Theoretische Physik 2: Elektrodynamik (Prof. A-S. Smith) Home assignment 11

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1 WiSe Prof. Dr. A-S. Smith Dipl.-Phys. Matthias Saba am Lehrstuhl für Theoretische Physik I Department für Physik Friedrich-Alexander-Universität Erlangen-Nürnberg Problem. Theoretische Physik 2: Elektrodynamik Prof. A-S. Smith Method of image charges Home assignment A point charge Q in vacuum is located a distance d away from a plane interface separating a semi-infinite dielectric medium from the vacuum. For definiteness choose coordinates such that z > corresponds to the vacuum and z < to the dielectric medium. a Give reasons for the existence of a scalar electrostatic potential ϕ with E ϕ. Derive the corresponding Laplace equations on both sides of the interface and formulate the appropriate matching conditions. b Introduce appropriate image charges to solve the electrostatic problem. Use symmetry arguments to choose a promising Ansatz. Determine the electrostatic potential ϕ, the corresponding electric field E and the displacement field D. c Show that the case of a metallic half space is included in the problem in a particular limit. Discuss the leading behavior of the fields far away from the point charge and distinguish the case of a dielectric and a metal. d Determine the induced surface charge σ ind and calculate the interaction energy of the external point charge Q and the surface charge distribution. What force does the dielectric medium exert on the external charge? Problem.2 Debye-Hückel theory In an electrolyte solution ions of opposite charges can freely float agitated by thermal fluctuations. To simplify consider only one species of cations of charge q + > and anions of charge q < of respective number density n + and n. Charge neutrality requires q + n + + q n. The constitutive equation of the Debye-Hückel electrolyte relates the charge densities of the cations ρ + x and anions ρ x to the electrostatic potential ϕ x via ρ + x q + n + exp q +ϕ x, ρ x q n exp q ϕ x. Here T denotes the temperature and k B is Boltzmann s constant. a Considering external charges ρ ext x in addition to the induced ones, ρ ind x ρ + x + ρ x, formulate the Poisson equation. Linearize the exponentials in ϕ x and show that 2 ϕ x λ 2 ϕ x 4πρext x, holds. Relate the Debye-Hückel screening length λ to the number densities n ±.

2 b Determine the electrostatic potential within the linearized theory for a point-like test charge ρ ext x Qδ x. Use the spherical symmetry of the problem and determine the solution of the differential equation for r x > that vanishes for r. Discuss the physical consequences of the result. Hint: The substitution ϕr ur/r simplifies the homogeneous differential equation. The constant of integration may be determined by matching to the Coulomb solution in vacuum close to the test charge. c Equivalently you may evaluate the Green function G x, y defined via λ 2 2 G x, y δ x y. Show that G k d 3 x e i k x y G x, y. satisfies an algebraic equation and perform the inverse Fourier transform; apply the residue theorem to perform the integration. d Consider a small spherical colloid of radius R suspended in the electrolyte. The colloid carries a charge Ze homogeneously distributed along the surface. There are no further charges inside of the colloid. Since the electrolyte cannot penetrate the colloidal particles, the usual Poisson equation holds in the inner region. Determine the electrostatic potential and compare your result to the point-like test charge. Due date:

3 WiSe Prof. Dr. A-S. Smith Dipl.-Phys. Matthias Saba am Lehrstuhl für Theoretische Physik I Department für Physik Friedrich-Alexander-Universität Erlangen-Nürnberg Theoretische Physik 2: Elektrodynamik Prof. A-S. Smith Solutions to Home assignment Solution of Problem. Method of image charges a The external charge distribution contains only the point charge at d,, d in front of the dielectric half space, ρ ext Qδ r d. Then Gauß s law reads in both half spaces for z > : div E 4πQδ r d for z < : div ε E. The electrostatic version of Faraday s law implies the existence of a scalar potential Boundary conditions: E E ϕ. i the tangential components of the electric field are continuous the electrostatic potential is continuous across the surface. ii the normal component of the displacement field is continuous ε z ϕ z z ϕ z+ b In vacuum the image charge is located at,, d, it is the mirror image of the real charge with respect to the interface as for the case of a metal compare the lecture. For the dielectric the image charge has to be located at,, d. The reasonable Ansatz is Q z > : ϕ + r r d + Q + r + d z < : ϕ r Q ε r d. The factor /ε for the potential is of course conventional; the potential is chosen such that in the dielectric it generates a displacement field D ε ϕ corresponding to an external charge Q. For ϕ to be continuous at z, one has to impose ϕ z+ + Q + Q + x 2 + y 2 + d Q 2 ε x 2 + y 2 + d ϕ 2 Q + Q + Q ε z

