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# Static electric fields unit 2-

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• 1. Static Electric Fields (UNIT-2)In this chapter we will discuss on the followings: Coulombs Law Electric Field & Electric Flux Density Gausss Law with Application Electrostatic Potential, Equipotential Surfaces Boundary Conditions for Static Electric Fields Capacitance and Capacitors Electrostatic Energy Laplaces and Poissons Equations Uniqueness of Electrostatic Solutions Method of Images Solution of Boundary Value Problems in Different Coordinate Systems.IntroductionIn the previous chapter we have covered the essential mathematical tools needed to study EMfields. We have already mentioned in the previous chapter that electric charge is a fundamentalproperty of matter and charge exist in integral multiple of electronic charge. Electrostatics can bedefined as the study of electric charges at rest. Electric fields have their sources in electriccharges.( Note: Almost all real electric fields vary to some extent with time. However, for manyproblems, the field variation is slow and the field may be considered as static. For some othercases spatial distribution is nearly same as for the static case even though the actual field mayvary with time. Such cases are termed as quasi-static.)In this chapter we first study two fundamental laws governing the electrostatic fields, viz, (1)Coulombs Law and (2) Gausss Law. Both these law have experimental basis. Coulombs law isapplicable in finding electric field due to any charge distribution, Gausss law is easier to usewhen the distribution is symmetrical.Coulombs LawCoulombs Law states that the force between two point charges Q1and Q2 is directly proportionalto the product of the charges and inversely proportional to the square of the distance betweenthem. Point charge is a hypothetical charge located at a single point in space. It is an idealisedmodel of a particle having an electric charge.PREPARED BY GADDALA JAYARAJU, M.TECH Page 1
• 2. Static Electric Fields (UNIT-2)Mathematically, ,where k is the proportionality constant.In SI units, Q1 and Q2 are expressed in Coulombs(C) and R is in meters.Force F is in Newtons (N) and , is called the permittivity of free space.(We are assuming the charges are in free space. If the charges are any other dielectric medium,we will use instead where is called the relative permittivity or the dielectric constantof the medium).Therefore .......................(2.1)As shown in the Figure 2.1 let the position vectors of the point charges Q1and Q2 are given byand . Let represent the force on Q1 due to charge Q2.Fig 2.1: Coulombs LawThe charges are separated by a distance of . We define the unit vectors as and ..................................(2.2) can be defined as . Similarly the force on Q1 due tocharge Q2 can be calculated and if represents this force then we can writeWhen we have a number of point charges, to determine the force on a particular charge due to allother charges, we apply principle of superposition. If we have N number of chargesPREPARED BY GADDALA JAYARAJU, M.TECH Page 2
• 3. Static Electric Fields (UNIT-2)Q1,Q2,.........QN located respectively at the points represented by the position vectors , ,...... , the force experienced by a charge Q located at is given by, .................................(2.3)Electric FieldThe electric field intensity or the electric field strength at a point is defined as the force per unitcharge. That is or, .......................................(2.4)The electric field intensity E at a point r (observation point) due a point charge Q located at(source point) is given by: ..........................................(2.5)For a collection of N point charges Q1 ,Q2 ,.........QN located at , ,...... , the electric fieldintensity at point is obtained as ........................................(2.6)The expression (2.6) can be modified suitably to compute the electric filed due to a continuousdistribution of charges.In figure 2.2 we consider a continuous volume distribution of charge r(t) in the region denoted asthe source region.For an elementary charge , i.e. considering this charge as point charge, we canwrite the field expression as:PREPARED BY GADDALA JAYARAJU, M.TECH Page 3
• 4. Static Electric Fields (UNIT-2) .............(2.7)Fig 2.2: Continuous Volume Distribution of ChargeWhen this expression is integrated over the source region, we get the electric field at the point Pdue to this distribution of charges. Thus the expression for the electric field at P can be writtenas: ..........................................(2.8)Similar technique can be adopted when the charge distribution is in the form of a line chargedensity or a surface charge density. ........................................(2.9) ........................................(2.10)Electric flux density:As stated earlier electric field intensity or simply ‘Electric field gives the strength of the field ata particular point. The electric field depends on the material media in which the field is beingconsidered. The flux density vector is defined to be independent of the material media (as wellsee that it relates to the charge that is producing it).For a linearisotropic medium under consideration; the flux density vector is defined as:PREPARED BY GADDALA JAYARAJU, M.TECH Page 4
• 5. Static Electric Fields (UNIT-2) ................................................(2.11)We define the electric flux Y as .....................................(2.12)Gausss Law: Gausss law is one of the fundamental laws of electromagnetism and it states thatthe total electric flux through a closed surface is equal to the total charge enclosed by the surface.Fig 2.3: Gausss LawLet us consider a point charge Q located in an isotropic homogeneous medium of dielectricconstant e. The flux density at a distance r on a surface enclosing the charge is given by ...............................................(2.13)If we consider an elementary area ds, the amount of flux passing through the elementary area isgiven by .....................................(2.14)But , is the elementary solid angle subtended by the area at the location of Q.Therefore we can writeFor a closed surface enclosing the charge, we can writePREPARED BY GADDALA JAYARAJU, M.TECH Page 5
• 6. Static Electric Fields (UNIT-2)which can seen to be same as what we have stated in the definition of Gausss Law.Application of Gausss LawGausss law is particularly useful in computing or where the charge distribution has somesymmetry. We shall illustrate the application of Gausss Law with some examples.1.An infinite line chargeAs the first example of illustration of use of Gausss law, let consider the problem ofdetermination of the electric field produced by an infinite line charge of density rLC/m. Let usconsider a line charge positioned along the z-axis as shown in Fig. 2.4(a) (next slide). Since theline charge is assumed to be infinitely long, the electric field will be of the form as shown in Fig.2.4(b) (next slide).If we consider a close cylindrical surface as shown in Fig. 2.4(a), using Gausss theorm we canwrite, .....................................(2.15)Considering the fact that the unit normal vector to areas S1 and S3 are perpendicular to theelectric field, the surface integrals for the top and bottom surfaces evaluates to zero. Hence wecan write,PREPARED BY GADDALA JAYARAJU, M.TECH Page 6
