electrostatics (xiiith)

52
Page-1 ELECTROST A TICS-1 JEE Syllabus : Coulomb’s law; Electric field and potential; Electrical potential energy of a system of point charges and of electrical dipoles in a uniform electrostatic field; Electric field lines; Flux of electric field; Gauss’s law and its application in simple cases, such as, to find field due to infinitely long straight wire, uniformly charged infinite plane sheet and uniformly charged thin spherical shell. Complete in 12 LECTURE Introduction A number of simple experiments demonstrate the existence of electric forces and charges. For example, after running a comb through your hair on a dry day, you will find that the comb attracts bits of paper. The attractive force is often strong enough to suspend the paper. Another simple experiment is to rub an inflated balloon with wool. The balloon then adheres to a wall, often for hours. When materials behave in this way, they are said to be electrified or to have become electrically charged. You can easily electrify your body by vigorously rubbing your shoes on a wool rug. Evidence of the electric charge on your body can be detected by lightly touching (and startling) a friend. Under the right conditions, you will see a spark when you touch, and both of you will feel a slight tingle. (Experiments such as these work best on a dry day because an excessive amount of moisture in the air can cause any charge you build up to “leak” from your body to the Earth. Electric and magnetic phenomenon are generally bracketed together, since both derive from charged particles. Magnetism, arises from charges in motion. However, in the frame of reference where all charges are at rest, the forces are purely electrical. The subject of electrostatics, as the name suggest deals with the physics of charges at rest. Electro (Related to charge) + statics (stationary). Hence it deals with stationary charges Properties of Electric Charges (i) Charge comes in two varieties, which are called “plus” and “minus”, Like charges repel each other and unlike charges attract each other. Only two kinds of electric charges exist because any unknown charge that is found experimentally to be attracted to a positive charge is also repelled by a negative charge. No one has ever observed a charged object that is repelled by both a positive and a negative charge. (ii) Charge is conserved : The charge of an isolated system is conserved. The algebraic sum of charges in any electrically isolated system does not change. (iii) Charge is quantized : Protons and electron are considered the only charge carriers. All desirable charges must be integral multiples of e. If an object contains n 1 protons and n 2 electrons, the net charge on the object is n 1 (e) + n 2 (– e) = (n 1 – n 2 ) e. Thus, the charge on any object is always an integral multiple of e and can be charged only in steps of e, i.e. charge is quantized. The step size e is usually, so small that we can easily neglect the quantization. Now, 1 C contains n units of basic charge e where n = C 10 6 . 1 C 1 19 ~ 6 × 10 12 The step size is thus very small as compared to the charges usually found on many cases we can assume a continuous charge variation. This was verified by millikan oil drop experiment. TEACHING NOTES

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  • Page-1

    ELECTROSTATICS-1JEE Syllabus :

    Coulombs law; Electric field and potential; Electrical potential energy of a system of point charges andof electrical dipoles in a uniform electrostatic field; Electric field lines; Flux of electric field; Gausss lawand its application in simple cases, such as, to find field due to infinitely long straight wire, uniformlycharged infinite plane sheet and uniformly charged thin spherical shell.

    Complete in 12 LECTUREIntroduction

    A number of simple experiments demonstrate the existence of electric forces and charges. For example,after running a comb through your hair on a dry day, you will find that the comb attracts bits of paper. Theattractive force is often strong enough to suspend the paper. Another simple experiment is to rub aninflated balloon with wool. The balloon then adheres to a wall, often for hours. When materials behavein this way, they are said to be electrified or to have become electrically charged. You can easily electrifyyour body by vigorously rubbing your shoes on a wool rug. Evidence of the electric charge on your bodycan be detected by lightly touching (and startling) a friend. Under the right conditions, you will see aspark when you touch, and both of you will feel a slight tingle. (Experiments such as these work best ona dry day because an excessive amount of moisture in the air can cause any charge you build up to leakfrom your body to the Earth.Electric and magnetic phenomenon are generally bracketed together, since both derive from chargedparticles. Magnetism, arises from charges in motion. However, in the frame of reference where all chargesare at rest, the forces are purely electrical. The subject of electrostatics, as the name suggest deals withthe physics of charges at rest.Electro (Related to charge) + statics (stationary). Hence it deals with stationary charges

    Properties of Electric Charges(i) Charge comes in two varieties, which are called plus and minus, Like charges repel each other

    and unlike charges attract each other.Only two kinds of electric charges exist because any unknown charge that is found experimentally to beattracted to a positive charge is also repelled by a negative charge. No one has ever observed a chargedobject that is repelled by both a positive and a negative charge.

    (ii) Charge is conserved : The charge of an isolated system is conserved. The algebraic sum of charges inany electrically isolated system does not change.

    (iii) Charge is quantized : Protons and electron are considered the only charge carriers. All desirablecharges must be integral multiples of e. If an object contains n1 protons and n2 electrons, the net chargeon the object is

    n1 (e) + n2 ( e) = (n1 n2) e.Thus, the charge on any object is always an integral multiple of e and can be charged only in steps of e,i.e. charge is quantized.The step size e is usually, so small that we can easily neglect the quantization. Now, 1C contains n unitsof basic charge e where

    n = C106.1

    C119

    ~ 6 1012

    The step size is thus very small as compared to the charges usually found on many cases we can assumea continuous charge variation. This was verified by millikan oil drop experiment.

    TEACHING NOTES

  • Page-2

    Meaning of a charged bodyA material is said to be charged if there are more of one kind of charge than the other. A negativelycharged body has excess electrons over protons, while a positively charged body has excess positivecharges over electrons. The protons are tightly bound in the nucleus, making them very difficult to remove.Charging a body therefore involves the removal, addition, and rearrangement of the orbital electrons. Abody becomes positively charged if it loses electrons, and negatively charged if electrons are added to it.Now we study ways in which this addition, removal or rearrangement is achieved in practice.

    WAYS OF CHARGING1. Charging by friction

    Charging by friction is the oldest form of charging. It was found that when an amber rod is rubbed withfur, the rod became negatively charged. The two bodies acquire opposite signs of electricity ; one getspositively charged, while the other becomes negatively charged. When two bodies are charged byfriction, they acquire the same magnitude of charge. Furthermore, the bodies retain these excess chargeseven when they are separated from each other.Note : Charging involves transformation of mass.

    2. Charging by conductionIn charging by contact, an uncharged body is brought into contact with a charged body. The unchargedbody acquires the same sign of charge as the charged body. The total charge is distributed between thetwo bodies.

    A B

    charged body

    uncharged body

    ++ +

    + ++++

    +++++

    ++ A B

    chargingby contact

    + +

    +

    +++

    +

    +

    +++

    +

    + +

    A B

    bodies retain chargeon separation

    +

    +

    +++

    +

    + +

    ++

    +

    +

    +

    +

    +

    +

    3. Charging by inductionIn charging by induction, an uncharged body is brought close to, but not touching, a charged body.Charging by induction is an example of the rearrangement of charges between bodies. The sign of theinduced charge is opposite to that of the inducing charge. Furthermore, the induced charges last onlywhile the inducing charge is present. The induced charged disappear when the inducing charge is takenaway. (Charge can be retained if use grounding)

    Asking questionThree objects are brought close to each other, two at a time. When objects A and B are broughttogether, they attract. When objects B and C are brought together, they repel. From this, we concludethat (a) objects A and C possess charge of the same sign. (b) objects A and C possess charges ofopposite sign. (c) all three of the objects possess charges of the same sign (d) one of the objects isneutral (e) we need to perform additional experiment to determine information about the charges on theobjects.

    Note : Neutral does not mean chargeless.Note : The bodies around is are almost neutral because there is microscopic balance of -ve and +ve

    charge.

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    Asking questionIf a negatively charged body is attracting a body of unknown nature of charged body What are thepossible signs of the unknown charged body.

    ?

    -ve charged body

    Answer It can have any charge as a ve charged body can definitely attract +ve, charged body, It canas well attract neutral or -ve charged body because of induction positive charge on a neutral body willbe closer to the -ve charge. Although the charge on +ve body should be small so that effect of thedistance overrule the slight effect of repulsion.

    Note : Attraction is not the proof the nature of charge on bodies.

    COULOMBS LAW

    From experimental observations on the electric force, Coulombs law can be expressed, giving themagnitude of the electric force of interaction between two point charges:

    Fe = ke 221

    r|q||q|

    where ke is a constantThe Coulomb constant ke in SI units has the value

    ke = 8.987 5 109 Nm2/C2 9 109 Nm2/C2This constant is also written in the form

    ke = 041

    where the constant 0 (lowercase Greek epsilon) is known as the permittivity of free space and has thevalue.

    0 = 8.854 2 1012 C2 / N m2The electric force is a conservative force.

    Ex.1 The electron and proton of a hydrogen atom are separated (on the average) by a distance of approximately5.3 1011 m. Find the magnitudes of the electric force and the gravitational force between the twoparticles.

    Sol. Charge and mass of the electron and protonParticle Charge (C) Mass (kg)Electron (e) 1.602 1917 1019 9.109 5 1031Proton (P) + 1.602 191 1019 1.672 61 1027

    From Coulombs law,

    Fe = ke 2r|e||e| = (8.99 109 N m2 / C2) 211

    219

    )m103.5()C1060.1(

    = 8.2 108 N

    Using Newtons law of universal gravitation

    Fg = G 2pe

    rmm

    = (6.67 1011 N m2 / kg2) 2112731

    )m103.5()kg1067.1()kg1011.9(

    = 3.6 1047 N

    The ratio Fe / Fg 2 1039.

    Gravitational force between charged atomic particles is negligible when compared with theelectric force.

