Coulomb's Laws Of Electrostatics

Coulomb's Laws Of Electrostatics :

First Law : Like charges repel each other and unlike charges attract each other. In other words, bodies having same charges repel each other and bodies oppositely charged attract each other.

The first law tells only about the nature of force i.e., repulsive or attractive. It does not tell about the magnitude of force. The magnitude of the force between two charged bodies is given by the second law.

Second Law : The force of attraction or repulsion between two charges is directly proportional to the product of magnitude of the two charges and inversely proportional to the square of distance between them.

If two point charges Q₁ and Q_{2 have distance d between them, then force  F between the charges can be mathematically expressed as :}

                             F = K × Q1Q2/d²

Where K is a constant of proportionality and its value depends upon the medium in which the charges are placed and the system of units used. In SI units force is measured in newton, charge in coulomb, distance in meter and the value of K is given as:
                 
                             K = 1/4πϵoϵr           (in S.I. system)
where   
ϵo = Absolute Permittivity of vacuum or prematurity of free space
ϵr = Relative Permittivity of the medium w.r.t. vacuum in which the charges are placed.
The value of  ϵo  = 8.854 x 10-12  farad/metre and the value of ϵr is different for different media, for air ϵr = 1. The value of 1/4πϵo = 9 x 109


                                          F = K × Q1Q2/d²
                                       F = (1/4πϵoϵr)  × Q1Q2/d²
                                     F = Q1Q2/4πϵoϵrd^{2  (in N)}

Units.   In the formula,     F = Q1Q2/4πϵoϵrd²  
F is measured in newton, Q1,Q2 in coulomb, d in metre and  ϵo in farad/metre.

Unit Charge :

(i) S.I.System : By unit charge, we mean one coulomb charge. By coulomb's second law,
                       F = Q1Q2/4πϵoϵrd²  newton
                      F =  9 x 109   newton 
and          Q1, Q2 = Q
                      ϵr = 1
                      ϵo = 8.854 x 10-12 
                      d = 1m
Put everything in above relation,
             9 x 109 =  ( Q ×  Q ) / 4π ×  8.854 x 10-12 ×1 ×1
                 Q2   =   ± 1 coulomb
Hence unit charge or one coulomb charge in S.I.system can be defined as:-

A unit charge or One coulomb is that much charge which when placed at a distance of one meter from an equal and similar charge in air, is repelled with a force of 9 x 109 newton from it. 

(ii) In C.G.S System : Unit of charge is stat coulomb
                                1 coulomb =  3 x 109  stat coulomb

Electrostatics

Introduction To Electrostatics : The branch of physics which deals with the study of charges at rest is called electrostatics.

If a glass rod is rubbed with silk cloth, electrons pass from glass rod to silk cloth and the glass rod becomes positively charged while the silk piece attains an equal negative charge. Since glass rod and silk cloth are both insulators, they retain charges on them because electrons cannot move.

In other words, electricity on them is static or at rest and hence the process comes under the Electrostatics.

Explanation of Charges : The number of electrons in an atom is equal to the number of protons, therefore, atom is neutral as whole. A body consists of atoms, therefore, the body is neutral under ordinary conditions. 

However, if from such a neutral body, electrons are detached (by rubbing etc.), there occurs a deficit of electrons in the body. Consequently the body no longer remains neutral due to the shortage of electrons. The result is that the body attains positive charge.

When a body is short of its due share of electrons or there is deficit of electrons in the body it is said to be positively charged.

On the other hand, a negatively charged body has excess of electrons from its normal due share. If a neutral body is supplied with electron, the body loses its neutrality and attains a negative charge.

Therefore, a negatively charged body has excess of electrons from its normal due share.

Total deficiency or excess of electrons in a body is known as charge.

To give a negative charge to any body, extra electrons must be supplied to it. To supply these extra electrons, work will have to be done, which is stored in the body in the form of energy. This makes the charged body capable of doing work.

Units of Charge : The charge on an electron is so small that it is not possible and convenient to take it as the unit. In practice, the charge is measured in coulomb (C). One coulomb of charge is equal to the charge on 625 × 10¹⁶ electrons. i.e.

Charge of 1 coulomb = charge on 625 × 10¹⁶  electrons.

∴ Practical unit of charge is coulomb. And a smaller unit of charge is micro-coulomb is also used.

1 micro-coulomb (i.e. 1 µC) = 10⁻⁶ C coulomb.

