Homework 7: Linear Dielectrics 1. [10 points] In class, we showed that the electric potential produced by a polarized dielectric for points outside of the dielectric was given by Prove that the electric potential produced by a polarized dielectric for points inside of the dielectric is also given by where the integral runs over the entire volume of the dielectric. [Hint: To show this, you need to use the result from Homework #6, Problem 2] Answer: Suppose we want to determine the macroscopic electric field at some point within a dielectric. By definition, the macroscopic field is the average electric field over a region large enough so that the polarization within the region remains constant. To define this macroscsopic field, suppose we place a Gaussian sphere within the dielectric that’s just large enough so that the polarization within this sphere is constant. Thus, the macroscopic field at some point within the dielectric consists of two parts: the average field over the sphere due to all charges outside the region and the average field over the sphere due to all the charges inside: ⃗ ⃗ ⃗ From Homework #6, Problem 2, the average field over a sphere, produced by charges outside the sphere, is equal to the electric field they produce at the center of the sphere. Therefore, ⃗ is the electric field due to dipoles outside of the sphere. These charges are far enough away that we can use the multipole expansion to determine the electric potential where the integration is performed over all space outside the sphere. For the dipoles within sphere, Homework #6, Problem 2 states that the average field over a sphere due to dipoles within the sphere is given by ⃗ ⃗ Since our Gaussian sphere is small enough that the polarization is essentially constant, this implies that average field over any sphere due to charges inside the sphere is the same as the electric field at the center of uniformly polarized sphere. Thus, the total macroscopic field is given by the potential where the integral is evaluated over the entire volume of the dielectric. 2. [10 points] A conducting spherical shell has inner radius a and outer radius c. The space between these two surfaces is filled with a dielectric for which the dielectric constant is between a and b, and between b and c, as shown below. Determine the capacitance of the system and the energy stored in the system. Answer: To find the electric field in the region apply Gauss’s law , we draw a Gaussian sphere of this radius and ∮⃗ ⃗ ̂ Based on linear dielectric theory, we have ⃗ In the region ⃗ ⃗ ⃗ ̂ , we draw a Gaussian sphere of this radius and apply Gauss’s law ∮⃗ ⃗ ̂ Based on linear dielectric theory, we have ⃗ ⃗ ⃗ ⃗ ̂ To find the potential difference, we have ∫ ⃗ ∫ ∫ [ ( ) ( )] Therefore, the capacitance is given by [ ( ) ( )] The energy of this configuration is given by ( ( ) [ ( ) ) ( )] [ ( ) ( )] 3. A point charge q is imbedded at the center of a sphere of linear dielectric material with susceptibility χe and radius R. a. [4 points] Find the electric field and the polarization. b. [4 points] Find the bound charge densities and . What is the total bound charge on the surface? c. [2 points] Where is the compensating negative bound charge located? Answer: (a) To find the electric field, we can apply Gauss’s law in the presence of a dielectric ∮⃗ ⃗ ̂ Based on linear dielectric theory, we have ⃗ ⃗ ⃗ ( ) ( ̂ ) The polarization is given by ⃗ ⃗ ( ) ̂ (b) The volume bound charge is given by ⃗ ( ) ̂ ( ) Recall the definition of the Dirac-Delta function ̂ ( ) ( ) Therefore, the volume bound charge is given by ⃗ ( ̂ ( ) ) ( The surface bound charge is given by ⃗ ̂ ( ) The total bound charge on the surface is given by ( ) ) ( ) Since the volume bound charge is zero everywhere except at the center, this implies that the compensating negative charge must be at the center. To show this, note that ∫ ∫ ( ) 4. [10 points] An uncharged conducting sphere of radius a is coated with a thick insulating shell of dielectric constant out to radius b. This object is now placed in a uniform electric field ⃗ ̂ . Find the electric field in the insulator and outside the dielectric sphere. [Hint: Note that for points far from the sphere ] Answer: To determine the electric field, we will use separation of variables in spherical coordinates. The azimuthally-symmetric general solution to Laplace’s equation is