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Chapter 4. Solution of Electrostatic Problems Poisson's equation

Chapter 4. Solution of Electrostatic Problems Poisson's equation Two governing differential equations for electrostatics : In a linear, isotropic medium ( D =  E ) : For a simple medium (linear, isotropic, and homogeneous) : Poisson's equation in coordinate systems RCS : CCS : SCS :.

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Chapter 4. Solution of Electrostatic Problems Poisson's equation

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  1. Chapter 4. Solution of Electrostatic Problems • Poisson's equation • Two governing differential equations for electrostatics : In a linear, isotropic medium (D = E) : For a simple medium (linear, isotropic, and homogeneous) : • Poisson's equation in coordinate systems • RCS : • CCS : • SCS : : Poisson's equations Laplacian operator

  2. Laplace's equation • If the medium under consideration contains no free charges, Poisson's equation reduces to • It is the governing equation for problems involving a set of conductors, such as capacitors, maintained at different potentials. • Example 4-1 : Parallel-plate capacitor (a) The potential at any point between the plates 1. Ignoring the fringing field  Electric field is uniform between the plates (as if the plates were infinitely large)  No variation of V in the x and z directions. 2. No free charges in the dielectric : Using Laplace's eq. (b) The surface charge densities on the plate : Laplace's equations B.C. : , At the lower plate : At the upper plate :

  3. Uniqueness of electrostatic solutions • Uniqueness theorem : A solution of Poisson's equation that satisfies the given boundary conditions is a unique solution. • A solution of an electrostatic problem satisfying its boundary solution is the only possible solution, irrespective of the method by which the solution is obtained. • A solution obtained by intelligent guessing is the only correct solution. (Proof) Assume that there are two solutions V1 and V2 satisfying Poisson's equation in . Integrating both sides over volume  : Considering very large S0 Since is nonnegative everywhere, should be zero. Vd is constant at all points in volume  including bounding surfaces S1, S2, …. Sn where Vd = 0. Vd = 0 throughout the volume  V1 = V2

  4. Method of images • Image charge method : In solving some electrostatic problems, the bounding surfaces can be replaced with appropriate "fictitious" charges (image charges) to satisfy the boundary conditions  This must be the only solution according to uniqueness theorem. • Example : A point charge Q above a very large grounded conducting plane  What is the potential at every point above the conducting plane (y > 0)? The potential above the conducting plane satisfies Laplace's equation except at the point charge. Boundary conditions : (1) V(x, 0, z) = 0, (2) At far point, V 0 What really happens is : (1) The point charge Q produces an electric field. (2) The field induces surface charges s on the grounded conducting plane.  In order to calculate V, we must find out s, but it is hard to determine s.

  5. [HW] Image charges must be located outside the region in which the filed is to be determined. Explain why. (Cont'd) (1) Potential above the conducting plate (y > 0) Another point charge –Q (image charge) is located at y = – d and the conducting plate is removed.  This situation satisfies the original boundary conditions.  This potential is the only solution of the original problem by uniqueness theorem. (2) Surface charge density on the conducting plate (0, d, 0) (0, -d, 0)

  6. Boundary-Value Problems (BVP) • BVP in electrostatics : the problem consisting of conductors maintained at specified potentialswith no isolated free charges  This kind of problems cannot be solved by the image method  We must solve these problems by solving Laplace's equation (it is a differential equation!) directly. • Types of BVP • Dirichlet problems : The value of potential is specified everywhere on the boundaries. • Neumann problems : The normal derivative of the potential is specified everywhere on the boundaries • Mixed BVP : The potential is specified over some boundaries, and the normal derivative of the potential is specified over the remaining ones. • How to solve the Laplace's equation in general? • One-dimensional problem (e.g. V is a function of one variable) : direct integration with the given boundary condition • Two- or three-dimensional (e.g. V is a function of more than one variables) : separation of variable method with the given boundary conditions

  7. Example 4-6 : Two grounded, semi-infinite parallel-plate electrodes by a distance and maintained at a specified potential at one end Sol) This is a BVP in Cartesian coordinate system  Finding V between the plates. separation constant : X(x), Y(y), and Z(z) can be expressed as a second-order ordinary differential equation.

  8. (Cont'd) • Because of the infinity in the z-direction, Boundary conditions in the x, y directions are (why?)

  9. (Cont'd) The solution should also satisfy the boundary condition (2) V(0, y) = V0 for all values of y from 0 to b. : Since Laplace's equation is a linear partial differential equation, the superposition of Vn(x, y) for the different values of n is also a solution  Fourier series expansion.

  10. Example 4-8 : The potential distribution between a long coaxial cable Sol) This is a BVP in cylindrical coordinate system. Laplace's equation in cylindrical coordinate system : General solution of the above equation is known as Bessel functions. (solution of Laplace’s equation in CCS)  Because the given structure is very long and has a symmetry in  -direction, the Laplace's equation in this problem must have only a r-dependence. Integrating both sides yields Boundary conditions : V(a) = 0, V(b) = V0

  11. Example 4-2 : Uniform spherical clouds of electrons Sol) Poisson's equation in spherical coordinate system : There are symmetries in the  and  directions : (a) Inside the clouds : Solve Poisson's equation. (b) Outside the clouds : Solve Laplace's equation . At the surface R = b, the E-field should be continuous Legendre functions. (solution of Laplace’s equation in SCS) , Since E must be finite at R = 0, C1 must be 0

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