r_{12}=r_1-r_2= -r_{21} $$

The magnitude of the vectors r21 and r12 is denoted by r21 and r12, respectively (r12 = r21). The direction of a vector is specified by a unit vector along the vector. To denote the direction from 1 to 2 (or from 2 to 1), we define the unit vectors.

$$ \hat r_{21} = \frac {r_{21}}{r_{21}}, \hat r_{12} =\frac {r_{12}}{r_{12}},\hat r_{21}= \hat r_{12}$$

Coulomb’s force law between two point charges q1 and q2 located at r1 and r2 is then expressed as:

$$ F_{21} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r_{21}^2} \cdot \hat{r}_{21} $$

If $q_1$ and $q_2$ are of opposite signs, $F_{21}$ is along – $r_{21} (= r_{ 12}$ ), which denotes attraction, as expected for unlike charges. Thus, we do not have to write separate equations for the cases of like and unlike charges.

$$F_{12} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r_{12}^2} \cdot \hat{r}{12} = -F{21}$$

This equation is valid for, both, positive and negative values of charge. Moreover, the equation is only valid for force between two charges in vacuum. In case there is an intervening medium, then the medium has to be accounted for by considering the permittivity of the medium. We will read this later in the chapter.

Questions from section 1.6:

1. State Coulomb’s law. From its mathematical expression, obtain the equation for force in the form: $F= \frac {1}{4πε_0 ε_r}. \frac {|q_1.q_2 |}{r^2} . $

2. State the limitations of Coulomb’s law.

3. Obtain the expression for dielectric constant in the form: $ \frac {F_a}{F_m }=ε_r $. Also, define dielectric constant.

4. Give the expression for coulomb’s force in vector form for like charges and unlike charges. Also give the relation between the two force expressions.

Example 1.4 Coulomb’s law for electrostatic force between two point charges and Newton’s law for gravitational force between two stationary point masses, both have inverse-square dependence on the distance between the charges/masses. (a) Compare the strength of these forces by determining the ratio of their magnitudes (i) for an electron and a proton and (ii) for two protons. (b) Estimate the accelerations of electron and proton due to the electrical force of their mutual attraction when they are 1 Å (= 10-10 m) apart? ($m_p$ = 1.67 × $10^{–27}$ kg, me = 9.11 × $10^{–31}$ kg)

Solution:
(a) (i) The electric force between an electron and a proton at a distance r apart is:
$$ F_e = – \frac {1}{4 \pi \epsilon_o} \frac {e^2}{r^2}$$
where the negative sign indicates that the force is attractive. The corresponding gravitational force (always attractive) is:
$$ F_G = – G \frac {m_pm_e}{r^2} $$
where $m_p$ and $m_e$ are the masses of a proton and an electron respectively.
$$ \bigg| \frac {F_e}{F_G} \bigg| = \frac {e^2}{4 \pi \epsilon_o G m_p m_e} = 2.4 \times 10^{39}$$
(ii) On similar lines, the ratio of the magnitudes of electric force to the gravitational force between two protons at a distance r apart is:
$$ \bigg| \frac {F_e}{F_G} \bigg| = \frac {e^2}{4 \pi \epsilon_o G m_p m_e} = 1.3 \times 10^{36}$$
However, it may be mentioned here that the signs of the two forces are different. For two protons, the gravitational force is attractive in nature and the Coulomb force is repulsive. The actual values of these forces between two protons inside a nucleus (distance between two protons is ~ $10^{-15}$ m inside a nucleus) are Fe ~ 230 N whereas $F_G ~ 1.9 × 10^{–34}$ N. The (dimensionless) ratio of the two forces shows that electrical forces are enormously stronger than the gravitational forces. (b) The electric force F exerted by a proton on an electron is same in magnitude to the force exerted by an electron on a proton; however, the masses of an electron and a proton are different. Thus, the magnitude of force is
$$ | F | =\frac {1}{4 \pu \epsilon_o }{e^2}{r^2} = 8.987 \times 10^9 Nm^2/C^2 \times (1.6 \times 10^{-19} C)^2 / (10^{-10}m)^2= 2.3 \times 10^{-8} N$$
Using Newton’s second law of motion, F = ma, the acceleration that an electron will undergo is a = 2.3×10$^{–8}$ N / 9.11 × 10${–31}$ kg = 2.5 × $10^{22}$ m/$s^2$ Comparing this with the value of acceleration due to gravity, we can conclude that the effect of gravitational field is negligible on the motion of electron and it undergoes very large accelerations under the action of Coulomb force due to a proton. The value for acceleration of the proton is
$$ 2.3 × 10^{–8} N / 1.67 × 10^{–27} \space kg = 1.4 × 10^{19} \space m/s^2$$

Example 1.5 A charged metallic sphere A is suspended by a nylon thread. Another charged metallic sphere B held by an insulating handle is brought close to A such that the distance between their centres is 10 cm, as shown in Fig. 1.5(a). The resulting repulsion of A is noted (for example, by shining a beam of light and measuring the deflection of its shadow on a screen). Spheres A and B are touched by uncharged spheres C and D respectively, as shown in Fig. 1.5(b). C and D are then removed and B is brought closer to A to a distance of 5.0 cm between their centres, as shown in Fig. 1.5(c). What is the expected repulsion of A on the basis of Coulomb’s law? Spheres A and C and spheres B and D have identical sizes. Ignore the sizes of A and B in comparison to the separation between their centres.

Solution: Let the original charge on sphere A be q and that on B be q′. At a distance r between their centres, the magnitude of the electrostatic force on each is given by

$ F = \frac {1}{ 4 \pi \epsilon_o}{qq’}{r^2}$ neglecting the sizes of spheres A and B in comparison to r. When an identical but uncharged sphere C touches A, the charges redistribute on A and C and, by symmetry, each sphere carries a charge q/2. Similarly, after D touches B, the redistributed charge on each is q′/2. Now, if the separation between A and B is halved, the magnitude of the electrostatic force on each is

$ F’ = \frac {1}{4 \pi \epsilon_o} \frac {(q/2)(q’/2)}{(r/2)^2} = \frac {1}{4 \pi \epsilon_o} \frac {(qq’)}{r^2} = F $

Thus the electrostatic force on A, due to B, remains unaltered.

