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. Gauss's Law We already know about electric field lines and electric flux. Electricflux through a closed surface S is which is the number of field lines passing through surface S. Statement of Gauss's Law “Electric flux through any surface enclosing charge is equal to q/ε0 , where q is the net charge enclosed by the surface” mathematically, where qenc is the net charge enclosed by the surface and E is the total electric field at each point on the surface under consideration. It is the net charge enclosed in the surface that matters in Gauss's law but the total flux of electric field E depends also on the surface chosen not merely on the charge enclosed. So if you have information about distribution of electric charge inside the surface you can find electric flux through that surface using Gauss's Law. Again if you have information regarding electric flux through any closed surface then total charge enclosed by that surface can also be easily calculated using Gauss's Law. Surface on which Gauss's Law is applied is known as Gaussian surface which need not be a real surface. Gaussian surface can be an imaginary geometrical surface which might be empty space or it could be partially or fully embedded in a solid body. Again consider equation 11 In left hand side of above equation E·da is scalar product of two vectors namely electric field vector E and area vector da. Area vector da is defined as the vector of magnitude |da| whose direction is that of outward normal to area element da. So, da=nˆda where nˆ is unit vector along outward normal to da. From above discussion we can conclude that, (1) If both E and surface area da at each points are perpendicular to each other and has same magnitude at all points of the surface then vector E has same direction as that of area vector as shown below in the figure. since E is perpendicular to the surface (2) If E is parallel to the surface as shown below in the figure E at all points on the surface. APPLICATIONS OF GAUSS'S LAW * Derivation of Coulomb’s Law. Coulomb’s law can be derived from Gauss's law. Consider electric field of a single isolated positive charge of magnitude q as shown below in the figure. Field of a positive charge is in radially outward direction everywhere and magnitude of electric field intensity is same for all points at a distance r from the charge. We can assume Gaussian surface to be a sphere of radius r enclosing the charge q. From Gauss's law since E is constant at all points on the surface therefore, surface area of the sphere is A=4πr2 thus, Now force acting on point charge q' at distance r from point charge q is This is mathematical statement of Coulomb’s law. * Electric field due to line charge Consider a long thin uniformly charged wire and we have to find the electric field intensity due to the wire at any point at perpendicular distance from the wire. If the wire is very long and we are at point far away from both its ends then field lines outside the wire are radial and would lie on a plane perpendicular to the wire. Electric field intensity has same magnitude at all points which are at same distance from the line charge. We can assume Gaussian surface to be a right circular cylinder of radius r and length l with its ends perpendicular to the wire as shown below in the figure. λ is the charge per unit length on the wire. Direction of E is perpandicular to the wire and components of E normal to end faces of cylinder makes no contribution to electric flux. Thus from Gauss's law Now consider left hand side of Gauss's law since at all points on the curved surface E is constant. Surface area of cylinder of radius r and length l is A=2πrl therefore, Charge enclosed in cylinder is q=linear charge density x length l of cylinder, or, q=λl From Gauss's law Thus electric field intensity of a long positively charged wire does not depends on length of the wire but on the radial distance r of points from the wire. Electric field due to charged solid sphere We'll now apply Gauss's law to find the field outside uniformly charged solid sphere of radius R and total charge q. In this case Gaussian surface would be a sphere of radius r>R concentric with the charged solid sphere shown below in the figure. From Gauss's law where q is the charge enclosed. Charge is distributed uniformly over the surface of the sphere. Symmetry allows us to extract E out of the integral sign as magnitude of electric field intensity is same for all points at distance r>R. Since electric field points radially outwards we have also as discussed magnitude of E is constant over Gaussian surface so, where 4πr2 is the surface area of the sphere. Again from Gauss's law we have Thus we see that magnitude of field outside the sphere is exactly the same as it would have been as if all the charge were concentrated at its center. * Electric field due to an infinite plane sheet of charge Consider a thin infinite plane sheet of charge having surface charge density σ (charge per unit area). We have to find the electric field intensity due to this sheet at any point which is distance r away from the sheet. We can draw a rectangular Gaussian pillbox extending equal distance above and below the plane as shown below in the figure. By symmetry we find that Eon either side of sheet must be perpandicular to the plane of the sheet, having same magnitude at all points equidistant from the sheet. No field lines crosses the side walls of the Gaussian pillbox i.e., component of E normal to walls of pillbox is zero. We now apply Gauss's law to this surface in this case charge enclosed is q=σA where A is the area of end face of Gaussian pillbox. E points in the direction away from the plane i.e., E points upwards for points above the plane and downwards for points below the plane. Thus for top and bottom surfaces, thus 2A|E|=σA/ε0 