Lecture 15 Perfect Conductors, Boundary Conditions, Method of Images

Transcription

1 Lecture 15 Perfect Conductors, Boundary Conditions, Method of Images Sections: 5.4, 5.5 Homework: See homework file

2 Perfect Conductors 1 metals such as Cu, Ag, Al are closely approximated by the concept of a perfect electric conductor (PEC) σ, τ charge relaxation is instantaneous charge is distributed in an infinitesimally thin layer on the surface of the conductor inside the conductor there is no charge (i.e., excess of electrons) due to Coulombic repulsion biased (charged) conductor isolated conductor in external field ( Q > ) ρ > ext E ρ s < ρ v = E = ( = ) ρ sq > V ρ s > s ρ v = E = 2

3 Perfect Conductors 2 underlying principle: charge arranges on surface so that resultant E field inside is zero; otherwise charge will keep moving until E = remember: E field is zero in the volume of a perfect conductor consequence: perfect conductors are equipotential and so are their surfaces B proof: V = E d L = V = V AB A B A (A and B belong to the volume or the surface of the conductor) LECTURE 15 slide 3

4 Shielding by Conducting Shell (Faraday Cage) in a cavity of a conductor E = (if there are no free charges inside) isolated conducting shell in external field ( Q = ) ext E ρ v = E = Γ E = V biased (charged) conducting shell ( Q > ) ρ v = E = E = Γ proof: take any closed contour Γ passing part through the cavity and part through the shell, apply the conservative property of E E d L = E d L E d L = E = Γ Γ shell Γcavity Is there any charge on the inner surface of the cavity? Hint: Gauss law LECTURE 15 slide 4

5 Faraday Cage: Illustrations every computer is enclosed in a metallic box, which protects it from EMI workers in the power industry are protected by Faraday cages when dealing with high voltage LECTURE 15 slide 5

6 Boundary Conditions at PEC Surfaces tangential components E= E N a N PEC is equipotential E t = D t = PEC tangential components are zero at PEC surface normal components lim D s N s s h S d = D S = Q = ρ S DN = ρs, C/m 2 E = ρ / ε N interface surface compare with D of planar charge: D= ± a ρ /2 s N = on PEC surfaces D DN N ρs a N LECTURE 15 slide 6 = = s a D =

7 Boundary Conditions at PEC Surface Example Voltage V = 1 V is applied to a coaxial cable whose inner wire has radius a = 1 mm and whose outer shield has radius b = 4 mm. The insulator is air (ε = ε ). (a) Find E as a function of the distance ρ from the center of the inner wire. (b) Find the surface and line charge densities at the inner wire and at the outer shield. LECTURE 15 slide 7

8 Fundamental Equation of Electrostatics D= ρ ( εe) = ρ v v ε V = ρ v in a uniform medium with sources (Poisson equation) V = ρ ε 2 v / in a uniform sourcefree medium (Laplace equation) 2 V = uniqueness theorem for the Poisson/Laplace equation if the following is given either V or V/ n at the boundary of the analyzed region (boundary conditions), and all charge densities in the analyzed region (source conditions) then the Poisson equation has one and only one solution LECTURE 15 slide 8

9 Method of Images 1 consider the two electrostatic problems Q h V b = ground Q h h V b = Q = Q source and BC are the same if analyzed region is upper hemisphere the field solution in the upper hemisphere identical in both cases we say that the two problems are equivalent LECTURE 15 slide 9

10 Method of Images 2 the method of images applies to any shape and orientation of the charge above ground plane [Ulaby&Ravaioli, Fundamentals of Applied Electromagnetics, 7 th ed.] LECTURE 15 slide 1

11 Method of Images Example A point charge Q = 1 12 C is located at S(,,3) cm above a ground plane at z =. Find ρ s at P(,4,) cm. LECTURE 15 slide 11

12 Method of Images 2 point charge at two orthogonal ground planes h 2 h 2 V b = h 1 Q V b = Q h 1 h 2 Q V b = h 1 V b = h 1 Q h 2 Q h 2 h 1 LECTURE 15 slide 12

13 Method of Images 3 point charge at two ground planes at an angle /n, n is integer a total of (2n 1) images required to build the equivalent problem Example: n = 3 V = b π /3 V b = Q V = Q b π /3 Q V b = if n is not integer, exact equivalent problem does not exist LECTURE 15 slide 13

14 You have learned: that the electric field is zero inside a perfect conductor the potential is constant everywhere inside and on a perfect conductor, i.e., the voltage between any two points is zero a closed conducting shell shields the inside volume from external electric fields the tangential E field is zero on the surface of a conductor the normal D component is equal to the surface charge density at the conductor s surface how the method of images can simplify solutions involving infinite PEC planes LECTURE 15 slide 14