1.7 Forces between multiples charges

The mutual electric force between two charges is given by Coulomb’s law. Consider a system of n stationary charges $q_1, q_2, q_3, …, q_n$ in vacuum with position vectors $r_1, r_2, r_3..r_n$ respectively.

Thus, force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to a number of other charges, taken one at a time. The individual forces are unaffected due to the presence of other charges. This is termed as the principle of superposition.

The principle of superposition says that in a system of charges $q_1, q_2, …, q_n$, the force on $q_1$ due to $q_2$ is the same as given by Coulomb’s law, i.e., it is unaffected by the presence of the other charges $q_3, q_4$, …, $q_n$.
$$ F_{12} = \frac{1}{4 \pi \varepsilon0} \cdot \frac{|q_1 q_2|}{r_{12}^2} \cdot \hat{r}_{12} $$

In the same way, the force on q1 due to q3, denoted by F13, is given by
$$ F_{13} = \frac{1}{4 \pi \varepsilon 0} \cdot \frac{|q_1 q_3|}{r_{13}^2} \cdot \hat{r}_{13}$$

Thus the total force F1 on q1 due to the two charges q2 and q3 is given as:
$$F_1 = F_{12} + F_{13} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{|q_1 q_2|}{r_{12}^2} \cdot \hat{r}{12} + \frac{1}{4 \pi \varepsilon_0} \cdot \frac{|q_1 q_3|}{r{13}^2} \cdot \hat{r}_{13}$$

The above calculation of force can be generalised to a system of charges more than three by using the principle of superposition.
$$ F_1=F_{12}+F_{13}+⋯.+F_{1n}= \frac { 1}{4πε0}.[\frac {|q_1.q_2 |}{r{12}^2}.\hat r_{12}+ .\frac { |q_1.q_3 |}{r_{13}^2}.\hat r_{13}+⋯…+ \frac { |q_1.q_m |}{r^2_{1n} }.\hat r_{1n}] $$
$$ F_1= \frac {q_n}{4πε0} ∑{i=2}^n \frac {q_i}{r_{1i}^2 }.\hat r_{1i} $$

The vector sum is obtained as usual by the parallelogram law of addition of vectors. All of electrostatics is basically a consequence of Coulomb’s law and the superposition principle.

Example 1.6 Consider three charges q1, q2, q3 each equal to q at the vertices of an equilateral triangle of side l. What is the force on a charge Q (with the same sign as q) placed at the centroid of the triangle, as shown in Fig. 1.7?

Solution:

In the given equilateral triangle ABC of sides of length l, if we draw a perpendicular AD to the side BC, AD = AC cos 30º = ( $ \sqrt 3 $/2 ) l and the distance AO of the centroid O from A is (2/3) AD = (1/ $ \sqrt 3 $ ) l. By symmatry AO = BO = CO.

Thus, Force $F_1$ on Q due to charge q at A = $ \frac{3 Q q}{4 \pi \varepsilon_0}{Qq}{l^2}$ along AO

Force $F_2$ on Q due to charge q at B = $ \frac {3}{4 \pi \varepsilon_o}{Qq}{l^2}$ along BO

Force $F_3$ on Q due to charge q at C = $ \frac {3}{4 \pi \varepsilon_o}{Qq}{l^2}$ along CO

The resultant of forces $F_2$ and $F_3$ is $\frac {3}{4 \pi \epsilon_o}{Qq}{l^2} $ along OA, by the parallelogram law. Therefore, the total force on Q = $\frac {3}{4 \pi \epsilon_o}{Qq}{l^2} (\hat r – \hat r) = 0$, where $ \hat r $ is the unit vector along OA. It is clear also by symmetry that the three forces will sum to zero. Suppose that the resultant force was non-zero but in some direction. Consider what would happen if the system was rotated through 60º about O.

Example 1.7 Consider the charges q, q, and –q placed at the vertices of an equilateral triangle, as shown in Fig. 1.8. What is the force on each charge?

Solution:

The forces acting on charge q at A due to charges q at B and –q at C are F12 along BA and F13 along AC respectively, as shown in Fig. 1.8. By the parallelogram law, the total force F1 on the charge q at A is given by
$F_1 = F \hat r_1 $ where $ \hat r_1 $ is a unit vector along BC.

The force of attraction or repulsion for each pair of charges has the same magnitude $ F = \frac {q^2}{4 \pi \epsilon_o l^2}$

The total force $F_2$ on charge q at B is thus $F_2 = F \hat r_2$, where $ \hat r_2$, is a unit vector along AC.

Similarly the total force on charge –q at C is $F_3$ = $ \sqrt 3 F \hat n $ , where $ \hat n $ is the unit vector along the direction bisecting the ∠BCA. It is interesting to see that the sum of the forces on the three charges is zero, i.e.,
$$F_1 + F_2 + F_3 = 0$$

The result is not at all surprising. It follows straight from the fact that Coulomb’s law is consistent with Newton’s third law. The proof is left to you as an exercise.

1.8 Electric Field

Recap: We know that if two-point charges are placed within a certain distance (say ‘r’), then there is a force between them in accordance with Coulomb’s law.

Going further, every charge produces an electric field. An electric field can be thought to be the area around a charge where it has influence over other charges. When another charge q is brought at some point P, the field there acts on it and produces a force. The electric field produced by the charge Q at a point r is given as:
$$ E(r ) = \frac {1}{4 \pi \epsilon_o } \frac {Q}{r^2} \hat r = \frac {1}{4 \pi \epsilon} \frac {Q}{r^2} \hat r $$

Now, considering the effect of the force exerted by a charge Q on another charge q, we can write:

$$ F = \frac {1}{4 \pi \epsilon_o } \frac {Qq}{r^2} \hat r $$

Comparing the above equations, it is evident that:

$F(r) = q.E(r)$

Note box: It is not just the charge Q that exerts a force on q. Even the charge q exerts an equal and opposite force on the charge Q.

