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Electrostatics

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Electrostatics PDF Notes, Important Questions and Formulas

 

Electrostatics-I

1. INTRODUCTION:

  1. Introduction:
    Electromagnetism is a science of the combination of electrical and magnetic phenomenon. Electromagnetism can be divided into 2 parts:
    1. Electrostatics: It deals with the study of charges at rest.
    2. Electrodynamics: It deals with the study of charges in motion (discusses magnetic phenomenon). In this chapter we will be dealing with charges at rest i.e. electrostatics.

  2. Structure of Atom:
    An atom consists of two parts (i) nucleus (ii) extra nuclear part. Nucleus consists of neutrons and protons and extra nuclear part has electrons revolving around nucleus. In a neutral atom. number of electrons = number of protons. charge of electrons = charge of protons = 1.602 × 10–19 coulomb. Normally positive charges are positron, proton and positive ions. In nature practically free existing positive charge are positive ions and negative charges are electrons.

  3. Electric Charge
    Charge of a material body or particle is the property (acquired or natural) due to which it produces and experiences electrical and magnetic effects. Some of naturally charged particles are electron, proton, particle etc.

  4. Types of Charge
    1. Positive charge: It is the deficiency of electrons compared to protons.
    2. Negative charge: It is the excess of electrons compared to protons.

  5. Units of Charge
    Charge is a derived physical quantity. Charge is measured in coulomb in S.I. unit. In practice we use mC (10–3C), C (10 –6 C), nC (10–9C) etc.
    C.G.S unit of charge = electrostatic unit = esu.
    1 coulomb = 3 × 109 esu of charge
    Dimensional formula of charge = [M°L°T1I1]

  6. Properties of Charge
    1. Charge is a scalar quantity: It adds algebraically and represents excess, or deficiency of electrons.
    2. Charge is transferable: Charging a body implies transfer of charge (electrons) from one body to another. Positively charged body means loss of electrons, i.e. deficiency of electrons. Negatively charged body means excess of electrons. This also shows that mass of a negatively charged body > mass of a positively charged identical body.
    3. Charge is conserved : In an isolated system, total charge (sum of positive and negative) remains constant whatever change takes place in that system.
    4. Charge is quantized: Charge on any body always exists in integral multiples of a fundamental unit of electric charge. This unit is equal to the magnitude of charge on electron (1e = 1.6 × 10–19 coulomb). So charge on anybody Q = ± ne, where n is an integer and e is the charge of the electron. Millikan's oil drop experiment proved the quantization of charge or atomicity of charge.
      Recently, the existence of particles of charge begin mathsize 12px style straight plus-or-minus 2 over 3 straight e thin space thin space and plus-or-minus 2 over 3 straight e end stylehas been postulated. These particles are called quarks but still this is not considered as the quantum of charge because these are unstable (They have very short span of life.)
    5. Like point charges repel each other while unlike point charges attract each other.
    6. Charge is always associated with mass, i.e., charge can not exist without mass though mass can exist without charge. The particle such as photon or neutrino which have no (rest) mass can never have a charge.
    7. Charge is relativistically invariant: This means that charge is independent of frame of reference, i.e., charge on a body does not change whatever be its speed. This property is worth mentioning as in contrast to charge, the mass of a body depends on its speed and increases with increase in speed.
    8. A charge at rest produces only electric field around itself; a charge having uniform motion produces electric as well as magnetic field around itself while a charge having accelerated motion emits electromagnetic radiation..


  7. Conductors and Insulators:
    Any object can be broadly classified in either of the following two categories:
    (i) Conductors
    (ii) Insulators
    (i)   Conductors: These are the materials that allow flow of charge through them. This category generally comprises of metals but may sometimes contain nonmetals too. (ex. Carbon in form of graphite.)
    (ii)Insulators: These are the materials which do not allow movement of charge through them.

  8. Charging of Bodies:
    An object can be charged by addition or removed of electrons from it. In general an object can either be a conductor or insulator. Thus we are going to discuss the charging of a conductor and charging of an insulator in brief.

