AND FIELDS

AND FIELDS on rubbing could attract light objects like straw, pith balls and bits of papers. You can perform the following activity at home to experience such an effect. Cut out long thin strips of white paper and lightly iron them. Take them near a TV screen or computer monitor.

You will see that the strips get attracted to the screen. In fact they remain stuck to the screen for a while. It was observed that if two glass rods rubbed with wool or silk cloth are brought close to each other, they repel each other [Fig. .

(a)]. The two strands of wool or two pieces of silk cloth, with which the rods were rubbed, also repel each other. However, the glass rod and wool attracted each other. Similarly, two plastic rods rubbed with cat’s fur repelled each other [Fig.

. (b)] but attracted the fur. On the other hand, the plastic rod attracts the glass rod [Fig. .

(c)] and repel the silk or wool with which the glass rod is rubbed. The glass rod repels the fur. If a plastic rod rubbed with fur is made to touch two small pith balls (now-a-days we can use polystyrene balls) suspended by silk or nylon thread, then the balls repel each other [Fig. .

(d)] and are also repelled by the rod. A similar effect is found if the pith balls are touched with a glass rod rubbed with silk [Fig. . (e)].

A dramatic observation is that a pith ball touched with glass rod attracts another pith ball touched with plastic rod [Fig. . (f )]. These seemingly simple facts were established from years of efforts and careful experiments and their analyses.

It was concluded, after many careful studies by different scientists, that there were only two kinds of an entity which is called the electric charge. We say that the bodies like glass or plastic rods, silk, fur and pith balls are electrified. They acquire an electric charge on rubbing. The experiments on pith balls suggested that there are two kinds of electrification and we find that (i) like charges repel and (ii) unlike charges attract each other.

The experiments also demonstrated that the charges are transferred from the rods to the pith balls on contact. It is said that the pith balls are electrified or are charged by contact. The property which differentiates the two kinds of charges is called the polarity of charge. When a glass rod is rubbed with silk, the rod acquires one kind of charge and the silk acquires the second kind of charge.

This is true for any pair of objects that are rubbed to be electrified. Now if the electrified glass rod is brought in contact with silk, with which it was rubbed, 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.

What can you conclude from these observations? It just tells us that unlike charges acquired by the objects FIGURE . Rods and pith balls: like charges repel and unlike charges attract each other. Interactive animation on simple electrostatic experiments: and Fields neutralise or nullify each other’s effect.

Therefore the charges were named as positive and negative by the American scientist Benjamin Franklin. We know that when we add a positive number to a negative number of the same magnitude, the sum is zero. This might have been the philosophy in naming the charges as positive and negative. By convention, the charge on glass rod or cat’s fur is called positive and that on plastic rod or silk is termed negative.

If an object possesses an electric charge, it is said to be electrified or charged. When it has no charge it is said to be neutral. and Fields by gaining electrons. When we rub a glass rod with silk, some of the electrons from the rod are transferred to the silk cloth.

Thus the rod gets positively charged and the silk gets negatively charged. No new charge is created in the process of rubbing. Also the number of electrons, that are transferred, is a very small fraction of the total number of electrons in the material body. Also only the less tightly bound electrons in a material body can be transferred from it to another by rubbing.

Therefore, when a body is rubbed with another, the bodies get charged and that is why we have to stick to certain pairs of materials to notice charging on rubbing the bodies. and Fields EXAMPLE . [This happens even when the light object is not a conductor. The mechanism for how this happens is explained later in Sections .

and . .] The centres of the two types of charges are slightly separated. We know that opposite charges attract while similar charges repel. However, the magnitude of force depends on the distance between the charges and in this case the force of attraction overweighs the force of repulsion.

As a result the particles like bits of paper or pith balls, being light, are pulled towards the rods. Example . How can you charge a metal sphere positively without touching it? Solution Figure .

(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. . (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. . (c). Disconnect the sphere from the ground.

The positive charge continues to be held at the near end [Fig. . (d)]. Remove the electrified rod.

The positive charge will spread uniformly over the sphere as shown in Fig. . (e). FIGURE .

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?

Interactive animation on charging a two-sphere system by induction: and Fields where n is any integer, positive or negative. This basic unit of charge is the charge that an electron or proton carries. By convention, the charge on an electron is taken to be negative; therefore charge on an electron is written as –e and that on a proton as +e. The fact that electric charge is always an integral multiple of e is termed as quantisation of charge.

There are a large number of situations in physics where certain physical quantities are quantised. The quantisation of charge was first suggested by the experimental laws of electrolysis discovered by English experimentalist Faraday. It was experimentally demonstrated by Millikan in . In the International System (SI) of Units, a unit of charge is called a coulomb and is denoted by the symbol C.

A coulomb is defined in terms the unit of the electric current which you are going to learn in a subsequent chapter. In terms of this definition, one coulomb is the charge flowing through a wire in s if the current is A (ampere), (see Chapter of Class XI, Physics Textbook , Part I). In this system, the value of the basic unit of charge is e = .602192 × – C Thus, there are about × electrons in a charge of –1C. In electrostatics, charges of this large magnitude are seldom encountered and hence we use smaller units μC (micro coulomb) = – C or mC (milli coulomb) = – C.

If the protons and electrons are the only basic charges in the universe, all the observable charges have to be integral multiples of e. Thus, if a body contains n1 electrons and n protons, the total amount of charge on the body is n × e + n1 × (–e) = (n – n1) e. Since n1 and n2 are integers, their difference is also an integer. Thus the charge on any body is always an integral multiple of e and can be increased or decreased also in steps of e.

