 Physics ELECTRIC CHARGES ,CONDUCTORS and INSULATORS AND COULOMB'S LAW

### TOPIC COVERED

color{blue}{star} Introduction
color{blue}{star} Electric Charges
color{blue}{star} Conductors and Insulators
color{blue}{star}Charging by Induction
color{blue}{star} BASIC PROPERTIES OF ELECTRIC CHARGE
color{blue}{star}Coulomb's Law

### INTRODUCTION

\color{blue} ✍️ All of us have the experience of seeing a spark or hearing a crackle when we take off our synthetic clothes or sweater, particularly in dry weather.

\color{blue} ✍️ The reason for these experiences is discharge of electric charges through our body, which were accumulated due to rubbing of insulating surfaces.

color{blue}ul text(Static) :  It means anything that does not move or change with time.

color{blue} ultext(Electrostatics) :  It deals with the study of forces, fields and potentials arising from static charges.

### ELECTRIC CHARGE \color{blue} ✍️The name electricity is coined from the Greek word text(elektron) meaning text(amber).

\color{blue} ✍️Many materials were known which on rubbing could attract light objects like straw, pith balls and bits of papers.

\color{blue} ✍️It was observed that if two glass rods rubbed with wool or silk cloth are brought close to each other, they repel each other .The two strands of wool or two pieces of silk cloth, with which the rods were rubbed, also repel each other as shown in fig 1.1.

\color{blue} ✍️We say that the bodies like glass or plastic rods, silk, fur and pith balls are electrified. They acquire an electric charge on rubbing.

\color{blue} ✍️ \color{blue} \mathbf(KEY \ CONCEPT)

color{red} ► There are two kinds of electrification and we find that

color{red} {(i)} Like charges repel

color{red} {(ii)} Unlike charges attract each other.

\color{blue} ✍️A simple apparatus to detect charge on a body is the color(green){text(Gold-Leaf Electroscope)} as shown in fig 1.2.

\color{blue} ✍️ 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.

\color{blue} {1 .}  Take a thin aluminium curtain rod with ball ends fitted for hanging the curtain.

\color{blue} {2 .} Cut out a piece of length about 20 cm with the ball at one end and flatten the cut end.

\color{blue} {3.}  Take a large bottle that can hold this rod and a cork which will fit in the opening of the bottle.

\color{blue} {4.}  Make a hole in the cork sufficient to hold the curtain rod snugly.

\color{blue} {5.}  Slide the rod through the hole in the cork with the cut end on the lower side and ball end projecting above the cork.

\color{blue} {6.}  Fold a small, thin aluminium foil in the middle and attach it to the flattened end of the rod by cellulose tape.

This forms the leaves of your electroscope.

### CONDUCTORS AND INSULATORS color(purple){ul{"CONDUCTORS"}}

\color{blue} {1.} Those which allow electricity to pass through them easily are called conductors.

\color{blue} {2.} They have electric charges (electrons) that are free to move inside the material.

\color{blue} {3.} When some charge is transferred to a conductor, it readily gets distributed over the entire surface of the conductor.

\color{blue} {4.} For Example: Metals, human and animal bodies and earth are conductors.

color(purple){ul{"INSULATORS"}}

\color{blue} {1.} Those which don't pass electricity through them are called insulators.

\color{blue} {2.} They don't have electric charges(electrons) that are free to move inside the material.

\color{blue} {3.} If some charge is put on an insulator, it stays at the same place.

\color{blue} {4.} For Example: The non-metals like glass, porcelain, plastic, nylon, wood.

\color{blue} ✍️  This property of the materials tells you why a nylon or plastic comb gets electrified on combing dry hair or on rubbing, but a metal article like spoon does not.

\color{blue} ✍️  When we bring a charged body in contact with the earth, all the excess charge on the body disappears by causing a momentary current to pass to the ground through the connecting conductor. This process of sharing the charges with the earth is called color(green){text(grounding)} or color(green){text(earthing)}.

\color{blue} ✍️ color(purple){text(Earthing)} provides a safety measure for electrical circuits and appliances. A thick metal plate is buried deep into the earth and thick wires are drawn from this plate; these are used in buildings for the purpose of earthing near the mains supply. The electric wiring in our houses has three wires: text(live, neutral) and text(earth).