• 7. Static Electric Fields (UNIT-2)Fig 2.4: Infinite Line Charge .....................................(2.16)2. Infinite Sheet of ChargeAs a second example of application of Gausss theorem, we consider an infinite charged sheetcovering the x-z plane as shown in figure 2.5.Assuming a surface charge density of for the infinite surface charge, if we consider acylindrical volume having sides placed symmetrically as shown in figure 5, we can write: ..............(2.17)PREPARED BY GADDALA JAYARAJU, M.TECH Page 7
• 8. Static Electric Fields (UNIT-2)Fig 2.5: Infinite Sheet of ChargeIt may be noted that the electric field strength is independent of distance. This is true for theinfinite plane of charge; electric lines of force on either side of the charge will be perpendicularto the sheet and extend to infinity as parallel lines. As number of lines of force per unit area givesthe strength of the field, the field becomes independent of distance. For a finite charge sheet, thefield will be a function of distance.3. Uniformly Charged SphereLet us consider a sphere of radius r0 having a uniform volume charge density of rv C/m3. Todetermine everywhere, inside and outside the sphere, we construct Gaussian surfaces ofradius r < r0 and r > r0 as shown in Fig. 2.6 (a) and Fig. 2.6(b).For the region ; the total enclosed charge will be .........................(2.18)PREPARED BY GADDALA JAYARAJU, M.TECH Page 8
• 9. Static Electric Fields (UNIT-2)Fig 2.6: Uniformly Charged SphereBy applying Gausss theorem, ...............(2.19)Therefore ...............................................(2.20)For the region ; the total enclosed charge will be ....................................................................(2.21)By applying Gausss theorem, .....................................................(2.22)Fig. 2.7 shows the variation of D for r0 = 1and .Electrostatic Potential and Equipotential SurfacesIn the previous sections we have seen how the electric field intensity due to a charge or a chargedistribution can be found using Coulombs law or Gausss law. Since a charge placed in thevicinity of another charge (or in other words in the field of other charge) experiences a force, themovement of the charge represents energy exchange. Electrostatic potential is related to the workdone in carrying a charge from one point to the other in the presence of an electric field.PREPARED BY GADDALA JAYARAJU, M.TECH Page 9
• 10. Static Electric Fields (UNIT-2)Let us suppose that we wish to move a positive test charge from a point P to another point Qas shown in the Fig. 2.8.The force at any point along its path would cause the particle to accelerate and move it out of theregion if unconstrained. Since we are dealing with an electrostatic case, a force equal to thenegative of that acting on the charge is to be applied while moves from P to Q. The workdone by this external agent in moving the charge by a distance is given by:Fig 2.8: Movement of Test Charge in Electric Field .............................(2.23)The negative sign accounts for the fact that work is done on the system by the external agent. .....................................(2.24)The potential difference between two points P and Q , VPQ, is defined as the work done per unitcharge, i.e. ...............................(2.25)It may be noted that in moving a charge from the initial point to the final point if the potentialdifference is positive, there is a gain in potential energy in the movement, external agentperforms the work against the field. If the sign of the potential difference is negative, work isdone by the field.PREPARED BY GADDALA JAYARAJU, M.TECH Page 10
• 11. Static Electric Fields (UNIT-2)We will see that the electrostatic system is conservative in that no net energy is exchanged if thetest charge is moved about a closed path, i.e. returning to its initial position. Further, the potentialdifference between two points in an electrostatic field is a point function; it is independent of thepath taken. The potential difference is measured in Joules/Coulomb which is referred to as Volts.Let us consider a point charge Q as shown in the Fig. 2.9.Fig 2.9: Electrostatic Potential calculation for a point chargeFurther consider the two points A and B as shown in the Fig. 2.9. Considering the movement of aunit positive test charge from B to A , we can write an expression for the potential difference as: ..................................(2.26)It is customary to choose the potential to be zero at infinity. Thus potential at any point ( rA = r)due to a point charge Q can be written as the amount of work done in bringing a unit positivecharge from infinity to that point (i.e. rB = 0). ..................................(2.27)Or, in other words,PREPARED BY GADDALA JAYARAJU, M.TECH Page 11
• 12. Static Electric Fields (UNIT-2) ..................................(2.28)Let us now consider a situation where the point charge Q is not located at the origin as shown inFig. 2.10.Fig 2.10: Electrostatic Potential due a Displaced ChargeThe potential at a point P becomes ..................................(2.29)So far we have considered the potential due to point charges only. As any other type of chargedistribution can be considered to be consisting of point charges, the same basic ideas now can beextended to other types of charge distribution also.Let us first consider N point charges Q1, Q2,.....QN located at points with position vectors ,,...... . The potential at a point having position vector can be written as: ..................................(2.30a)or, ...........................................................(2.30b)For continuous charge distribution, we replace point charges Qn by corresponding chargeelements or or depending on whether the charge distribution is linear, surfaceor a volume charge distribution and the summation is replaced by an integral. With thesemodifications we can write:PREPARED BY GADDALA JAYARAJU, M.TECH Page 12
• 13. Static Electric Fields (UNIT-2)For line charge, ..................................(2.31)For surface charge, .................................(2.32)For volume charge, .................................(2.33)It may be noted here that the primed coordinates represent the source coordinates and theunprimed coordinates represent field point.Further, in our discussion so far we have used the reference or zero potential at infinity. If anyother point is chosen as reference, we can write: .................................(2.34)where C is a constant. In the same manner when potential is computed from a known electricfield we can write: .................................(2.35)The potential difference is however independent of the choice of reference. .......................(2.36)We have mentioned that electrostatic field is a conservative field; the work done in moving acharge from one point to the other is independent of the path. Let us consider moving a chargefrom point P1 to P2 in one path and then from point P2 back to P1 over a different path. If thework done on the two paths were different, a net positive or negative amount of work wouldhave been done when the body returns to its original position P1. In a conservative field there isno mechanism for dissipating energy corresponding to any positive work neither any source ispresent from which energy could be absorbed in the case of negative work. Hence the questionof different works in two paths is untenable, the work must have to be independent of path anddepends on the initial and final positions.Since the potential difference is independent of the paths taken, VAB = - VBA , and over a closedpath,PREPARED BY GADDALA JAYARAJU, M.TECH Page 13
