1. Electrostatic field in dielectric media When a material has no free charge carriers or very few charge carriers, it is known as dielectric. For example glass, quartz, when a dielectric material is placed in an electric field, etc. The free electron rearrange themselves, sitting in to a new equilibrium configuration. The electronic charge on each atom is displaced with respect to the positive charge as shown in diagram. The net result that each atom molecule behaves like an electrical dipole. Under such condition, the materials is said to be polarized. we can summarize the effect of the electric field on a dielectric material by saying that dipoles are product in the material. 1

2. Of course, since what really is involved here is a charge separation, not only dipoles but also quadruple and higher multiple will be produced. If we place an atom in an electric field, the pulls of this electric field in the electrons and on the nucleus are in opposite directions, this causes a small separation between centroid of the negative and positive charge distributions of atom and gives the atom an induced dipole moment. Wherever we have an electrical dipole of dipole moment, then the potential due to this at any point is given by Let us suppose that the dipole moment of one molecule the dipole moment of all molecules in the medium therefore average dipole moment per unit volume 2

3. 1- To find out the potential and electric field outside a dielectric medium (polarized) Let there be a dielectric medium of volume, let there be a point p at a distance r from an elementary volume as shown in the diagram. Let the dipole moment per unit volume, when the dipole moment of the volume is potential du at point p is given by We know that From eq. (2), eq. (1) becomes We know that. 3

4. Applying divergence theorem we get, Eq. (3) has two term, both are a scalar quantities, the first term has units of charge per unit area, the second term has unit of charge per unit volume eq. (3) can be written as, Where is charge per unit area and is the charge per unit volume. From eq. (4) we can write 4

5. From eq. (5) we conclude that a polarized material behaves as if it has a charge and this charge can be called charge due to polarization. This charge due to polarization is Q: show that a polarized material behaves like a free charge, find out the expression for the polarized charge. The electric field E is given by 5

6. 3-2 Electric field inside a dielectric: Consider a dielectric medium PQRS as shown in diagram. Imagine a fine cavity AB in this medium. Let ABCD be a closed loop since Following the loop ABCD we can write Where and are electric field in the dielectric and vacuum respectively. If the side AD and BC of the loop are infinitely small then become nearly zero Eq. (1) becomes or Now if, then Where stand for tangential component. Eq. (2) tell us that the tangential component of electric field inside a dielectric is equal to the tangential component of the electric field outside the dielectric. 6

7. 3-3Gauss law in a dielectric (electric displacement) Let us suppose a dielectric medium as PQRS. Let there be free charge on the surface. Let the polarization charge be, image a surface S. Applying Gauss law to this surface we can write Where is the electric field inside the dielectric. Now 7

8. Putting value of in eq. (1) we get Let Therefore eq. (3) becomes; This equation is the Gauss law in a dielectric. The vector quantity is known electric displacement. From eq. (5), in the vacuum, This is the usual Gauss law. From eq. (5) we write, applying divergence theorem, we get Let be the charge density in the volume V, then Combining eq. (6) & (7) we get or Eq. (8) is the differential equation of the Gauss law in a dielectric medium. 8

9. 3-4 A point charge in a dielectric: A point charge q is placed at O in a medium of dielectric constant k, and permittivity. A point p at a distance r from q is given as shown in the diagram. Let us imagine a spherical surface of radius (r) passing through the point p. If is the electric displacement at p, then applying Gauss theorem we can write, From eq. (2) Or 9

10. But Substituting this value in above equation we get, From eq. (4) we find that the presence of dielectric change’s the effective value of the point charge (q) to the value. This mean that the dielectric medium reduces the effective value of the free charge. Q: show that the dielectric medium reduces the effective value of the free charge. From eq. (4) we can write And from eq. (2) we have, or 10

11. Q: prove that the effective charge in a dielectric is equal to. Let us find out the value of in the dielectric. We know that Let us apply Gauss law to a spherical surface of radius r which is located concentrically about q. for convince q will be located at the origin, then Because Eq. (4) show that the total charge this means that in the dielectric, the point charge q becomes an effective charge. 3-5 Properties of dielectric: 1- Susceptibility: is the property of the dielectric medium and is equal to the ratio of the dipole moment/unit volume and the electric field Approximation, the right side can be regarded as the first term in a Tailor-series expansion of P in term of E for strong electric filed. 11

