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Dielectrics - II Lecture 21: Electromagnetic Theory Professor D. K. Ghosh, Physics Department, I.I.T., Bombay In the previous lecture we obtained a multipole expansion of the potential due to a continuous charge distribution. We discussed in some detail, the dipole part of this expansion and will be doing so in further detail later in this lecture. Let us make some comments on the quadrupoleterm. The most elementary quadrupole consists of two dipoles oriented anti-parallel. A linear dipole is one where the four charges are lined up with the two dipoles being end to end but directed oppositely, so that there is a charge 2q at the centre and โq each at the ends. _ ++ + ++ _ _ + Recall that the expression for the quadrupole contribution to the potential is given by 3 ๐๐๐ข๐๐๐๐ข๐๐๐๐ 3 3 3 ๐ฅ๐ ๐ฅ๐ ๐ฅ๐ฅ 1 โโ ) ๐3 ๐ โฒ = โ โ ๐๐๐ ๐ ๐ = โโ 5 โซ (3๐ฅ๐โฒ ๐ฅ๐โฒ โ ๐ฟ๐๐ ๐ โฒ2 ) ๐ (๐โฒ 4๐๐0 2๐ ๐๐๐๐ข๐๐ 2๐ 5 ๐=1 ๐=1 ๐=1 ๐=1 where the quadrupole moment tensor is given by ๐๐๐ = โซ ๐๐๐๐ข๐๐ โโ ) ๐3 ๐ โฒ (3๐ฅ๐โฒ ๐ฅ๐โฒ โ ๐ฟ๐๐ ๐ โฒ2 ) ๐ (๐โฒ Note that since I and j take three values each, the quadrupole moment is a tensor of rank two with 9 components. A few properties of quadrupoleare : 1. The tensor is traceless. (Recall that the trace of a matrix is sum of its diagonal components. This can be easily proved by taking i=j and summing over i. 1 2. At large distances quadrupolefield decreases as inverse fourth power of the distance from the source. (This is easily proved by inspection as the potential goes as inverse cube of distance). We now return back to the dipole term which is the most important term for the case of a neutral dielectric. Observe that the dipole moment defined by ๐ = โซ ๐๐(๐)๐3 ๐ depends, in general, on our choice of origin, for, if we let ๐ โ ๐ + ๐, we would have ๐ โ โซ(๐ + ๐)๐(๐)๐3 ๐ = ๐ + ๐ โซ ๐(๐)๐3 ๐ = ๐ + ๐๐ Thus only for the case of the dielectric not having a net charge will it be independent of the choice of origin. We had seen that the potential due to the dipole contribution to the charge distribution is the one that we would obtain if we have a volume charge density and a surface charge density, ๐๐๐๐๐๐๐ (๐) = 1 ๐๐ ๐๐ โซ ๐๐ โฒ + โซ ๐3 ๐ โฒ โฒ โโโ 4๐๐0 ๐๐ข๐๐๐๐๐ |๐ โ โโโ ๐| ๐๐๐๐ข๐๐ |๐ โ ๐ | where the charge densities are related to the polarization vector by ๐๐ = โโ โ ๐โ ๐๐ = ๐โ โ ๐ฬ It may be emphasized that the bound charges, whether in volume or on the surface, are not fictitious charges; they are very real and their only difference with the free charges which we have in conductors is that they are bound to their parent atoms or molecules. โ is an auxiliary field defined by The โdisplacement vectorโ ๐ท โ = ๐0 ๐ธโ + ๐โ ๐ท Using the relationship of the polarization vector with the bound charge densities, we get, โ = ๐0 โ โ ๐ธโ + โ โ ๐โ โโ ๐ท ๐๐ + ๐๐ = ๐0 โ ๐๐ = ๐๐ ๐0 โ is given by mentally isolating the bound charges from free charges. There is really no Thus the field ๐ท physical way of doing it but defining such a displacement vector leads to substantial simplification. โ , We can easily derive the integral form of the equation satisfied by ๐ท โ ๐3 ๐ = โฎ๐ท โ โ ๐๐ = โซ ๐๐ ๐3 ๐ = ๐๐ โซโโ ๐ท ๐ 2 โ โ ๐๐ = ๐๐ โฎ๐ท ๐ where๐๐ is the total free charges in the sample. Boundary Conditions on the surface of a dielectric ฬ ๐ We recall that in deriving the boundary condition On the surface of a conductor, we had taken a Gaussian Pillbox of infinitesimal height, half inside the conductor โ๐ท โ ฬ = โ๐ ฬ ๐โฒ And half outside. The field inside the conductor being zero, the contribution to the flux came only from the top face of the pillbox, since the thickness of the pillbox being negligible the curved surface did not contribute. In the present case, there is a difference because the field inside is not zero. Further, one has to remember that at the interface, the normal is outward for the outside face and inward for the inside face. Let us assume that there no free charges on the surface. The volume integral of the electric field over the pillbox is given by 1 โซ ๐๐ ๐3 ๐ ๐0 1 = โ โซ โ โ ๐โ๐3 ๐ ๐0 โซ โ โ ๐ธโ ๐3 ๐ = Using the divergence theorem on both sides, โฎ ๐ธโ โ ๐๐ = โ 1 โฎ ๐โ โ ๐๐ ๐0 The contribution to the surface integral on both sides, as mentioned before, are from the surfaces only with the directions of the normal being opposite, (๐ธโ๐๐ข๐ก โ ๐ธโ๐๐ ) โ ๐ฬฮ๐ = 1 ๐๐ ฮ๐ ๐โ โ ๐ฬฮ๐ = ๐0 ๐0 Where we have taken the direction of normal in this equation as going away from the surface (i.e. outward). Thus we have, (๐ธโ๐๐ข๐ก โ ๐ธโ๐๐ ) โ ๐ฬ = ๐๐ ๐0 This is precisely the relationship that we had earlier for the boundary condition for a conductor except that the surface charge density was due to the free charges. Adding the free charge density, the discontinuity in the normal component of the electric field is given by 3 (๐ธโ๐๐ข๐ก โ ๐ธโ๐๐ ) โ ๐ฬ = ๐๐ + ๐๐ ๐0 We can obtain the discontinuity in the normal component of the displacement vector from ๐ท๐ = ๐0 ๐ธ๐ + ๐๐ = ๐0 ๐ธ๐ + ๐๐ We have, โ ๐๐ข๐ก โ ๐ท โ ๐๐ ) โ ๐ฬ = ๐0 (๐ธโ๐๐ข๐ก โ ๐ธโ๐๐ ) โ ๐ฬ โ ๐๐ (๐ท the reason for the minus sign is that outside the material ๐๐ = 0. Thus โ ๐๐ข๐ก โ ๐ท โ ๐๐ ) โ ๐ฬ = ๐๐ (๐ท โ to be continuous, Thus, if there are no free surface charges we have the normal component of ๐ท ๐ท1๐ = ๐ท2๐ The tangential components can be shown to be continuous by use of Stokeโs theorem. We take a rectangular contour across the surface. ๐ธ1๐ก = ๐ธ2๐ก In the above, the outside medium is denoted as 1 and the inside by 2. We define a linear dielectric as one where the polarization at a point in the dielectric is proportional to the electric field at the point. ๐โ = ๐0 ๐๐ ๐ธโ The quantity ๐๐ is known as the electric