Difference between Transverse Waves and Longitudinal Waves

Transverse waves

Longitudinal waves

1.       In transverse waves the particles of the medium vibrate in a direction at right angles to the direction of propagation of the wave. 2.       Transverse waves travel in the form of crests and troughs. One crest and one trough constitute a wave. 3.       Transverse waves are possible in media which possess the properties of elasticity of shape or they have  a free surface i.e., they are possible in solids and liquids. 4.       Transverse waves can be polarized. 5.       Transverse displacement travel with the wave but there is no change in pressure of medium. 6.       Transverse waves can be represented by sine curve directly. 7.       Examples : Waves in ropes, ripples on the surface of liquids and electromagnetic waves are transverse waves.

1.       In longitudinal waves the particles of the medium vibrate back and forth parallel to the direction of propagation of the wave. 2.       Longitudinal waves travel in the form of compressions and rarefactions. One compression and one rarefaction constitute one wave. 3.       Longitudinal waves are possible in media which possess the properties of elasticity of volume  i.e., they are possible in solids, liquids and gases but in gases, this is the only wave motion possible. 4.       Longitudinal waves can not be polarized. 5.       Pressure of medium changes when longitudinal displacement passes through it, but there is no transvere displacement in the medium. 6.       Longitudinal waves cannot be represented by sine curve directly. 7.       Examples : Sound waves in air, gases and waves in springs are longitudinal waves.

Microscope

A microscope is a device used to see small object much magnified at the least distance of distinct vision.

Least distance of distinct vision : The minimum distance from the eye at which the objects are clearly visible is called the least distance of distinct vision. It is denoted by D. For a normal eye it is about 25 cm.

The two types of microscopes are discussed below :
1. Simple microscope or magnifying glass,
2. Compound microscope

1. Simple microscope or Magnifying glass
The size of an object depends upon the angle subtended by the object at the eye. The angle subtended depends upon the dimensions of the object and its distance from the eye. If the object is  brought nearer, the angle subtended at the eye increases and the object appears bigger and more distinct. Thus to see san object distinctly, it should be moved very near to the eye. But if it is brought near the eye at a distance less than the least distance of distinct vision, it becomes indistinct. Thus to see the object distinctly, It should be moved very close to the eye but its image should be formed at the least distance of distinct vision.

An ordinary convex lens of small focal length kept close to the eye can be used as a simple microscope or magnifying glass.

Watch makers use a single convex lens to get a magnified view of the fine parts of the watch.

Magnifying Power : The magnifying power of a simple microscope is the ratio of the angle subtended at the eye by the image as seen through the lens to the angle subtended by the object at the unaided eye, when both are placed at the least distance of distinct vision, it is denoted by M.

The mathematical formula for Magnifying Power, M is

                                            M = D/u
            
                                               = 1+D/f

Where, D is the least distance of distinct vision and is equal to v.
D being constant, the magnifying power depends upon the focal length of the lens.
       Smaller the focal length, greater will be the magnifying power of the lens.

2. Compound microscope : The magnification produced by simple microscope is small and can only be increased by decreasing the focal length of the lens. But there is a practical limit to it. Large magnification can be obtained by using compound microscope in which magnification is obtained in two stages by using two convex lenses.

Magnifying Power : Magnifying power of the microscope is defined as the ratio of the angle subtended by the image at the eye as seen through the microscope to the angle subtended by object at the unaided eye when both are placed at the least distance of distinct vision.

The mathematical formula for Magnifying Power, M is       

fe = focal length of the eye-piece
Fo = focal length of the objective

v = L = length of the microscope tube,
and u = Fo

Hence,                  M = v/u (1+D/fe)

                                 = L/Fo(1+ D/fe)

The above relation shows that magnifying power is inversely proportional to the focal length of the objective and the focal length of the eye-piece. Therefore, magnifying power can be increased by
1. taking the objective of short focal length
2. taking the eye-piece of short focal length.


Practice Problems :
1. A simple magnifier (convex lens) has a focal length of 10 cm. Find its magnifying power.
Solution :
                  f = 10 cm
Magnifying power of a simple microscope(convex lens) M = 1+D/f 
              
⇒                M = 1+25/10     
                        = 1 + 2.5    
                        = 3.5                         { D = 25 cm for normal eye}

Magnifying power of simple convex lens of f = 10 cm is 3.5

2. A simple microscope is made of a combination of two lenses in contact of powers +15D and +5D. Calculate the magnifying power of the microscope, if the image is formed at 0.25m, the least distance of distinct vision.
Solution :   
               Powers of the two lenses are
                p1 = + 15 D
and           p2 = + 5D 
                D  = distance of distinct vision = 0.25 m
∴ Power of the combination, P = + 15 + (+5)
                                            = + 20 D 
∴ Focal length of the lens, f = 1/(+20)
                                        = 0.05 m
Now, magnifying power of simple microscope,

                                  M = (1+ D/f)
                                      = 1 + 0.25/0.05
                                      = 1 + 5 = 6 (Ans.)

