1ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityLecture 8Boundary Value ProblemsIn this lecture you will learn:• How to solve some interesting boundary value problemsECE 303 – Fall 2007 – Farhan Rana – Cornell UniversitySolutions of Laplace Equation in Spherical Coordinates()BrAr +=rφSpherically Symmetric SolutionA Constant Uniform Electric Field Solution()()zArAr −=−=θφcosrzθ() ()zArrEˆ=−∇=rrrφA Dipole Oriented Along z-Axis Solution()()2cosrArθφ=rzθ+q-qFor all the following solutions()02=∇ rrφ+q2ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityA Non-Uniformly Charged Spherical Shell - IConsider a charged spherical sheet where the surface charge density is fixed and is given by:θz()θσσcoso=+++++++-------Looking from outside the potential sort of looks like that of a dipole. So try a dipole-like solution for r > aa()()CrArout+=2cosθφrFor r > a:For r < a: Try something that does not go to infinity at r = 0()()DrBrin+=θφcosr00Why??Why??ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityA Non-Uniformly Charged Spherical Shell - IIθz+++++++-------aHow would you know if your solution is the right one??Uniqueness Theorem:There is only one right solution and If you have a solution that satisfies all the boundary conditions then you have the right solution x()( )0.00=∫=∞=−=∞=rsdErrrrrrrφφsdr∞For r < a:()()DrBrin+=θφcosr0Why??3ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityA Non-Uniformly Charged Spherical Shell - IIIθz+++++++-------a()θσσcoso=()()2cosrAroutθφ=r()()θφcosrBrin=rBoundary Conditions:(1) Potential inside and outside must be continuous at the surface of the shell()()()()θθφφcoscos2aBaArraroutarin=⇒===rr()() ()()()()()oooaroutoaroutoarrinarroutoBaArrrrEEεθσθθθσφεφεσεcoscoscos2cos3,,=+⇒=∂∂+∂∂−⇒=−====rr(2) Discontinuity in the radial electric field at the surface must be related to the local surface charge densityECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityA Non-Uniformly Charged Spherical Shell - IVz+++++++-------()θσσcoso=A surface charge density of the form:produces a uniform z-directed E-field inside the charged shell and a dipole-like E-field outside the charged shellSolution:()()23cos3raroooutθεσφ⎟⎟⎠⎞⎜⎜⎝⎛=r()()zrrooooin⎟⎟⎠⎞⎜⎜⎝⎛=⎟⎟⎠⎞⎜⎜⎝⎛=εσθεσφ3cos3r4ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityA Perfect Metallic Sphere in a Uniform E-Field - IConsider a perfect metal sphere placed inside a constant and uniform z-directed E-fieldθzaThe field lines shown on the right cannot be the right picture – there cannot be any E-field inside a perfect metalQuestion: So what happens when a metal sphere is placed inside a constant and uniform electric field ??Answer: Induced surface charges appear on the metal sphere that screen out the outside E-fieldxEoECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityA Perfect Metallic Sphere in a Uniform E-Field - IIQuestion: What kind of surface charge density gets induced on the metal sphere that produces a uniform E-field inside the sphere which completely cancels the applied E-field??Answer: Induced surface charge density on the metal sphere must look like:()θσcos∝θzax+++++++-------Eo5ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityA Perfect Metallic Sphere in a Uniform E-Field - IIISolution:The outside potential must look like a combination of:(i) The potential corresponding to a uniform z-directed E-field, and(ii) The dipole-like potential from the induced surface charge density on the metal sphereSo try the following solution for r > a:()()()θθφcoscos2rErAroout−=rFor r > a:For r < a: Metal sphere is equipotential with zero potential()0=rinrφθzax+++++++-------EoECE 303 – Fall 2007 – Farhan Rana – Cornell University()()()θθφcoscos2rErAroout−=rFor r > a:For r < a: Metal sphere is an equipotentialwith zero potential()0=rinrφA Perfect Metallic Sphere in a Uniform E-Field - IVBoundary Condition:(i) The potential at the surface of the metal must be continuous (i.e. the same inside and outside) and, therefore, must be zero() ()()()0coscos2=−⇒===θθφφaEaArroaroutarinrrSolution:()()()θθφcoscos23rEraErooout−=rθzax+++++++-------Eo6ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityA Perfect Metallic Sphere in a Uniform E-Field - VSurface Charge Density:Now lets calculate the induced surface charge density on the metal sphere surface()()()θεσσφεσεcos30,,ooaroutoarrinarroutoErrEE=⇒=−∂∂−⇒=−===rAs expected, the induced surface charge density is just what is needed to completely cancel the applied E-field inside the metal sphere to give a net zero E-fieldθzax+++++++-------EoECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityParallel Plate Capacitor with a Dielectric - IConsider the parallel plate capacitor now with a dielectric between the platesV+-dφ = Vφ = 0area = Ax0Need to find the potential between the plates()xφNeed to solve:0222=∂∂=∇xφφWith the two boundary conditions:Vdx =⎟⎠⎞⎜⎝⎛−=2φ02=⎟⎠⎞⎜⎝⎛+=dxφSolution is:() ()()dVxxxEdxVxx=∂∂−=⇒⎟⎠⎞⎜⎝⎛−=φφ21ε7ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityParallel Plate Capacitor with a Dielectric - IIV+-dφ = Vφ = 0area = Ax0Solution is:()()()dVxxxEdxVxx=∂∂−=⇒⎟⎠⎞⎜⎝⎛−=φφ21εSurface Charge Densities on Metal PlatesOn the left plate (σL)• The field inside the metal is zero• The field just outside the metal is V/ddVdVLLεσσε=⇒=⎟⎠⎞⎜⎝⎛− 0On the right plate (σR)• The field just outside the metal is V/d• The field inside the metal is zerodVdVRRεσσε−=⇒=⎟⎠⎞⎜⎝⎛−0uEEσεε=−1122ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityParallel Plate Capacitor with a Dielectric - IIIV+-dφ = Vφ = 0area = AεSurface Charge Densities on Metal PlatesdVLεσ=dVRεσ−=++++++++--------dAdVdQVQCVdAQεε====Total charge on the positive plate: Capacitance:8ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityParallel Plate Capacitor with a Non-Uniform Dielectric - I2dφ = Vφ = 0area = Ax01ε2εV+-2d12In region 1 need to solve:0222=∂∂=∇xφφassume a solution:()BxAx+=1φIn region 2 need to solve:0222=∂∂=∇xφφassume a solution:()DCxx+=2φECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityParallel Plate Capacitor