• 14. Static Electric Fields (UNIT-2) .................................(2.37)Applying Stokess theorem, we can write: ............................(2.38)from which it follows that for electrostatic field, ........................................(2.39)Any vector field that satisfies is called an irrotational field.From our definition of potential, we can write .................................(2.40)from which we obtain, ..........................................(2.41)From the foregoing discussions we observe that the electric field strength at any point is thenegative of the potential gradient at any point, negative sign shows that is directed fromhigher to lower values of . This gives us another method of computing the electric field, i. e. ifwe know the potential function, the electric field may be computed. We may note here that thatone scalar function contain all the information that three components of carry, the same ispossible because of the fact that three components of are interrelated by the relation .Example: Electric DipoleAn electric dipole consists of two point charges of equal magnitude but of opposite sign andseparated by a small distance.PREPARED BY GADDALA JAYARAJU, M.TECH Page 14
• 15. Static Electric Fields (UNIT-2)As shown in figure 2.11, the dipole is formed by the two point charges Q and -Q separated by adistance d , the charges being placed symmetrically about the origin. Let us consider a point P ata distance r, where we are interested to find the field.Fig 2.11 : Electric DipoleThe potential at P due to the dipole can be written as: ..........................(2.42)When r1 and r2>>d, we can write and .Therefore, ....................................................(2.43)We can write, ...............................................(2.44)The quantity is called the dipole moment of the electric dipole.Hence the expression for the electric potential can now be written as: ................................(2.45)PREPARED BY GADDALA JAYARAJU, M.TECH Page 15
• 16. Static Electric Fields (UNIT-2)It may be noted that while potential of an isolated charge varies with distance as 1/r that of anelectric dipole varies as 1/r2 with distance.If the dipole is not centered at the origin, but the dipole center lies at , the expression for thepotential can be written as: ........................(2.46)The electric field for the dipole centered at the origin can be computed as ........................(2.47) is the magnitude of the dipole moment. Once again we note that the electric field ofelectric dipole varies as 1/r3 where as that of a point charge varies as 1/r2.Equipotential SurfacesAn equipotential surface refers to a surface where the potential is constant. The intersection of anequipotential surface with an plane surface results into a path called an equipotential line. Nowork is done in moving a charge from one point to the other along an equipotential line orsurface.In figure 2.12, the dashes lines show the equipotential lines for a positive point charge. Bysymmetry, the equipotential surfaces are spherical surfaces and the equipotential lines are circles.The solid lines show the flux lines or electric lines of force.PREPARED BY GADDALA JAYARAJU, M.TECH Page 16
• 17. Static Electric Fields (UNIT-2)Fig 2.12: Equipotential Lines for a Positive Point ChargeMichael Faraday as a way of visualizing electric fields introduced flux lines. It may be seen thatthe electric flux lines and the equipotential lines are normal to each other.In order to plot the equipotential lines for an electric dipole, we observe that for a given Q and d,a constant V requires that is a constant. From this we can write to be theequation for an equipotential surface and a family of surfaces can be generated for various valuesof cv.When plotted in 2-D this would give equipotential lines.To determine the equation for the electric field lines, we note that field lines represent thedirection of in space. Therefore, , k is a constant .................................................................(2.48) .................(2.49)For the dipole under consideration =0 , and therefore we can write, .........................................................(2.50)Integrating the above expression we get , which gives the equations for electric fluxlines. The representative plot ( cv = c assumed) of equipotential lines and flux lines for a dipole isshown in fig 2.13. Blue lines represent equipotential, red lines represent field lines.PREPARED BY GADDALA JAYARAJU, M.TECH Page 17
• 18. Static Electric Fields (UNIT-2)Boundary conditions for Electrostatic fieldsIn our discussions so far we have considered the existence of electric field in the homogeneousmedium. Practical electromagnetic problems often involve media with different physicalproperties. Determination of electric field for such problems requires the knowledge of therelations of field quantities at an interface between two media. The conditions that the fieldsmust satisfy at the interface of two different media are referred to as boundary conditions .In order to discuss the boundary conditions, we first consider the field behavior in some commonmaterial media.In general, based on the electric properties, materials can be classified into three categories:conductors, semiconductors and insulators (dielectrics). In conductor , electrons in the outermostshells of the atoms are very loosely held and they migrate easily from one atom to the other.Most metals belong to this group. The electrons in the atoms of insulators or dielectrics remainconfined to their orbits and under normal circumstances they are not liberated under theinfluence of an externally applied field. The electrical properties of semiconductors fall betweenthose of conductors and insulators since semiconductors have very few numbers of free charges.The parameter conductivity is used characterizes the macroscopic electrical property of amaterial medium. The notion of conductivity is more important in dealing with the current flowand hence the same will be considered in detail later on.If some free charge is introduced inside a conductor, the charges will experience a force due tomutual repulsion and owing to the fact that they are free to move, the charges will appear on thesurface. The charges will redistribute themselves in such a manner that the field within theconductor is zero. Therefore, under steady condition, inside a conductor .From Gausss theorem it follows that = 0 .......................(2.51)The surface charge distribution on a conductor depends on the shape of the conductor. Thecharges on the surface of the conductor will not be in equilibrium if there is a tangentialcomponent of the electric field is present, which would produce movement of the charges. Henceunder static field conditions, tangential component of the electric field on the conductor surfaceis zero. The electric field on the surface of the conductor is normal everywhere to the surface .Since the tangential component of electric field is zero, the conductor surface is an equipotentialsurface. As = 0 inside the conductor, the conductor as a whole has the same potential. We mayfurther note that charges require a finite time to redistribute in a conductor. However, this time isvery small sec for good conductor like copper.PREPARED BY GADDALA JAYARAJU, M.TECH Page 18