  • Page-4

    Coulombs law in vector formRemember that force is a vector quantity and must be treated accordingly. In vector form the force oncharge q1 due to charge q2 is expresed as

    F12 = ke 221

    rqq

    r {q1 & q2 should be substituted with sign.}

    r is a unit vector directed from q2 toward q1

    r +

    +

    q1

    q2r^

    F21

    F12

    (a)

    +F21

    F12

    (b)

    When the charges are of the same sign, the force is repulsive. When the charges are of opposite signs,the force is attractive.

    The electric force obeys Newtons third law, the electric force exerted by q2 and q1 is equal in magnitudeto the force exerted by q1 on q2 and in the opposite direction ; that is, F21 = F12.If q1 and q2 are of opposite sign, as shown in figure (b), the product q1q2 is negative. A negative productindicates an attractive force, so that the charges each experience a force toward the other.

    Coulomb forces follows principle of Superposition. i.e. the resultant force on any one of the charge isequal to the vector sum of the forces exerted by the various individual charges. For example, if fourcharges are present, then the resultant force exerted by charge 2, 3 and 4 on charge 1 is

    1413121 FFFF

    Charges in medium : Consider two point charges q1 and q2 kept in a medium of permittivity r (r 1and is 1 for vacuum).The medium will be consisting of atoms which are neutral but not chargeless. If any substance is presentin vicinity of a charge, the positive and negative charges of atom (nuclei and electrons) experienceelectric force, which in turn leads to a partial separation of these charges. These partially separatedcharges present on atoms apply additional electric force on charges which in combination with the forceof interaction between q1 and q2 gives the resultant field. If we know the force of interaction and theadditional electric force due to induced charges, we can forget about the presence of the substance itselfwhile calculating the resultant force, since the role of the substance has already been taken into accountwith the help of induced charges.Thus, the resultant force in the presence of a substance is determined simply as the superposition of theexternal field and the field of induced charges.

    r

    q1 q2

    + (medium is )

    F1 = F12 + F force on 1 due to polarisation.

    F1 represents net force on 1 and = 0r41 2

    21

    rqq

    But force on q1 due to q2 is still F12 = 2 2/1021

    r4qq

  • Page-5

    Eg. Very simple Eg Consider three point charges located at the corners of a right triangle as shown in figure,q1 = q3 = 5.0 C, q2 = = 2.0C, and a = 0.10m. Find the resultant force exerted on q3.

    F23F13

    q3+aq2

    a

    q1 +

    2a

    x

    y

    Sol. F23 = 9.0 N in the coordinate system shown in figure, the attractive force F23 is to the left (in the negativex direction). The magnitude of the force F13 exerted by q1 on q3 is F3 = 11N the repulsive force F23makes an angle of 45 with the x axis. Combing F13 with F23 by the rules of vector addition. To demostrate the principle of superposition.

    Eg. Two identical small charged spheres, each having a mass of 3.0 102 kg, hang in equilibrium as shownin figure. The length of each string is 0.15m, and the angle is 5.0. Find the magnitude of the charge oneach sphere. ////////////////////////////////

    aq

    L

    L

    L=0.15m = 5.0

    q

    Sol. (1) Fx = T sin Fe = 0(2) Fy = T cos mg = 0

    |q| = 4.4 108CThere is no way we could find the sign of the charge from the information given. In fact, the sign of thecharge is not important. The situation will be exactly the same whether both spheres are positivelycharged or negatively charged.

    Asking questionSuppose we propose solving this problem without the assumption that the charges are of equal magnitude.We claim that the symmetry of the problem is destroyed if the charges are not equal, so that the stringswould make two different angles with the vertical, and the problem would be much more complicated.How would you respond?

    Ans. You should argue that the symmetry is not destroyed and the angles remain the same. Newtons third lawrequires that the electric forces on the two charges be the same, regardless of the equality or none qualityof the charges. The solution of the example remains the same. The symmetry of the problem would bedestroyed if the masses of the spheres were not the same. In this case, the strings would make differentangle with the vertical and the problem would be more complicated.

  • Page-6

    The electric fieldThe concept of a field was developed by Michael Faraday. An electric field is said to exist in the regionof space around a charged object. When another charged object, qo (the test charge) enters this space,we say the test charge experiences an electric force, Fe due to this field.We define the electric field due to the source charge at the location of the test charge to be theelectric force on the test charge per unit charge.

    E 0

    e

    qF

    + ++ +

    + +

    + +

    +

    +

    +

    +

    +

    Qq0

    E

    The vector E has the SI units of newtons per coulomb (N/C). The direction of E, as shown figure, is thedirection of the force a positive test charge experiences when placed in the field It is important toremember that Electric field E is produced by charge Q, which is separate from the field produced bythe test charge itself.Also, note that the an electric field is a property of its source, the presence of the test charge is notnecessary for the field to exist. The test charge is used to detect the electric field.

    When using E 0

    e

    qF

    , we must assume that the test charge q0 is small enough that it does not disturb the

    charge distribution responsible for the electric field. If vanishingly small test charge q0 is placed near auniformly charged metallic sphere, as in figure (a) the charge on the metallic sphere, which produces theelectric field, remains uniformly distributed. If the test charge is great enough (q0 >> q0), as in figure (b)the charge on the metallic sphere is redistributed and the ratio of the force to the test charge is different: (Fe / q0 Fe/q0). That is, because of this redistribution of charge on the metallic sphere, the electricfield it sets up is different from the field it sets up in the presence of the much smaller test charge q0..

    + +q0 q >> q0 0

    (a) (b)

    Why concept of electric field necessary?Modern understanding of electric interaction between two charges is visualized in terms of the electricfield concept. A charge produces an electric field around itself ; this field then exerts force on the othercharge. Thus, the interaction between two charges is a two step process.

    For two charges, the measurable quantity is the force on a charge which can be directly determined usingCoulombs law . Why then introduce this intermediate quantity called the electric field?

    When charges are stationary, the concept of electric field is convenient, but not really necessary. Electricfield in electrostatics is an elegant way of characterising the electrical environment of a system of charges.

    The true physical significance of the concept of electric field, however,emerges only when we gobeyond electrostatics and deal with time-dependent electromagnetic phenomena.

    Suppose we consider the force between two distant charges q1, q2 in accelerated motion. The greatest

  • Page-7

    speed with which a signal or information can go from one point to another is c, the speed of light. Thus,the effect of any motion of q1 on q2 cannot arise instantaneously. There will be some time delay betweenthe effect (force on q2) and the cause (motion of q1). It is precisely here that the notion of electric field(strictly, electromagnetic field) is natural and very useful. The field picture is this: the accelerated motionof charge q1 produces electromagnetic waves, which then propagate with the speed c, reach q2 andcause a force on q2. The notion of field elegantly accounts for the time delay. Thus, even though electricand magnetic fields can be detected only by their effects (forces) on charges, they are regarded asphysical entities, not merely mathematical constructs. They have an independent dynamics of their own,i.e., they evolve according to laws of their own. They can also transport energy. Thus concept of field isnow among the central concepts in physics.

    Field due to a point chargeAccording to Coulombs law, the force exerted by point charge q on the test charge q0 is :

    rr

    qqkF 20

    ee

    +q r

    P

    Fq0

    r^

    (a)+q r

    P

    F

    r^

    (b)

    where r is a unit vector directed from q toward q0. By E = Fe/q0, the electric field created by q is :

    rrqkE 2ee

    The source charge sets up an electric field at point P, directed away from q.

    +q F

    P

    q0

    r^

    (c)q

    Er^

    (d)

    Pr

    If q is negative, as in figure (c), the force on the test charge is toward the source charge, so the electricfield at P is directed toward the source charge, as in figure (d)

    Superposition of electrostatic fieldsIf we are dealing with many charges then electric field at a point p is the vector sum

    ii

    21

    ie rr

    qkE

    where ri is the distance from the ith source charge q, to the point P and ir is a unit vector directed fromqi toward P. If some more charge are added, more terms are added to the summation. However, thereis no change to the terms that were already there, provided that the original charges do not move. If weknow the electric fields generated by two different sets of charges separately, the electric field generatedby both together is simply the vector sum of the two separate fields. The two fields, which each occupythree-dimensional space, are superimposed on one another. Because it has this property, the electricfield is said to satisfy the principle of superposition.

  • Page-8

    IllustrationsEx. Two identical positive point charges q are placed on the x-axis at

    x = a and x = + a, as shown in figure. (i) Plot the variation of E along the x-axis.(ii) Plot the variation of E along the y-axis.

    Sol. (i)

    E

    +q+q xx = a x = + a

    The variation of along the x-axis. Electric field directed along the positive x-axis is taken as positive

    E on point charge is undefined

    (ii) The direction of electric field along the positive y-axis is taken as positive.

    Ey = 2/322 )ay(kqy2

    Emax = 278

    2akq

    a / 2a / 2

    y

    E

    occurs at

    y = 2

    a

    Ex. Three point charges lie along the x axis as shown in figure. The positive charge q1 = 24.0 C is atx = 3.00m, the positive charge q2 = 6.00 C is at the origin, and the resultant force acting on q3 is zero.What is the x coordinate of q3?

    + F23q2

    x

    +F13 q1

    x

    3.00m

    3.00m-x

    q3

    Sol. 232

    e x|q||q|k = ke 2

    31

    )x00.3(|q||q|

    Noting that ke and |q3| are common to both sides and so can be dropped, we solve for x and find that(3.00 x)2 |q2| = x2 |q1|

    This can be reduced to the following quadratic equation :solving this quadratic equation for x, we find that the positive root is x = 1m. There is also a second root,x = 3m. This is another location at which the magnitudes of the forces on q3 are equal, but both forcesare in the same direction at this location.

  • Page-9

    Asking questionSuppose charge q3 is constrained to move only along the x axis. From its initial position at x = 1m, it ispulled a very small distance along the x axis. When released, will it return to equilibrium or be pulledfurther from equilibrium? That is, is the equilibrium stable or unstable?

