Chapter 3 Boundary-Value Problems in Electrostatics

1. Chapter 3 Boundary-Value Problems in Electrostatics Differential Equations for Electric Potential Method of Images Method of Separation of Variables 1.Differential Equations for Electric Potential 2.Method of Image 3.Method of Separation of Variables in Rectangular Coordinates 4.Method of Separation of Variables in Cylindrical Coordinates 5.Method of Separation of Variables in Spherical Coordinates

2. 1.Differential Equations for Electric Potential The relationship between the electric potential  and the electric field intensity E is Taking the divergence operation for both sides of the above equation gives In a linear, homogeneous, and isotropic medium, the divergence of the electric field intensity E is

3. In infinite free space, the electric charge density confined to in V produces the electric potential given by The differential equation for the electric potential is which is called Poisson’s equation. In a source-free region, and the above equation becomes which is called Laplace’s equation. The solution of Poisson’s Equation. which is just the solution for Poisson’sEquation in free space.

4. Applying Green’s function gives the general solution of Poission’s equation For infinite free space, the surface integral in the above equation will become zero, and Green’s function becomes In the source-free region, the volume integral in the above equation will be zero. Therefore, the second surface integral is considered to be the solution of Poisson’s equation in source-free region, or the integral solution of Laplace’s equation in terms of Green’s function.

5. An equation in mathematical physics is to describe the changes of physical quantities with respect to space and time. For the specified region and moment, the solution of an equation depends on theinitial condition and theboundary condition, respectively, and both are also called the solving condition. Usually the boundary conditions are classified into three types: 1. Dirichet boundary condition: The physical quantities on the boundaries are specified. 2. Neumann boundary condition: The normal derivatives of the physical quantities on the boundaries are given. 3. Mixed boundary-value condition: The physical quantities on some boundaries are given, and the normal derivatives of the physical quantities are specified on the remaining boundaries.

6. For any mathematical physics equation, the existence, the stability, and the uniqueness of the solutions need to be investigated. The existence of the solution is that whether the equation has a solution or not for the given condition of the solution. The stability of the solution refers to whether the solution is changed substantially when the condition or the solution is changed slightly. The uniqueness of the solution is whether the solution is uniqueor not for the prescribed condition of the solution. Electrostatic fields exist in nature, and the existence of the solution of the differential equations for the electric potential is undoubted.

7. The stability of Poisson’s and Laplace’s equations have been proved in mathematics, and the uniqueness of the solution of the differential equations for the electric potentialcan be proved also. In many practical situations, the boundary for the electrostatic field is on a conducting surface. In such cases, the electric potential on the boundary is given by the first type of boundary condition, and the electric charge is given by the second type of boundary condition. Therefore, the solution for the electrostatic field is unique when the charge is specified on the surface of the conducting boundary. For electrostatic fields with conductors as boundaries, the field may be given uniquely when the electric potential , its normal derivative, or the charges is given on the conducting boundaries. That is the uniqueness theorem for solutions to problems on electrostatic fields.

8. 2. Method of Image Essence: The effect of the boundary is replaced by one or several equivalent charges, and the original inhomogeneous region with a boundary becomes an infinite homogeneous space. Basis：The principle of uniqueness. Therefore, these charges should not change the original boundary conditions. These equivalent charges are at the image positions of the original charges, and are called image charges, and this method is called the method of images. Key：To determine the values and the positions of the image charges. Restriction：These image charges may be determined only for some special boundaries and charges with certain distributions.

9. P r P q r q Dielectric h Dielectric Dielectric Conductor h Considering the electric potential of an infinite conducting plane is zero, we have . （1）A point electric charge and an infinite conducting plane The effect of the boundary is replaced by a point charge at the image position, while the entire space becomes homogeneous with permittivity , then the source of electric potential at any pointP will be due to the charges qand q',

10. z   The distribution of the electric field lines and the equipotential surfaces are the same as that of an electric dipole in the upper half- space. 动画 The electric field lines are perpendicular to the conducting surface everywhere, which has zero potential.