• 19. Static Electric Fields (UNIT-2)Let us now consider an interface between a conductor and free space as shown in the figure 2.14.Fig 2.14: Boundary Conditions for at the surface of a ConductorLet us consider the closed path pqrsp for which we can write, .................................(2.52)For and noting that inside the conductor is zero, we can write =0.......................................(2.53)Et is the tangential component of the field. Therefore we find thatEt = 0 ...........................................(2.54)In order to determine the normal component En, the normal component of , at the surface ofthe conductor, we consider a small cylindrical Gaussian surface as shown in the Fig.12. Letrepresent the area of the top and bottom faces and represents the height of the cylinder. Onceagain, as , we approach the surface of the conductor. Since = 0 inside the conductor iszero, .............(2.55) ..................(2.56)Therefore, we can summarize the boundary conditions at the surface of a conductor as:Et = 0 ........................(2.57)PREPARED BY GADDALA JAYARAJU, M.TECH Page 19
• 20. Static Electric Fields (UNIT-2) .....................(2.58)Behavior of dielectrics in static electric field: Polarization of dielectricHere we briefly describe the behavior of dielectrics or insulators when placed in static electricfield. Ideal dielectrics do not contain free charges. As we know, all material media are composedof atoms where a positively charged nucleus (diameter ~ 10-15m) is surrounded by negativelycharged electrons (electron cloud has radius ~ 10-10m) moving around the nucleus. Molecules ofdielectrics are neutral macroscopically; an externally applied field causes small displacement ofthe charge particles creating small electric dipoles.These induced dipole moments modifyelectric fields both inside and outside dielectric material.Molecules of some dielectric materials posses permanent dipole moments even in the absence ofan external applied field. Usually such molecules consist of two or more dissimilar atoms and arecalled polar molecules. A common example of such molecule is water molecule H2O. In polarmolecules the atoms do not arrange themselves to make the net dipole moment zero. However, inthe absence of an external field, the molecules arrange themselves in a random manner so thatnet dipole moment over a volume becomes zero. Under the influence of an applied electric field,these dipoles tend to align themselves along the field as shown in figure 2.15. There are somematerials that can exhibit net permanent dipole moment even in the absence of applied field.These materials are called electrets that made by heating certain waxes or plastics in the presenceof electric field. The applied field aligns the polarized molecules when the material is in theheated state and they are frozen to their new position when after the temperature is brought downto its normal temperatures. Permanent polarization remains without an externally applied field.As a measure of intensity of polarization, polarization vector (in C/m2) is defined as: .......................(2.59)n being the number of molecules per unit volume i.e. is the dipole moment per unit volume.Let us now consider a dielectric material having polarization and compute the potential at anexternal point O due to an elementary dipole dv.PREPARED BY GADDALA JAYARAJU, M.TECH Page 20
• 21. Static Electric Fields (UNIT-2)Fig 2.16: Potential at an External Point due to an Elementary Dipole dv.With reference to the figure 2.16, we can write: ..........................................(2.60)Therefore, ........................................(2.61)........(2.62)where x,y,z represent the coordinates of the external point O and x,y,z are the coordinates of thesource point.From the expression of R, we can verify that .............................................(2.63) .........................................(2.64)Using the vector identity, ,where f is a scalar quantity , we have, .......................(2.65)PREPARED BY GADDALA JAYARAJU, M.TECH Page 21
• 22. Static Electric Fields (UNIT-2)Converting the first volume integral of the above expression to surface integral, we can write .................(2.66)where is the outward normal from the surface element ds of the dielectric. From the aboveexpression we find that the electric potential of a polarized dielectric may be found from thecontribution of volume and surface charge distributions having densities ......................................................................(2.67) ......................(2.68)These are referred to as polarisation or bound charge densities. Therefore we may replace apolarized dielectric by an equivalent polarization surface charge density and a polarizationvolume charge density. We recall that bound charges are those charges that are not free to movewithin the dielectric material, such charges are result of displacement that occurs on a molecularscale during polarization. The total bound charge on the surface is ......................(2.69)The charge that remains inside the surface is ......................(2.70)The total charge in the dielectric material is zero as ......................(2.71)If we now consider that the dielectric region containing charge density the total volumecharge density becomes ....................(2.72)Since we have taken into account the effect of the bound charge density, we can writePREPARED BY GADDALA JAYARAJU, M.TECH Page 22
• 23. Static Electric Fields (UNIT-2) ....................(2.73)Using the definition of we have ....................(2.74)Therefore the electric flux densityWhen the dielectric properties of the medium are linear and isotropic, polarisation is directlyproportional to the applied field strength and ........................(2.75) is the electric susceptibility of the dielectric. Therefore, .......................(2.76) is called relative permeability or the dielectric constant of the medium. is calledthe absolute permittivity.A dielectric medium is said to be linear when is independent of and the medium ishomogeneous if is also independent of space coordinates. A linear homogeneous andisotropic medium is called a simple medium and for such medium the relative permittivity is aconstant.Dielectric constant may be a function of space coordinates. For anistropic materials, thedielectric constant is different in different directions of the electric field, D and E are related by apermittivity tensor which may be written as: .......................(2.77)For crystals, the reference coordinates can be chosen along the principal axes, which make offdiagonal elements of the permittivity matrix zero. Therefore, we havePREPARED BY GADDALA JAYARAJU, M.TECH Page 23