    Ans. If the charge is moved to the right, F13 becomes larger and F23 becomes smaller. This results in a netforce to the right, in the same direction as the displacement. Thus, the equilibrium is unstable.Note that if the charge is constrained to allowed to move up and down along y -axis in figure, theequilibrium is stable. In this case, if the charge is pulled upward (or downward) and released, will moveback toward the equilibrium position and undergo oscillationExplain using graph of E vs x.}To learn stable, unstable and nuetral equilibrium.

    Ex. Four identical charges are fixed at the corners of a square of side a. Find electric field at point p which isat a distance z lying on the line perpendicular to the plane of the square passing through the centre ofsquare.

    q

    q

    q

    q

    p

    Ans. 30aq

    22.qz

    Electric field of a continuous charge distribution :The total electric field at P due to all elements in the charge distribution is approximately.

    ii

    21

    ie rr

    qkE

    Considering the charge distribution as continuous, the total field at P in the limit qi 0 is

    i

    i21

    i0qe rr

    qlimkEi

    = ke rr

    dq2

    OPTIONALProblem solving tactics for calculating the electric field from continuous chargedistributions

    1. Identify the type of charge distribution and compute the charge density or .2. Divide the charge distribution into infinitesimal charges dq, each of which will act as a tiny point charge.3. The amount of charge dq, i.e., within a small element dl, dA or dV is

    dq = dl (charge distributed in length)dq = dA (charge distributed over a surface)dq = dV (charge distributed throughout a volume)

  • Page-10

    4. Draw at point P the dE vector produced by the charge dq. The magnitude of dE is

    dE = 04

    1 2r

    dq

    Vector dE is along radial line joining dq to P, dE is directed away for positive charge dq while directedtowards dq for negative dq.

    5. Resolve the dE vector into its components. Identify any special symmetry features to show whether anycomponent(s) of the field that are not cancelled by other components.

    6. Write the distance r and any trigonometric factors in terms of given coordinates and parameters.7. The electric field is obtained by summing over all the infinitesimal contributions.

    EdE

    = 20r4dq

    8. Perform the indicated integration over limit of integration that include all the source charges.

    Ex. A thin rod of length has a uniform positive charge per unit length . Calculate the electric field at a pointP that is located along the long axis of the rod and a distance a from one end

    xE

    y

    Pa l

    xdx

    dq= dx

    Ans. )( alalke

    Asking questionSuppose we move to a point P very far away from the rod. What is the nature of the electric field at sucha point ?

    Ans. If P is far from the rod (a >> ), then in the denominator of the final expression for E can be neglected,and E kel/a2.

    Electric field due to finite rod at perpendicular distance x from the wire.y = x tan dy = x sec2 d

    dEx = 2)secx(cosdyK

    P

    dy

    y

    x

    dEy = 2)secx(sindyK

    so Ex = xK

    (sin 2 sin 1)

    Ey = xK

    (cos 1 cos 2)

    1 and 2 are to be used with sign.

    For eg in this figure 1 = 3

    ,2 = 4

    /3/4 P

  • Page-11

    Results:(i) For semi infinite wire

    E x = xK

    E y = xK

    O

    E = xK2

    at 45 with OP.

    (ii) For infinite wire

    Ex = 2 xK

    Ey = 0

    O

    Electric field due to arc = charge distribution / unit length. Find field at the centre due to arc.

    drRr

    d

    dRd

    So only the horizontal components are cancelled and only verticalcomponents are added.

    dE = 2/

    02R

    cos)dR(K

    netE

    = R2sinK2

    Electric field due to the ring(at a point P lying a distance x from its center along the central axis perpendicular to the plane of the ring)

    +

    +

    ++

    ++ +

    +++

    +

    +++

    +

    PdEx

    dEdE

    x

    r

    dq

    a

    The magnitude of the electric field at P due to the segment of charge dq is

    dE = ke 2rdq

    This field has an x component dEx = dE cos along the x-axis and a component dE perpendicular tothe x-axis. The resultant field at P must lie along the x-axis because the perpendicular components of thefield created by any charge element is canceled by the perpendicular component created by anelement on the opposite side of the ring.

  • Page-12

    +

    +

    ++

    ++ +

    ++

    +

    +

    +++

    +

    dE2

    dE1

    1

    2

    dEx = dE cos =

    2e r

    dqkrx

    = dq)ax(xk

    2/322e

    All elements of the ring make the same contribution to the field at P because they are all equidistant fromthis point. Thus, we can integrate to obtain the total field at P

    Ex = dq)ax(xk

    2/322e = 2/322

    e

    )ax(xk

    Q

    +

    +

    ++

    ++ +

    +++

    +

    +++

    +

    PdEx

    dEdE

    x

    r

    dq

    a

    ++

    ++

    ++ +

    ++

    +

    +

    +++

    +

    dE2

    dE1

    1

    2

    Ex. A thread carrying a uniform charge per unit length has the configurations shown in Fig. a and b.Assuming a curvature radius R to be considerably less than the length of the thread, find the magnitude ofthe electric field strength at the point O.

    RO

    (a)R

    O

    (b)

    [Ans. (a) E = R42

    0

    ; (b) E = 0 ]

    Ex. Suppose a negative charge is placed at the center of the ring in and displaced slightly by a distancex

  • Page-13

    Asking questionWhat will be the changes in the result if the ring is non uniformly charged.

    Ex. A system consists of a thin charged wire ring of radius R and a very long uniformly charged threadoriented along the axis of the ring, with one of its ends coinciding with the centre of the ring. The totalcharge of the ring is equal to q. The charge of the thread (per unit length) is equal to . Find the interactionforce between the ring and the thread.

    [Ans. F = R4q

    0

    ]

    Electric field due to disk (at a point P that lies along the central perpendicular axis of the disk and adistance x from the center of the disk.Let disk has radius R has a uniform surface charge density .

    dq

    x Pdr

    r

    R

    The ring of radius r and width dr shown in has a surface area equal to 2rdr. The charge dq on this ringis 2r dr. Using this result of field due to the ring

    dEx = 2/322e

    )rx(xk

    (2r dr)

    To obtain the total field at P, we integrate this expression over the limits r = 0 to r = R. x is a constant

    Ex = kex R

    02/322 )rx(

    drr2

    =

    2/122e )Rx(

    x1k2

    This result is valid for all values of x > 0 and x> x ; thus, the expression in bracket reduces to unity to give usthe near-field approximation.

    Ex = 2ke = 02

    * Explain that the field created by a uniformly charged infinite sheet is same..

  • Page-14

    ELECTRIC FIELD LINES and FLUX

    Concept of field linesWe have already studied electric field in the few lectures. It is a vector quantity and can be representedas we represent vectors. Let us try to represent E due to a point charge pictorially. Let the point chargebe placed at the origin. Draw vectors pointing along the direction of the electric field with their lengthsproportional to the strength of the field at each point. Since the magnitude of electric field at a pointdecreases inversely as the square of the distance of that point from the charge, the vector gets shorter asone goes away from the origin, always pointing radially outward. Figure. shows such a picture.

    In this figure, each arrow indicates the electric field, i.e., the force acting on a unit positive charge, placedat the tail of that arrow. Connect the arrows pointing in one direction and the resulting figure representsa field line. We thus get many field lines, all pointing outwards from the point charge.

    Relative density of field lines represent magnitude of electric fieldHave we lost the information about the strength or magnitude of the field now, because it was containedin the length of the arrow? No. Now 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. Another person may draw more lines. But the number of lines is not important. Infact, an infinite number of lines can be drawn in any region. It is the relative density of lines indifferent regions which is important. We draw the figure on the plane of paper, i.e., in two dimensionsbut we live in three-dimensions. So if one wishes to estimate the density of field lines, one has to considerthe number of lines per unit cross-sectional area, perpendicular to the lines. Since the electric fielddecreases as the square of the distance from a point charge and the area enclosing the charge increasesas the square of the distance, the number of field lines crossing the enclosing area remains constant,whatever may be the distance of the area from the charge. We started by saying that the field lines carryinformation about the direction of electric field at different points in space. Having drawn a certain set offield lines, the relative density (i.e., closeness) of the field lines at different points indicates the relativestrength of electric field at those points. The field lines crowd where the field is strong and are spacedapart where it is weak. Figure shows a set of field lines.

    Relative density of field lines is inversely propotional to square of distanceWe can imagine two equal and small elements of area placed at points R and S normal to the field linesthere. The number of field lines in our picture cutting the area elements is proportional to the magnitudeof field at these points. The picture shows that the field at R is stronger than at S. To understand thedependence of the field lines on the area, or rather the solid angle subtended by an area element, let us

  • Page-15

    try to relate the area with the solid angle, a generalization of angle to three dimensions. Recall how a(plane) angle is defined in two-dimensions. Let a small transverse line element l be placed at a distancer from a point O. Then the angle subtended by l at O can be approximated as = l/r. Likewise, inthree-dimensions the solid angle subtended by a small perpendicular plane area S, at a distance r, canbe written as = S/r2. We know that in a given solid angle the number of radial field lines is the same.In Fig., for two points P1 and P2 at distances r1 and r2 from the charge, the element of area subtendingthe solid angle is r12 at P1 and an element of area r22 at P2, respectively. The number of lines(say n) cutting these area elements are the same. The number of field lines, cutting unit area element istherefore n / (r12 ) at P1 and n / (r22 ) at P2, respectively. Since n and are common, thestrength of the field clearly has a 1 / r2 dependence.