susceptibility. Using the relation between the displacement vector and the electric field vector, we have, โ = ๐0 ๐ธโ + ๐โ ๐ท = ๐0 (1 + ๐๐ )๐ธโ = ๐0 ๐ ๐ธโ โก ๐๐ธโ The constant ๐ = (1 + ๐๐ ) is known as the dielectric constant of the medium and ๐ = ๐0 ๐ is known as the permittivity of the medium. Example 1 : Image Problem with Dielectric Consider a point charge in a semi-infinite dielectric medium with permittivity ๐1 occupying the space ๐ง > 0, the remaining half space ๐ง < 0 has a permittivity ๐2 . Depending on the problem either of the media could be vacuum with permittivity ๐0 . 4 ๏ฅ๏ฒ๏ P r rโ ๏ฅ๏ฒ๏ ๏ฅ๏ฑ๏ O qโ dO ๏ฅ๏ฑ๏ d q qโ z d The interface is the z=0 plane and has no free charges on it. Based on our experience of solving similar problem using images, we will attempt to write down an expression for the potential at the point P. Since there are no free charges on the surface ๐ท๐ = ๐ท๐ง is continuous across the surface. Thus ๐1 ๐ธ๐ง |๐๐ = ๐2 ๐ธ๐ง |๐๐ข๐ก In addition, of course, the tangential component of ๐ธโ is continuous. Since the surface is in the x-y plane, we have ๐ธ๐ฅ |๐๐ = ๐ธ๐ฅ |๐๐ข๐ก ๐ธ๐ฆ |๐๐ = ๐ธ๐ฆ |๐๐ข๐ก In solving this problem, we fall back on the uniqueness theorem and attempt to solve the problem by intuitive method. Assume that in the medium 1,i.e., in the region z>0, the solution is the same as one would obtain by filling up the entire space by a dielectric of permittivity ๐๐ and put an image charge ๐ โฒ at ๐ง = โ๐. The potential at a point P (the entire space is now a uniform medium) is given by ๐1 = 1 ๐ ๐โฒ [ ] 1 + 4๐๐1 [(๐ง โ ๐)2 + ๐2 ]2 [(๐ง + ๐)2 + ๐2 ]12 Where we have used polar coordinates for the x-y plane, ๐2 = ๐ฅ 2 + ๐ฆ 2 . For the potential in medium 2, we assume that the solution would be the same as one would obtain by filling up the entire space by a dielectric of permittivity ๐๐ and put an image charge at the location of the charge q, i.e. at ๐ง = ๐ . We denote the sum of the original charge plus its image by ๐". Thus the potential at the point P is due to a single charge ๐" and is given by ๐2 = 1 ๐" 4๐๐2 [(๐ง โ ๐)2 + ๐2 ]12 Using the continuity of the normal component of the displacement vector, 5 ๐ท1๐ = ๐ท2๐ โ โ๐1 ๐๐1 ๐๐2 = โ๐2 | | ๐๐ง ๐ง=0 ๐๐ง ๐ง=0 (Remember that the direction of the outward normal at the interface was taken as uniquely defined). We have ๐๐1 1 โ๐(๐ง โ ๐) โ๐โฒ (๐ง + ๐) = [ ] 3 + ๐๐ง 4๐๐1 [(๐ง โ ๐)2 + ๐2 ]2 [(๐ง + ๐)2 + ๐2 ]32 ๐๐2 1 โ๐(๐ง โ ๐) = ๐๐ง 4๐๐2 [(๐ง โ ๐)2 + ๐2 ]32 Thus we get, (๐ โ ๐ โฒ )๐ ๐"๐ (1) 3 [๐2 + ๐2 ] [๐2 + ๐2 ]2 The tangential component of the electric field is obtained by differentiating the expressions for the potential with respect to x or y, 3 2 = ๐๐1 1 โ๐๐ฅ โ๐ โฒ ๐ฅ = + [ ] ๐๐ฅ 4๐๐1 [(๐ง โ ๐)2 + ๐2 ]32 [(๐ง + ๐)2 + ๐2 ]32 ๐๐2 1 โ๐"๐ฅ = ๐๐ฅ 4๐๐2 [(๐ง โ ๐)2 + ๐2 ]32 So that we have, ๐ + ๐โฒ 3 2 ๐" = ๐1 [๐2 + ๐2 ] 3 (2) ๐2 [๐ 2 + ๐2 ]2 From (1) and (2), we obtain, ๐1 โ ๐2 ๐ ๐1 + ๐2 2๐2 ๐" = ๐ ๐1 + ๐2 ๐โฒ = In both media Laplaceโs equation and the boundary conditions are satisfied and the solution is unique. 