Some practice problems :

1. An object is placed at a distance of 15 cm from a convex lens of focal length 30 cm. The size of image formed, in comparison to size of object is
(a) same                                   (b) double
(c) half                                      (d) 4 times

2. A simple magnifier(convex lens) has  a focal length of 10 cm. Its magnifying  power is :
(a) 0.1                                       (b) 2.5
(c) 3.5                                       (d) 10

3. Distance of distinct vision for a normal eye is :
(a) 5 cm                                     (b) 25 cm
(c) 50 cm                                   (d) infinity

4. In a simple microscope, the object is placed 
(a) between F and its lens            (b) at F
(c) between f and 2f                    (d) beyond 2f

5. Which of the following is not true ?
(a) Microscope is used to see small objects.
(b) Telescope  is used to see distant objects.
(c) In a microscope objective is larger than eye piece.
(d) In a telescope objective is larger than eye piece.

6. For a simple microscope if the final image is located at the least distinct vision D from the eye placed close to the lens, the magnifying power is :
(a) D/f                                   (b) 1 + D/f
(c) f/D                                   (d) f × D

7. Ratio of focal length of the objective to the focal length of the eye piece is greater than one for :
(a) telescope                                      (b) microscope
(c) both telescope and microscope       (d) neither telescope nor microscope

8. The final image produced by a simple microscope is :
(a) virtual and erect                           (b) virtual and inverted
(c) real and erect                               (d) real and inverted

9. A convex lens acts as a simple microscope when the object is placed
(a) between F and 2F                        (b) between optical centre and focus
(c) at 2F                                           (d) at F

10. The linear magnification of an image is m. The magnification for area will be
(a) m                                              (b) m/2
(c) m²                                                                  (d) m^1/2
  
11. As an object is moved from infinity towards the pole of a convex lens, the magnification of image will
(a) remain same                              (b) decrease 
(c) increase                                     (d) depend upon the presence of medium

12. Which of the following formula do not represent magnification by a lens– (where terms has usual meaning).
(a) I/O                                           (b) v/u
(c) (f – v)/f                                     (d) f/u

13. When the length of the tube of a compound microscope is increased, its magnifying power will
(a) increase                                  (b) decrease
(c) remain unchanged                    (d)become infinity

14. The magnifying power of a compound microscope in terms of magnification mo due to objective and magnifying power me by the eye piece is given by
(a) mo/me                                     (b) mo × me
(c) mo+me                                    (d) me/mo

15.If the least distance of distinct vision is 25 cm, then the convex lens of focal length 5 cm acts as a magnifier of magnifying power,
(a) 5                                          (b) less than 5
(c) 6                                          (d) more than 6

Optical Instruments

An optical instrument is an arrangement of lenses, prisms or mirrors, which enables us to see better than what we can see with the naked eye.

Optical Projection :
The process of obtaining distinct images on a screen or in a light sensitive device (the eye, photographic films etc.) of large distant objects, small nearly objects or fine details of large nearly object is known as  optical Projection. The devices used for getting the images in the above cases are called optical instruments.

The principle of optical instruments is to increase the angle of view for the image as compared to the viewing angle of the object. The angle of view is the angle at which rays from the extreme points of the object or its image converge at the optical centre of the eye.

Optical instruments provide a two dimensional (plane) image of a three dimensional object.

Types of Optical Instruments :
There are two types of optical instruments :
1. Optical instruments in which a real image is formed on the screen.
Examples : 
               Photographic camera, Magic lantern, Epidiascope, Cinema projector, Over head projector, Human eye etc. are the examples of this type of optical instruments.
2. Optical instruments in which a virtual image is formed and is directly seen with the eye.
Examples :
               Telescope, Microscope, Prism, Binoculars etc. are the examples of this type of optical instruments.

Defects In Image Formation By Lenses And Their Correction

Images formed by lenses suffer from the following defects:
1. Spherical aberration
2. Chromatic aberration

1. Spherical aberration : It is the inability of a lens to bring all the rays to meet to one point after suffering refraction.

                              
Spherical aberration is due to the greater deviation of the rays from outer edges of the lens as compared to that of the rays incident on the central part.

Removal Of Spherical Aberration : Spherical aberration can be removed or minimized by the following methods :
1. By using a stop or making the diameter of the lens small.
2. By using Lenses of large Focal lengths.
3. By selecting suitable curvature to the two surfaces of the lens.
4. By Minimising the deviation.
5. By suitable combination of lenses.