    Drawing field linesThe picture of field lines was invented by Faraday to develop an intuitive non- mathematical way ofvisualizing electric fields around charged configurations. Faraday called them lines of force. This term issomewhat misleading, especially in case of magnetic fields. The more appropriate term is fieldlines (electric or magnetic) that we will use. Electric field lines are thus a way of pictorially mappingthe electric field around a configuration of charges. An electric field line is, in general, a curve drawn insuch a way that the tangent to it at each point is in the direction of the net field at that point. An arrow onthe curve is obviously necessary to specify the direction of electric field from the two possible directionsindicated by a tangent to the curve. A field line is a space curve, i.e., a curve in three dimensions. Figureshows the field lines around some simple charge configurations.

    As mentioned earlier, the field lines are in 3-dimensional space, though the figure shows them only in aplane. The field lines of a single positive charge are radially outward while those of a single negativecharge are radially inward. The field lines around a system of two positive charges (q, q) give a vividpictorial description of their mutual repulsion, while those around the configuration of two equal andopposite charges (q, q), a dipole, show clearly the mutual attraction between the charges.

    Properties of field lines:The field lines follow some important general properties:

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

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

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    (iv) Electrostatic field lines do not form any closed loops. This follows from the conservative nature ofelectric field.

    Illustration 1. Figure shows the sketch of field lines for two point charges 2Q and Q.

    E = 0

    2Q Q

    The pattern of field lines can be deduced by considering the following points:(a) Symmetry : For every point above the line joining the two charges there is an equivalent point belowit. Therefore, the pattern must be symmetrical about the line joining the two charges.(b) Near field : Very close to a charge, its own field predominates. Therefore, the lines are radial andspherically symmetric.(c) Far field : Far from the system of charges, the pattern should look like that of a single point charge ofvalue (2Q Q) = + Q, i.e., the lines should be radially outward.(d) Null point : There is one point at which E = 0. No lines should pass through this point.(e) Number of lines : Twice as many lines leave + 2Q as entre Q.Figure (a) shows the incorrectly and (b)shows the correctly drawn field lines for a collection of fourcharges Q, 4Q, 2Q and +Q.

    -Q +4Q

    -2Q

    -Q

    P

    (a)

    -Q +4Q

    -2Q

    -Q(b)

    ELECTRIC FLUXAnalogy with flow of water and concept of flux

    Consider flow of a liquid with velocity v, through a small flat surface dS, in a direction normal to thesurface. The rate of flow of liquid is given by the volume crossing the area per unit time v dS andrepresents the flux of liquid flowing across the plane. If the normal to the surface is not parallel to thedirection of flow of liquid, i.e., to v, but makes an angle with it, the projected area in a plane perpendicularto v is v dS cos . Therefore the flux going out of the surface dS is n.v dS.For the case of the electric field, we define an analogous quantity and call it electric flux.We should however note that there is no flow of a physically observable quantity unlike the case of liquidflow. In the picture of electric field lines described above, we saw that the number of field lines crossinga unit area, placed normal to the field at a point is a measure of the strength of electric field at that point.This means that ifwe place a small planar element of area S normal to E at a point, the number of field lines crossing it isproportional to ES. Now suppose we tilt the area element by angle . Clearly, the number of field linescrossing the area element will be smaller. The projection of the area element normal to E is S cos.Thus, the number of field lines crossing S is proportional to ES cos . When = 90, field lines willbe parallel to S and will not cross it at all (Figure).

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    Area as a vectorThe orientation of area element and not merely its magnitude is important in many contexts. For example,in a stream, the amount of water flowing through a ring will naturally depend on how you hold the ring. Ifyou hold it normal to the flow, maximum water will flow through it than if you hold it with some otherorientation. This shows that an area element should be treated as a vector. It has a magnitude and also adirection. How to specify the direction of a planar area? Clearly, the normal to the plane specifies theorientation of the plane. Thus the direction of a planar area vector is along its normal. How to associatea vector to the area of a curved surface? We imagine dividing the surface into a large number of verysmall area elements. Each small area element may be treated as planar and a vector associated with it, asexplained before.Notice one ambiguity here. The direction of an area element is along its normal. But a normal can pointin two directions. Which direction do we choose as the direction of the vector associated with the areaelement? This problem is resolved by some convention appropriate to the given context. For the case ofa closed surface, this convention is very simple. The vector associated with every area element of aclosed surface is taken to be in the direction of the outward normal. This is the convention used in Fig.

    Thus, the area element vector S at a point on a closed surface equals S n where S is the magnitudeof the area element and n is a unit vector in the direction of outward normal at that point.

    Vectorial definition of fluxWe now come to the definition of electric flux. Electric flux through an area element S is defined by

    = E.S = E S coswhich, as seen before, is proportional to the number of field lines cutting the area element. The angle here is the angle between E and S. For a closed surface, with the convention stated already, is theangle between E and the outward normal to the area element. Notice we could look at the expression ES cos in two ways: E (S cos ) i.e., E times the projection of area normal to E, or ES i.e.,component of E along the normal to the area element times the magnitude of the area element. The unitof electric flux is N C1 m2.

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    The basic definition of electric flux given by Eq. = E.S = E S cos can be used, in principle, tocalculate the total flux through any given surface. All we have to do is to divide the surface into small areaelements, calculate the flux at each element and add them up. Thus, the total flux through a surface Sis

    ~ E SThe approximation sign is put because the electric field E is taken to be constant over the small areaelement. This is mathematically exact only when you take the limit S 0 and the sum in Eq. ~ E S is written as an integral.

    Gausss LawLet us again consider a positive point cahrge q located at the centre of a sphere of radius r, as shown infigure

    +

    r

    qdA

    E

    Gaussiansurface

    from equation E = ke 2rq

    r we know that the magnitude of the electric field everywhere on the surface

    of the sphere is E = keq/r2. The field lines are directed radially outward and hence are perpendicular tothe surface at every point on the surface. That is, at each surface point, E is parallel to the vector Airepresenting a local element of area Ai srrounding the surface point. Therefore,

    E Ai = EAiand from equation E = dAE = dAEn we find that net flux through are gaussian surface is

    E = dAE = dAE = E dAwhere we have moved E outside of the integral because, by symmetry, E is constant over the surface andgiven by E = keq / r2. Furthermore, because the surface is spherical dA = A = 4r2. Hence, the net fluxthrough the gaussian surface is

    E = 2e

    rqk

    (4r2) = 4keq

    we know that ke = 1/4 0, we can write this equation in the form

    E = 0

    q

    Note from equation E = 0

    q that the net flux through the spherical surface is proportional to the charge

    inside. The flux is indepdent of the radius r because the area of the spherical surface is proportional to r2,whereas the electric field is proportional to 1/r2. Thus, in the product of area and electric field, thedependence on r cancels.

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    Asking questionWhat if the charge in figure were not located at the center of the spherical gaussian surface ?

    +

    r

    qdA

    E

    Gaussiansurface

    Ans. In this case, the situation does not possess enough symmetry to evaluate the electric field. Because thecharge is not at the center, the magnitude of E

    would vary over the surface of the sphere and the vector

    E would not be everywhere perpendicular to the surface. But still the flux through the spherical surface

    is q / 0Now consider several closed surfaces surrounding a charge q, as shown in.

    S3S2

    S1

    Surface S1 is spherical, but surface S2 and S3 are not. From equation E = 0

    q , the flux that passes

    through S1 has the value q / 0. As we discussed in the preceding section, flux is proportional to thenumber f electric field lines passing through a surface. The construction shown in figure shows that thenumber of lines through S1 is equal to the number of lines through the nonspherical surfaces S2 and S3.Thefore, we conclude that the net flux through any closed surface surrounding a point charge q isgiven by q / 0 and is independent of the shape of that surface.

    Flux through closed surface for a charge kept outside the surfaceNow consider a point charge located outside a closed surface of arbitary shape, as shown.

    q

    As you can see from this construction, any electric field line that enters the surface leaves the surface atanother point. The number of electric field lines entering the surface equals the number leaving the surface.Therefore, we conclude that the net electric flux through a closed surface that surrounds no charge iszero.

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    Gauss Law for multiple chargesLet us calculate flux for many charges. We once again use the superposition principle, which states thatthe electric field due to many charges is the vector sum of the electric fields produced by the individualcharges. Therefore, we can express the flux through any closed surface as :

    Ad.....).EE(Ad.E 21

    where E

    is the total electric field at any point on the surface produced by the vector addition of theelectric fields at that point due to the individual charges. Consider the system of charges shown in figure.

    q2

    q3 S

    q1

    S q4

    S

    The surface S surrounds only one charge, q1 ; hence, the net flux through S is q1 / 0. The flux throughS due to charges q2, q3 and q4 outside it is zero because each electric field line that enters S at one pointleaves it at another. The surface S surrounds charges q2 and q3 ; hence, the net flux through it is (q2 +q3)/ 0. Finally, the net flux through surface S is zero because there is no charge inside this surface. Thatis, all the electric field lines that enter S at one point leave at another. Notice that charge q4 does notcontribute to the net flux through any of the surfaces because it is outside all of the surface.Gausss law, which is a generalization of what we have just described, states that the net flux throughany closed surface is :

    E = 0inqAd.E

    where qin represents the net charge inside the surface and E represents the electric field at any point on

    the surface.Some points to be emphasized about the gauss law :

    (i) It is true for any closed surface no matter what its shape is size.

    (ii) The q includes sum of all charges enclosed by the surface.

    (iii) In situations when the surface is so chosen that there are some charges inside and some outside, the E

    (whose flux appear in the equation) is due to all charges, just term q in the law represents only totalcharge inside.

    (iv) The gaussian surface should not pass through any discrete charge however, it can pass through acontinuous charge distribution.

    Application of gauss lawDefinition of a Gaussian surface : While applying Gausss law we are interested in evaluating the integral

    E = dAEThe closed surface for which the flux is calculated is generally an imaginary or hypothetical surface,called a Gaussian surface. Whenever we apply Gausss law we may choose a surface of any size andshape as our Gaussian surface. But selecting a proper size and shape for a Gaussian surface is a keyfactor for determining flux. Here are the list of different types of the Gaussian surfaces to be chosen fora given charge distribution.