### CHARGING BY INDUCTION \color{blue} ✍️ When we touch a pith ball with an electrified plastic rod, some of the negative charges on the rod are transferred to the pith ball and it also gets charged. Thus the pith ball is charged by contact. It is then repelled by the plastic rod but is attracted by a glass rod which is oppositely charged.

\color{blue} ✍️ Let us try to understand what could be happening by performing the following experiment.

color{purple}{(i)} Bring two metal spheres, A and B, supported on insulating stands, in contact as shown in Fig. 1.4(a).

color{purple}{(ii)} Bring a positively charged rod near one of the spheres, say A, taking care that it does not touch the sphere. The free electrons in the spheres are attracted towards the rod.

\color{blue} ✍️ This leaves an excess of positive charge on the rear surface of sphere B. Both kinds of charges are bound in the metal spheres and cannot escape. They, therefore, reside on the surfaces, as shown in Fig. 1.4(b). The left surface of sphere A, has an excess of negative charge and the right surface of sphere B, has an excess of positive charge.

\color{blue} ✍️ However, not all of the electrons in the spheres have accumulated on the left surface of A. As the negative charge starts building up at the left surface of A, other electrons are repelled by these.

\color{blue} ✍️ In a short time, equilibrium is reached under the action of force of attraction of the rod and the force of repulsion due to the accumulated charges. Fig. 1.4(b) shows the equilibrium situation. The process is called induction of charge and happens almost instantly.

\color{blue} ✍️ The accumulated charges remain on the surface, as shown, till the glass rod is held near the sphere. If the rod is removed, the charges are not acted by any outside force and they redistribute to their original neutral state.

color{purple}{(iii)} Separate the spheres by a small distance while the glass rod is still held near sphere A, as shown in Fig. 1.4(c). The two spheres are found to be oppositely charged and attract each other.

color{purple}{(iv)} Remove the rod. The charges on spheres rearrange themselves as shown in Fig. 1.4(d). Now, separate the spheres quite apart. The charges on them get uniformly distributed over them, as shown in Fig. 1.4(e).

\color{blue} ✍️ In this process, the metal spheres will each be equal and oppositely charged. This is charging by induction. The positively charged glass rod does not lose any of its charge, contrary to the process of charging by contact.
Q 3270780616 How can you charge a metal sphere positively without
touching it?
Class 11 Chapter 1 Example 1 Solution:

Figure 1.5(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. 1.5(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. 1.5(c). Disconnect
the sphere from the ground. The positive charge continues to be
held at the near end [Fig. 1.5(d)]. Remove the electrified rod. The
positive charge will spread uniformly over the sphere as shown in
Fig. 1.5(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.

### BASIC PROPERTIES OF ELECTRIC CHARGE

\color{blue} ✍️ We have seen that there are two types of charges, namely positive and negative and their effects tend to cancel each other. Here, we shall now describe some other properties of the electric charge.

\color{blue} ✍️ If the sizes of charged bodies are very small as compared to the distances between them, we treat them as point charges. All the charge content of the body is assumed to be concentrated at one point in space.

color{purple}ulbb{"Additivity of charges : "}

\color{blue} ✍️ We have not as yet given a quantitative definition of a charge; we shall follow it up in the next section. We shall tentatively assume that this can be done and proceed.

\color{blue} ✍️ If a system contains two point charges q_1 and q_2, the total charge of the system is obtained simply by adding algebraically q_1 and q_2 , i.e., charges add up like real numbers or they are scalars like the mass of a body. If a system contains n charges q_1, q_2, q_3, …, q_n, then the total charge of the system is q_1 + q_2 + q_3 + … + q_n .

\color{blue} ✍️ Charge has magnitude but no direction, similar to the mass. However, there is one difference between mass and charge. Mass of a body is always positive whereas a charge can be either positive or negative. Proper signs have to be used while adding the charges in a system. For example, the total charge of a system containing five charges +1, +2, –3, +4 and –5,
in some arbitrary unit, is (+1) + (+2) + (–3) + (+4) + (–5) = –1 in the same unit.

color{purple}ulbb{"Charge is conserved : "}

\color{blue} ✍️ We have already hinted to the fact that when bodies are charged by rubbing, there is transfer of electrons from one body to the other; no new charges are either created or destroyed. A picture of particles of electric charge enables us to understand the idea of conservation of charge.