• 24. Static Electric Fields (UNIT-2) .......................(2.78)Media exhibiting such characteristics are called biaxial. Further, if then the medium iscalled uniaxial. It may be noted that for isotropic media, .Lossy dielectric materials are represented by a complex dielectric constant, the imaginary part ofwhich provides the power loss in the medium and this is in general dependant on frequency.Another phenomenon is of importance is dielectric breakdown. We observed that the appliedelectric field causes small displacement of bound charges in a dielectric material that results intopolarization. Strong field can pull electrons completely out of the molecules. These electronsbeing accelerated under influence of electric field will collide with molecular lattice structurecausing damage or distortion of material. For very strong fields, avalanche breakdown may alsooccur. The dielectric under such condition will become conducting.The maximum electric field intensity a dielectric can withstand without breakdown is referred toas the dielectric strength of the material.Boundary Conditions for Electrostatic Fields:Let us consider the relationship among the field components that exist at the interface betweentwo dielectrics as shown in the figure 2.17. The permittivity of the medium 1 and medium 2 are and respectively and the interface may also have a net charge density Coulomb/m.Fig 2.17: Boundary Conditions at the interface between two dielectricsWe can express the electric field in terms of the tangential and normal components ..........(2.79)PREPARED BY GADDALA JAYARAJU, M.TECH Page 24
• 25. Static Electric Fields (UNIT-2)Boundary Conditions for Electrostatic Fields:Let us consider the relationship among the field components that exist at the interface betweentwo dielectrics as shown in the figure 2.17. The permittivity of the medium 1 and medium 2 are and respectively and the interface may also have a net charge density Coulomb/m.Fig 2.17: Boundary Conditions at the interface between two dielectricsWe can express the electric field in terms of the tangential and normal components ..........(2.79)i.e., .......................(2.81c)Thus we find that the normal component of the flux density vector D is discontinuous acrossan interface by an amount of discontinuity equal to the surface charge density at theinterface.ExampleTwo further illustrate these points; let us consider an example, which involves the refraction of Dor E at a charge free dielectric interface as shown in the figure 2.18.Using the relationships we have just derived, we can write .......................(2.82a) .......................(2.82b)In terms of flux density vectors,PREPARED BY GADDALA JAYARAJU, M.TECH Page 25
• 26. Static Electric Fields (UNIT-2) .......................(2.83a) .......................(2.83b)Therefore, .......................(2.84)Fig 2.18: Refraction of D or E at a Charge Free Dielectric InterfaceCapacitance and CapacitorsWe have already stated that a conductor in an electrostatic field is an Equipotential body and anycharge given to such conductor will distribute themselves in such a manner that electric fieldinside the conductor vanishes. If an additional amount of charge is supplied to an isolatedconductor at a given potential, this additional charge will increase the surface charge density .Since the potential of the conductor is given by , the potential of the conductorwill also increase maintaining the ratio same. Thus we can write where the constant ofproportionality C is called the capacitance of the isolated conductor. SI unit of capacitance isCoulomb/ Volt also called Farad denoted by F. It can It can be seen that if V=1, C = Q. Thuscapacity of an isolated conductor can also be defined as the amount of charge in Coulombrequired to raise the potential of the conductor by 1 Volt.Of considerable interest in practice is a capacitor that consists of two (or more) conductorscarrying equal and opposite charges and separated by some dielectric media or free space. Theconductors may have arbitrary shapes. A two-conductor capacitor is shown in figure 2.19.PREPARED BY GADDALA JAYARAJU, M.TECH Page 26
• 27. Static Electric Fields (UNIT-2)Fig 2.19: Capacitance and CapacitorsWhen a d-c voltage source is connected between the conductors, a charge transfer occurs whichresults into a positive charge on one conductor and negative charge on the other conductor. Theconductors are equipotential surfaces and the field lines are perpendicular to the conductorsurface. If V is the mean potential difference between the conductors, the capacitance is given by . Capacitance of a capacitor depends on the geometry of the conductor and thepermittivity of the medium between them and does not depend on the charge or potentialdifference between conductors. The capacitance can be computed by assuming Q(at the sametime -Q on the other conductor), first determining using Gauss’s theorem and thendetermining . We illustrate this procedure by taking the example of a parallel platecapacitor.Example: Parallel plate capacitorFig 2.20: Parallel Plate CapacitorFor the parallel plate capacitor shown in the figure 2.20, let each plate has area A and a distanceh separates the plates. A dielectric of permittivity fills the region between the plates. ThePREPARED BY GADDALA JAYARAJU, M.TECH Page 27
• 28. Static Electric Fields (UNIT-2)electric field lines are confined between the plates. We ignore the flux fringing at the edges ofthe plates and charges are assumed to be uniformly distributed over the conducting plates withdensities and - , .By Gauss’s theorem we can write, .......................(2.85)As we have assumed to be uniform and fringing of field is neglected, we see that E is constantin the region between the plates and therefore, we can write . Thus, for a parallelplate capacitor we have,........................(2.86)Series and parallel Connection of capacitorsCapacitors are connected in various manners in electrical circuits; series and parallel connectionsare the two basic ways of connecting capacitors. We compute the equivalent capacitance for suchconnections.Series Case: Series connection of two capacitors is shown in the figure 2.21. For this case wecan write, .......................(2.87)Fig 2.21: Series Connection of CapacitorsPREPARED BY GADDALA JAYARAJU, M.TECH Page 28
• 29. Static Electric Fields (UNIT-2)Fig 2.22: Parallel Connection of CapacitorsThe same approach may be extended to more than two capacitors connected in series.Parallel Case: For the parallel case, the voltages across the capacitors are the same.The total chargeTherefore, .......................(2.88)Electrostatic Energy and Energy DensityWe have stated that the electric potential at a point in an electric field is the amount of workrequired to bring a unit positive charge from infinity (reference of zero potential) to that point.To determine the energy that is present in an assembly of charges, let us first determine theamount of work required to assemble them. Let us consider a number of discrete charges Q1,Q2,......., QN are brought from infinity to their present position one by one. Since initially there isno field present, the amount of work done in bring Q1 is zero. Q2 is brought in the presence of thefield of Q1, the work done W1= Q2V21 where V21 is the potential at the location of Q2 due to Q1.Proceeding in this manner, we can write, the total work done.................................................(2.89)Had the charges been brought in the reverse order,.................(2.90)Therefore,PREPARED BY GADDALA JAYARAJU, M.TECH Page 29
• 30. Static Electric Fields (UNIT-2)................(2.91)Here VIJ represent voltage at the Ith charge location due to Jth charge. Therefore,Or, ................(2.92)If instead of discrete charges, we now have a distribution of charges over a volume v then we canwrite, ................(2.93)where is the volume charge density and V represents the potential function.Since, , we can write .......................................(2.94)Using the vector identity, , we can write ................(2.95)In the expression , for point charges, since V varies as and D varies as , theterm V varies as while the area varies as r2. Hence the integral term varies at least as andthe as surface becomes large (i.e. ) the integral term tends to zero.PREPARED BY GADDALA JAYARAJU, M.TECH Page 30