Now, if q = 1C in the above equation, we have Force exerted by the charge Q on q equal to the Electric field of Charge Q. We can derive the definition of Electric field from this as follows:

Definition box: “The electric field due to a charge at a point in space may be defined as the force that a unit positive charge would experience if placed at that point.”

Because we are using an external charge to understand the concept of electric field of a charge Q, we need to take into consideration the effect of electric field of the charge q also. We need to make the effect of electric field of charge q as small as possible so that it does not affect the measurement of Electric field of charge Q. This is done by using a very small value of charge (smallest value of charge possible is the unit charge: 1.602 x $10^{-19}$ C). We can then take limits as follows:

$$ \vec E = \lim_{q→0} \bigg( \frac {F}{q} \bigg) $$

As q tends to zero, the electric field of the test charge q will tend to zero hence causing minimum distortion in measure the electric field strength.

Recap:

1) The space surrounding a charge in which another charge comes under the influence of the said charge and experiences an electrical force is known as the Electric field of the charge.

2) Coulomb’s force between two point charges can be viewed as a result of the interaction between the electric fields of two charges.

Questions from sections 1.7 and 1.8:

1. Obtain the expression for force when n stationary charges interact in vacuum, in the form:
$F_1 = \frac{q_n}{4 \pi \varepsilon_0} \sum_{i=2}^{n} \frac{q_i}{r_{1i}^2} \cdot \hat{r}_{1i}$

2. Define Electric field due to a charge.

3. What is value of the smallest value of charge possible?

Example 1.8 An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude 2.0 × 104 N $C^{–1}$ [Fig. 1.10(a)]. The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance [Fig. 1.10(b)]. Compute the time of fall in each case. Contrast the situation with that of ‘free fall under gravity’.

Solution:

In Fig. 1.10(a) the field is upward, so the negatively charged electron experiences a downward force of magnitude eE where E is the magnitude of the electric field. The acceleration of the electron is $ a_e = eE/m_e$ where me is the mass of the electron.

Starting from rest, the time required by the electron to fall through a distance h is given by
$$ t_e = \sqrt { \frac {2h}{a_e} } = \sqrt { \frac {2hm_e}{eE} }$$

For e = 1.6 × $10^{–19}$C, $m_e$ = 9.11 × $10^{–31}$ kg, E = 2.0 × $10^4$ N $C^{–1}$, h = 1.5 × 10$^{–2}$ m, $t_e$ = 2.9 × $10^{–9}$s

In Fig. 1.10 (b), the field is downward, and the positively charged proton experiences a downward force of magnitude eE. The acceleration of the proton is $ a_p = eE/m_p$ where $m_p$ is the mass of the proton; $m_p$ = 1.67 × $10^{–27}$ kg. The time of fall for the proton is
$$ t_p = \sqrt { \frac {2h}{a_p}} = \sqrt { \frac {2h m_p}{eE}} = 1.3 \times 10^{-7} s$$

Thus, the heavier particle (proton) takes a greater time to fall through the same distance. This is in basic contrast to the situation of ‘free fall under gravity’ where the time of fall is independent of the mass of the body. Note that in this example we have ignored the acceleration due to gravity in calculating the time of fall. To see if this is justified, let us calculate the acceleration of the proton in the given electric field:

$$a_p = \frac{e E}{m_p}$$

$$ a_p = \frac{(1.6 \times 10^{-19} \, C) \times (2.0 \times 10^4 \, N\,C^{-1})}{1.67 \times 10^{-27} \, kg}$$

$$ = 1.9 \times 10^{12} \, m\,s^{-2} $$

which is enormous compared to the value of g (9.8 m s–2), the acceleration due to gravity. The acceleration of the electron is even greater. Thus, the effect of acceleration due to gravity can be ignored in this example.

Example 1.9 Two-point charges $q_1$ and $q_2$, of magnitude $+10^{–8}$ C and –$10^{–8}$ C, respectively, are placed 0.1 m apart. Calculate the electric fields at points A, B and C shown in Fig. 1.10

Solution:

The electric field vector $E_{1A}$ at A due to the positive charge q1 points towards the right and has a magnitude

$$ E_{1A} = \frac {( 9 \times 10^9 Nm^2C^{-2}) \times (10^{-8}C)}{(0.05 m^2)}= 3.6 \times 10^4 \space N C^{-1}$$

The electric field vector $E_{2A}$ at A due to the negative charge q2 points towards the right and has the same magnitude. Hence the magnitude of the total electric field $E_A$ at A is $ E_{1A} = \frac {(9 \times 10^9Nm^2C^{-2}) \times (10^{-8}C)}{(0.05m^2)}= 3.6 \times 10^4 \space NC^{-1}$

$E_A$ is directed toward the right.

The electric field vector $E_{1B}$ at B due to the positive charge q1 points towards the left and has a magnitude
$$ E_{1B} = \frac {( 9 \times 10^9 Nm^2C^{-2})\times (10^{-8}C)}{(0.05m)^2} = 3.6 \times 10^4 NC^{-1}$$

The electric field vector $E_{2B}$ at B due to the negative charge $q_2$ points towards the right and has a magnitude
$$ E_{2B} = \frac {( 9 \times 10^9 Nm^2C^{-2})\times (10^{-8}C)}{(0.15m)^2} = 4 \times 10^3 NC^{-1}$$

The magnitude of the total electric field at B is
$$ E_B = E_{1B} – E_{2B} = 3.2 × 10^4 N C^{–1}$$
$ E_B$ is directed towards the left. The magnitude of each electric field vector at point C, due to charge $q_1$ and $q_2$ is
$$ E_{1C} =E_{2C}= \frac {( 9 \times 10^9 Nm^2C^{-2})\times (10^{-8}C)}{(0.10m)^2} = 9 \times 10^3 NC^{-1}$$

The directions in which these two vectors point are indicated in Fig. 1.11. The resultant of these two vectors is
$$ E_C = E_1 cos \frac {\pi}{3} + E_2 cos \frac {\pi}{3} = 9 \times 10^3 NC^{-1}$$
$E_C$ points towards the right.