6 Dielectrics -II Lecture 21: Electromagnetic Theory Professor D. K. Ghosh, Physics Department, I.I.T., Bombay Tutorial Assignment 1. Find the first few multipole moments of a linear quadrupole aligned along the z direction. 2. Consider a quadrupole in the xy plane as shown. Determine the components of the quadrupole moment tensor in Cartesian coordinates. y +q โ๐ a a a x a โ๐ +q q 3. Two dielectrics having permittivity ๐1 and ๐2 have an interface which has no free charges. The โโโโ1 in medium 1 makes an angle ๐1 with the interface while ๐ธ โโโโ2 , the field in medium electric field ๐ธ 2 makes an angle ๐2 . Find the relationship between the two angles. ๐ฬ ๏ฑ๏ฒ๏ ๏ฑ๏ฑ๏ โโโโ ๐ธ2 โโโโ ๐ธ1 Solutions to Tutorial assignment 1. In spherical polar coordinates, the charges +q are at (๐, 0, 0) and (๐, ๐, 0). The charge densities of the distribution of charges is ๐(๐, ๐, ๐) = +๐ ๐ฟ(๐ โ ๐)๐ฟ(๐)๐ฟ(๐) + ๐๐ฟ(๐ โ ๐)๐ฟ(๐ โ ๐)๐ฟ(๐) โ 2๐๐ฟ 3 (๐) The multipole moment of this charge distribution is (for ๐ > 0, ) 7 โ (๐, ๐๐๐ = โซ ๐(๐)๐ ๐ ๐๐๐ ๐)๐3 ๐ โ (0,0) โ (๐, = ๐๐๐ [๐๐๐ + ๐๐๐ 0)] 9 (๐ โ ๐)! ๐ = ๐โ ๐ [๐๐๐ (1) + ๐๐๐ (โ1)] 4๐(๐ + ๐)! 9 (๐ โ ๐)! ๐ = ๐โ ๐ [๐๐๐ (1) + (โ1)๐ ๐๐๐ (1)] 4๐(๐ + ๐)! y The multipole moment vanishes for all odd ๐. For even ๐, q 9 (๐ โ ๐)! ๐ ๐๐๐ = 2๐โ ๐ ๐๐๐ (1) 4๐(๐ + ๐)! -2q q Further, ๐๐๐ (1) = 0 except for m=0. Thus ๐๐๐ x 9 = 2๐โ ๐๐ ๐๐ (1)๐ฟ๐.0 4๐ z (The above formula is valid for even values of ๐ > 0, for ๐ = 0, ๐00 = 0 because the total charge of the quadrupole is zero) 2. The charge density can be written in terms of delta functions ๐(๐ฅ, ๐ฆ, ๐ง) = ๐๐ฟ(๐ฅ โ ๐)๐ฟ(๐ฆ โ ๐)๐ฟ(๐ง) โ ๐๐ฟ(๐ฅ โ ๐)๐ฟ(๐ฆ + ๐)๐ฟ(๐ง) +๐๐ฟ(๐ฅ + ๐)๐ฟ(๐ฆ + ๐)๐ฟ(๐ง) โ ๐๐ฟ(๐ฅ + ๐)๐ฟ(๐ฆ โ ๐)๐ฟ(๐ง) ๐๐๐ = โซ (3๐ฅ๐ ๐ฅ๐ โ ๐ฟ๐๐ ๐ 2 )๐(๐ฅ, ๐ฆ, ๐ง)๐๐ฅ๐๐ฆ๐๐ง ๐๐ฅ๐ฅ = โซ (3๐ฅ 2 โ ๐ 2 )๐(๐ฅ, ๐ฆ, ๐ง)๐๐ฅ๐๐ฆ๐๐ง = 0 = ๐๐ฆ๐ฆ ๐๐ฅ๐ฆ = โซ 3๐ฅ๐ฆ๐(๐ฅ, ๐ฆ, ๐ง)๐๐ฅ๐๐ฆ๐๐ง = 12๐๐2 = ๐๐ฆ๐ฅ All other components are zero because the quadrupole is in the xy plane. 3. At the interface between two dielectrics, the tangential component of ๐ธโ is continuous ๐ธ1 cos ๐1 = ๐ธ2 cos ๐2 โ is continuous, In the absence of free surface charge at the interface the normal component of ๐ท ๐ท1 sin ๐1 = ๐ท2 sin ๐2 ๐1 ๐ธ1 sin ๐1 = ๐2 ๐ธ2 sin ๐2 Thus, ๐1 tan ฮธ1 = ๐2 tan ๐2. 8 Self Assessment Quiz 1. Find the first few multipole moments of a linear quadrupole aligned along the x direction and also when it is aligned along y direction. 2. Three linear quadrupoles are arranged along the x, y and z axes with the central charge of each being at the origin and the other two charges at a distance a each from the origin. Show that the quadrupole moment of this superposition of the quadrupoles is zero. What is the lowest non-vanishing multipole of this configuration? 