• 31. Static Electric Fields (UNIT-2)Thus the equation for W reduces to ................(2.96) , is called the energy density in the electrostatic field.Poisson’s and Laplace’s EquationsFor electrostatic field, we have seen that ..........................................................................................(2.97)Form the above two equations we can write ..................................................................(2.98)Using vector identity we can write, ................(2.99)For a simple homogeneous medium, is constant and . Therefore, ................(2.100)This equation is known as Poisson’s equation. Here we have introduced a new operator, ( delsquare), called the Laplacian operator. In Cartesian coordinates, ...............(2.101)Therefore, in Cartesian coordinates, Poisson equation can be written as: ...............(2.102)In cylindrical coordinates, ...............(2.103)PREPARED BY GADDALA JAYARAJU, M.TECH Page 31
• 32. Static Electric Fields (UNIT-2)In spherical polar coordinate system, ...............(2.104)At points in simple media, where no free charge is present, Poisson’s equation reduces to ...................................(2.105)which is known as Laplace’s equation.Laplace’s and Poisson’s equation are very useful for solving many practical electrostatic fieldproblems where only the electrostatic conditions (potential and charge) at some boundaries areknown and solution of electric field and potential is to be found throughout the volume. We shallconsider such applications in the section where we deal with boundary value problems.Uniqueness TheoremSolution of Laplace’s and Poisson’s Equation can be obtained in a number of ways. For a givenset of boundary conditions, if we can find a solution to Poisson’s equation ( Laplace’s equation isa special case), we first establish the fact that the solution is a unique solution regardless of themethod used to obtain the solution. Uniqueness theorem thus can stated as:Solution of an electrostatic problem specifying its boundary condition is the only possiblesolution, irrespective of the method by which this solution is obtained. To prove this theorem, asshown in figure 2.23, we consider a volume Vr and a closed surface Sr encloses this volume. Sr issuch that it may also be a surface at infinity. Inside the closed surface Sr , there are chargedconducting bodies with surfaces S1 , S2 , S3,....and these charged bodies are at specified potentials.If possible, let there be two solutions of Poisson’s equation in Vr. Each of these solutions V1 andV2 satisfies Poisson equation as well as the boundary conditions on S1 , S2 ,....Sn and Sr. Since V1and V2 are assumed to be different, let Vd = V1 -V2 be a different potential function such that, ........................(2.106) and ...............(2.107) ............................(2.108)PREPARED BY GADDALA JAYARAJU, M.TECH Page 32
• 33. Static Electric Fields (UNIT-2)Thus we find that Vd satisfies Laplace’s equation in Vr and on the conductor boundaries Vd iszero as V1 and V2 have same values on those boundaries. Using vector identity we can write: ...............(2.109)Using the fact that , we can write ..........................(2.110)Here the surface S consists of Sr as well as S1 , S2, .....S3.We note that Vd =0 over the conducting boundaries, therefore the contribution to the surfaceintegral from these surfaces are zero. For large surface Sr , we can think the surface to be aspherical surface of radius . In the surface integral, the term varies as whereasthe area increases as r2, hence the surface integral decreases as . For , the integralvanishes.Therefore, . Since is non negative everywhere, the volume integral can bezero only if , which means V1 = V2. That is, our assumption that two solutions aredifferent does not hold. So if a solution exists for Poisson’s (and Laplace’s) equation for a givenset of boundary conditions, this solution is a unique solution.Moreover, in the context of uniqueness of solution, let us look into the role of the boundarycondition. We observe that, , which, can be written as . Uniqueness of solution is guaranteed everywhere if or =0on S.The condition means that on S. This specifies the potential function on S andis called Dirichlet boundary condition. On the other hand, means onS. Specification of the normal derivative of the potential function is called the Neumannboundary condition and corresponds to specification of normal electric field strength or chargedensity. If both the Dirichlet and the Neumann boundary conditions are specified over part of Sthen the problem is said to be over specified or improperly posed.PREPARED BY GADDALA JAYARAJU, M.TECH Page 33
• 34. Static Electric Fields (UNIT-2)Method of ImagesForm uniqueness theorem, we have seen that in a given region if the distribution of charge andthe boundary conditions are specified properly, we can have a unique solution for the electricpotential. However, obtaining this solution calls for solving Poisson (or Laplace) equation. Aconsequence of the uniqueness theorem is that for a given electrostatics problem, we can replacethe original problem by another problem at the same time retaining the same charges andboundary conditions. This is the basis for the method of images. Method of images is particularlyuseful for evaluating potential and field quantities due to charges in the presence of conductorswithout actually solving for Poisson’s (or Laplace’s) equation. Utilizing the fact that aconducting surface is an equipotential, charge configurations near perfect conducting plane canbe replaced by the charge itself and its image so as to produce an equipotential in the place of theconducting plane. To have insight into how this method works, we consider the case of pointcharge Q at a distance d above a large grounded conducting plane as shown in figure 2.24.Fig 2.24For this case, the presence of the positive charge +Q will induce negative charges on the surfaceof the conducting plane and the electric field lines will be normal to the conductor. For the timebeing we do not have the knowledge of the induced charge density . Fig 2.25o apply the method of images, if we place an image charge -Q as shown in the figure 2.26, thesystem of two charges (essentially a dipole) will produce zero potential at the location of theconducting plane in the original problem. Solution of field and potential in the region plane(conducting plane in the original problem) will remain the same even the solution is obtained bysolving the problem with the image charge as the charge and the boundary condition remainsunchanged in the region .PREPARED BY GADDALA JAYARAJU, M.TECH Page 34
• 35. Static Electric Fields (UNIT-2)Fig 2.26We find that at the point P(x, y, z), ...............(2.111)Similarly, the electric field at P can be computed as: ...............(2.112)The induced surface charge density on the conductor can be computed as: ...............(2.113)The total induced charge can be computed as follows: ...............(2.114)Using change of variables,PREPARED BY GADDALA JAYARAJU, M.TECH Page 35