1.9 Electric field Strength or Electric Intensity

The strength of the Electric field is also known as Electric Intensity.

Definition box:   The force experienced by a unit positive charge placed at any point in an electric field is known Electrical Intensity.

Electrical intensity is different at different points in an Electric field.

Electrical intensity is denoted by $ \vec E $ as it is a vector quantity.

The SI unit is: $NC^{-1}$ or $Vm^{-1}$

If $ \vec F $ is the force on a test charge q at some point in an electric field, then Electrical intensity $ \vec E $ can be written as:

$$ \vec E= \frac {\vec F}{q} $$

The direction of Electrical Intensity $ \vec E $ is same as the direction of force $ \vec E $

Note box:   The measured value of Electrical Intensity will be less than the actual value of Electrical Intensity for a positive charge. For a negative charge, the measured value of Electrical Intensity will be greater than the actual value of Electrical Intensity.

Electrical Intensity due to a point charge

We have a charge Q (Source charge) and we have another point charge qo (Test chare). They are separated by a distance ‘r’.
Now, applying Coulomb’s law, we get the Electrical Force acting on the test charge due to the source charge as:

$$ F= \frac {1}{4πε_0}. \frac { |q.q_o |}{r^2} $$

Rearranging this equation,

$$ F/q_o= \frac {1}{4πε_0 }. \frac { |q|}{r^2} $$

Recap: We know that Electrical intensity is:

$$ F/q_o= \frac {1}{4πε_0 }. \frac { |q|}{r^2} $$

Substituting this:

$$ E = \frac {1}{4πε_0 }. \frac { |q|}{r^2} $$

Now this force is along OP as shown in the diagram as the charge is positive.
If the charge would have been negative, the magnitude of the Electrical Intensity would have been but the direction would have been along PO.

Note: A point where electric field intensity within an electric field is zero is known as a neutral point.

Electric field due to a system of charges

Figure 1.12 Electric field at a point due to a system of charges is the vector sum of the electric fields at the point due to individual charges.

Definition box:   Electric field at a point due to a system of charges is the vector sum of the electric fields at that point due to individual charges.

Consider a system of charges $q_1, q_2… q_n$ with position vectors $r_1, r_2… r_n$.
Electric field $E_1$ at r due to $q_1$ at $r_1$ is given by:
$$ E_1 = \frac {1}{4 \pi \epsilon_o} \frac {q_1}{r^2_{1P}} \hat r_{1P}$$
where $ \hat r_{1P} $ is a unit vector in the direction from $q_1$ to P,
and $r_{1p}$ is the distance between $q_1$ and P.
In the same manner, electric field $E_2$ at r due to $q_2$ at $r_2$ is:
$$ E_2 = \frac {1}{4 \pi \epsilon_o} \frac {q_2}{r^2_{2P}} \hat r_{2P}$$
where $ \hat r_{2P} $ is a unit vector in the direction from $q_2$ to P,
and $r_{2p}$ is the distance between $q_2$ and P.

For the $n^{th}$ charge:

Recap: Principle of Superposition
According to the principle of superposition, the net electric field at a point due to a system of charges is equal to the vector sum of all the electric fields produced by the individual charges at the point.
$$ \ce { E(r) = E1 (r) + E2 (r) + … + E_n(r) } $$
$$= \frac {1}{4 \pi \epsilon_o } \frac {q_1}{r^2_{1P}} \hat r_{1P} + \frac {1}{4 \pi \epsilon_o} \frac {q_2}{r^2_{2P}} \hat r_{2P}+……..+\frac {1}{4 \pi \epsilon_o} \frac {q_n}{r^2_{nP}} \hat r_{nP}$$
$$ E(r)= \frac {1}{4 \pi \epsilon_o} \Sigma_{i=1}^{n} \frac {q_i}{r^2_{iP}} \hat r_{iP}$$

Physical Significance of Electric field

The term field in physics generally refers to a quantity that is defined at every point in space and may vary from point to point. It was Michael Faraday who introduced the concept of Electric field.

Recap: Electric field is a property of a charge (or a system of charges). The test charge has no effect on the Electric field of a charge (or system of charges).

The true physical significance of the concept of electric field emerges when we go beyond electrostatics and into (time-dependent) electromagnetic phenomena.

Let us understand how:
Now we have two charges $q_1$ and $q_2$. These are not static charges (All charges considered till now and for the rest of this chapter are all static charges). These are charges in motion.

Now, a motion of $q_1$ will produce a force on $q_2$ and vice-versa. However, the motion of q1 will not immediately produce a force on $q_2$. There is a time delay. It is the concept of Electric field that accounts for this delay. How?

“The accelerated motion of charge $q_1$ produces electromagnetic waves, which then propagate with the speed c, reach $q_2$ and cause a force on $q_2$.”

Speed ‘c’ is the speed of light. The maximum speed at which any information, signal or energy can be transferred form one point to another is at the speed of light. Hence it has been considered to demonstrate that even at the maximum speed possible there is going to be a delay; and that we have to account for it.

Hence, this time delay taken for the force to be experienced by $q_2$ is accounted for by the concept of electric field.

Questions from section 1.9:

1. Define electric intensity. Give its vector notation.
2. Give the expression for Electrical force acting on the test charge due to the source charge.
3. Define Electric field at a point due to a system of charges.
4. Explain the physical significance of electric field.

1.10 Electric field lines

Electric field needs to be represented. This representation is done using lines and these lines are known as Electric field lines.

Electric field lines represent the electric field due to a charge. They are only a visual representation tool. They can be thought of as imaginary lines along which a unit positive charge moves or tends to move.

Field of a point charge

Electric field lines are as shown in the figure. They are vector representations. Here, a positive charge is considered and hence you can see the lines of force radiating outwards. (This has already been discussed previously). The magnitude of the field is represented by the density of field lines. E is strong near the charge, so the density of field lines is more near the charge and the lines are closer. Away from the charge, the field gets weaker and the density of field lines is less, resulting in well-separated lines. The number of lines is not important. In fact, an infinite number of lines can be drawn in any region. It is the relative density of lines in different regions that is important.