3. A rectangular parallelepiped having dimension ๐ × ๐ × ๐ is filled with a dielectric in which the polarization is given by ๐๐, where ๐ is the position vector of a point in the dielectric with respect to the centre of the parallelepiped. Find the bound charge densities in the dielectric and show that the total bound charge is zero. 4. A line charge with a linear charge density ๐ is in the vacuum, at a distance d from the surface of a semi-infinite dielectric of permittivity ๐. Obtain the image charge that must be put at a distance d inside the dielectric to simulate the charge density at the boundary when the entire space has permittivity ๐0 . 5. At the interface between two dielectrics with relative permittivity 2 and 3, there is a free charge density 3.24 × 10โ5 C/m2. The electric field strength in the first medium has a magnitude 3 × 3 4 106 V/m and its direction makes an angle of tanโ1 ( ) with the interface. Find the electric field in the second medium. Solutions to Self Assessment Quiz 1. The method is the same as that adopted for Problem 1 of the tutorial assignment. When the ๐ ๐ 2 2 ๐ 3๐ ). 2 2 charges are along the x-axis, the coordinates of the charges +q are at (๐, , ) , (๐, , multipole moments are given by, for ๐ > 0 โ (๐, ๐๐๐ = โซ ๐(๐)๐ ๐ ๐๐๐ ๐)๐ 3 ๐ ๐ ๐ ๐ 3๐ โ โ = ๐๐๐ [๐๐๐ ( , ) + ๐๐๐ ( , )] 2 2 2 2 ๐๐๐ 3๐๐๐ 9 (๐ โ ๐)! ๐ = ๐โ ๐ [๐๐๐ (0)๐ 2 + ๐๐๐ (0)๐ 2 ] 4๐(๐ + ๐)! 9 The 9 (๐ โ ๐)! ๐ = ๐โ ๐ ๐๐๐ (0)[(๐)๐ + (โ๐)๐ ] 4๐(๐ + ๐)! R.h.s. is zero for odd values of m. For the quadrupole located along the y axis, ๐ ๐ โ โ ๐๐๐ = ๐๐๐ [๐๐๐ ( , 0) + ๐๐๐ ( , ๐)] 2 2 9 (๐ โ ๐)! ๐ = ๐โ ๐ [๐๐๐ (0) + (โ1)๐ ๐๐๐ (0)] 4๐(๐ + ๐)! 9 (๐ โ ๐)! ๐ = ๐โ ๐ ๐๐๐ (0)[1 + (โ1)๐ ] 4๐(๐ + ๐)! Once again, the multipole moment is zero for odd values of m. 2. To find the resulting multipole moments, we add the contribution to the multipole moment from the three cases (Problem 1 of Tutorial and Problem 1 of the Quiz) ,For ๐ > 0 9 (๐ โ ๐)! ๐ ๐๐๐ = ๐ โ ๐ [2๐๐ (1)๐ฟ๐.0 + ๐๐๐ (0)[๐ ๐ + (โ๐)๐ + 1 + (โ1)๐ ]] 4๐(๐ + ๐)! We need to only compute even m. Take ๐ = 1, i.e. the dipole moment. For ๐ = 1, ๐ = 0. Recall that the first term is zero for odd value of ๐. Further since ๐10 (0) = 0, the dipole moment also vanishes. Let us compute quadrupole moments for which ๐ = 2. Since ๐20 (๐) = 3๐2 โ1 , 2 We have ๐20 (1) = 1, ๐20 (0) = โ1/2, ๐20 = 0. For calculating ๐22 note that the first term is zero because of Kronecker delta function while the second term valishes. Thus all components of the quadrupole moment vanishes. The next moment is for ๐ = 3. The first term only contributes for even ๐, and, for the second term we have ๐3๐ (1) = 0. Thus the octupole moments also vanish. The first non-zero moment is ๐ = 4, which is calledโ hexadecupoleโ moment. 