or, |E|=σ/2ε0 Here one important thing to note is that magnitude of electric field at any point is independent of the sheet and does not decrease inversely with the square of the distance. Thus electric field due to an infinite plane sheet of charge does not falls of at all. ………………………………………………………………. Electric Potential and Electric Field Consider a charge placed in an electric field generated by fixed charges. Let us chose some arbitrary reference point in the field. At this point, the electric potential energy of the charge is defined to be zero. This uniquely specifies the electric potential energy of the charge at every other point in the field. For instance, the electric potential energy to at some point along any path. As, we can calculate depends both on the particular charge is simply the work done in moving the charge from . It is clear, from (from previous knowledge of , that which we place in the field, and the magnitude and direction of the electric field along the chosen route between points and . However, it is also clear that is directly proportional to the magnitude of the charge is . Thus, if the electric potential energy of a charge then the electric potential energy of a charge at the same point is at point . We can exploit this fact to define a quantity known as the electric potential. The difference in electric potential between two points and in an electric field is simply the work done in moving some charge between the two points divided by the magnitude of the charge. Thus, (80) where denotes the electric potential at point potential between points and , etc. This definition uniquely defines the difference in electric , but the absolute value of the potential at point therefore, without loss of generality, set the potential at point of a charge at some point potential at that point: remains arbitrary. We can equal to zero. It follows that the potential energy is simply the product of the magnitude of the charge and the electric (81) The electric potential at point (relative to point ) is solely a property of the electric field, and is, therefore, the same for any charge placed at that point. We shall see exactly how the electric potential is related to the electric field later on. The dimensions of electric potential are work (or energy) per unit charge. The units of electric potential are, therefore, joules per coulomb ( ). A joule per coulomb is usually referred to as a volt (V): i.e., (82) Thus, the alternative (and more conventional) units of electric potential are volts. The difference in electric potential between two points in an electric field is usually referred to as the potential difference, or even the difference in ``voltage,'' between the two points. A battery is a convenient tool for generating a difference in electric potential between two points in space. For instance, a twelve volt (12V) battery generates an electric field, usually via some chemical process, which is such that the potential difference between its positive and negative terminals is twelve volts. This means that in order to move a positive charge of 1 coulomb from the negative to the positive terminal of the battery we must do 12 joules of work against the electric field. (This is true irrespective of the route taken between the two terminals). This implies that the electric field must be directed predominately from the positive to the negative terminal. More generally, in order to move a charge work through a potential difference we must do , and the electric potential energy of the charge increases by an amount the process. Thus, if we move an electron, for which minus 1 volt then we must do an electron volt (eV): i.e., in C, through a potential difference of joules of work. This amount of work (or energy) is called (83) The electron volt is a convenient measure of energy in atomic physics. For instance, the energy required to break up a hydrogen atom into a free electron and a free proton is eV. We have seen that the difference in electric potential between two arbitrary points in space is a function of the electric field which permeates space, but is independent of the test charge used to measure this difference. Let us investigate the relationship between electric potential and the electric field. Consider a charge which is slowly moved an infinitesimal distance along the difference in electric potential between the final and initial positions of the charge is change in the charge's electric potential energy is given by -axis. Suppose that the . By definition, the (84) From Eq. (76), the work which we perform in moving the charge is (85) where is the local electric field-strength, and the -axis. By definition, Energy conservation demands that on the charge), or is the angle subtended between the direction of the field and , where is the -component of the local electric field. (i.e., the increase in the charge's energy matches the work done (86) which reduces to (87) We call the quantity the gradient of the electric potential in the -direction. It basically measures how fast the potential varies as the coordinate is changed (but the coordinates and are held constant). Thus, the above formula is saying that the -component of the electric field at a given point in space is equal to minus the local gradient of the electric potential in the -direction. According to Eq. (87), electric field strength has dimensions of potential difference over length. It follows that the units of electric field are volts per meter ( per coulomb: i.e., . Of course, these new units are entirely equivalent to newtons (88) Consider the special case of a uniform -directed electric field generated by two uniformly charged parallel planes normal to the -axis. It is clear, from Eq. (87), that if is to be constant between the plates then must vary linearly with in this region. In fact, it is easily shown that (89) where is an arbitrary constant. According to Eq. (89), the electric potential decreases continuously as we move along the direction of the electric field. Since a positive charge is accelerated in this direction, we conclude that positive charges are accelerated down gradients in the electric potential, in much the same manner as masses fall down gradients of gravitational potential (which is, of course, proportional to height). Likewise, negative charges are accelerated up gradients in the electric potential. According to Eq. (87), the -component of the electric field is equal to minus the gradient of the electric potential in the -direction. Since there is nothing special about the -direction, analogous rules must exist for the and -components of the field. These three rules can be combined to give - (90) Here, the derivative is taken at constant and , etc. The above expression shows how the electric field , which is a vector field, is related to the electric potential , which is a scalar field. We have seen that electric fields are superposable. That is, the electric field generated by a set of charges distributed in space is simply the vector sum of the electric fields generated by each charge taken separately. Well, if electric fields are superposable, it follows from Eq. (90) that electric potentials must also be superposable. Thus, the electric potential generated by a set of charges distributed in space is just the scalar sum of the potentials generated by each charge taken in isolation. Clearly, it is far easier to determine the potential generated by a set of charges than it is to determine the electric field, since we can sum the potentials generated by the individual charges algebraically, and do not have to worry about their directions (since they have no directions). Equation (90) looks rather forbidding. Fortunately, however, it is possible to rewrite this equation in a more appealing form. Consider two neighboring points displacement of point relative to point . Let points. Suppose that we travel from along the move from along the -axis, and finally moving to to and . Suppose that is the vector be the difference in electric potential between these two by first moving a distance along the as we move along the -axis, then moving -axis. The net increase in the electric potential is simply the sum of the increases -axis, and along the as we move along the -axis, as we as we move -axis: (91) But, according to Eq. (90), , etc. So, we obtain (92) which is equivalent to (93) where is the angle subtended between the vector and the local electric field . Note that attains its most negative value when . In other words, the direction of the electric field at point corresponds to the direction in which the electric potential decreases most rapidly. A positive charge placed at point is accelerated in this direction. Likewise, a negative charge placed at is accelerated in the direction in which the potential increases most rapidly (i.e., ). Suppose that we move from point to a neighboring point in a direction perpendicular to that of the local electric field (i.e., ). In this case, it follows from Eq. (93) that the points and lie at the same electric potential (i.e., ). The locus of all the points in the vicinity of point which lie at the same potential as is a plane perpendicular to the direction of the local electric field. More generally, the surfaces of constant electric potential, the so-called equipotential surfaces, exist as a set of non-interlocking surfaces which are everywhere perpendicular to the direction of the electric field. Figure 14 shows the equipotential surfaces (dashed lines) and electric field-lines (solid lines) generated by a positive point charge. In this case, the equipotential surfaces are spheres centred on the charge. Figure 14: The equipotential surfaces (dashed lines) and the electric field-lines (solid lines) of a positive point charge. In Sect. 4.3, we found that the electric field immediately above the surface of a conductor is directed perpendicular to that surface. Thus, it is clear that the surface of a conductor must correspond to an equipotential surface. In fact, since there is no electric field inside a conductor (and, hence, no gradient in the electric potential), it follows that the whole conductor (i.e., both the surface and the interior) lies at the same electric potential. ……………………………………………………………………………… MULTIPOLE EXPANSION A multipole expansion is a series expansion of the effect produced by a given system in terms of an expansion parameter which becomes small as the distance away from the system increases. Therefore, the leading one or terms in a multipole expansion are generally the strongest. The first-order behavior of the system at large distances can therefore be obtained from the first terms of this series, which is generally much easier to compute than the general solution. Multipole expansions are most commonly used in problems involving the gravitational field of mass aggregations, the electric and magnetic fields of charge and current distributions, and the propagation of electromagnetic waves. Electric Multipole Expansion For a given charge distribution, we can write down a multipole expansion, which gives the potential as a series in powers of , where is the distance from the origin to the observation point. We know that the potential in general is In the integral, where is the position of charge element is the angle between and . From the law of cosines . We can rewrite this as From the theory of Legendre polynomials, it is known that the last factor in this expression is a generating function for the polynomials. That is, if we write the square root as an power series, we get The coefficient of in the series is the Legendre polynomial . This can be verified for the first few terms by calculating the Taylor series