  • Page-21

    Charge distribution Gaussian surface Electric fieldPoint charge Spherical RadialSpherical charge distribution Spherical RadialLine of charge Cylindrical RadialPlanar charge Cylindrical Normal to surface

    Ex Consider a cube of edge a, kept in a uniform electric field of magnitude E, directed along x-axis asshown in Figure.

    1

    AdA1

    l

    Ez FdA

    4l 2

    x

    dA2G

    C E

    dA3 3D

    B

    Sol. The net flux is algebraic sum of the flux through all the faces of the cube.Note that flux through faces ABFE, BCGE, ADHE, CDHG is zero because E is normal to area vectoron these faces.Flux through EFGH,

    (E)EFGH = E(a2) cos 180

    = Ea2Area vector and electric field vector are opposite to each other.Flux through ABCD,

    (E)ABCD = E(a2) cos 0

    = + Ea2Area vector and electric field vector are parallel to each other.Net flux over all the six faces is

    (E) = (E)ABFE + (E)BCGF + (E)ADHE+ (E)CDHG + (E)ABCD + (E)EFHG= 0 + 0 + 0 + 0 + ( Ea2) + Ea2= 0

    Ex. Find electric flux through square of side a, due to charge placed at distance a/2 from centre of a square.

    a/2

    Sol. Let us enclose the charge q by a cubical gaussian surface with q at its centre.

    q

    a

    a/2

    By symmetry all face have equal flux

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    = 06

    q

    Objective : Flux from a point charge is distributed symmetrically

    Ex. A point charge q is placed on the apex of a cone of semi-vertex angle . Show that the electric flux

    through the base of the cone is 02

    )cos1(q

    .

    Sol. Consider a Gaussian sphere with its centre at the apex and radius the slant length of the cone. The fluxthrough the whole sphere is q / 0. Therefore, the flux through the base of the cone,

    E =

    0AA

    0

    q

    Here, A0 = area of whole sphere = 4R2

    and A = area of sphere below the base of the cone.Consider a differential ring of radius r and thickness dr.

    dA = (2r) R d

    (b)

    x

    R

    C

    Rdr

    = (2R sin ) R d as r = R sin = (2R2) sin d

    A =

    0

    2 dsin)R2(

    A = 2R2 (1 acos )

    The desired flux is E =

    0AA

    0

    q

    = )R4()cos1()R2(

    2

    2

    0

    q = 02

    )cos1(q

    Objective : Introduction to application of solid angle

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    OPTIONALEx. A charge is placed at the centre of a cylindrical surface. Find total flux passing through lateral curved

    surface.

    q

    2R

    45R

    Sol. Total flux linked can be divided into parts(i) Flux through lateral surface L(ii) Flux through end caps AIf end cap substends solid angle at centre.

    L + 2A = 0

    q

    A = 0

    q

    4

    Note : Solid angle is given by = 2 (1 cos )

    = 0

    q

    24

    2112

    = 2

    211q

    0

    L = 02q ]

    Calculation of electric field using Gauss Law :I. Shell

    A charged shell with total charge Q distributed uniformly on the surface of shell find field at x.(i) x > RStudent will attemp it like thisLet gaussian surface of r = x

    E

    dA = 0

    Q

    x

    QR

    Possible fielddirections

    E 4x2 = 0

    Q

    E = 20x4

    Q

    {Emphasize this point}Note : But this soln is incomplete because, this law only gives E

    along dA, and how do we know that E

    is along Ad

    .

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    Proper LogicField cant be along tangent no tangential component can exist because of symmetry of charge distribution.Only direction possible is radial.

    AdE

    = 0

    Q

    AdE

    = 0

    Q

    Also by symmetry field at each point an gaussian surface has same magnitude is same, so Ecan be taken out of integral

    E dA = 0Q

    E 4x2 = 0

    Q

    E = 2

    0 x4Q

    ]

    (ii) x < RIn this case gaussian surface does not include any charge

    AdE

    = 0 [as charge in gaussian surface is zero]Student can interpret it in this manner E

    is 0

    RQ

    x

    Gaussian surface

    Emphasize that AdE

    = 0 may also means E

    to Ad

    or E is 0 or AdE

    = 0

    Emphasize proper method

    AdE

    = 0

    By symmetry E

    is along Ad

    Also dAE

    0 as all will be either +ve or ve by symmetry

    AdE

    = 0

    Now dA 0 0E

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    II. Uniformly charged sphere with const volumetric charge density.(i) field at outside point x > RBy symmetry we see that tangentical components of E is cancelled and net field is radial only.

    AdE

    = 0

    3R34

    Also there is uniform charge distribution E at every point has same magnitude and directed radially.

    dAE = 0

    3R34

    x

    R E

    Gaussian surface E 4x2 = 0

    3

    3R4

    E = 20

    3

    x4

    R34

    x > R ; Hence proved.

    (iii) Field at inside pointsx < R

    AdE

    = 0

    3x34

    x

    Gaussian surface

    Again by symmetry E is only along radial direction.

    E Ad

    = 0

    3x34

    E 4x2 = 0

    3

    3

    x34

    E = 03x

    x < R

    Hence proved.

    OPTIONALEx. Charge is distributed throughout a spherical region of space in such a manner that its volume charge

    density is given by = ar2, 0 r R

    where a s is a constant. Find the field at distance r from the center.

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    dr

    dv = 4 r dr 2r

    Y

    R

    Z X

    Sol. Since charge distribution is volumetric we choose a small volume element dV within the sphere. Becausethe charge distribution is spherically symmetric, we choose a thin spherical shell of radius r and thicknessdr. We can think of entire spherical charge to be made up of concentric shells.Area of sphere is 4r2.

    Therefore, dV = 4r2dr

    q = dV = R

    0

    22 )drr4()ar(

    = 5ar54

    E = 04

    1 . 5

    4 . ar3

    Ex. Field inside a cavity :Consider a spherically symmetric charge distribution of charge density , with a cavity inside it. Fieldinside cavity can be obtained by super position of electric field due to complete sphere and electric fielddue to cavity part i.e. at the given point subtract the contribution of electric field due to cavity part fromfield of complete sphere.

    Sol. Contribution of complete sphere sphereE

    = 03r

    Ecavity = 03x

    rl xResultant field E

    = sphereE

    + cavityE

    E

    = 03

    )xr(

    cavityinsideE

    = l

    03

    So field inside cavity is uniform.

    Objective : (i) Cavity can be considered having equal amount of positive and negative charge(ii) Cavity of uniform charged non conducting sphere has uniform E.

    III. Infinite sheet, constant charge density We choose a small closed cylinder whose axis is perpendicular to plane as our gaussian surface. Fromsymmetry we expect E to be directed perpendicular to the plane on both sides as shown, and to beuniform over the end caps of the cylinder.

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    x

    AE dA

    E

    Gaussion surface

    AdE

    = 0

    A

    Ad.E

    + Ad.E

    + Ad.E

    = 0

    A

    curved left cap right cap

    Now if the consider a plane through x, charge above and below is same, by symmetry again E is onlyalong Ad

    .

    through lateral surface is zero as E is along x-axis and here Ad

    is perpendicular to it.

    Ad.E

    = 0curved

    Thus, E

    = 02

    Asking question :Why we should consider the cylindrical gaussian surface in front as well as behind the sheet?

    IV. Hollow cylinder (infinite length)What type of charge distribution cylinder can have?(i) Field outside cylinder x > RLinear charge distribution and surface charge distributionLet us consider a coaxial cylindrical gaussian surface.

    AdE

    = 0l

    By previous proved reason l

    x

    E x 2 x l = 0l

    E = 0x2

    (ii) Field inside cylinderx < Rq enclosed = 0

    AdE = 0

    Now E

    is uniform throughout and radialAlso E

    is along Ad

    in region x < R

    2

    3

    1

    3

    so this shows that E = 0 E 0 x < R

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    ELECTRIC POTENTIAL

    The electrostatic Potential Energy

    The concepts of work and potential energy were discussed in class XI. The work done by a force wasdefined as

    Work done = force distance moved by the point of application of force.

    Examples considered in class XI included the work done against the gravitational force in lifting a mass,and against the restoring force of a spring when it is stretched. In both cases energy must be expendedto do the work, but this energy does not disappear. It is stored as potential energy, which may later bereleased : gravitational potential energy may, for example, be released by allowing an object to fall.

    Work done by electric charges

    The concepts of work and potential energy also apply when the forces are electrical. Consider the twopositive charges q and q1 as shown in figure.

    ri

    rf

    A B Cqq1

    Lets say the charge q1 is fixed, but q may be moved. There is a repulsive force between the two chargesand when they are separated by a distance r the magnitude of the force is given by equation

    F = 20

    1

    r4qq

    The force on q acts in the direction AB. If q moves a distance dr along the line BC, the work done by theforce is Fdr. The amount of work done when q moves from B to C,

    W = f

    i

    r

    rFdr =

    f

    i

    r

    r2

    0

    1

    r4qq

    dr = 0

    1

    4qq

    f

    i

    r

    rr1

    = 0

    1

    4qq

    fi r1

    r1

    This work represents the difference in the electrical potential energy of the system of the two chargeswhen q moves from B to C. It is natural to choose the potential energy to be zero at rf = , and with thischoice the total potential energy U of the two charges when they are at A and B, separated by a distanceri, is

    U = W = i01

    r4qq

    Equation U = W = i01

    r4qq is still valid, but if q and q1 have different signs the right-hand side is

    negative as the work must be done to pull the charges apart.The potential energy of q depends only on its distance from q1 and not on the direction.