11. z   Electric charge conservation：When a point charge qis above an infinite conducting plane, the induced opposite charge will be distributed on the conducting surface, and the magnitude of the image charge should be equal to the total induced charge. We can also prove the claim by making use of the relationship between the density of the charge and the electric field intensityor the derivative of the electric potential on the conducting surface. The above equivalence is established only for the upper half-space for which the source and the boundary condition are both unchanged.

12. For the semi-infinite wedge conducting boundary, the method of images is also applicable. However, the images can be found only for conducting wedges with angle given by where nis an integer. In order to keep the wedge boundary at zero-potential, several image charges are required.  q q    /3 /3    When an infinite line charge is nearby an infinite conducting plane, the method of images can be applied as well, based on the principle of superposition.

13. P r a r' q O q d f （2）A point charge and a conducting sphere. To replace the effect of the boundary of the conducting sphere, let an imagepoint charge q'be placed on the line segment between the point charge qand the center of the sphere. Then the electric poten-tial on the surface of the sphere is then given by Requiring that the electric potential at any point on the surface of the sphere be zero, the image charge must be

14. The ratio must be constant for any point on the surface of the sphere to obtain an image charge with a fixed value. If △OPq'~ △OqP , then . Thus the quantity of the image charge should be P r a r' q O q d f The distance d is The electric field intensity outside the sphere can be found out from qandq' .

15. P r a r' q O q d f If the conducting sphere is ungrounded, then the opposite charges will be induced on the side of the conducting sphere facing the point charge, while the induced charge on the other side of the sphere is positive. The total induced charge on the surface of the conducting sphere should be zero. If the image charge q' is put in, then another image charge q" is needed in order to satisfy the neutrality condition.

16. P r a r' q O q d f The image charge must be at the center of the sphere to ensure that the surface of the sphere is an equipotential surface. In fact, since the sphere is ungrounded, the electric potential is non-zero. Since qand q' produce a zero potential on the surface of the sphere, the second image charge q" is present to produce a certain electric potential.

17. P r a -l O l d An image line charge is put in to represent the charge on the cylinder and placed parallel to it, at a distance dfrom the axis. f The electric field intensity produced by an infinite line charge of density given by The electric potential with respect to a reference point at a distance r from the line electric charge is （3）A line charge and a charged conducting cylinder.

18. If is also taken as the reference point for the electric potential produced by the image line charge , then and produce an electric potential at a point P on the surface of the cylinder given by P r a -l O l d f Since the conducting cylinder is an equipotential body, in order to satisfy this boundary condition the ratio must be a constant. Let , we find

19. q" q E' r0 Et 2 1 1 2   q E" q' En E et 1 2 en （4）A point charge and an infinite dielectric plane. + = To find the field in the upper half-space, the image charge q'can be used to replace the effect of the bound charges on the interface, and the entire space becomes a homogeneous medium with permittivity1. For the lower half-space, the function of the point charge q and the bound charges on the interface can be replaced by the image point charge q" at the position of the original point charge q, and the entire space becomes a homogeneous medium with permittivity 2.

20. The fields must satisfy the boundary condition, i.e., the tangential components of the electric fields are continuous, and the normal components of the electric flux density are equal on the other side. We have therefore, q" q E' r0 Et 2 1 1 2   q E" q' En E et 1 2 en The electric field intensities produced by the point charges q, q', and q" are, respectively, To satisfy the boundary condition, we find the image charges are, respectively, + =

21. a V O b We have Example. Given the radius of the internal conductor of a coaxial line is a, and its electric potential is U. The external conductor is grounded, and its internal radius is b. Find theelectric potential and the electric field intensity between the internal and the external conductors. Solution:The method of images cannot be used, and we have to solve the differential equation for electric potential. The cylindrical coordinate system is selected. Since the field depends only on the variable r, the Laplace’s equation for the electric potential only involves the variable r, as given by

22. Considering We have Solving for the constants C1 and C2 gives a V O b We obtain

23. For the given boundary-value problem it is very importantto select the coordinate system in order to determine the integration constants from the given boundary conditions. In general, the boundary of the region of interest should conform with the coordinate system to be used. In practice, the boundary-value problems for electrostatic fields are related to three coordinate variables. One efficient method to solve three-dimensional Laplace’s equation is the method of separation of variables. This method reduces a three-dimensional partial differential equation to three ordinary differential equations, and the method of separation of variables is established for 11 coordinate systems.