expansion of the square root term about . This is tedious to do by hand, but using Maple, we get, defining : It is important to note that the angle is equivalent to the angle in spherical coordinates only if the observation point lies on the axis, since that is the only configuration where the angle between the observation vector and a charge element corresponds to the spherical coordinate angle . (A more general multipole expansion uses spherical harmonics rather than just Legendre polynomials, but that’s a topic for a more advanced post.) With this restriction, we can substitute the series expansion back into 1 to get The first few terms in this series have special names. The term is where is the total charge. This is called the monopole term, and shows that to a first approximation, the potential of any charge distribution is just the potential of a point charge with the same total charge. The next term in the series is This is called the dipole term. For Finally, for , we get the quadrupole term we get the octopole term As an example, consider a solid sphere with a charge density We can use the integrals above to find the first non-zero term in the series, and thus get an approximation for the potential. Note that we can do this only for points on the z axis. By direct calculation, we have for the monopole term: since the integral over gives zero. Thus the monopole term vanishes, as it always does if the total charge is zero. For the dipole term, we get This time, the integral over For the quadrupole term gives zero, since the term is odd relative to the interval . The octopole term comes out to zero, since the terms in are again odd relative to the interval . ………………………………………………………………………………. Work and energy – point charges Since charges exert forces on each other through their electric fields, it will require the expenditure of energy, or work, to assemble any configuration of charges. Here we’ll have a look at how much energy is required to assemble, and thus how much energy is stored, in a collection of discrete charges. The force on a charge due to an electric field is . From elementary physics, we know that the work done when an object is moved against a force is the negative (since we’re opposing the force) of (force) times (distance). In general, if the force varies as a function of position, we get where the integral is taken over the path through which the object is moved. The minus sign is an indication that we are opposing the force ; if we work instead with the force that we must exert to move the object, then the minus sign is omitted. For the electric field, then, we get To get the last line, we’ve used the fact that, in electrostatics, the line integral of the electric field is independent of the path; it depends only on the endpoints and . We’ve seen earlier that the line integral of the field is the negative of the potential difference between the two endpoints. So, in other words, the potential difference between two points is the work per unit charge required to move a charge between those two points. If we’ve set the reference point for the potential at infinity (that is, at infinity), then the work required to bring in a charge from infinity to a point is We can apply this formula to find out how much energy is required to assemble a collection of point charges. To place a single charge at a location takes no work, since there are no fields to work against. Bringing in a second charge requires working against the field due to . The potential due to is , so if we want to place at position the work required is Before we go any further, it’s worth noting that this formula gives rise to a bit of a problem. What if we want to assemble a point charge itself? That is, suppose we want to build up a point charge of a certain size by bringing together other point charges and, in effect, gluing them together. This seems to be a valid procedure, since after all, if a charge is truly a mathematical point, we should be able to pile as many of these point charges on top of each other as we like without increasing the volume (that is, zero) occupied by the sum of all the charges. However, if we try that, the above formula says this will require an infinite amount of work (since ). This is, in fact, a recognized problem in electrodynamics, and the problems don’t go away even in the quantum mechanical theory. In fact, we can’t even get out of the problem by saying that there is no such thing as a point charge, since a lot of physicists think that the electron might actually be a point charge (at least its diameter, if it’s non-zero, is so small that nobody has actually measured it yet). With that caution in mind, let’s ignore the problem and carry on. If we assume that the existence of point charges is possible (without taxing our minds as to how they are built), we can continue to add more point charges to our distribution. Adding a third charge at location requires work The total work required to assemble all three charges is then The general pattern should be fairly obvious by now. To assemble charges, the total work is If we extend the second sum to cover 1 through ), we get where (excluding is the potential due to all the charges in the collection except . As an example, suppose we have arranged two charges of at the ends of a diagonal in a square, and a charge of on one of the other two corners of the square. How much work is required to bring in another charge of from infinity and place it at the remaining corner of the square? To work this out, we need to find the potential at this corner due to the existing three charges. If the length of each side of the square is , then we get Once all four charges have been assembled, the total energy stored in the collection is In the second line, the potential at the location of one of the charges is . Multiplying this by gives the first