12. 3-7 Poisson’s and Laplace’s equation in a dielectric: From Gauss theorem for a dielectric, we know that Where D is the electric displacement and is the charge density per unit volume. If is the permittivity of the medium and is the electric field, then Now we know that. Put this value in eq. (2) and (1) we gets Or Eq. (3) can also be written as follows 12

13. When a dielectric is placed in an electric field, the charge in the dielectric rest only on the surface of the dielectric. Therefore, for any point inside the dielectric the value of is equal to zero. For each points eq. (3) becomes Eq. (5) is Laplace’s equation and can be applied to problem dealing with dielectric. 13

14. 3-8 Dielectric sphere in a uniform field Let us suppose that a dielectric sphere of radius ‘a’ is placed in a uniform electric field, as shown in the diagram. Let its dielectric constant be k. Laplace’s equation gives Where is the potential at any point. The general solution of the equation is In this problem, there is no point or spherical charge on the sphere, therefore in eq. (2) For this problem will also be zero. ( negligent higher order) eq. (2) becomes For simplicity the constant can be made=0 14

15. Eq. (4) can be divided in to two equations, one for outside the sphere and other for inside the sphere, these equation are outside (5) inside The potential inside sphere eq. (5) should give where r=0.When we put r=0 in eq. (5) for, the second term makes which is wrong, therefore should be zero. Hence (1) At r=a Also at r=a or Therefore applying first boundary condition we get, 15

16. If the z axis is taken along the field. For outside points, we know at Therefore eq. (8) gives us that in eq. (5) should be therefore And Applying second boundary condition we get, Now 16

17. Multiplying eq. (7) by 2 we get Placing value of in eq. (7) we get 17

18. The electric intensity inside can be found from eq. (14) by finding and And Q: Show that electric field inside the dielectric sphere has the direction of and is given by. 18

19. 3-9 Microscopic theory of the dielectric Both the electric susceptibility χ and the dielectric constant k are determined by the atomic structure of the dielectric. The atom consist of an electron bound to a spherically symmetric core by an elastic force. The induced dipole moment is then due to the displacement of the electron from its equilibrium position (the effect of electric field on an atom). If an electric field E is applied to the electron, the amount by electric force displace the electron from equilibrium is The corresponding dipole moment is If there are n atoms per unit volume, then the dipole moment per unit volume is The susceptibility and dielectric constant are therefore and (, ) 19

20. Have the same unit. Expressed in term of the mass of the electron and the vibration frequency, so that For helium, the typical resonant frequency is with density EXPERMINTAL VALUE = 20

21. 3-10 Polar molecules (the langevin - Debye formula) If a gas consist of molecules with a permanent dipole moment, then the polarization depend on the alignment of these molecules. If the dipole moment of one molecule is then the potential energy of this molecule in an electric field is where is the angle between and E. In the absence of thermal motion, all the molecules would settle in the configuration of least energy, with p exactly aligned along E. the thermal motion destroy this perfect alignment; however, there remains some partial alignment, and on the average, p has a positive component along E. According to statistical mechanics, the probability that a molecule is oriented at an angle between and with respect to the electric field is proportional to, where T is the absolute temperature and k is Boltzmann's constant,. 21

22. The potential energy of a permanent dipole in an electric field is Probability for orientation between and The effective dipole moment of a molecular dipole is the component along the field direction, i.e. Using the above principle, the average value of this quantity is found to be, then eq.(4) becomes Solving eq.(5) we get Eq.(6) is called Langevin equation 22

23. The polarization has been define as If there N molecules per unit volume the polarization and has the direction of E `The displacement of molecule per unit polarization field is called polarizabilty, This result have been derived by negligent include dipole moment, let the deformation polarizabilty, The total polarizabilty which is known langevin- Debye equation Consequently, the susceptibility and the dielectric constant are n: number of molecule per unit volume 23