Did you know?   The visual representation of field lines was also invented by Faraday to develop a non- mathematical way of visualizing electric fields around charged configurations. Faraday called them lines of force. This term is somewhat misleading, especially in case of magnetic fields. The more appropriate term is field lines (electric or magnetic).

Note box:   Electric Intensity is greater near the charge and decreases with distance

Properties of Electric field lines

1) Field lines start from positive charges and end at negative charges. If there is a single charge, they may start or end at infinity.

2) In a charge-free region, electric field lines can be taken to be continuous curves without any breaks.

3) 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. However two tangents can be drawn from the point of intersection which would imply that the Electric field vector has two directions at the point. This would still not be true since the direction of the Electric field vector is always unique.)

4) Electrostatic field lines do not form any closed loops. This follows from the conservative nature of electric field.

5) The Electric field vector, $ \vec E $is tangential to a line of force at each point.

6) Electric field lines are denser in areas where the field in stronger and are less dense and spread out when the field is weaker.

7) The electric field lines do not pass through a conductor. Hence the electric field inside a conductor is always zero. They pass through non-conductors.

8) Electric field lines cannot be closed loops.

9) In a uniform electric field, electric field lines are equidistant and parallel to one another.

The Area Vector, $ \vec { ds}$

The very notion of area being a vector can be confusing as Area is a scalar quantity. Area continues to be a scalar quantity. It does not change into a vector. It is only treated as a vector to help us work with certain problems in electrostatics and electromagnetism. However, at the end, treating area as a vector does not distort the results derived.

From a surface S, consider a small element, ds. The vector representation would be $ \vec { ds}$.The magnitude of the vector is represented by ds. Its direction is perpendicular to the area element and is directed outwards.

$ = \vec {ds} = \vec n.ds$
$ \vec n $– unit vector taken perpendicular to the surface in an outward direction.

Electric Flux

Definition box:   Electric Flux of a closed surface is defined as the total number of electric field lines passing through the surface along the normal.

Note box:   The electric flux through a closed surface in an electric field can also be defined mathematically. Electric flux is nothing but the electric field lines passing over a given surface. i.e $$ i.e \space φ= \int dφ= ∫ \vec E.\vec {ds} =E.ds.cosθ $$ Explanation   A surface is subjected to an Electric field as shown in the diagram. Consider a small element area,  $ \vec {ds}$ The flux through tis elementary surface area is now given by: Δφ = E. ΔS Cosθ Θ – is the angle between the surface and the direction of Electric field. Flux is given by: Now this result can be applied to the entire surface which shall be split into a number of elemental surfaces. Hence, Φ = E1ds1 Cosθ1  + E2ds2 Cosθ2   +……..+ EndsnCosθn   $$ \ce { Φ = E1ds1 Cosθ1  + E2ds2 Cosθ2   +……..+ E_nds_nCosθ_n    } $$ or Φ ≈ ∑ E. ΔS Cos θ This entity can be considered to be fairly constant as the Electric field over the small area is considered to be constant.

Questions from section 1.10:

1. What are electric field lines? Represent the field lines emerging from a positive charge.
2. List the properties of Electric field lines
3. Define electric flux. Give the expression for electric flux.

1.11 Electric Dipole

Definition box: An electric dipole is a pair of equal and opposite point charges q and –q, separated by a distance 2a. The line connecting the two charges defines a direction in space.

By convention, the direction from –q to q is said to be the direction 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 obviously zero. This does not mean that the field of the electric dipole is zero. Since the charge q and –q are separated by some distance, the electric fields due to them, when added, do not exactly cancel out. However, at distances much larger than the separation of the two charges forming a dipole (r >> 2a), the fields due to q and –q nearly cancel out.

The field of an electric dipole

The electric field of the pair of charges (–q and q) at any point in space can be found out from Coulomb’s law and the superposition principle. The results are simple for the following two cases:

(i) when the point is on the dipole axis,

(ii) when it is in the equatorial plane of the dipole, i.e., on a plane perpendicular to the dipole axis through its centre.

The electric field at any general point P is obtained by adding the electric fields $E_{–q }$ due to the charge –q and $E_{+q}$ due to the charge q, by the parallelogram law of vectors.

1) For points on the axis
$$ E_{-q} = – \frac {q}{4 \pi \epsilon_o(r+a)^2} \hat p$$

Where $ \cap p $is the unit vector along the dipole axis (from –q to q). Also

$$ E_{+q} = – \frac {q}{4 \pi \epsilon_o(r-a)^2} \hat p$$

The total field at P is:

$$ E = E_{+q} + E_{-q} = \frac {q}{4 \pi \epsilon_o } \bigg [ \frac {1}{(r-a)^2} – \frac {1}{(r+a)^2} \bigg] \hat p $$
$$ = \frac {q}{4 \pi \epsilon_o} \frac {4ar}{ (r^2-a^2)^2 } \hat p$$
For r >>a
$$ E = \frac {4qa}{4 \pi \epsilon_o r^3 } \hat p $$ (r >>a)
2) For points on the equatorial plane
The magnitude of the electric fields due to the two charges +q and –q are given by:
$$ E_{+q} = \frac {q}{4 \pi \epsilon_o} \frac {1}{r^2+a^2} $$
$$ E_{-q} = \frac {q}{4 \pi \epsilon_o} \frac {1}{r^2+a^2} $$
$$E = -(E_{+q} + E_{-q}) cos \theta \space \hat p$$
$$ =- \frac {2qa}{4 \pi \epsilon_o (r^2 +a^2)^{3/2}} \hat p$$
At large distance (r>>a),this reduces to
$$ E = – \frac {2qa}{4 \pi \epsilon_o r^3} \hat p $$ (r>>a)
At a point on the dipole axis:
$$ E = \frac {2p}{4 \pi \epsilon_o r^3} \hat p $$ (r>>a)
At a point on the equatorial plane:
$$ E = \frac {p}{4 \pi \epsilon_o r^3} \hat p $$ (r>>a)

Note box:   1) The electric field due to a dipole for a single charge decreases at a rate of $1/r^2$. At large distance, the electric field due to dipole decreases as $1/r^3$. The magnitude and the direction of the dipole field depend not only on the distance r but also on the angle between the position vector r and the dipole moment p.   2) When the dipole size 2a approaches zero, the charge q approaches infinity in such a way that the product p = q × 2a is finite. Such a dipole is referred to as a point dipole.