2. Chromatic aberration  or colour defect :  The inability of the lens to focus all the colours at one point is called chromatic aberration.

Achromatism : The removal  of chromatic aberration or color defect is called achromatism. 
It is possible to remove chromatic aberration by having a combination of a suitable convex lens with a concave lens. Such a combination is known as achromatic combination.


Some Multiple Choice Questions for practice :

1. Which lens is used to correct the defect of long sightedness :
(a)   Concave                                        (b) convex
(c)   double concave                              (d) plano concave

2. Aberration in a lens is due to:
(a) nature of light                       
(b) defect in lens material
(c) deviation from simple lens equation   
(d) defect  in experiment arrangement

3. chromatic aberration in alens is due to the;
(a) inability of the lens to focus axial and marginal rays at the same point 
(b) inability of the lens to focus longer and shorter wavelengths of the light at the same point
(c) total internal reflection
(d) the fact that two faces of the lens may have different curvatures

4. Where is the image of the distant object formed by a short sighted eye ?
inability of the lens to focal axial and marginal rays at the same point 
(a)  At yellow spot                          (b) At blind spot
(c)  In front of retina                      (d) Behind the retina

5. Chromatic aberration of a lens results, because :
(a) the marginal and paraxial rays do not meet at the same point                
(b) of the different radii of curvature of the two surfaces
(c) of the prismatic action of the lens                                              
(d) of the defect in manufacturing

6. Ability of the eye to see objects at all distances is called
(a)  Binocular vision                       (b) myopia
(c)  hypermetropia                         (d) accomodation

7.Defect of hypermetropia can be remedied by using a

(a) convex lens                            (b) concave lens
(c) cylinderical lens                      (d) bi-focal length lens

8. Defect of color blindness can be cured by using a 
(a) contact lens                           (b) cylinderical lens
(c) bi-focal length lens                 (d) None of these

9. Spherical aberration can be reduced by
(a) using stops
(b) using lenses of large focal lengths
(c) using plano-convex lenses
(d) All the above methods

10. A plano-convex lens is used as objective lens in a telescope, then to minimise spherical aberration its

(a) convex surface should face the parallel rays  
(b) plane surface should face the parallel rays
(c) centre is covered with black paper
(d) edges are covered with black paper

11. chromatic observation in the formation of images by a lens arises because

(a) of non paraxial rays                                              
(b) the focal length varies with wavelength 
(c) of the defect in grinding                                               
(d)  radii of curvature of the two sides are not same

12. The chromatic aberration in a lens is because of 
(a) the large size of object                                                
(b) the object being very close to object
(c) manufacturing defect               
(d) dispersion of light in the material of lens

13.In order to reduce the spherical aberration in optical instruments , one should use
(a) plano convex lenses               (b) convex lenses
(c) plane mirrors                         (d) concave lenses

14. How should people wearing their spectacles work with a microscope ?
(a) They should take off the spectacles                                          
(b) They should keep on wearing the spectacles
(c) It hardly matters whether they take off or put on their spectacles 
(d) Special lenses of more power are used in the spectacles

15. The defects in the formation of image arising due to marginal and paraxial rays undergoing different deviation in a lens is known as
(a) Chromatic aberration              (b) Coma
(c)  Distortion                              (d) Spherical aberration

Total Internal Reflection (TIR)

It is the process of reflection of light into a denser medium from an interface of the denser medium and a rarer medium.

Essential conditions for TIR : There are two essential conditions for the process to occur.

(i) Light should travel from denser medium to rarer medium

(ii) Angle of incidence in denser medium should be greater than the critical angle for the pair of media in contact.

Critical Angle : Critical angle for a pair of media in contact is the angle of incidence in the denser medium corresponding to which angle of refraction in the rarer medium is 90°. It is denoted by C and its value depend upon the nature of media in contact.

Let we have an interface XY separating a rarer medium (Air) and a denser medium (water). Let medium (a) is air and medium (b) is water as shown in fig.

Let O be an object lying in the denser medium and the rays starting from it are as shown in fig. below,

                                                                             

A ray of light from O falling perpendicular to the interface XY passes without any deviation along AB. A second ray OA1, making some angle with the normal on interface will refract along A1B1 shown. Third ray OA3 is falling at a particular angle of incidence called critical angle (i2= C) at point A3 and the ray after striking point A3 bends towards the interface A3B3, in the case the angle of refraction is 90°.

                        Another ray OA4 falling at the interface XY at an angle greater than the critical angle (i3 > C), the rays after striking at point A4 will be reflected back in the medium (II) along A4B4 as shown in fig. This process is called Total internal Reflection (TIR).