    Note : Electric force is a Central force.Central force : Any force satisfying

    )r(f|F|

    i.e. function of distance from a fixed point

    rF i.e. directed along line joiningAll Central forces are conservative force.

  • Page-29

    The Electrostatic Potential

    The potential energy per unit charge U/q0 is independent of the value of q0 and has a value at every pointin an electric field. This quantity U/q0 is called the electric potential (or simply the potential) V. Thus, theelectric potential at any point in an electric field is :

    V = 0q

    U

    Since potential energy is a scalar quantity so the electric potential also is a scalar quantity.If the test charge is moved between two positions A and B in an electric field, the charge-field systemexperiences a change in potential energy. Potential is defined as the change in potential energy of thesystem when a test charge is moved between the points divided by the test charge q0.

    V = 0qU

    = B

    Ad.E l

    1. Units : Joule / coulomb = volt

    2. Dimension : qw

    = AT

    TLM 221 = M1L2T3A1

    3. It is a scalar quantity. As work can be +ve or ve, the electric potential can also be +ve or ve.

    4. Just as with potential energy, only differences in electric potential are meaningful. We often take the valueof the electric potential to be zero at some convenient point in an electric field.The most general referencepoint is infinite & potential at is assumed to be zero. (This fails for line charge & sheet).

    5. Electric potential is a scalar characteristic of an electric field, independent of any charges that may beplaced in the field.

    Electric fields lines always point in the direction of heighest decreasing electric potential

    d

    qB

    A

    E

    d

    mB

    A

    g

    can be explained using analogy with gravity..

    Now suppose that a test charge q0 moves from A to B. We can calculate the change in the potentialenergy of the charge-field system from Equations :

    U = q0V = q0Ed

    From this result, we see that if q0 is positive, then U is negative.

  • Page-30

    Ex. Calculate Potential difference between points A and B in front of Infinite charge sheet.

    VB VA = V = B

    A

    sd.E

    = B

    A

    ds)0cosE( = B

    A

    sd.E

    Because E is constant, we can remove it from the integral sign ; this gives

    V = E B

    A

    Edds

    The negative sign indicates that the electric potential at point B is lower than at point A ; that is, VB < VA.

    Ex. If j4i3E

    N/C and potential at origin is zero find potential at (2, 4) and find work required to takea particle of charge 10C from (1,1) to (2,2).

    Asking QuestionsQ.1 In figure, two points A and B are located within a region in which

    there is an electric field. The potential difference V = VB VA is(a) positive (b) negative (c) zero.

    Q.2 In figure. a negative charge is placed at A and then moved to B. Thechange in potential energy of the charge-field system for this processis (a) positive (b) negative (c) zero.

    B

    A

    E

    Equipotential surfaces :

    For an isolated charge q1 the electrostatic potential at a distance r fromq1 is q1 / (40r). All points on the surface of a sphere of radius r areat the same potential. Spherical surfaces centred on q1 are equipotentialsurfaces.

    q1

    Electric field is perpendicular to the equipotential surface. This is obvious for a single charge, for whichthe field lines are radial and the equipotential are spherical. Electric field is always perpendicular toequipotential surfaces, no matter what the distribution of charge. This is easily proved by considering asmall movement of a test charge on an equipotential surface. No work is done, and it follows that theelectric field does not have a component lying in the surface, that is, the field is perpendicular to thesurface.

    Properties of equipotential surfaces.

    1. These are imaginary surfaces where potential of all points are equal.

    2. W.D. to take a charge from one point on equipotential surface to other is zero.

    3. Two equipotential surface do not intersect each other.

    4. Direction of electric field at any point on equipotential surface is to the surface and in direction in whichv is decreasing. (if v = 0 at ).

  • Page-31

    Asking question Q.1 Arrange ER, EP, ER in order of magnitude.

    10V 20V 30V 40V

    A B C D

    P Q R

    E > E > ER Q P

    Equi Potential surfaces

    Q.2Find the direction of field at P.A

    B D20V10V

    5V

    C

    C is direction of E field

    P

    Q.3

    l

    Calculate shape of equipotential surfaces.

    Potential due to combination of charges : (superposition principle)If a point is located in field of more than one charge then potential of that point is summation of individualpotential with sign, where all potentials are calculated with same reference point.

    Methods of calculation of potential(1) Find E

    in space and calculate potential at a point using V = - sd.E

    (2) Calculate potential due to elements and add them up.

    Potential on the axis of ring

    +

    +

    ++

    ++ +

    ++

    +

    +

    +++

    +

    Px

    r

    dq

    a

    (1) Potential at P using relation between E

    & V

    Eeff = 04q 2/322 )xr(

    x =

  • Page-32

    V =

    x

    .eff sdE

    = 04

    q

    dx)xr(

    xx2/322

    = 220 xr4

    q

    (2) Potential at P using superposition

    dV = 04

    1

    q

    022 xr

    dq = 220 xr4

    q

    Asking question Q.1 What happens if the ring is non uniformly charged?

    Q.2 Draw graph of V versus r and relate it with the graph of E

    (i) Centre at x = 0

    E = 0 ; V = r4

    q

    0 (maximum)

    (ii) InfiniteE = 0 ; V = 0

    (iii) E is maximum at x = 2

    R & Emax = 332

    20R4

    Q

    R / 2R / 2

    x

    E

    r

    V

    Potential difference due to Infinite charged wire

    r1

    +ve

    r2

    Potential difference in moving from r1 r2.

    So V2 V1 = 2

    1

    r

    r 0r2(dr)

    V2 V1 = 02

    (nr1 nr2)

    * So here it is clear that V1 0 as r1 * Also here as V 0

    V = dV = xkdq

    will not work as when we write potential due to an element as xkdq

    we

    assume potential at infinite to be zero.

  • Page-33

    Potential due to solid uniformly charged non conducting sphere

    Rr

    V = 04

    1 r

    Qr > R

    V = r4

    r34

    0

    3

    +

    R

    r 0

    2

    x4dxx4

    r < R

    V = 0

    2rR

    21r

    31 222

    = 30

    22

    R8)rR3(q

    Results :(i) Centre r = 0

    E = 0 ; V = 23

    04

    1 R

    Q

    (ii) Surface r = R

    E = 20 R4q ; V = R4

    q

    0(Graph of field and potential due to shell)

    O

    Inverse square~ I/r2

    Linear~ r

    a r

    E

    O

    Inverse square~ I/r2

    Linear~ r

    a r

    E

    (iii) V

    parabola

    r=Rr

    V 1r V

    r=Rr

    V 1r

    Ex. A fixed solid sphere of radius R is made of a material dielectric constant K = 1. Uniform charge density

    . A bullet of mass m and having charge q is fired towards its centre. From a distance x (x > R) fromcentre, calculate min. velocity so it can cross the sphere. Assume no other force acting on the bulletexcept the electrostatic force.

    Finding E from V

    VE

    = grad (V)

  • Page-34

    grad =

    k

    zj

    yi

    x (x, y, z co-ordinate system)

    Ex. V = x + y + z find E

    xV

    = 1 ; yV

    = 1 ; zV

    = 1

    E

    = )kji(

    Ex. V = xyz find E at (1, 1, 1)

    xV

    = yz ; yV

    = xz ; zV

    = xy

    E

    = )kxyjxziyz(

    = )kji( Q. The electric potential varies in space according to the relation V = 3x + 4y. A particle of mass 10Kg

    starts from rest from point (2, 3.2) under the influence of this field. Find the velocity of the particle whenit crosses the x-axis. The charge on the particle is +1C. Assume V and (x, y) are in S.I. units.

    [Ans. 2mm/s]

    III. Concept of Self energy / potential energy of system energy required to createa system is called self energy.

    1. Electric potential energy of a systemDefinition : It is the work done to assemble the charges from infinite separation to present configurationw/o change in K.E. of any particle.

    Methods of calculation

    (i) Keep all charges at separation from each other (ii) Find PE of each charge dues to field and then bring them one by one in present configuration other charges. and calculate the work done.

    PEsys = Wi PEsys = 2.........PEPEPE 321

    WherePE1 = PE12 + PE13 + ...........PE2 = P21 + PE23useful for symmetric arrangements.

    A Bq1 q2

    q1C

  • Page-35

    Method I Work done to bring A w1 = 0

    Work done to bring B w2 = rqkq 21

    Work done to bring C w3 = rqkq 31 + r

    qkq 32

    Self energy = w1 + w2 + w3

    Method II : U = 21 C/AB/AA VV(q + qB )VV( C/BA/B + qC (VC/A + VC/B)]

    U = 21

    n

    1i

    n

    1rr/ii Vq

    Ex. Find energy to the break. the system

    Sol. U = 21

    7

    1i

    7

    1

    /r

    i rVq

    q

    q

    qaq

    q

    q

    q

    q

    7

    1

    /r

    i rV =

    3a)q(k

    2akq3

    a)q(k3

    as because of (q) is also same.

    Work done to distroy = U = - q821

    3akq

    2aka3

    aka3

    Calculation of self energy due to continuous charge distributions1. Shell

    Method I : let say that x charge has been brought and if further dx charge from inifinity.

    U = Q

    0 Rdxkx R

    Q

    U = R2

    KQ2

    Method II : Here V due to other charges is

    U = 21

    Q

    0 RkQ

    dq

    Qdq

    multiplied by 21

    because every interaction counted two times

    U = R2

    kQ2

  • Page-36

    2. Solid sphereOnly by Method I : Let us consider a shell of radius x has ben created and further dx them created.

    U = R

    0 0

    3

    x4x3/4

    (4 x2 dx))

    U = R5

    kQ3 2

    Ex. Find self energy of system

    r

    Q R1 1 Q R2 2

    Sol. Because of three parts to create both sphere and then their interacles.

    U = rQkQ

    R5kQ3

    R5kQ3 21

    2

    22

    1

    21

    Energy density of electric field* Energy / unit volume where field is

    202

    1 E

    This is the energy required to create electric field per unit volume.