Physical significance of dipoles

In most molecules, the centres of positive charges and of negative charges lie at the same place. Therefore, their dipole moment is zero. Such molecules are called non-polar molecules.

Ex: $CO_2, CH_4$

They develop a dipole moment when an electric field is applied. But in some molecules, the centres of negative charges and of positive charges do not coincide. Therefore, they have a permanent electric dipole moment, even in the absence of an electric field. Such molecules are called polar molecules.

Ex: $H_2O, NH_3$, HCL, CO

Dipole in a uniform electric field

Fig 1.14, insert title: Electric force on a dipole: (a) E parallel to p, (b) E antiparallel to p.

Consider a permanent dipole of dipole moment p in a uniform external field E. A permanent dipole is one in which p exists irrespective of E; it has not been induced by E.

Force in q = q.E

Force on –q = -q.E

Since it is a uniform electric field (E), the forces cancel each other out and the net force on the dipole is zero.

However, this does not mean that there are no forces acting on the dipole. It is only the net force that is zero. Forces, however, continue to act at different points in space. This creates a torque on the dipole.
If a net force exists, then the torque is dependent on the origin. However, in this case, as the net force is zero, the torque is independent of the origin.

Recap: Torque(τ) = F.d

The magnitude of this torque, hence, will be a product of the magnitude of each force and the arm of the couple; because the points are located in 3d space and are at an angle with respect to the dipole, this distance can be written as 2asinθ.

From the triangle AOB,

Sinθ = BC / AB

BC = Sinθ. AB

Magnitude of the torque (τ) = F.d
= (qE)(2aSinθ)
= 2qaE.Sinθ

Its direction is normal to the plane of the paper, coming out of it.

τ = PE. Sinθ
because, (P = q. 2a)

This torque will tend to align the dipole with the field E. When p is aligned with E, the torque is zero.

Note box:   1) Torque is maximum when θ=$90^0$­ (τmax = P.E) 2) Torque is minimum when θ=$0^0$­ or θ=$180^0$­ 3) The dipole moment is in stable equilibrium when θ=$0^0$­ 4) The dipole moment is in stable equilibrium when θ=$180^0$­

In a non-uniform electric field

The net force now will obviously have some value. Now, a torque also exists on the system. However, the net torque on the system is zero.

Looking at the figures above, we have two cases:

a) Dipole moment p is parallel to Electric field E.
Net force in the direction of increasing Electric field.

b) Dipole moment p is anti-parallel to Electric field E.
Net force in the opposite direction of increasing Electric field.

Questions from section 1.11:

1. Give the expression for electric fields due to charges, +q and –q acting at a point P on:

a) the axis

b) the equatorial plane

2. What are polar and non-polar molecules?

3. Explain what a permanent dipole is.

Example 1.10 Two charges ±10 μC are placed 5.0 mm apart. Determine the electric field at (a) a point P on the axis of the dipole 15 cm away from its centre O on the side of the positive charge, as shown in Fig. 1.21(a), and (b) a point Q, 15 cm away from O on a line passing through O and normal to the axis of the dipole, as shown in Fig. 1.16(b).

Solution:

(a) Field at P due to charge +10 μC
$$ = \frac {10^{-5}}{4 \pi (8.854 \times 10^{-12} C^2N^{-1}m^{-2}}\times \frac {1}{(15-0.25)^2 \times 10^{-4}m^2}$$

= 4.13 × $10^6$ N $C^{–1}$ along BP

Field at P due to charge –10 μC

$$ = \frac {10^{-5}}{4 \pi (8.854 \times 10^{-12} C^2N^{-1}m^{-2}}\times \frac {1}{(15+0.25)^2 \times 10^{-4}m^2}$$

= 3.86 × $10^6$ N $C^{–1}$ along PA The resultant electric field at P due to the two charges at A and B is = 2.7 × $10^5$ N $C^{–1}$ along BP

In this example, the ratio OP/OB is quite large (= 60). Thus, we can expect to get approximately the same result as above by directly using the formula for electric field at a far-away point on the axis of a dipole. For a dipole consisting of charges ± q, 2a distance apart, the electric field at a distance r from the centre on the axis of the dipole has a magnitude $ E = \frac {2p}{4 \pi \epsilon_o r^3} (r/a >> 1)$

where p = 2a q is the magnitude of the dipole moment. The direction of electric field on the dipole axis is always along the direction of the dipole moment vector (i.e., from –q to q). Here, p =$10^{–5}$ C × 5 × $10^{–3}$ m = 5 × $10^{–8}$ Cm

$$ E = \frac {2 \times 5 \times 10^{-8} Cm}{4 \pi (8.854 \times 10^{-12} C^2N^{-1}m^{-2}}\times \frac {1}{(15)^3 \times 10^{-6}m^3} = 2.6 \times 10^5 NC^{-1}$$

along the dipole moment direction AB, which is close to the result obtained earlier.

(b) Field at Q due to charge + 10 μC at B

(c) $E = \frac {10^{-5} C}{4 \pi (8.854 \times 10^{-12} C^2N^{-1}m^{-2}}\times \frac {1}{15^2 + (0.25)^2 \times 10^{-4}m^2} = 3.99 \times 10^6 NC^{-1} $ Along BQ

Field at Q due to charge –10 μC at A

${1}{15^2 + (0.25)^2 \times 10^{-4}m^2} = 3.99 \times 10^6 NC^{-1} $ Along QA

Clearly, the components of these two forces with equal magnitudes cancel along the direction OQ but add up along the direction parallel to BA.