4 Thus at this point it is obvious that placing the image charges at a different location would not lead to a solution. For the displacement to be continuous across the surface, we calculate Qd Q + d D + z+ z ϕ + z+ x 2 + y 2 + d 2 3/2 D z ε z ϕ z thus the electrostatic problem is solved indeed provided with solution Q Q + Q, Q d x 2 + y 2 + d 2 3/2 ; Q + ε ε + Q, Q 2ε ε + Q. Introducing β ε /ε +, the potential takes a compact form, z > : ϕ + r Q r d βq r + d βq z < : ϕ r r d. The electric fields are trivially obtained from E ϕ and D ε E, z > : z < : D+ r E + r Q r d r d 3 βq r + d r + d 3 ε D r E r βq r d r d 3. c In a metal, there cannot be a static electric field. Thus we try β, i.e. ε. Obviously, one recovers the solution of the lecture. The image charge attains the value Q + Q whereas with Q /ε the potential inside the metal vanishes. The far field behavior is found by expanding the potentials for r d, it is convenient to perform the Taylor expansion for small d, r d r + d r r 3 + O d/r 2. Then, for a dielectric body the far field behavior is given by z > : ϕ + r Q β r + Or 2, z < : ϕ r Q β r + Or 2. i.e., far away the electric field looks like that of a point charge Q β. The factor β quantifies the dielectric shielding. In the case of a metal, the shielding is perfect. Inside the metal, there is no field at all. In the half-space z >, the leading contribution is z > : ϕ + r Q r d r + d r 3 r 2Q d + Or 3, i.e. a dipolar pattern with dipole strength p 2Q d. 2

5 d The induced surface charges are connected with a jump in the polarization. Consider a sheet of width h containing the surface, then σ ind lim h hρ ind lim h h div P. There are no external charges at the surface, then by Coulomb s law div D and one concludes Thus with r x, y, σ ind 4π n βqd 2π 4πhρ ind h div E n E + E Q r d r d 3 βq r + d r + d 3 x 2 + y 2 + d 2 3/2, 4π n βq r d r d 3 where I have observed n r, n d d and r + d r d. The interaction energy between the induced surface charges and the point charge is σ ind r Q U r d d 2 r βq2 d dxdy 2π x 2 + y 2 + d 2 2 d 2πRdR βq2 2π R 2 + d 2 2, substitute s 2 R 2 + d 2, sds RdR, U βq 2 sds d d s 4 Q2 β 2d. This result is interpreted as interaction of the original charge Q with the image charge βq separated by a distance 2d. Correspondingly there is a force F U d β Q2 2d 2 and the dielectric attracts the point charge. It is amusing to note that the interaction energy may be interpreted in terms of image charges. For the force one would come to the wrong conclusion F βq 2 /4d 2... Solution of Problem.2 Debye-Hückel theory a In the presence of cations, anions, and external charges, Gauß s law of electricity reads E 4πρ x 4πρ + x + 4πρ x + 4πρ ext x. Faraday s law of induction reduces to E implying the existence of a electrostatic potential, E ϕ. Then the Poisson equation yields 2 ϕ x 4πρ x 4πρ + x 4πρ x 4πρ ext x. Linearizing the constitutive law and allowing for charge neutrality, q + n + + q n, yields 2 ϕ x 4πq + n + q +ϕ x 4πq n q ϕ x 4πρ ext x 4π q 2 + n + + q n 2 ϕ x 4πρ ext x, 3