    Ex. Self energy of shellAns. Now let us consider at x, energy required to make field at x.

    U = 2

    20R

    0 xQ

    41

    21

    4x2 dx

    = dxx4xQk

    21 2

    2

    22

    R0

    =

    R

    22

    0 42Qk

    21

    = R2

    kQ2

    Note : We dont write self energy of point charge.

  • Page-37

    Ex. Solid sphere :-Here since E in sphere varies in different ways for x > R and x < R

    x U = dxx43

    x21 2

    R

    0

    2

    00

    + dxx4xQ

    41

    21 2

    R

    2

    20

    0

    U = R5

    kQ3 2

    {Optional} Irodov 3.140. A point charge q is located at the Fig. centre O of a spherical uncharged conductinglayer provided with a small orifice (Fig. ) The inside and outside radii of the layer are equal to a and brespectively. What amount of work has to be performed to slowly transfer the charge q from the pointO through the orifice and into infinity?

    Ans. Electric Dipole

    Definition : When two charges of equal magnitude and opposite sign are separated by a very small distance,then the arrangement is called electric dipole.

    d

    d qq

    (i) Dipole moment : P = q d

    ; d

    is always taken reason -ve to +ve

    (ii) Axis of dipole : is line joining -q to +q(iii) Mid point of axis of dipole is centre of dipole(iv) bisector of line joining +q to -q is equatorial line of dipole.(v) All the distances are measured from centre of dipole.

    Electric potential due to dipole

    OA B

    P

    q q

    r

    for d

  • Page-38

    VP =

    cos

    2dr

    kq

    cos

    2dr

    kq =

    22

    2 cos4

    dr

    )cosd(kq ~ 2rcosdqk

    = 2rr.pk

    = 2rkp

    cos

    Electric field due to dipole :)V(E

    =

    rr

    r

    2r

    coskp

    = rrcoskP23

    +

    rsinkP3

    +

    r

    P

    ErEnetE

    = 3rkP

    )sinrcos2(

    E = 04

    1 3

    2

    rcos31P

    where (r >> d)

    direction tan = 2tan

    where is angle of Enet with r

    Er = 30r4cosP2

    ; E = 30r4

    sinP

    (i) On the axis : Maximum potential and electric field at = 0 i.e. on axis for given distance r

    +

    Formulanotapplicable

    P = 0V +ve

    = V ve

    E EE

    E = 04

    1 3r

    P2

    V = 04

    1 2r

    P

    E

    is in direction of P

    (ii) Equatorial line : Minimum potential and field at = 90 i.e. on equitorial plane for given distance r.

    E = 04

    1 3r

    P+

    E

    P

    V = 0Ex. Find equivalent dipole moment of a ring having linear charge densities + and on its two halves.

    .

  • Page-39

    Electric dipole in uniform field :

    (A) Force 21 FF

    = 0

    0Fnet

    qE q

    +q

    qE

    E

    P

    E

    (B) Torque

    22

    sinqE||

    l

    In the vector form Ep

    (C) Potential Energy

    The work done by an external agent in rotating the dipole withoutchange in its kinetic energy is stored an potential energy in thefield and dipole system.It is given by

    U

    O /2

    U = p.E.or U = p E cos

    where is the angle between the dipole and the electric field vector, as shown in figure.There potential energy U as a function of the angle is plotted in figure.

    Note : the potential energy is minimum at = 0 the potential energy is maximum at =

    the dipole is in stable equilibrium at = 2

    .

    Potential energy in this posn. is zerodU = d

    = PE

    0

    dcos

    90

    +

    E

    = PE sin = PE cos = EP

    +

    (D) It will undergo SHM if turned by small angle

    PE sin = 22

    dtd.I

    for small

    +

    E

    stable

    2

    2

    dtd

    = IPE

    = I

    PE

    +

    unstable

    t

  • Page-40

    Electrostatic Pressure {Optional}

    If a small piece of radius b is removed from a charged spherical shell of radius a (>>b), calculate electricintensity at the midpoint of the aperture, assuming the density of charge to be .

    PED

    ER

    EDER

    a

    [Sol. Consider the shell to be made up of a disc of radius b and the remainder. If ED and ER are the intensitiesdue to disc and the remainder respectively at P, then for a charged spherical shell (or conductor)

    Eout = 0

    and Ein = 0

    Now as for outside the shell both ED and ER will be directed outwards while inside ER will be outwardswhile ED inwards so that :

    Eout = ER + ED and Ein = ER ED ........ (2)And hence equating Eqs. (1) and (2),

    ER + ED = 0

    and ER ED = 0

    Solving these for ER and ED :

    ER = ED = 02

    i.e., field at the aperture will be (/20) directed outwards.

    Note : As intensity on the disc (element) the to remainder is (/20), electric force on it will be,

    dF = dq E = ( ds)

    02 =

    0

    2

    2 ds

    So force per unit area on a charged conductor due to its own charge

    dsdF

    = 0

    2

    2

    = 210E2 [as for a conductor E = 0

    ]

    This force is called mechanical force or electrostatic pressure. ].

  • Page-41

    CONDUCTORS

    A conductor means infinite no. of free charges which move in random direction so the lattice becomespositively charged. Conductors contain charge carriers, these charge carriers are electrons. In a metal,the outer (valence) electrons part away from their atoms and are free to move. These electrons are freewithin the metal but not free to leave the metal. The free electrons form a kind of gas; they collide witheach other and with the ions, and move randomly in different directions. The positive ions made up of thenuclei and the bound electrons remain held in their fixed positions.

    1. Inside a conductor, electrostatic field is zero

    Consider a conductor, neutral or charged. There may also be an external electrostatic field. In the staticsituation, when there is no current inside or on the surface of the conductor, the electric field is zeroeverywhere inside the conductor. This fact can be taken as the defining property of a conductor. Aconductor has free electrons. As long as electric field is not zero, the free charge carriers would experi-ence force and drift. In the static situation, the free charges have so distributed themselves that theelectric field is zero everywhere inside. Electrostatic field is zero inside a conductor.

    Further Explanation

    What happens if conductor is placed in an exterrnal field.Lets keep a positive charge q charge near a neutral or acharged conductor then the electrons goes closeto q.

    The redistribution of electrons inside conductor takes place which generates an internal electric field intE

    .

    intEEE extnet

    +++

    ++

    +

    Eint Eext

    neutral conductor

    So an e experience netE

    If Eint Eext , then the e move such that they will create a stronger intE

    which will tend to cancel Eext.This constitutes current and therefore energy conservation is not valid.

    netE

    has to be O instantaneously..

    0EEE intextnet

    2. The interior of a conductor can have no excess charge in the static situationA neutral conductor has equal amounts of positive and negative charges in every small volume or surfaceelement. When the conductor is charged, the excess charge can reside only on the surface in the staticsituation.

  • Page-42

    ExplanationThis follows from the Gausss law. Consider any arbitrary volume element v inside a conductor. If weconsider any small gaussian surface inside

    Q

    AdE

    = 0 [as E = 0]

    On the closed surface S bounding the volume element v, electrostatic field is zero. Thus the total electricflux through S is zero. Hence, by Gausss law, there is no net charge enclosed by S.

    qenclosed = 0Since the surface S can be made as small as you like, i.e., the volume v can be made vanishingly small.This means there is no net charge at any point inside the conductor, and any excess charge mustreside at the surface.

    Note : Thus Solid conducting sphere is same as a shell. Note : You may emphasise again but qin = 0 does not imply that E = 0 from gauss law.

    3. At the surface of a charged conductor, electrostatic field must be normal to the surface atevery pointIf E were not normal to the surface, it would have some non-zero component along the surface. Freecharges on the surface of the conductor would then experience force and move. In the static situation,therefore, E should have no tangential component. Thus electrostatic field at the surface of a chargedconductor must be normal to the surface at every point. (For a conductor without any surface chargedensity, field is zero even at the surface.)

    Asking questionConsider a neutral conducting sphere placed near a infinite non conducting uniform sheet of charge.Draw the field near the infinite sheet.

    Sol.

    +

    +

    ++

    +

    Emphasize following facts from figure(i) Field is normal to surface conductor(ii) Field inside conductor is zero

    Further discussionWhen we place an ideal conductor in an electric field E, the free electrons experience a force in theopposite direction of the field and migrate to one side of the conductor as shown in figure (a).

  • Page-43

    E+++

    ++

    (a)

    (a) The accumulation of electrons leaves one side positively charged and the other negative. This chargeddistribution creates an electric field in a direction opposite to the applied field. Theredistribution of charge takes place till net field inside the conductor is zero. Therefore, in electrostaticequilibrium, the electric field inside an ideal conductor is zero.If we place a Gaussian surface, an infinitesimal distance below the surface, the electric field is zero atevery point on this Gaussian surface because it is inside the conductor [figure (b)]. Gausss law thenimplies that the net charge contained within the Gaussian surface is zero. In electrostatic equilibrium,excess charge on an ideal isolated conductor must reside on the conductors surface. No free chargecan exist anywhere within the electrostatic conductor.

    Note: Also if some external field is present then charge distribution on a Solid conducting sphere anda conducting shell will be non uniform.

    OPTIONAL :Electric field at the surface of a charged conductor

    E = n0

    where is the surface charge density and n is a unit vector normal to the surface in the outwarddirection.To derive the result, choose a pill box (a short cylinder) as the Gaussian surface about any point P on thesurface, as shown in figure.