  9. Charging of Conductors:
    Conductors can be charged by
    (a)  Rubbing or frictional electricity
    (b)  Conduction & Induction (will be studied in later sections)
    (c)   Thermionic emission (will be study the topic "heat")
    (d)  Photo electric emission (will be studied under the topic modern physics)

  10. Charging of Insulators:
    Since charge cannot flow through insulators, neither conduction nor induction can be used to charge, insulators, so in order to charge an insulator friction is used. Whenever an insulator is rubbed against a body exchange of electrons takes place between the two. This results in appearance of equal and opposite charges on the insulator and the other body. Thus the insulator is charged. For example rubbing of plastic with fur, silk with glass causes charging of these things.
    To charge the bodies through friction one of them has to be an insulator.

 

 2.     COULOMB’S LAW

Coulombs, through his experiment found out that the two charges ‘q1’ and ‘q2’ Kept at distance ‘r’ in a medium as shown in figure-1 exert a force ‘F’ on each other. The value of force F is given by

 begin mathsize 12px style straight F equals fraction numerator Kq subscript 1 straight q subscript 2 over denominator straight r squared end fraction end style

 

This law gives the net force experienced by q1 and q2 taking in account the medium surrounding them. Where F gives the magnitude of electrostatic force q1 and q2 are the magnitudes of the two interacting charges. K is electrostatic constant which depends upon the medium surrounding the two charges. This force F acts along the line joining the two charges and is repulsive if q1 and q2 are of same sign and is attractive if they are of opposite sign. Let us take some examples on application of coulomb's Law.

 

Electrostatics-II


Chapter Cover: Questions Based on electric flux for various configuration, Question on solid angle concept, Gauss law Based Questions.

  1. ELECTRIC FLUX:
    Any group of electric lines of forces passing through a given surface, we call electric flux and it is denoted by ϕ.

  • Area as a Vector :
    Till now we have considered area of a surface as a scalar quantity but for further analysis we treat area of a surface as a vector quantity whose direction is along the normal to the surface. The area vector S of a surface which has surface area S can be written as S = Sň
    Where ň is the unit vector in the direction along normal to the surface.



    If a surface is three dimensional we consider a small elemental area dS on this surface and direction of this elemental area vector is along the local normal of the surface at the point where elemental area is chosen as shown. Thus

     begin mathsize 12px style dS with rightwards arrow on top equals dS straight a with hat on top end style
    Here â is the unit vector in the direction along the normal at elemental area dS


  • Electric Field Strength in Terms of Electric Flux:
    Earlier we’ve defined that the density of electric lines gives the magnitude of electric field strength. Mathematically the numerical value of electric field strength at a point in the region of electric field can be give as the electric flux passing through a unit normal area at that point.

    Flux = ϕ=∫E. begin mathsize 12px style dA with rightwards arrow on top end style, If E is constant, ϕ=E.A
    In a uniform electric field shown in figure. If ϕ be the flux passing through an area S which is normal to the electric field lines, the value of electric field strength at this surfaces can be given as

    begin mathsize 12px style straight E equals straight ϕ over straight S end style



    Or flux through the surface can be given as

    ϕ=ES

    If in an electric field, surface is not normal as shown in figure. Here the are ABCD is inclined at an angle Ѳ from the normal to electric field. Here we resolve the area ABCD in two perpendicular components as shown in figure. One is S cos Ѳ, which is area ABC’D’ normal to electric field direction and other is S sin Ѳ, which is area CDC’D’ along the direction of electric field.