• 36. Static Electric Fields (UNIT-2) ...............(2.115)Thus, as expected, we find that an equal amount of charge having opposite sign is induced on theconductor.Method of images can be effectively used to solve a variety of problems, some of which areshown in the next slide.1.Point charge between semi-infinite conducting planes.Fig 2.27 :Point charge between two perpendicular semi-infinite conducting planesFig 2.28: Equivalent Image charge arrangementPREPARED BY GADDALA JAYARAJU, M.TECH Page 36
• 37. Static Electric Fields (UNIT-2)In general, for a system consisting of a point charge between two semi-infinite conducting planesinclined at an angle (in degrees), the number of image charges is given by .2.Infinite line charge (density C/m) located at a distance d and parallel to an infiniteconducting cylinder of radius a.Fig 2.29: Infinite line charge and parallel conducting cylinderFig 2.30: Line charge and its imageThe image line charge is located at and has density . See Appendix 1 for details3.A point charge in the presence of a spherical conductor can similarly be represented in terms ofthe original charge and its image.Electrostatic boundary value problemIn this section we consider the solution for field and potential in a region where the electrostaticconditions are known only at the boundaries. Finding solution to such problems requires solvingLaplace’s or Poisson’s equation satisfying the specified boundary condition. These types ofproblems are usually referred to as boundary value problem. Boundary value problems forpotential functions can be classified as:• Dirichlet problem where the potential is specified everywhere in the boundaryPREPARED BY GADDALA JAYARAJU, M.TECH Page 37
• 38. Static Electric Fields (UNIT-2)• Neumann problem in which the normal derivatives of the potential function are specifiedeverywhere in the boundary• Mixed boundary value problem where the potential is specified over some boundaries and thenormal derivative of the potential is specified over the remaining ones.Usually the boundary value problems are solved using the method of separation of variables,which is illustrated, for different types of coordinate systems.Boundary Value Problems In Cartesian Coordinates:Laplaces equation in cartesian coordinate can be written as, ...............(2.116)Assuming that V(x,y,z) can be expressed as V(x,y,z) = X(x)Y(y)Z(z) where X(x), Y(y) and Z(z) arefunctions of x, y and z respectively, we can write: ..............(2.117)Dividing throughout by X(x)Y(y)Z(z) we get .In the above equation, all terms are function of one coordinate variably only, hence we can write: ...........................................................................(2.118)kx, ky and kz are seperation constant satisfyingAll the second order differential equations above has the same form, therefore we discuss thepossible form of solution for the first one only.Depending upon the value of kx , we can have different possible solutions.PREPARED BY GADDALA JAYARAJU, M.TECH Page 38
• 39. Static Electric Fields (UNIT-2) .........................(2.119)A,B,C and D are constants.Similarly we have the possible solutions for other two differential equations. The particularsolutions for X(x), Y(y) and Z(z) to be used will be chosen based on the nature of the problemsand the constants A,B etc are evaluated from the boundary conditions. We illustrate the methoddiscussed above with some examples.Example 1: As shown in the figure 2.31, let us consider two electrodes, one grounded and theother maintained at a potential V. The region between the plates is filled up with two dielectriclayers. Further, we assume that there is no variation of the field along x or y and there is no freecharge at the interface.Fig 2.31The Laplace equation can be written as: for which we can write the solution in general form as V= Az+B.Considering the two regions, .......................(2.120)Applying the boundary conditions V=0 for z=0 and V=V0 for z = d, we can writePREPARED BY GADDALA JAYARAJU, M.TECH Page 39
• 40. Static Electric Fields (UNIT-2) ..................(2.121)Both the solution should give the same potential at z =a . Therefore, ...........................................(2.122a)or, ...................................(2.122b)Another equation relating to A1 and A2 is required to solve for the two constants A1 and A2.We note that at the dielectric boundary at z = a , the normal component of electric flux densityvector is constant. Noting that , we can write ...........................(2.123a) or, ..................(2.123b)Solving for A1 and A2, we find that ..................(2.124a) .................(2.124b)Using the above expressions for A1 and A2, we can find and .Example-2: In the earlier example, we considered a situation where the potential function was afunction of only one coordinate. In the figure 2.32 we consider a problem where the potential is afunction of two coordinate variables. We consider two number of grounded semi-infinite parallelplate electrodes separated by a distance d. A third electrode maintained at potential V0 andinsulated from the grounded conductors is placed as shown in the figure 2.32. The potentialfunction is to be determined in the region enclosed by the electrodes.PREPARED BY GADDALA JAYARAJU, M.TECH Page 40
• 41. Static Electric Fields (UNIT-2)Fig 2.32: Potential Function Enclosed by the ElectrodesAs discussed above, we can write V(x,y,z) = X(x) Y(y) Z(z). We now construct the solution for Vsuch that it satisfies the stated conditions at the boundary.We find that V is independent of x. Therefore, X(x) = C1 , where C1is a constant.Therefore, ..................(2.125)We observe that the potential has to be equal to V0 at y = 0 and as the potential functionbecomes zero. Therefore, the y dependence can be expressed as an exponential function.Similarly, since V = 0 for z = 0 and z = d, the z dependence will be a sinusoidal function with kz= k, k being a real number.Therefore, .....................(2.126) ...............(2.127) .........................(2.128)Combining all these terms we can write: , where C is a constant.Using the condition that V(y,d) =0, we can write , n=1,2,3,....For a particular n, . Therefore we can write the general solution asPREPARED BY GADDALA JAYARAJU, M.TECH Page 41
• 42. Static Electric Fields (UNIT-2) , where Cns are yet to be determined.Given that . Therefore . Multiplying both sides by onboth sides and integrating from z =0 to z = d we can write, .....................(2.129)Using orthogonality conditions and evaluating the integrals we can show that ..................................................(2.130)The general solution for the potential function can therefore be written as .........................(2.131)n the region y > 0 and 0< z< d .Boundary Value Problem in Cylindrical and Spherical Polar CoordinatesAs in the case of Cartesian coordinates, method of separation of variables can be used to obtainthe general solution for boundary value problems in the cylindrical and spherical polarcoordinates also. Here we illustrate the case of solution in cylindrical coordinates with a simpleexample where the potential is a function of one coordinate variable only.Example 3: Potential distribution in a coaxial conductorLet us consider a very long coaxial conductor as shown in the figure 2.33, the inner conductor ofwhich is maintained at potential V0 and outer conductor is grounded.PREPARED BY GADDALA JAYARAJU, M.TECH Page 42