Therefore, the resultant electric field at Q due to the two charges at A and B is

$ = 2 \times \frac {0.25}{\sqrt {15^2+ (0.25)^2}} \times 3.99 \times 10^6 N \space C^{-1}$ along BA

$= 1.33 \times 10^5 NC^{-1}$ Along BA

As in (a), we can expect to get approximately the same result by directly using the formula for dipole field at a point on the normal to the axis of the dipole:

$$ E = \frac {p}{4 \pi \epsilon_o r^3} \space (r/a >> 1) $$

$$ E = \frac {5 \times 10^{-8} Cm}{4 \pi (8.854 \times 10^{-12} C^2N^{-1}m^{-2}} \times \frac {1}{(15)^3 \times 10^{-6}m^3} = 1.33 \times 10^5 NC^{-1}$$

The direction of electric field in this case is opposite to the direction of the dipole moment vector. Again the result agrees with that obtained before.

1.12 Continuous charge distribution

Considering charge distribution as a discrete element and not as continuous helped greatly in mathematical calculations. Charge distribution is termed discrete because charge is always considered to exist as an integral number of the basic charge (e).

However, it is not always accurate and practical to work in terms of discrete charges. For some problems and applications, we need to look at charge distribution as continuous. For example, on the surface of a charged conductor, it is impractical to specify the charge distribution in terms of the locations of the microscopic charged constituents. It is even better to consider an area element, ΔS and specify the charge Δq on that element.

This are ΔS is large enough to include a large number of electrons but is small on a macroscopic scale. Surface density charge is then defined as:

$$ σ= \frac {Δq}{Δs} Cm^{-2}$$

The surface density at a point is the charge per unit area around that point.

Now, the linear charge density is the charge per unit length.

$$ λ= \frac {Δq}{ΔL} Cm^{-1} $$

where ΔL is a small line element of wire.

The volume charge density (ρ) is the charge per unit volume.

$$ ρ= \frac {Δq}{ΔV} Cm^{-3}$$

where ΔV is a small line element of wire.

The electric field due to a continuous charge distribution can be found as follows:

Consider a continuous charge distribution in space which has volume charge density ρ

$$ ∆E= \frac{1}{4πε0}.ρ \frac{ΔV}{r^2} .\hat r’ $$

$$ E= \frac{1}{4πε_0}.∑{all ∆V} ρ \frac{ΔV}{r^2} .\hat r’ $$

Questions from section 1.12:

1. Why is charge distribution said to be discrete?

2. Define and give the mathematical expressions of:

a) Surface density

b) Linear charge density

c) Volume charge density.

3. Obtain the expression for electric field due to a continuous charge distribution.

1.13 Gauss’s Theorem

As a simple application of the notion of electric flux, 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, as shown in the Figure.

Flux through a sphere enclosing a point charge q at its centre

The flux through an area element ΔS is:

$$ Δϕ=E.ΔS= \frac {q}{4πε_0 r^2}.r ̂ΔS $$

where we have used Coulomb’s law for the electric field due to a single charge q. The unit vector $ \vec r$ is along the radius vector from the centre to the area element. Now, since the normal to a sphere at every point is along the radius vector at that point, the area element ΔS and have the same direction. Therefore,

$$ Δϕ= \frac {q}{4πε_0 r^2}.ΔS $$
since the magnitude of a unit vector is 1.

The total flux through the sphere is obtained by adding up flux through all the different area elements:

$$ ϕ=∑_{dl ds} \frac {q}{4πε_0 r^2}.ΔS $$
Since each area element of the sphere is at the same distance r from the charge,

$$ ϕ= \frac {q}{4πε0 r^2 } ∑{dl ds}ΔS $$

Now, S, the total area of the sphere, equals $4πr^2$. Thus,

$$ ϕ= \frac {q}{4πε_0 r^2 }.4πr^2 $$

The above equation is a simple illustration of a general result of electrostatics called Gauss’s law.

We state Gauss’s law without proof:

“Electric flux through a closed surface S is: “$ϕ= \frac {q}{ε_0} $

q = total charge enclosed by S.

The law implies that the total electric flux through a closed surface is zero if no charge is enclosed by the surface.
Here the electric field is uniform and we are considering a closed cylindrical surface, with its axis parallel to the uniform field 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 E, so by definition of flux, $φ_3$ = 0. Further, the outward normal to 2 is along E while the outward normal to 1 is opposite to E.

Therefore,

If q=0, then

$$ϕ_1=-E.S_1$$

$$ϕ_2=-E.S_2$$

$$ S_1=S_2=S$$

Where S is the area of circular cross-section.

Example 1.11 The electric field components in Fig. 1.18 are $E_x = αx^{1/2}$, $E_y = E_z$ = 0, in which α = 800 N/C $m^{1/2}$. Calculate

(a) the flux through the cube, and (b) the charge within the cube. Assume that a = 0.1 m

Solution:

Solution (a) Since the electric field has only an x component, for faces perpendicular to x direction, the angle between E and ΔS is ± π/2. Therefore, the flux φ = E.ΔS is separately zero for each face of the cube except the two shaded ones. Now the magnitude of the electric field at the left face is
$$E_L = αx^{1/2} = αa^{1/2}$$
(x = a at the left face).
The magnitude of electric field at the right face is
$$ E_R = α x^{1/2} = α (2a)^ ½ $$
(x = 2a at the right face).
The corresponding fluxes are
$$ φ_L= E_L .ΔS = ΔS E_L ⋅ \hat n=E_L ΔS cosθ = –E_L ΔS, since θ = 180° $$
$$ = –Ela^2 $$
$$φ_R= E_R .ΔS = E_R ΔS cosθ = E_R ΔS,$$ since θ = 0°
$$ = E_Ra^2 $$
Net flux through the cube
$$ = φ_R + φ_L = E_Ra^2 – E_La^2 = a^2 (E_R – E_L) = αa^2 [(2a)^ 1/2 – a^1/2]$$
$$ = αa^{5/2} (\sqrt 2 – 1)$$
$$ = 800 (0.1)^{5/2} (\sqrt 2–1)$$
$$ = 1.05 N m^2 C^{-1}$$
(b) We can use Gauss’s law to find the total charge q inside the cube. We have φ = q/$ε_0$ or q = φ$ε_0$. Therefore,
$$q = 1.05 × 8.854 × 10^{–12} C = 9.27 × 10^{–12} C.$$

Example 1.12 An electric field is uniform, and in the positive x direction for positive x, and uniform with the same magnitude but in the negative x direction for negative x. It is given that E = 200 $ \hat i$ N/C for x > 0 and E = –200 $ \hat I $ N/C for x < 0. A right circular cylinder of length 20 cm and radius 5 cm has its centre at the origin and its axis along the x-axis so that one face is at x = +10 cm and the other is at x = –10 cm (Fig. 1.28).