    Here the total flux passing through the given area ABCD is same which is passing through its normal components S cos Ѳ, thus here the flux ϕ through the area can be given as
    ϕ=ES cosѲ     [S cos Ѳ=area of ABC’D’]

    If we consider the direction of area vector normal of the area surface, as shown in figure, Ѳ would be the angle between S and E. Thus flux through the surface ABCD can be given as
    ϕ=E.S

 

  1. Electric Flux in Non-uniform Electric Field
    In non-uniform electric field, we can calculate electric flux through a given surface by integrating the above expression for elemental surface area of the given surface. For this consider the situation shown in figure. If we wish to calculate electric flux through the surface M shown in figure. For this we consider an elemental area dS on the surface M as shown. At this position if electric field is E then the electric flux through this elemental area dS can be given

    dϕ = Ed cos θ

    The flux through the surface M can be given as

    begin mathsize 12px style straight ϕ equals integral dϕ equals integral for straight M of EdScosθ end style

  2. Electric flux through a circular Disc:



    Figure shows a point charge q placed at a distance? From a disc of radius R. Here we wish to find the electric flux through the disc surface due to the point charge q. We know a point charge q originates electric flux in radically upward direction. The flux of q which is originated in cone shown in figure passes through the disc surface.


    To calculate this flux, we consider an elemental ring on disc surface of radius X and width dx as shown.

    Area of this ring (strip) is

    dS=2πx dx

    The electric field due to q at this elemental ring is given as
    begin mathsize 12px style straight E equals fraction numerator Kq over denominator left parenthesis straight x squared plus straight space squared right parenthesis end fraction end style
    If dϕ is the flux passing through this elemental ring we have

    dϕ= EdScosѲ

    begin mathsize 12px style table attributes columnalign left end attributes row cell equals fraction numerator Kq over denominator left parenthesis straight X squared plus straight space squared right parenthesis end fraction cross times 2 πx space dx cross times fraction numerator blank over denominator square root of straight X squared plus straight space squared end root end fraction end cell row cell equals fraction numerator 2 πKqlxdx over denominator left parenthesis straight l squared plus straight X squared right parenthesis to the power of 3 divided by 2 end exponent end fraction end cell end table end style

    Total flux through the disc surface can be given by integrating this expression over the whole area of disc thus total flux can be given as

    begin mathsize 12px style table attributes columnalign left end attributes row cell straight ϕ equals integral dϕ equals integral from straight O to straight R of fraction numerator straight q over denominator 2 straight space straight epsilon subscript 0 end fraction fraction numerator straight x straight space dx over denominator left parenthesis straight space squared plus straight X squared right parenthesis to the power of 3 divided by 2 end exponent end fraction end cell row cell straight equals fraction numerator straight q over denominator 2 straight space straight epsilon subscript 0 end fraction integral from straight O to straight R of fraction numerator straight x space dx over denominator left parenthesis straight space squared plus straight X squared right parenthesis to the power of 3 divided by 2 end exponent end fraction end cell row cell straight equals fraction numerator straight q over denominator 2 straight space straight epsilon subscript 0 end fraction left square bracket negative fraction numerator 1 over denominator square root of straight space squared plus straight X squared end root end fraction right square bracket subscript straight O superscript straight R end cell row cell straight equals fraction numerator straight q over denominator 2 straight space straight epsilon subscript 0 end fraction left square bracket 1 over blank minus fraction numerator 1 over denominator square root of straight space squared plus straight R squared end root end fraction right square bracket end cell end table end style

    The above result can be obtained in a much simpler way by using the concept of solid angle and Gauss’s Law, shortly we’ll discuss it.


  3. Electric Flux Through the Lateral Surface of a Cylinder due to a Point Charge:

    Figure. Shows a cylindrical surface of length L and radius R. On its axis at its centre a point charge q is placed. Here we wish to find the flux coming out from the lateral surface of this cylinder due to the point charge q. For this we consider an elemental strip of width dx on the surface of cylinder as shown. The area of this strip is

    dS = 2πR. Dx

    The electric field due to the point charge on the strip can be given as  begin mathsize 12px style straight E equals fraction numerator Kq over denominator left parenthesis straight X squared plus straight R squared right parenthesis end fraction end style