\color{blue} ✍️ When we rub two bodies, what one body gains in charge the other body loses.
Within an isolated system consisting of many charged bodies, due to interactions among the bodies, charges may get redistributed but it is found that the total charge of the isolated system is always conserved.Conservation of charge has been established experimentally.

\color{blue} ✍️ 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

color{purple}ulbb{"Quantisation of charge "}

\color{blue} ✍️ Experimentally it is established that all free charges are integral multiples of a basic unit of charge denoted by e. Thus charge q on a body is always given by

bb{q = n e}

\color{blue} ✍️ 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.

\color{blue} ✍️ 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.

\color{blue} ✍️ 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 1912.

\color{blue} ✍️ 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 1 s if the current is 1 A (ampere), (see Chapter 2 of Class XI, Physics Textbook , Part I). In this system, the value of the basic unit of charge is e = 1.602192 × 10^(–19) C

\color{blue} ✍️ Thus, there are about 6 × 1018 electrons in a charge of –1C. In electrostatics, charges of this large magnitude are seldom encountered and hence we use smaller units 1 μC (micro coulomb) = 10^(–6) C or 1 mC (milli coulomb) = 10^(–3) 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.

\color{blue} ✍️ Thus, if a body contains n_1 electrons and n_2 protons, the total amount of charge on the body is n_2 × e + n_1 × (–e) = (n_2 – n_1) e. Since n_1 and n_2 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

\color{blue} ✍️ 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.

\color{blue} ✍️ 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.

\color{blue} ✍️ At the macroscopic level, one deals with charges that are enormous compared to the magnitude of charge e. 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.

\color{blue} ✍️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., 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.
Q 3280580417 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?
Class Chapter 1 Example 2 Solution:

In one second 10^9 electrons move out of the body. Therefore the charge given out in one second is 1.6 xx 10^(–19) xx 10^9 C = 1.6 xx 10^(–10) C. The time required to accumulate a charge of 1 C can then be estimated to be 1 C ÷ (1.6 xx 10^(–10) C//s) = 6.25 xx 10^9 s = 6.25 xx 109 ÷ (365 xx 24 xx 3600) years = 198 years. Thus to collect a charge of one coulomb, from a body from which 109 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 xx 10^(24) electrons.
Q 3200580418 How much positive and negative charge is there in a cup of water?
Class Chapter 1 Example 3 Solution:

Let us assume that the mass of one cup of water is 250 g. The molecular mass of water is 18g. Thus, one mole (= 6.02 xx 10^(23) molecules) of water is 18 g. Therefore the number of molecules in one cup of water is (250//18) xx 6.02 xx 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) xx 6.02 xx 10^(23) xx 10 xx 1.6 xx 10^(–19) C = 1.34 xx 10^7 C.

### Coulomb's law

\color{purple}ul(★ \color{purple} " DEFINITION ALERT")

color(blue){ul("Coulomb’s law")}

If two point charges q_1, q_2 are separated by a distance r in vacuum, the magnitude of the force (F) between them is given by:

\color{red}ul(★ \color{red} " FORMULA ALERT")

F=k ((|q_1q_2|)/r^2)

\color{green} ✍️ \color{green} \mathbf(KEY \ CONCEPT)

\color{blue} ✍️ When the separation between two spheres is much larger than the radius of each sphere, the charged spheres may be regarded as point charges.

\color{blue} ✍️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//2. Repeating this process, we can get charges q//2, q//4, etc.

\color{blue} ✍️ 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 9 × 10^9. q_1 = q_2 = 1 C, r = 1 m , then:

F = 9 × 10^9 N

That is, 1 C is the charge that when placed at a distance of 1 m from another charge of the same magnitude in vacuum experiences an electrical force of repulsion of magnitude 9 × 10^9 N.

One coulomb is evidently too big a unit to be used. In practice, in electrostatics, one uses smaller units like 1 mC or 1 μC.