    The pill box is partly inside and partly outside the surface of the conductor. It has a small area of crosssection S and negligible height.Just inside the surface, the electrostatic field is zero; just outside, the field is normal to the surface withmagnitude E. Thus, the contribution to the total flux through the pill box comes only from the outside(circular) cross-section of the pill box. This equals ES (positive for > 0, negative for < 0), sinceover the small area S, E may be considered constant and E and S are parallel or antiparallel. Thecharge enclosed by the pill box is S.By Gausss law

  • Page-44

    ES = 0

    S||

    E = 0

    ||

    Including the fact that electric field is normal to the surface, we get the vector relation, Eq. E = n0

    ,

    which is true for both signs of . For > 0, electric field is normal to the surface outward; for < 0,electric field is normal to the surface inward.

    4. Electrostatic potential is constant throughout the volume of the conductor and has the samevalue (as inside) on its surfaceSince E = 0 inside the conductor and has no tangential component on the surface, no work is done inmoving a small test charge within the conductor and on its surface. That is, there is no potential differencebetween any two points inside or on the surface of the conductor. Hence, the result. If the conductor ischarged, electric field normal to the surface exists; this means potential will be different for the surfaceand a point just outside the surface. In a system of conductors of arbitrary size, shape and chargeconfiguration, each conductor is characterised by a constant value of potential, but this constant maydiffer from one conductor to the other.

    Charge density on a conductors surface in abscence of an external field

    ++++

    +++ +

    ++ +

    + +++

    +++++

    ++

    Rr

    1r

    Now consider a single conductor with a nonspherical shape. If a charge is given to this conductor figure,the charge density will not be uniform on the entire surface. A portion where the surface is more flatmay be considered as part of a sphere of larger radius. The charge density at such a portion will besmaller from equation (1). At portions where the surface is more curved, the charge density will belarger. More precisely, the charge density will be larger where the radius of curvature is small. Thus thedistribution of charge Q on surface of conductor is Non-uniform

    and curvatureR

    1

    Charged conductor in external field :Figure shows a charge Q placed infront of a neutral conductor.

    Here the charge density also depends on external field and

    int.indout EE

    = 0

    ++

    ++

    +

    Q1

    Q

    Note : So here charge Q1 must be distributed non uniformly unrelated to Radius of curvature.

  • Page-45

    Ex. A point charge q is kept at a distance l from neutral sphere of radius R. Find electric potential V at P,solid conducting sphere.

    xP

    r

    R

    C

    Q

    l

    +++

    +q

    Sol. Potential due to induced charges is zero as centreC is equidistant from all induced chargesVolume of solid conductor is equipotential volume

    Vc = Vp = RKQKq

    l

    Also at PVP = Vdue to q + Vdue to induced charges + Vdue to Q

    lKq

    + RKQ

    = xKq

    + RKQ

    + Vdue to induced charges

    Vinduced charge = lKq

    xKq

    externalE

    + conductorE

    + inducedE

    = 0Resultant electric field inside material is zero

    Ex. Find electric field due to induced charges at P

    | pE

    | by induced charges = 2xKq

    Cavity inside a conductor :Cavity is aplace surrounded from all sides by the conductor such that without touching the body we cantreach cavity.

    Electrostatic shieldingConsider a conductor with a cavity, with no charges inside the cavity. A remarkable result is that theelectric field inside the cavity is zero, whatever be the size and shape of the cavity and whatever be thecharge on the conductor and the external fields in which it might be placed. We have proved a simplecase of this result already: the electric field inside a charged spherical shell is zero. The proof of the resultfor the shell makes use ofthe spherical symmetry of the shell. But the vanishing of electric field in the(charge-free) cavity of a conductor is, as mentioned above, is a general result. A related result is thateven if the conductor is charged or charges are induced on a neutralconductor by an external field, allcharges reside only on the outer surface of a conductor with cavity.Whatever be the charge and field configuration outside, any cavity in a conductor remains shielded fromoutside electric influence: thefield inside the cavity is always zero. This is known as electrostatic shielding.The effect can be made use of in protecting sensitive instruments from outside electrical influence. Figuregives a summary of the important electrostatic propertiesof a conductor.

  • Page-46

    1. If there is no charge present inside the cavity than field inside it is zero.(This does not mean that we are implying no net charge. If there is some charge present hereand there in cavity such that there sum total is zero then the following disscussion will not bevalid)

    Proof wrong sol.Let us consider a gauesion surface just near to cavity.

    AdE

    = 0 q in = 0

    Sol. From this we dont have proved that no charge resides on cavity but have proved that net charge oncavity surface is O.Now suppose a conductor of arbitrary shape contains a cavity as shown in figure.

    B

    A

    Let us assume that no charges are inside the cavity. In this case, the electric field inside the cavity must bezero regardless of the charge distribution on the outside surface of the conductor. Furthermore, the fieldin the cavity is zero even if an electric field exists outside the conductor.To prove this point, we use the fact that every point on the conductor is at the same electric potential, andtherefore any two points A and B on the surface of the cavity must be at the same potential. Now imaginethat a field E exists in the cavity and evaluate the potential difference VB VA defined by equation.

    V 0qU

    = B

    AdsE

    VB VA = B

    AdsE

    Because VB VA = 0, the integral of E ds must be zero for all paths between any two points A and Bon the conductor. The only way that this can be true for all paths is if E is zero everywhere in the cavity.Thus, we conclude that a cavity surrounded by conducting walls is a field-free region as along no chargesare inside the cavity.

  • Page-47

    2. Equal and opposite charge is induced on the inner surface of cavity.

    Figure shows a conductor with a cavity inside it. A charge q is placed insidecavity. In electrostatic equilibrium charge distribution will be as shown in figure q

    q

    +q

    charge on inner surface of cavity is q since material of conductor is initially neutral, equal and oppositecharge appears on outer surface.

    B

    Aq

    p

    C+

    ++ +

    +

    +

    +++

    +

    Gaussian surface

    Proof: Let us consider a gaussian surface just outside cavity inside material of conductor. As E in material of

    conductor is zero.

    dAE = 0 qenclosed = 0

    Thus equal and opposite charge is induced on the inner surface of cavity.

    3. The electric field due to charges on the outer surface of conductor is zero for all the pointsinside the conductor seperately

    Consider a charged conductor having charge +q1 and Q is kept inside the cavity. Lets call charge Qinside cavity as A, the induced charge -Q on the surface of the cavity as B and the charge on the surfaceof the conductor Q + q1 as C.

    A

    B

    C

    Now field inside the conductor is, netE

    and AE

    , CB EandE

    are fields due to charge A,B and C inside theconductor.

    and, netE

    = AE

    + CB EE

    = 0

    Now the electric field due to charges on the outer surface of conductor is zero for all the pointsinside the conductor seperately and the AB EE

    is zero seperately..

  • Page-48

    Special case : When spherical cavity in present inside spherical conductor and charge is at the centre.Then field inside the cavity is just due to charge A only as new field lines are already to B and E due toB = 0. [Symmetrical distribution]

    AqB

    C

    Note : When if we displace the charge q inside the cavity, it will only affect the charge distribution at B thecharge distribution at C will remain unaffected.

    Ex. Find field at P, the conductor is neutral. and cavityis having a charge q inside the cavity.

    qA

    R

    rx

    P

    BC +

    + ++

    +

    ++

    +

    Sol. Ep = 2rRQ

    as EA + EB = O

    Ex. A hollow, uncharged spherical conductor has inner radius a and outer radius b. A positive point charge+q is in the cavity at the centre of the sphere. Make the graph E and potential V(r) everywhere, assumingthat V = 0 at r = .

    a ab bZ r

    kqr2

    kqr2

    kq b

    Er

    kq r +

    kq b

    kq a

    kq r

    V

    +aq

    b

    + +

    +++

    +

  • Page-49

    Ex. Find charge on all surfaces :Sol. Applying Gauss law on opposite faces.

    sdE

    = 0

    P

    A C EB D F

    5Q 3Q4Q

    x x 3Q-x

    3Q+x4Qx

    2Q+x

    as through two lids E = 0 as part of conductor and through lateral surface sdE

    = 0 qenclosed = 0

    Important : (i) Thus facing surfaces have equal and opposite charges.(ii) Using this also prove that outer faces of the two last plates have equal charges.

    02

    Area/)xQ4(

    02Area/)xQ2(

    = 0 [as no field because of equal is cancelled]

    4Q x 2Q x = 0 2x = 2Q x = Q.

    Ex. A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conductinghollow spherical shell. Let the potential difference between the surface of the solid sphere and that of theouter surface of hollow shell be V. What will be the new potential difference between the same twosurfaces if the shell given a charge 3Q ?

    b

    +

    +

    +++

    +

    +

    +

    +++

    ++

    + ++ Q

    Sphere

    Shell

    [Sol. In case of a charged conducting sphere,

    Vin = VC = VS = Rq

    41

    0

    and Vout = rq

    41

    0

    So if a and b are the radii of sphere and spherical shell respectively, potential at their surfaces will be,

    Vsphere = aQ

    41

    0 and Vshell = b

    Q4

    1

    0

    And so according to given problem

    V = Vsphere Vshell =

    b

    1a1

    4Q

    0........... (i)

    Now when the shell is given a charge (3Q) the potential at its surface and also inside will change by V0

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    =

    b

    Q34

    1

    0

    So that now

    Vsphere = 041

    aQ

    + VV0

    and Vshell = 00V

    bQ

    41

    And hence

    Vsphere Vshell = 04Q

    b1

    a1

    = V [from Eq.(1)]

    i.e., if any charge is given to external shell, the potential difference between sphere and shell will notchange. This is because by presence of charge on outer shell, potential everywhere inside and on thesurface of shell will change by same amount and hence potential difference between sphere and shell willremain unchanged.

    (I) When two conductors are connected :When two conductors are connected charge redistributes on the connected conductors till theirpotential becomes equal

    Ex. A metal sphere A of radius a is charged to potential V. What will be its potential if it is enclosed by aspherical conducting shell B of radius b and the two are connected by a wire ?

    Sol. If the charge on spher