    If dϕ is the electric flux through the strip, we can write

    dϕ=EdScosѲ



    begin mathsize 12px style table attributes columnalign left end attributes row cell equals fraction numerator Kq over denominator left parenthesis straight X squared plus straight R squared right parenthesis end fraction cross times 2 straight pi space Rdx cross times fraction numerator straight R over denominator square root of straight X squared plus straight R squared end root end fraction end cell row cell equals 2 straight pi straight space KqR squared cross times dx over left parenthesis straight X squared plus straight R squared right parenthesis to the power of 3 divided by 2 end exponent end cell end table end style

    Total flux through the lateral surface of cylinder can be given by integrating the above result for the complete lateral surface, which can be given as

    begin mathsize 12px style table attributes columnalign left end attributes row cell straight ϕ equals integral dϕ equals fraction numerator qR squared over denominator 2 straight space straight epsilon subscript 0 end fraction integral from negative straight L divided by 2 to plus straight L divided by 2 of dx over left parenthesis straight X squared plus straight R squared right parenthesis to the power of 3 divided by 2 end exponent end cell row cell Or      straight ϕ equals straight q over straight epsilon subscript 0. fraction numerator blank over denominator square root of straight space squared plus 4 straight R squared end root end fraction end cell end table end style

    The solution of above integration is left for students as exercise. This situation can also be easily handled by using the concepts of Gauss’s Law, we’ll discuss in next section.

  4. Electric flux Produced by a Point Charge

    The figure. shows a point charge placed at the centre of a spherical surface of radius R from which electric lines are originated and coming out of the surface of sphere. For clarity and convenience only lower half of sphere is drawn in the picture. As the charge q is inside the sphere, whatever flux it originates will come out from the spherical surface. To find the total flux, we consider an elemental area dS on surface. The electric field on the points on surface of sphere can be given as

    begin mathsize 12px style straight E equals Kq over straight R squared end style

    The electric flux coming out from the surface dS is

    begin mathsize 12px style table attributes columnalign left end attributes row cell dϕ equals straight E.. dS with rightwards arrow on top equals EdS end cell row cell left square bracket As straight space straight theta equals 0 straight space shown straight space in straight space figure right square bracket end cell row cell Thus    dϕ equals Kq over straight R squared dS end cell end table end style

    Total flux coming out from the spherical surface is

    begin mathsize 12px style straight ϕ equals integral dϕ straight equals straight integral Kq over straight R squared dS end style

    At every point of spherical surface, magnitude of electric field remains same hence we have

    begin mathsize 12px style table attributes columnalign left end attributes row cell straight ϕ equals Kq over straight R squared integral dS end cell row cell Or    straight ϕ equals Kq over straight R squared cross times 4 πR squared end cell row cell left square bracket As straight space integral dS equals 4 πR squared right square bracket end cell row cell straight ϕ equals straight q over straight epsilon subscript 0 end cell end table end style

    Thus total flux, the charge q originates is begin mathsize 12px style straight q over straight epsilon subscript 0 end style. Similarly a charge –q absorbs begin mathsize 12px style straight q over straight epsilon subscript 0 end style 

    electric lines (flux) into it.

    Figure shows a charge q enclosed in a closed surface S of random shape. Here we can say that the total electric flux emerging out from the surface S is the complete flux which charge q is originating, hence flux emerging from surface is

    begin mathsize 12px style straight ϕ subscript straight s equals straight q over straight epsilon subscript 0 end style



    The above result is independent of the shape of surface if only depends on the amount of charge enclosed by the surface.

  5. Flux Calculation in the Region of Varying Electric field:

    In a region electric field depends on x direction as

    E=E0x2

    In the cube of edge a shown in figure from front face electric

    Flux goes in which can be given as

    ϕin=E0 (2a) 2.a2=4E0a4

    From the other surface flux coming out can be given as

    ϕout = E0(3a2).a2=9E0 a4

    Here ϕoutin for the cubical surface hence net

    Flux=ϕoutin=5 E0a4



 

 

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