Electrostatics: Energy and Forces on Multipoles and Conducting Spheres in External Fields , Assignments of Physics

Download Electrostatics: Energy and Forces on Multipoles and Conducting Spheres in External Fields and more Assignments Physics in PDF only on Docsity! PHYS606, Spring ‘09, Homework 4 Solutions Instructor: Prof. Bedaque Notation/Convention: • Gaussian units with c 6= 1 are used throughout. (Let the grader know if this is violated, so that it can be corrected.) • The metric used is as follows: gµν =     1 0 0 0 0 −1 0 0 0 0 −1 0 0 0 0 −1     . • The coordinate x0 = ct. • For derivatives ∂0 = ∂ ∂x0 = 1 c ∂ ∂t and ∂j = ∂ ∂xj . Care needs to be taken since ∂j = −∂j with the metric we are using. • Summation over repeated Lorentz indices is done when one is up and the other is down. Summation over Cartesian coordinates is done when two are repeated no matter their location. 1 PHYS606, Spring ‘09, Homework 4 Solutions Instructor: Prof. Bedaque Problem 1. Energy, forces, and torques on multipoles Consider a static charge distribution characterized by a charge q, and electric dipole p and an electric quadrupole Q in and external electric field. Assuming that the external field varies slowly within the size of the charge distribution: a. Show that the energy of the charge distribution due to the interaction with the external field is given by U = qφ(0) + #p ·E + #Qij∂iEj + · · · , (1) where # are numerical coefficients. Determine the #’s. b. Using the result above, calculate the energy of one electric dipole in the electric field of another dipole (assuming their separation is much larger than their sizes). c. What is the analogue expression for the force acting on the charge distribution? Consider now a localized steady current distribution immersed in an external magnetic field varying litle within the size fo the current distribution. d. What is the expression for the energy of the magnetic dipole? e. What is the force exerted by the external field on the dipole? f. What is the torque exterted on the dipole? Solution Part a. Recall that the energy density for the electromagnetic field is u = 1 8π (Eext + E) 2, where E is the field due to the static charge distribution and Eext is the external field. Upon expanding this, the only energy term dependent on the interaction between the fields is just uint = 1 4π Eext · E. If we integrate the interaction energy density, we get the energy (using Eext = φext and ∇ ·E = 4πρ and integration by parts) U = 1 4π ∫ d3xEext · E = 1 4π ∫ d3xφext∇ · E = ∫ d3xφext(x)ρ(x). 2 PHYS606, Spring ‘09, Homework 4 Solutions Instructor: Prof. Bedaque Problem 2. Jackson, 4.10 Two concentric conducting spheres of inner and outer radii a and b, respectively, carry the charges ±Q. The empty space between the spheres is half-filled by a hemispherical shell of dielectric (of dielectric constant ǫ), as shown in the figure. Figure 1: Problem 4.10 (a) Find the electric field everywhere between the spheres. (b) Calculate the surface-charge distribution on the inner sphere. (c) Calculate the polarization-charge density induced on the surface of the dielectric at r = a. Solution Part (a) For this part we make the ansatz that E = E(r)r̂ since the potential at r = a and r = b is constant. Of course, we will check to see if this is a valid solution. Now, we can use the Gauss’s theorem for the displacement field D which tells us ∮ S D · n̂ dA = 4πQfree (2) where Qfree is the free charge inclosed by S. We now take S to be a sphere around zero at a radius r in between a and b. We then obtain ∮ S D · n̂ dA = 4πQ ∫ left E · r̂ dA + ∫ right ǫE · r̂ dA = 4πQ 2πr2(1 + ǫ)E(r) = 4πQ. By our ansatz. This then gives E(r) = 2Q (1 + ǫ)r2 . (3) 5 PHYS606, Spring ‘09, Homework 4 Solutions Instructor: Prof. Bedaque Does this actually work though? Clearly in the left and right regions separately ∇ · D = 0 since the same is true for E and ǫ is constant in both regions. Clearly this is derivable from a potential which makes the potential constant on both the inner sphere and the outer sphere, so that boundary condition is satisfied. Lastly, we need to check the boundary between the vacuum and the dielectric material. Recall that in the abscence of free charge that the normal component of D must be contin- uous. Well, the normal component of D is zero (between the two mediums), so it must be continuous in this case. Lastly, we need the tangential E field to be continuous, and this can be clearly seen by the fact that for a given r, E is constant. Thus, we have satisified all boundary conditions and have a unique solution to the problem! Part (b) With the E field we can easily get the surface charge distribution by noting E · n̂|r=a = 4πσ, where σ is on the inner sphere. Thus, we have that σ = Q 2πa2(1 + ǫ) . This is the surface charge distribution on the inner sphere. Part (c) Now, we can use the D field to obtain the free charge density. In the region without a dielectric this is the same condition as in (b), so we can focus on the region with dielectric. Here we have that D · n̂|r=a = 4πσfree. So in the region of dielectric D = ǫE so we obtain σfree = Q 2πa2(1 + ǫ) . Now we know that σfree + σinduced = σ, so we can solve for σinduced to get σinduced = − (ǫ − 1)Q 2πa2(ǫ + 1) . As expected, it induces a negative charge on the surface. 6