3. Let us position the dielectric so that the end faces are parallel to the three Cartesian axes. 10 The volume density of bound charges is โ๐ป โ ๐โ = โ๐(๐ป โ ๐) = โ3๐. Since the density is constant, the total volume charge is obtained by multiplying this with the volume, i.e., ๐๐ = โ3๐ ๐๐๐. Now we need to calculate the bound charges on the surface. Note that ๐ โ ๐ฬ on each face is given by the distance of that face from the centre. For instance, the top face (or the bottom face) which is at a distance b from ๐๐ the origin has ๐๐ = ๐โ โ ๐ฬ = . The total bound charge on this face is obtained by multiplying the 2 constant density with the area of the face ac giving a contribution of ๐ ๐๐๐ . 2 Each of the six faces contribute the same, giving a total of 3๐๐๐๐ which is exactly the equal and opposite of the total volume charge. 4. Let the charge density at the interface be . If the entire region had permittivity ๐0 , the electric field at the interface can be calculated by taking a Gaussian cylinder perpendicular to the sheet and its ends equidistant from the sheet. The normal component of the electric field (directed ๐ outward from the respective face)due to the induced charge is given by 2๐ . Let the normal 0 component of the electric field at the point P on the sheet be due to the line charge be ๐ธ๐ . At a point P, ๐ ๐ ๐ ๐ธ๐ = cos ๐ = 2 2๐๐0 ๐ 2๐๐0 ๐ฅ + ๐2 Since the normal components are oppositely directed, the resultant normal component is P ๐ 2๐0 x โ ๐ธ๐ . The normal component of the net electric field on the other side of the interface is given by ๐ 2๐0 d ๏ฌ๏ + ๐ธ๐ . Since the normal component of D field is continuous, we have (recall that the outward normal direction is uniquely defined) ๐ ๐ โ ๐ธ๐ = โ๐ ( + ๐ธ๐ ) 2๐0 2๐0 Solving, we get, ๐โ1 ๐โ1 ๐ ๐ ๐=โ 2๐0 ๐ธ๐ = โ 2๐0 2 ๐+1 ๐+1 2๐๐0 ๐ฅ + ๐2 This is the induced charge. The field due to this at P is ๐ ๐โ1 ๐ ๐ =โ 2 2๐0 ๐ + 1 2๐๐0 ๐ฅ + ๐2 Comparing this with the expression for the normal component of the field, we see that this field is equivalently produced by locating an image line with line charge density ๐โ1 ๐โฒ = โ๐ ๐+1 11 at a distance d from the interface, inside the dielectric. 5. At the interface between two dielectrics, the tangential component of ๐ธโ is continuous ๐ธ1 cos ๐1 = ๐ธ2 cos ๐2 . Since the interface has free charges, we have ๐ท2 sin ๐ โ ๐ท1 sin ๐ = ๐๐ . The last relation gives ๐2 ๐ธ2 sin ๐2 โ ๐1 ๐ธ1 sin ๐1 = ๐๐ . 4 Thus ๐ธ2 cos ๐2 = ๐ธ1 cos ๐1 = 3 × 106 × (5) = 2.4 × 106 . Taking ๐0 = 9 × 10โ12 F/m , we get from the second equation, ๐ธ2 sin ๐2 = ๐๐ ๐1 ๐ธ1 sin ๐1 + ๐2 ๐2 Substituting values, 2 3 3.24 × 10โ5 ๐ธ2 sin ๐2 = × 3 × 106 × + = 1.2 × 106 + 1.2 × 106 = 2.4 × 106 3 5 3 × 9 × 10โ12 Thus tan ๐2 = 1, i.e. the electric field makes an angle of 450 with the interface. The field 4 strength in the second medium is ๐ธ2 = ๐ธ1 cos ๐1 / cos ๐2 = 3 × 106 โ2 = 3.39 × 106 V/m. 5 12