\color{green} ✍️ \color{green} \mathbf(KEY \ CONCEPT)

The constant k = 1/(4πε_0) ,

So that Coulomb’s law is written as:

color{ green} {F=1/(4πε_0)((|q_1q_2|)/r^2)}

where,
ε_0 =permittivity of free space .
ε_0 = 8.854 × 10^(–12) C^2 N^(–1)m^(–2).

### VECTOR FORM OF COULOMB'S LAW \color{blue} ✍️. Let the position vectors of charges q_1 and q_2 be vec r_1 and vec r_2 . We denote force on q_1 due to q_2 by vec F_(12) and force on q_2 due to q_1 by vec F_(21). The two point charges q_1 and q_2 have been numbered 1 and 2 for convenience and the vector leading from 1 to 2 is denoted by r_(21):

vec r_(21) = vec r_2 –vec r_1.

In the same way, the vector leading from 2 to 1 is denoted by r_(12):

vec r_(12) = vec r_1 –vec r_2 = –vec r_(21)

\color{green} ✍️ \color{green} \mathbf(KEY \ CONCEPT)

color {blue}●The magnitude of the vectors vec r_(21) and vec r_(12) is denoted by r_(21) and r_(12), respectively (r_(12) = r_(21)).

color {blue}● 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)= vec r_(21)/r_(21) , hat r_(12)=vec r_(12)/r_(12) ,hat r_(21)= - hat r_(12)

Coulomb’s force law between two point charges q_1 and q_2 located at r_1 and r_2 is then expressed as

vec F_(21)=(1/(4πε_0))((q_1q_2)/r_(21)^2)hat r_(21)

\color{green} ✍️ \color{green} \mathbf(KEY \ CONCEPT)

color {blue}● Above equation is valid for any sign of q_1 and q_2 whether positive or negative.

color {blue}{1.} If q_1 and q_2 are of the same sign (either both positive or both negative), vec F_(21) is along hat r_(21), which denotes repulsion, as it should be for like charges.

color {blue}{2.} If q_1 and q_2 are of opposite signs, vec F_21 is along – hat r_ (21)(= hat 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. takes care of both cases correctly.

color {blue}● The force vec F_(12) on charge q_1 due to charge q_2, is obtained by simply interchanging 1 and 2, i.e.,

vec F_(12)=(1/(4πε_0))((q_1q_2)/r_(12)^2)
Thus, Coulomb’s law agrees with the Newton’s third law.

color {blue}● Coulomb’s law gives the force between two charges q_1 and q_2 in vacuum. If the charges are placed in matter or the intervening
space has matter, the situation gets complicated due to the presence of charged constituents of matter.
Q 3210580419 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? (mp = 1.67 xx
10^(–27) kg, me = 9.11 xx 10^(–31) kg)
Class Chapter 1 Example 4 Solution:

(a) (i) The electric force between an electron and a proton at a distance
r apart is:

 F_e = - (1 )/( 4 pi ε_0) e^2/r^2

where the negative sign indicates that the force is attractive. The
corresponding gravitational force (always attractive) is:

 F_G = - G (m_p m_e)/r^2

where m_p and m_e are the masses of a proton and an electron
respectively.

 | F_e/F_G| = e^2/( 4 pi ε_0 G m_p m_e) = 2.4 xx 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 :

 | F_e/F_G| = e^2/( 4 pi ε_0 Gm_p m_e) = 1.3 xx 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 F_e ~ 230 N
whereas F_G ~ 1.9 xx 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| = (1 )/( 4 pi ε_0) e^2/r^2 = 8.987 xx 10^9 Nm^2//C^2 xx (1.6 xx 10^(–19) C)^2 // (10^(–10) m)^2

= 2.3 xx 10^(–8) N

Using Newton’s second law of motion, F = m a, the acceleration
that an electron will undergo is

a = 2.3 xx 10^(–8) N // 9.11 xx 10^(–31) kg = 2.5 xx 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 xx 10^(–8) N // 1.67 xx 10^(–27) kg = 1.4 xx 10^(19) m//s^2
Q 3210680510 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. 1.7(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. 1.7(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. 1.7(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.
Class Chapter 1 Example 5 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 = 1/( 4 pi ε_0) = (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' = 1/( 4 pi ε_0) = ((q//2)(q'//2))/(r//2)^2 = 1/( 4 pi ε_0) (qq')/r^2 = F

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