The step size e is, however, very small because at the macroscopic level, we deal with charges of a few μC. At this scale the fact that charge of a body can increase or decrease in units of e is not visible. The grainy nature of the charge is lost and it appears to be continuous. This situation can be compared with the geometrical concepts of points and lines.

A dotted line viewed from a distance appears continuous to us but is not continuous in reality. As many points very close to each other normally give an impression of a continuous line, many small charges taken together appear as a continuous charge distribution. At the macroscopic level, one deals with charges that are enormous compared to the magnitude of charge e. Since e = .

× – C, a charge of magnitude, say μC, contains something like 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. At the microscopic level, where the charges involved are of the order of a few tens or hundreds of e, i.e., EXAMPLE .

EXAMPLE . they can be counted, they appear in discrete lumps and quantisation of charge cannot be ignored. It is the scale involved that is very important. Example .

If electrons move out of a body to another body every second, how much time is required to get a total charge of C on the other body? Solution In one second electrons move out of the body. Therefore the charge given out in one second is . × – × C = .

× – C. The time required to accumulate a charge of C can then be estimated to be C ÷ ( . × – C/s) = . × s = .

× ÷ ( × × ) years = years. Thus to collect a charge of one coulomb, from a body from which electrons move out every second, we will need approximately 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 cm contains about . × electrons. Example . How much positive and negative charge is there in a cup of water?

Solution Let us assume that the mass of one cup of water is g. The molecular mass of water is 18g. Thus, one mole (= . × molecules) of water is g.

Therefore the number of molecules in one cup of water is ( / ) × . × . Each molecule of water contains two hydrogen atoms and one oxygen atom, i.e., electrons and protons. Hence the total positive and total negative charge has the same magnitude.

and Fields spheres. When the separation between two spheres is much larger than the radius of each sphere, the charged spheres may be regarded as point charges. However, the charges on the spheres were unknown, to begin with. How then could he discover a relation like Eq.

( . )? Coulomb thought of the following simple way: Suppose the charge on a metallic sphere is q. If the sphere is put in contact with an identical uncharged sphere, the charge will spread over the two spheres.

By symmetry, the charge on each sphere will be q/ *. Repeating this process, we can get charges q/ , q/ , etc. Coulomb varied the distance for a fixed pair of charges and measured the force for different separations. He then varied the charges in pairs, keeping the distance fixed for each pair.

Comparing forces for different pairs of charges at different distances, Coulomb arrived at the relation, Eq. ( . ). Coulomb’s law, a simple mathematical statement, was initially experimentally arrived at in the manner described above.

While the original experiments established it at a macroscopic scale, it has also been established down to subatomic level (r ~ – m). Coulomb discovered his law without knowing the explicit magnitude of the charge. In fact, it is the other way round: Coulomb’s law can now be employed to furnish a definition for a unit of charge. In the relation, Eq.

( . ), k is so far arbitrary. We can choose any positive value of k. The choice of k determines the size of the unit of charge.

In SI units, the value of k is about × . The unit of charge that results from this choice is called a coulomb which we defined earlier in Section . . Putting this value of k in Eq.

( . ), we see that for q1 = q2 = C, r = m F = × N That is, C is the charge that when placed at a distance of m from another charge of the same magnitude in vacuum experiences an electrical force of repulsion of magnitude × N. One coulomb is evidently too big a unit to be used. In practice, in electrostatics, one uses smaller units like mC or μC.

The constant k in Eq. ( . ) is usually put as k = /4πε0 for later convenience, so that Coulomb’s law is written as q q F ( . ) ε0 is called the permittivity of free space .

The value of ε0 in SI units is ε = . × – C2 N–1m– * Implicit in this is the assumption of additivity of charges and conservation: two charges (q/ each) add up to make a total charge q. Charles Augustin de Coulomb ( – ) Coulomb, a French physicist, began his career as a military engineer in the West Indies. In , he returned to Paris and retired to a small estate to do his scientific research.

He invented a torsion balance to measure the quantity of a force and used it for determination of forces of electric attraction or repulsion between small charged spheres. He thus arrived in at the inverse square law relation, now known as Coulomb’s law. The law had been anticipated by Priestley and also by Cavendish earlier, though Cavendish never published his results. Coulomb also found the inverse square law of force between unlike and like magnetic poles.

and Fields EXAMPLE . Example . 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 Å (= - m) apart? (mp = . × – kg, me = . × – kg) Solution (a) (i) The electric force between an electron and a proton at a distance r apart is: e e F = − where the negative sign indicates that the force is attractive.

The corresponding gravitational force (always attractive) is: p e G F G = − where mp and me are the masses of a proton and an electron respectively. . e G p e F e F Gm m (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 : e G p p F e F Gm m . × 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 ~ - m inside a nucleus) are Fe ~ N whereas FG ~ . × – 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| = e = . × Nm2/C2 × ( . × –19C) / ( –10m) = .

× – N Using Newton’s second law of motion, F = ma, the acceleration that an electron will undergo is a = . × – N / . × – kg = . × m/s2 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 . × – N / . × – kg = . × m/s2 Interactive animation on Coulomb’s law: EXAMPLE .

Example . 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 cm, as shown in Fig. .

(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. .

(b). C and D are then removed and B is brought closer to A to a distance of . cm between their centres, as shown in Fig. .

(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.

FIGURE . and Fields EXAMPLE . 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 qq F ′ neglecting the sizes of spheres A and B in comparison to r.