(a) What is the net outward flux through each flat face?

(b) What is the flux through the side of the cylinder?

(c) What is the net outward flux through the cylinder?

(d) What is the net charge inside the cylinder?

Solution:

(a) We can see from the figure that on the left face E and ΔS are parallel. Therefore, the outward flux is
$$ φ_L= E.ΔS = – 200 \hat i . S $$
$$ = + 200 ΔS, since $ \hat i $ . S iΔ = – ΔS
= + 200 × π $(0.05)^2$ = + 1.57 N $m^2 C^{–1}$
On the right face, E and ΔS are parallel and therefore
$$ φ_R = E.ΔS = + 1.57 N m^2 C^{–1 }$$
(b) For any point on the side of the cylinder E is perpendicular to ΔS and hence E.ΔS = 0. Therefore, the flux out of the side of the cylinder is zero.
(c) Net outward flux through the cylinder φ = 1.57 + 1.57 + 0 = 3.14 N $m^2$ $C^{-1}$

(d) The net charge within the cylinder can be found by using Gauss’s law which gives
$$ q = ε_0φ$$
$$ = 3.14 × 8.854 × 10^{–12} C $$
$$ = 2.78 × 10^{–11} C$$

Applications of Gauss’s Law

Field due to an infinitely long straight uniformly charged wire

Flux through the Gaussian surface = flux through the curved cylindrical part of the surface
= E × 2πrl
The surface includes charge equal to λ l. Gauss’s law then gives
$ E × 2πrl = λl/ε_0$
$$ E= \frac {λ}{2πε_o r}$$
$$ E= \frac {λ}{2πε_o r}.\hat n$$
Field due to a uniformly charged infinite plane sheet
$$ 2 EA = σA/ε_0$$
$$ or, E = σ/2ε_0$$
Vectorically,
$$ E= \frac {σ}{2ε_o}.n ̂$$
Field due to a uniformly charged thin spherical shell
$$ E \times 4πr^2= \frac {σ}{ε_o} .4πr^2 $$
$$ E= \frac {σr^2}{ε_o r^2}= \frac {q}{4πε_o r^2} $$
$$ E= \frac {q}{4πε_o r^2 }.\hat r ̂$$

Field inside the shell

The flux through the Gaussian surface, calculated as before, is E × $4πr^2$. However, in this case, the Gaussian surface encloses no charge. Gauss’s law then gives
E × $4πr^2$= 0
i.e., E = 0 (r < R )

Example 1.13 An early model for an atom considered it to have a positively charged point nucleus of charge Ze, surrounded by a uniform density of negative charge up to a radius R. The atom as a whole is neutral. For this model, what is the electric field at a distance r from the nucleus?

The charge distribution for this model of the atom is as shown in Fig. 1.20. The total negative charge in the uniform spherical charge distribution of radius R must be –Z e, since the atom (nucleus of charge Z e + negative charge) is neutral. This immediately gives us the negative charge density ρ, since we must have
$$ \frac {4 \pi R^3}{3} \rho = 0 -Ze $$
$$ or \rho = \frac {3Ze}{4 \pi R^3}$$

To find the electric field E(r) at a point P which is a distance r away from the nucleus, we use Gauss’s law. Because of the spherical symmetry of the charge distribution, the magnitude of the electric field E(r) depends only on the radial distance, no matter what the direction of r. Its direction is along (or opposite to) the radius vector r from the origin to the point P. The obvious Gaussian surface is a spherical surface centred at the nucleus. We consider two situations, namely, r < R and r > R

(i) r < R : The electric flux φ enclosed by the spherical surface is φ = E (r ) × 4 π $r^ 2$ where E (r ) is the magnitude of the electric field at r. This is because the field at any point on the spherical Gaussian surface has the same direction as the normal to the surface there, and has the same magnitude at all points on the surface. The charge q enclosed by the Gaussian surface is the positive nuclear charge and the negative charge within the sphere of radius r,

$$ i.e., q = Ze + \frac {4 \pi r^3}{3} \rho $$

Substituting for the charge density ρ obtained earlier, we have $ q = Ze – Ze \frac {r^3}{R^3}$ Gauss’s law then gives

$$E(r) = \frac {Ze}{4 \pi \epsilon_o } \bigg( \frac {1}{r^2}- \frac {r}{R^3} \bigg); r < r $$

The electric field is directed radially outward.

(ii) r > R: In this case, the total charge enclosed by the Gaussian spherical surface is zero since the atom is neutral. Thus, from Gauss’s law, E (r ) × 4 π $r^ 2$ = 0 or E (r ) = 0; r > R At r = R, both cases give the same result: E = 0.

Questions from section 1.13:

1. Obtain the expression for flux through a sphere enclosing a point charge q at its centre, in the form:
   . $ ϕ= \frac {q}{4πε_0 r^2 }.4πr^2. $
2. State Gauss’s law.
3. Obtain the expression for field due to an infinitely long straight uniformly charged wire in the form:
   . $ E= \frac {λ}{2πε_o r}.\hat n ̂$
4. Obtain the expression for field due to a uniformly charged infinite plane sheet in the form:
   . $ E= \frac {σ}{2ε_o }. \hat n $
5. Obtain the expression for field due to a uniformly charged thin spherical shell in the form:. $ E= \frac {q}{4πε_o r^2 }. \hat r $.What would be the field inside the shell?