Hence in short, we can conclude that when  a ray of light traveling from an optically denser medium to an optically rarer medium is incident at an angle greater than the critical angle for the pair of medium in contact, the ray is totally reflected back into denser medium. This process is called Total Internal Reflection.

Relation between refractive index and critical angle :

               ^Iµ_{II  = 1/sin C}

As is seen from fig., if the angle of incidence (i3 >C) greater than critical angle, the light ray is completely reflected back into the water (the denser medium).

Applications of Total Internal Reflection :

1. The Brilliance of Diamond : Refractive index of diamond is approximately 2∙45

⇒            Sin C = 1/µ = 1/2∙45 = 0∙408

⇒                  C = 1/sin (0∙408) = 24∙1°

i.e., critical angle for diamond is 24∙1. 

Diamond is cut in such a way that a ray of light enters the diamond at an angle greater than critical angle i.e., 24∙1°, hence light ray suffers multiple total internal reflections at different faces and remains within the diamond. This makes the diamond sparkle.

2. Total Internal Reflection in Prisms : 

For a glass prism,      µ = 1∙5

                          Sin C = 1/µ            

                                  = 1/1∙5 = 0∙667

                               C = 1/sin (0∙667) =42°

Thus, in a right angled isosceles prism, a ray of light incident at an angle of incidence 45° (i.e., greater than critical angle 42°), the light ray will be totally internally reflected.

                          

                          

                                       

3. Mirage :It is an optical illusion observed in deserts due to total internal reflection of light. Mirage disappoints a traveller in a sandy desert because a traveller on a look out for some supply of water gets feeling of presence of a pond of water as he sees the image of tree down below but did not get any water.
                    

4. Optical Fibres : Optical fibers are made up of glass or plastic may be as thin as a human hair. Optical fibers have two parts (i) Core (ii) Cladding

(i) A Core is the central part or main part of an optical fibre. Diameter of core varies from 6µm to 250 µm. It is made from a pure glass of high refractive index (approximately 1⋅7)

(ii) Cladding: Core is surrounded by a cladding of glass or plastic of relatively small refractive index (approximately 1⋅5). Cladding is usually between 10 µm to 150 µm thick. The interface between core and cladding acts as a cylindrical mirror. A ray of light falling at this interface at an angle more than critical angle gets totally internally reflected.

                               

                       

Optical fibres help in communication of signals in the form of light through many kilometers without loss in energy or signal. Optical fibers are also used in medical instruments like fibrescope to examine internal body cavities such as the stomach, galbladder etc.

5. Shinning of air bubble in water.


Practice Problem:

If refractive index of water and glass are 4/3 and 5/3 respectively and the light is going from glass  to water. Find the critical angle (C).
Solution:

             Given,            ^aµ_w =  4/3,    ^aµ_g =  5/3

                                 ^wµ_g  = ^aµ_{g /} ^aµ_w     

                                       =  5/3 X 3/4   

                                       = 5 /4  = 1⋅25
                                             
We know that,           ^wµ_g  =  1/Sin C

∴                      1/Sin C   = 1⋅25
⇒                        Sin C   = (1/1⋅25) = 0⋅8
                              C     =  (1/Sin) (0⋅8)
                              C      = 53⋅13° (Ans.)


Some Multiple choice Practice Problems :
1. The critical angle of light passing from glass to air is minimum for 
(a) red                               (b) green
(c) yellow                           (d)violet

2. A well cut diamond appears bright due to 
(a) it emits bright light 
(b) scattering of light
(c) total internal reflection
(d) it emits luminous particles

3.Total internal reflection will not take place, when light ray travels from
(a) glass to air
(b) glass to water
(c) water to air
(d) water to glass

4. µ = Sin C, where C is the angle of incidence in optically
(a) rarer medium              (b) denser medium
(c) air                              (d) transparent

5. Critical angle for a pair of media in contact is the angle of incidence in the denser medium corresponding to which angle of refraction in the rarer medium is
(a) 180°                          (b) 30°
(c) 45°                            (d) 90°

6.Condition for Total Internal Reflection is
(a) ∠i in denser medium should be greater than critical angle
(b) Light should travel from denser medium to rarer medium
(c)  ^Iµ_{II = 1/ Sin C
(d) All the above are true}

7. If the critical angle for total internal reflection from a medium to vacuum is 30°, then velocity of light in medium is
(a) 1⋅5 x 10⁸ m/s                   (b) 0⋅75 x 10⁸ m/s    

(c) 3 x 10⁸ m/s                      (d) 2 x 10⁸ m/s