24. Let Let these constants be , and we have 3. Method of Separation of Variables in Rectangular Coordinates In rectangular coordinate system, Laplace’s Equation for electric potential is Substituting it into the above equation, and dividing both sides by X(x)Y(y)Z(z), we have Where each term involves only one variable. The only way the equation can be satisfied is to have each term equal to a constant.

25. or where kx，ky，kz are called the separation constants, and they could be real or imaginary numbers. The three separation constants are not independent of each other, and they satisfy the following equation The three-dimensional partial differential equation is separated to three ordinary differential equations, and the solutions of the ordinary differential equations are easier to obtain. The solution of the equation for the variable x can be written as where A, B, C, D are the constants to be determined.

26. The separation constants could be imaginary numbers. If is an imaginary number, written as , then the equation becomes or The solutions of the equations for the variables y and z have the same forms. The product of these solutions gives the solution of the original partial differential equation. It is very important to select the forms of the solutions, which depend on the given boundary conditions. The constants in the solutions are also related to the boundary conditions.

27. y  = 0  = 0 d x O  = 0 Example. Two semi-infinite, grounded conducting planes are parallel to each other with a separation of d. The finite end is closed by a conducting plane held at electric potential  0, and is isolated from the semi-infinite grounded conducting plane with a small gap. Find the electric potentialin the slot constructed by the three conducting planes. Solution: Select rectangular coordinate system. Since the conducting plane is infinite in the z-direction, the potential in the slot must be independent of z, and this is a two-dimensional problem. The Laplace’s Equation for the electric potential becomes

28. y  = 0  = 0 d In order to satisfy the boundary conditions and , the solution of Y(y) should be selected as x O  = 0 From the boundary condition , we have  = 0 at y = 0 , and the constant B = 0. In order to satisfy , the constant ky should be Using the method of separation of variables, and let The boundary conditions for the electric potential in the slot can be expressed as

29. We find Since ，we obtain Then The constant kx is an imaginary number, and the solution of X(x) should be Since    atx = 0 , the constantC = 0, and Where the constantC = AD.

30. Since  =  0atx = 0 , and we have The right side of the above equation is variable, since Cand n are not fixed. To satisfy the requirement at x = 0, one needs to take the linear combination of the equation as the solution, leading to In order to satisfy the boundary condition x = 0,  = 0, and we have

31. y  = 0  = 0 x  = 0 Equipotential surfaces d Electric field lines 0 The right side of the above equation is Fourier series. By using the orthogonality between the terms of a Fourier series, the coefficients Cn can be found as Finally, we find the electric potential in the slot as

32. Let And we have 4.Method of Separation of Variables in Cylindrical Coordinates In cylindrical coordinate system, Laplace’s equation has the form where the second term is a function of the variable only, while the first and the third are independent of , leading to

33. or where kis the separation constant, and it could be real or imaginary number. The domain of the variable is , in this case the change of the field with must be a periodic function with the period of 2. Let (m is integer), then the solution of the above equation is Considering and the above equation for variable  , the previous equation can be rewritten as where A and Bare the constants to be determined.

34. or The first term of the left side is a function of the variable ronly, and the second is a function of the variable zonly, they should be given by a constant. Let where the constant kz could be real or imaginary number, so that trigonometric functions, hyperbolic functions, or exponential functions can be applied. If kz is a real number, we can take where C and D are constants to be determined. Substituting the equation for the variable z into the previous equation gives

35. If we let , then the above equation becomes where Eand F are constants to be determined, with being the first kind of Bessel function of order m, and being the second kind of Bessel function of order m. Since at r = 0 , we can only take the first kind of Bessel’s function as the solution if the region of the field includes the point r = 0 . which is called a Bessel equation, and its solution is a Bessel function, given as The solution of the above quation should be the linear combination of products of the solutions R(r),() , Z(z).