• 43. Static Electric Fields (UNIT-2)Fig 2.33: Potential Distribution in a Coaxial ConductorFrom the symmetry of the problem, we observe that the potential is independent of variation.Similarly, we assume the potential is not a function of z. Thus the Laplace’s equation can bewritten as ...................................................(2.132)Integrating the above equation two times we can write .....................................................(2.133)where C1 and C2 are constants, which can be evaluated from the boundary conditions ............................................................(2.134)Evaluating these constants we can write the expression for the potential as .......................(2.135)Furthure, .......................(2.136)Therefore, ........................(2.137)If we consider a length L of the coaxial conductor, the capacitancePREPARED BY GADDALA JAYARAJU, M.TECH Page 43
• 44. Static Electric Fields (UNIT-2) ....................................(2.138)Example-4: Concentric conducting spherical shellsAs shown in the figure 2.34, let us consider a pair of concentric spherical conduction shells,outer one being grounded and the inner one maintained at potential V0.Considering the symmetry, we find that the potential is function of r only. The Laplace’sequation can be written as: ............(2.139)From the above equation, by direct integration we can write, ...............................(2.140)Applying boundary conditions ................................(2.141)we can find C1 and C2 and the expression for the potential can be written as: ...........................(2.142)The electric field and the capacitor can be obtained using the same procedure described inExample -3.PREPARED BY GADDALA JAYARAJU, M.TECH Page 44
• 45. Static Electric Fields (UNIT-2)Fig 2.34: Concentric conducting spherical shellsADDITIONAL PROBLEMS SOLVED: 1. Two point charges Q 1and Q2 are located at (3,0,0) and (1,2,0). Find the relation between Q1 and Q2 such that x and y component of the total force on a test charge qi placed at (1,- 1,0) are equal.Solution:The force F1 on q0 due to Q1 is given bySince x and y component of the total force are equal Or, Or, Or,PREPARED BY GADDALA JAYARAJU, M.TECH Page 45
• 46. Static Electric Fields (UNIT-2) 2. A line charge of charge density lies along z -axis and a point charge Q0 lies at (0,9,0). Find Q0 so that net force on a test charge located at (0,3,0) is zero. Solution: We know that for an infinite line charge along z, the field at any radial distance is given by In the present case, the test charge is located on the y-axis. Thus the force F1 on the test charge q0 due to the line charge isThe force F2 on q0 due to Q0 is Given C/m 3. (a) The figure Figure P2.1shows an arc AB of radius a and carrying uniform charge density C/m lies on the xy plane. Calculate the electric field intensity at point P(0,0,h). (b) What is the electric field at P due to a ring of radius a carrying C/m if the ring lies on the x y plane with its centre at the origin.PREPARED BY GADDALA JAYARAJU, M.TECH Page 46
• 47. Static Electric Fields (UNIT-2)Figure P2.1 Solution : The electric field at P due to the elementary charge dQ (b) In the case of the circular ring, we have from part (a) =0 , =PREPARED BY GADDALA JAYARAJU, M.TECH Page 47
• 48. Static Electric Fields (UNIT-2)Alternatively,Figure P2.2 Using symmetry , for any elementary charge dQ , a symetrical charge dQ exists , so that and 4. A circular disc of radius a carries a uniform surface charge density as shown in Figure P2.3. Determine the potential and electric field at a point P(0,0,z) on the axis (z > 0) .PREPARED BY GADDALA JAYARAJU, M.TECH Page 48
• 49. Static Electric Fields (UNIT-2)Figure P2.3Solution : Let us consider an elementary area The charge dQ on this elementary area is given byThe potential at the point P due this elementary charge is given by The electric fieldPREPARED BY GADDALA JAYARAJU, M.TECH Page 49
• 50. Static Electric Fields (UNIT-2)5.Determine the work done in carrying charge from P1((1,1,-3) to P2(4,2,-3) in an electricfield .Solution :(a) x = y2 dx = 2y.dy E.dl = y.2y.dy + y2.dy = 3 y2dy(b) Both points P1 and P2 are on z=-3 plane.Euation of the line joining P1 and P2PREPARED BY GADDALA JAYARAJU, M.TECH Page 50
• 51. Static Electric Fields (UNIT-2) 6. Assume that a circular tube of radius a and height h is placed on the xy plane with its axis coinciding with z axis. A total charge Q is uniformly distributed on the tube. Determine V and E at a point z>h on the z axis.Figure P2.4Total charge = QLet us consider a ring of charge consisting of a thin strip of width dz at a height z. If weconsider a small length of the strip at an angle . Then the chargeThe potential due to the ringPREPARED BY GADDALA JAYARAJU, M.TECH Page 51
• 52. Static Electric Fields (UNIT-2) The potential at P(0,0,z)Substituting t = z - z 7. Dipole moment is located at the point (1, 1,-2). Determine the potential at a point (3, 4, 2). Solution : We know that for a dipole moment p, the potential V is given by Where is the position vector of the dipole centre. Here we havePREPARED BY GADDALA JAYARAJU, M.TECH Page 52
• 53. Static Electric Fields (UNIT-2)8.The interface between a dielectric medium having relative permittivity 4 and free space ismarked the y = 0 plane. If the electric field next to the interface in the free space region is givenby , determine E field on the other side of the interface.Solution :Let represents the electric field in the free space and represents the electric fieldin the dielectric region.Since x & z component of the electric fields are parallel to the interface at y = 0, by continuity oftangential field componentBy continuity of the normal component of the flux density vector, we can write in the dielectric region is given by V/m. 9. A parallel plate capacitor has a plate area of 1 cm2 and 2mm separation between the plates. The space between the plates is filled with a dielectric whose relative permittivity varies linearly from 2 to 4. Neglecting the fringing effect, what would be the capacitance?PREPARED BY GADDALA JAYARAJU, M.TECH Page 53
• 54. Static Electric Fields (UNIT-2) Ans: 10. Consider a cylindrical capacitor as shown in the Figure P2.5. Determine the capacitance neglecting the fringing given that a dielectric material of dielectric constant fills the region and another dielectric material of dielectric constant fills the region .Figure P2.5 Solution : We find that electric filed has only component and therefore If we consider a total charge Q then where & Similarly,PREPARED BY GADDALA JAYARAJU, M.TECH Page 54
• 55. Static Electric Fields (UNIT-2) Solution : Let us consider the parallel plate capacitor with area A, separation between the platesd & thepermittivity varies from to . The plates are placed at z = 0 and z = d. Given : A = 1 cm2 = 10 -4 m 2 -3 d = 2 mm = 10 m At any height z from the bottom plate, The charge density on the plate is The electric field can be written as The potential The capacitance C is given byPREPARED BY GADDALA JAYARAJU, M.TECH Page 55
• 56. Static Electric Fields (UNIT-2)Substituting the values, The total potential V = V 1 + V 2 = =The capacitanceIt may be noted that, the same relation can also be derived by considering the series connectionof two capacitors formed by the two dielectric regions.PREPARED BY GADDALA JAYARAJU, M.TECH Page 56