given by ( ) ) ( ∑( ) We note that since there is a zero net charge inside the dielectric sphere, this implies that for . Since for points far outside the sphere, this implies that ( ) ( ∑ ) Therefore the general solution with these boundary conditions is ( ) ) ( ∑( ∑ { ) ( ) To determine the coefficients, we note that the potential must be continuous at implies that ( ) We note that the potential must be continuous at . This implies that ( ) ( ( ) ) ( ) . This Finally, we note that the normal derivative of the potential undergoes a discontinuity at ∑( For ( ) ( ) ) ) ∑( ( ) . ( ) , we have [using (1)] ( ( [( ) ) ) ( ) ] ( ) Comparing (4) with (3) gives [( ) ] ( ) Using these results, we can write (2) as ( We can now simplify (4) for ( ) ( ) as ( ) ) Using (5) gives us [ ( ) [ ] ( ) ] ( [ ) ( ) ] Therefore, the electric potential inside the dielectric is given by ( ) ) ( ∑( ) ( ( [ ( ) ] [ ( ) ] ) ) The electric field inside the insulator is given by ⃗ ( ) [ ( ) ] [ {( ) ( ) ̂ ( ) ( ) ] The electric potential outside the sphere is given by ( ) ( ∑ ( ) ( ) ) ( [ Since ( ) ] ) [ ( ) ] outside the dielectric, we have ( ) ( ( ) ) ( ) The electric field outside the sphere is given by ⃗ ( ) [( ) ̂ ( ) ̂] ̂} 5. The space between the plates of a parallel-plate capacitor is filled with two slabs of linear dielectric material, as shown below. Each slab has a thickness a so the total distance between the plates is 2a. Slab 1 has a dielectric constant of 2, and slab 2 has a dielectric constant of 1.5. The free charge density on the top plate is and the free charge density on the bottom plate is – . a. [5 points] Find ⃗ , ⃗ , and ⃗ in each slab. b. [3 points] Find the location and amount of all bound charge. c. [2 points] Find the potential difference between the plates Answer: (a) The electric displacement can be obtained from Gauss’s law ∮⃗ By symmetry, the electric displacement only points in the z-direction and thus, there is no contribution from the curved wall of our Gaussian surface. Therefore, we have ∮⃗ The electric displacement from the top plate is ⃗ ( ) ̂ above the top plate and is ⃗ ( ) ( ̂ ) below the top plate. Similarly, the contribution from the bottom plate is ⃗ ( ) ̂ below the bottom plate and ⃗ ( ) ( ̂ ) above the bottom plate. Since the displacement is a constant, the sum of the contributions cancels all displacement outside the capacitor and gives ⃗ ̂ inside the capacitor (pointing downward). Therefore, the displacement has the same value and direction in either dielectric. Using linear dielectric theory, the electric field in the top slab is given by ⃗ ̂ ⃗ ̂ ̂ where it is assumed that ̂ points downward. The polarization in each slab is given by ⃗ ⃗ ( )⃗ ̂ ⃗ ⃗ ( )⃗ ̂ (b) Since the polarization is uniform in each slab, then the volume bound charge will be zero in each slab. This means that there will be surface bound charge on the top and bottom of both slabs. The surface bound charge for slab 1 is given by { { The surface bound charge for slab 2 is given by { { (c) The potential difference between both plates is due solely to the bound charge and is given by ( ) Bonus [7 points]: Consider a linear dielectric material composed of nonpolar molecules. Here, you are going to derive a relationship between the atomic polarizability and the electric susceptibility . Here are the steps to this derivation: a. First, we need to examine the total macroscopic electric field in the medium. Suppose that the space allotted to each atom is a sphere of radius R. Show that the macroscopic field for the dielectric material is given by ⃗ ⃗ ⃗ )⃗ ( where N is the number of atoms per unit volume and ⃗ sphere due to the atom itself. is the average field over the b. Use the result from (a) to conclude that ( ) This result is known as the Clausius-Mossotti formula. Answer: The macroscopic field is due to everything except the particular atom under consideration and due to atom itself. This gives ⃗ ⃗ ⃗ We know that the electric field outside of the atom will be related to the polarization ⃗ ⃗ ⃗ Where N is the density of atoms. Based on Homework #6, Problem 2, we know that the average field over a sphere due to charges inside the sphere is given by ⃗ ⃗ Therefore, the electric field is given by ⃗ ⃗ ⃗ ⃗ ⃗ )⃗ ( The polarization is given by ⃗ ⃗ ⃗ ( Using linear dielectric theory, we have ) ( )⃗ ⃗ ⃗ ⃗ ( Solving for ) ( ) gives ( )