6 or 2 λ 2 ϕ x 4πρ ext x with the screening length λ determined as λ 2 4π q 2 + n + + q 2 n. b The field equation for ϕ x in the presence of an external point charge reads 2 λ 2 ϕ x 4πQδ x; taking advantage of the spherical symmetry, it simplifies for r x > to Substituting one obtains ϕr ur r d 2 dr 2 ϕr + 2 d r dr ϕr λ 2 ϕr. ϕ u r u r 2, ϕ u r 2u r 2 + 2u r 3, u λ 2 u ur Ce r/λ + Ce +r/λ. The solution shall remain finite vanish at infinity, C, and ϕr C r exp r/λ. We determine the constant by integrating the Poisson equation over a small sphere with radius R λ, 2 ϕ x d 3 x 4π Qδ x d 3 x + λ 2 ϕ x d 3 x Gauß s theorem With one finds x <R x R x <R ϕ x ndf 4πQ + 4π λ 2 ϕ x n ϕr r Thus, the potential is determined as R r 2 dr ϕr. x <R C r 2 e r/λ + r C λ r 2 + Or, 4πC + OR 2 4πQ + OR 2 C Q. ϕr Q r exp r. λ Note that the potential decays much faster to zero than the Coulomb solution. In fact the electric field decays exponentially rather than algebraically. Reversely, by Gauß s law the total charge of a sphere of radius R λ is electrically neutral. Thus, the electrolyte screens the external charge. The screening is almost perfect on length scales r λ. For a positive test charge, anions accumulate close to the charge, whereas the cations are repelled. The situation is of course reversed for a negative test charge. 4

7 c Obviously the Green function is translationally invariant and from the result for ϕ x one concludes exp x y /λ G x, y G x y. 4π x y To perform the Fourier transform, use a coordinate system such that k ê z, then k x k r cos ϑ. G k d 3 x e i k x G x G k 2π 2π 2ik 2ik π r 2 dr k 2 + λ 2 Hence the algebraic equation is r 2 dr e r/λ 4πr which is of course the Fourier transform of 2π sin ϑdθ ikr cos ϑ e r/λ dϕ e 4πr du e ikru r 2 dr e r/λ e ikr e ikr 4πr ikr dr [e ] rλ ik e rλ +ik λ ik λ + ik k 2 + λ 2 G k, 2 λ 2 G k δ x. λ + ik λ ik 2ik λ ikλ + ik For the inverse transformation, choose k-coordinates such that x ê k z, k x kr cos ϑ k G x d k 2π 3 k 2 + λ 2 2π 3 2π 2 2π 2 2π 2 e i k x k 2 dk π k 2 dk k 2 + λ 2 dϑ k 2π du e ikru dϕ k e ikr cos ϑ k k 2 + λ 2 k 2 dk e ikr k 2 + λ 2 e ikr ikr k 2 dk k 2 + λ 2 ikr eikr 2π 2 kdk k 2 + λ 2 ir eikr Now we apply the residue theorem. There are two poles k ±iλ. Since r >, we close the contour by a circle in the upper complex plane. Then the exponential rapidly makes the 5

8 contribution of the arc negligibly small. Thus G x 2πi Res kiλ k 2π 2 k 2 + λ 2 ir eikr kk iλ 2πi lim k iλ 2π 2 k 2 + λ 2 k 2πi lim k iλ 2π 2 k + iλ ir eikr G x e r/λ 4πr, r x. ir eikr d For the colloid the constitutive equation holds only for r R; inside we have the usual Poisson equation, 2 ϕ x 4πρ ext x + ϕ x Θ x R λ2 Thus ϕ x const inside the colloid maximum principle for harmonic functions, see Problem.6. Outside the sphere, the above result for ϕ x holds. The constant C is determined by integrating over a sphere of radius R; only the first term on the r.h.s. contributes to the volume integral. Rewriting the l.h.s. using Gauß s theorem, ϕ x ndf 4πZe x R 4πR 2 C R 2 e R/λ + R λ 4πZe, thus ϕr Z e r e r/λ with the effective charge Z ZeR/λ + R/λ. 6