36. If the electric fields are independent of z, then we have . The above equation becomes The solutions of the above equation are exponential functions, If the field is independent of z, and also independent of , then m = 0 . The solution is given by Considering all of the above situations, the solution can be written as

37. Example. An infinitely long conducting cylinder of radius a is placed in a homogeneous electrostatic field . The direction of is perpendicular to the axis of the conducting cylinder, as shown in the figure. Find the electric fields intensity inside and outside the cylinder. Solution: Select the cylindrical coordinate system. Let z-axis be the axis of the cylinder, and is aligned with the x-axis, so that. y a x O E0 When the conducting cylinder is in an electrostatic equilibrium state, the electric field intensity inside the cylinder is zero, with the cylinder being an equipotential body. The tangential component of the electric field intensity on the surface of the cylinder is zero.

38. y a x O E0 which states that if , the electric potential is independent of the function , but proportional to r and . Hence, one may conclude that the coefficients , and m = 0. Since the electric potential outside the cylinder should be independent of z, the solution becomes the general form and it should satisfy the following two boundary conditions: (a) The tangential component of electric field intensity on the surface of the cylinder is zero, and has no component in the z-direction due to symmetry. This leads to the result (b) The electric potential at infinity should be the same as that required by the original field as given by

39. y a x O E0 Hence, the electric potential function now reduces to Based on the given boundary conditions, we find the coefficients B1 and D1 as And the electric potential outside the cylinder as The electric field intensity outside the cylinder can be obtained as

40. y   a   x Electric field lines     Surface charges E0 Equipotential surfaces The electric field lines, the equipotential surfaces outside the cylinder, and the charges on the surface are shown as follows:

41. Let We have 5. Method of Separation of Variables in Spherical Coordinates In spherical coordinate system, Laplace’s equation becomes For the same reason as before,the solution for  should be

42. And The first term of the above equation is the function of r only, while the second is independent of r. Hence, the first term should be a constant. It is more convenient to write the constant as n(n+1) so that where n is an integer. This is Euler’s equation, and the general solution being given by

43. And Let , then we have The above equation is the associated Legendre’s equation, and the general solution is the sum of the first kind of associated Legender’s functions and the second kind of associated Legender’s functions , where m < n . If n is an integer, and are polynomials with finite number of terms, and n was required to be an integer.

44. From the properties of the second kind of associated Legender’s functions , we know that as . Thus, if the region includes or , for which , we only can take the first kind of associated Legender’s functions as the solution. In view of this, we have The field is independent of , so that m = 0 . In this case, which is called Legendre’s function of the first kind, and the general solution is Therefore, we take the following linear combination as the general solution

45. Solution: Select spherical coordinate system, and let the direction of E0 coincide with the z-axis, i.e. . Obviously, in this case the field is rotationally symmetrical to the z-axis, and independent of . y a z  E0 0 Example. Assume a dielectric sphere with radius aand permittivity  is placed in free space, and with an originally uniform electrostatic field E0, as shown in the figure. Find the electric field intensity in the dielectric sphere. In this way, the solution of the distribution functions of the electric potential inside and outside the sphere as follows, respectively,

46. ①The electric potential atthe center of the sphere should be finite. y a z  E0 0 The distribution functions of the electric potentials inside and outside the sphere should satisfy the following boundary conditions: ②The electric potential at infinity should be ③ The electric potential should be continues at the surface of the sphere, i.e. ④The normal derivatives of electric potentials at the surface should satisfy

47. In order to satisfy the boundary condition ②, all of the coefficients Anexcept A1 should be zero, and , so that where Considering the boundary condition ①, the coefficient Dn should be zero, hence Reconsidering the boundary condition ③ , we have To satisfy the boundary condition ④, we obtain

48. Since the above equations hold for any, the corresponding coefficients on both sides should be equal. Hence, we find Finally, we find the electric potentials inside and outside the sphere are, respectively,

49. Since , we find the electric field intensity inside the sphere is If there is a spherical air bubble in an infinite homogeneous dielectric with the permittivity , then the electric field intensity is y a z  E0 0 The field inside the sphere is still uniform, and the field intensity inside the sphere is less than that outside the sphere. The electric field intensity inside the sphere is greater than that outside the sphere.