two terms in the square brackets. Similar logic for one of the charges gives the last two terms in the brackets. The factor of 2 in the numerator arises from the fact that there are two each of and charges. Work and energy – continuous charge For discrete point charges, the work required to assemble a collection of where is the potential at the location of charge For a continuous charge density charges is due to all the other charges (excluding ). we can write this as an integral over the volume containing the charges: Note, however, that there is a subtle distinction between the discrete and continuous formulas. In the discrete formula, the potential term in the sum excludes the charge , but in the integral form, the potential is the complete potential due to the entire charge distribution. In the continuous case, we don’t talk about point charges (unless we write the density as a sum of delta functions), so in that sense the continuous formula is more accurate. However, if point charges such as electrons do truly exist, then we can’t either build them or take them apart, so it seems fair enough to exclude the energies involved in doing so. However, as we’ve seen in the post on discrete charges, the energy associated with a point charge is actually infinite, so it’s dodgy to just ignore it. This problem plagues both classical and quantum electrodynamics, but most books just ignore it. The integral formula can be expressed in terms of the electric field by using a bit of vector calculus. We know from Gauss’s law for electrostatics that so we can write the work as A theorem from vector calculus says In the last line, we used the relation . Therefore, the work becomes We can now use the divergence theorem on the first term, and convert it from a volume integral to a surface integral if we select some surface that encloses all the charge (we’re assuming that we’re dealing with realistic systems so that all the charge is at a finite distance, and not with things like infinite planes of charge). That is, we can say Since all we require is that the surface encloses all the charges, we can let the surface tend to infinity. In that case, since the charges are all at finite distances, and we are taking the potential to be zero at infinity, and the electric field falls off as , the surface integral will go to zero at infinity. The volume integral is always positive (since we’re integrating the square of the field, which is always positive), so what happens is that as we include more volume, the surface integral decreases and the volume integral increases in such a way as to keep the total work constant. That is, we get where the integral now covers all space. Note the distinction between 2 and 10: in the first case, we need to integrate only over that volume where ; in the second case we need to integrate over all space, since in general the electric field is always non-zero over any finite distance, even for a localized charge distribution. As an example, we can work out the energy stored in a uniformly charged solid sphere of radius We’ll do it four different ways to show how each of the above methods works. and charge . Example 1. We can use 2. We found the potential of the sphere earlier (Example 1 in this post). The charge density in terms of the total charge is and the potential inside the sphere (all we need here, since we need to integrate only over that volume where ) is Therefore, we get Example 2. Same problem, but now we use 10. We worked out the field earlier (Example 2 in this post). In this case, we will need the field for both inside and outside the sphere, since we must integrate over all space. We have, for inside: and for outside: The energy is then Example 3. This time we use the formula 9, so the integral is split between a volume integral and a surface integral. If we use a surface of radius , then the volume component can be worked out using the same integrals as in Example 2, but changing the limit on the second integral. Note that as , this integral tends to the total energy as worked out in the previous two examples: . The surface integral uses the potential and field at a distance sphere, which is The field at is Both of these are constants over the bounding sphere, so we get The total energy is . This time we need the potential outside the Example 4. Finally, we can build up the solid sphere by adding successive layers of charge of thickness have, when the sphere has intermediate radius : This amount of charge is brought from infinity and added to a spherical volume of charge from above that the potential of such a sphere at its outer boundary is and radius . We . We know so the amount of work needed to add this charge to the sphere is We can then integrate this to find the total work: ………………………………………………………….. Dielectric Constant? It is also known as Relative permittivity. If two charges q 1 and q 2 are separated from each other by a small distance r. Then by using the coulombs law of forces the equation formed will be ———— 2.6 In the above equation is the electrical permittivity or you can say it, Dielectric constant. If we repeat the above case with only one change i.e. only change in the separation medium between the charges. Here some material medium must be used. Then the equation formed will be. ——2.7 Now after division of equation 2.6 with equation 2.7 we get… ——2.8 In the above figure is the Relative Permittivity. Again one thing to notice is that the dielectric constant is represented by the symbol(K) but permittivity by the symbol Now after the division third equation is created. Its we describe this equation then the new definition can be made from this. i.e. It is the ratio of the force of communication between the two point charges separated from each by a distance using air/Vacuum as a medium to the force of communication between the same charges placed at the same distance and using some material medium. Faraday has done many experiments on this topic. He found that When we insert some insulating material in the space present between the plates of a capacitor , having its plates fully charged then the capacitance of the capacitor will increase definitely. Values of dielectric constant varies with the change in materials. Some of the examples of materials is shown below in the table Values of K Material Air( 1 atm) 1.0006 Amber 2.8 Alcohol 25 Water 81 Hydrogen 1.00026 Germanium 16 Vaccum/Air 1 Metals ∞ Polar and Non Polar Dielectrics? Those materials which have the ability to transfer the electric effects without conducting. Dielectrics exists basically in two types. 1.Polar Dielectrics 2. Non polar Dielectrics Polar Dielectrics: Polar dielectrics are those in which the possibility of center coinciding of the positive as well as negative charge is almost zero i.e. they don’t coincide with each other. The reason behind this is their shape. They all are of asymmetric shape. Some of the examples of the polar dielectrics is NH3, HCL, water etc. Non Polar dielectrics: In case of non polar dielectrics the centres of both positive as well as negative charges coincide. Dipole moment of each molecule in non polar system is zero. All those molecules which belong to this category are symmetric in nature. Examples of non polar dielectrics are: methane , benzene etc. …………………………… What is Dielectric Polarization? In this case a non polar dielectric like methane is taken and placed in some external electric field. Center of positive charge of individual molecules is pulled automatically in the same direction as that of electric field towards the plate having negative charge. Similarly, the centre of the negative charged electrons is dragged in the opposite direction of the electric field, towards the plate having positive charge. So the centres of positive as well as negative charges are set apart. Due to this the molecules are deformed from their original shape and hence separate at last. So, due to the above process each molecule gets some dipole moment . After sometime these molecules will get polarized when the forces of attraction between the centres of positive and negative charges and the force due to electric field will reach at some stable state. Now the individual molecules will exist as separate tiny dipoles. ————–------------------------------------ 2.22 In the above equation alpha is proportionality constant. Its is also called as Atomic polarizability. By using the equation 2.22 = Cm / (C2 N-1m-2 ) * (NC-1) =m3 Now from the above result it is cleared that the Alpha i.e. atomic polarizability has similar dimensions as that of volume. Mostly values of alpha vary from 10-29 to 10-30. We are going to made some experiments on a rectangular slab. Let’s name it as ABCD. Keep it in the influence of the electric field present between the two plates. For this we have to think that all the atoms must be polarized in a uniform manner. Let the displacement between the charges be x. So the equation will be: p=qx If the total number of atoms present per unit volume be taken as N. But N is equal to the total dipole moment density . New equation formed will be: P=Np Or we can write it as P=Nqx Here P is the density of the dipole. Its other name is Electric polarization. Its units are C -m-2 Lets us think about the interior of the slab. The cancellation of the charges due to their equal magnitude of positive and negative charge, the volume charge density will be Zero. Some amount of positive charge will generate in the surface CD.It can be seen in the figure below. The equation of Effective electric field present inside a dielectric which is polarized is given by: E=Eo – E p ---------------------------------------—-2.23 Eo= = p/ i / = Qi/a * ------------------------------------—-2.24 The nature of the dielectric slab will affect the value of E. Now it is proved from the definition and the equation given below. The charge in external electric field divided by the reduced value of electric field gives us the constant i.e. k , which is the constant for material of the rectangular slab ABCD. /E=K ----------------------------------------------—2.25 (b) Electric Susceptibility: It states that the reduced value of electric field effects the electric polarization P due to a relation of direct proportion between them. P is directly proportional to E. So P= * *E --------------------------------——2.26 In the above equation is the electric susceptibility. The main purpose of to make the dimensionless. Its main purpose is to tell about the complete electric manner of acting of a dielectric. Value of Susceptibility varies with the variation of dielectrics. The value of Electric susceptibility for vacuum is equal to zero. E= Eo – P / – * *E/ By using the equation no 2.25 Or E = Eo – E Or E o= E+E=E (1+ K=1+ ) ---------------------------------——– 2.27 Dielectric constant and the susceptibility are related to each other according to the equation 2.27 In polar dielectrics individual molecules have their individual dipole moments. But when the External electric field’s influence remains inactive then the net dipole moment of the elements of dielectric becomes zero. Lets consider the case in which external electric fields influence is present. Due to this value of dipole moment in the individual elements increase by a little amount. Then another affect will take place, a twisting force will be generated which will try to change the alignment of electric dipoles parallel to that of the electric field. The alignment depends upon the amount of the electric field passed. Stronger the electric field higher will be the alignment. This also depends upon the temperature. The alignment of the dipoles increases if we increase the temperature. ……………………. See pdf file “CM & Multipole”