1. Electron and proton and other particles ( charge particle ) exert a long – range force on one another, like gravitation, this force is inversely proportional to the square of the distance between the particle ; but unlike gravitation, this force is attractive or repulsive, depending on what particle are involved. Coulomb’s law : The electric force that a point charge exerts on a point charge q at a distance r has magnitude 1

2. The force is directed along the line joining the charges. the force is repulsive if the charge have the same sign, and it is attractive if the charge have opposite sign. If the charge at the origin of our coordinate system, then The potential energy associated with the coulomb force is The electric field ; Gauss’ law : For a charge placed at the origin of coordinates, we define the electric field that this charge produces at some distance r as 2

3. According to eq.(2), we can then express the force on a second charge as The net electric field of several charges if the charges are with position, then the net electric field In the calculation of electric field, to describe the distribution of charges by a continuous charge density. Since is mount of charge in the volume near the point. 3

4. Gauss’s law says that the total normal outward flux over a closed surface in an electric field is equal to times the charge enclosed by the surface. returning to the div. theorem we can write, From div. theorem we have Therefore Let us now suppose that charge uniformly distribution in space and charge density then Gauss's equation become this is differential equation of Gauss’ law. 4

5. Application : 1-We wont to find out the electric field at a point p due to a point charge q placed at o as shown in the diagram. Sol. Consider a sphere of radius r passing through the point p. let the electric field at p be. then by Gauss’ law Or. 5

6. 2-To find out the electric field at a point p due to an infinite line of charge : Consider a cylindrical surface through the point p as shown in the diagram and let be the electric field at the point p. by Gauss’ law 6

7. Ex. Find the electric field to the electric dipole consisting of two point like charge at z ? Sol. For 2 then 7

8. a t θ=0 p location on the z direction (r) For θ=90,ф=0 8

9. The electrostatic potential (Poisson’s equation) : We know that a vector field of zero curl is conservative and can be express as the gradient of scalar function. the electrostatic field satisfied this condition, it is therefore conservative, for any arbitrary closed path And it can be expressed as the gradient of scalar function From Gauss’ law, we have that Or This equation is known as Poisson equation and solution of this equation can be used for solving any electrostatic problem. 9

10. For electrostatic problem in which the charge density at most of point is zero, we can write the Poisson equation as (charge free region) This equation is known as Laplace’s equation in conductors, the charge is on the surface and therefore for any point other than surface the charge density, hence Laplace’s equation can be applied rectangular coordinates cylindrical coordinate Spherical coordinates 10

11. Theorem No.1: State that if are different solutions of Laplace’s equation, then Is also a solution of Laplace’s equation. Where c 1,c 2, …… c n are different constants Theorem No.2: States that if and are the two solution of Laplace’s equation with same conditions, then either or. 11

21. 2-4 Solutions of Laplace’s equation : 1-when the potential is dependent on only one variable: 1-a:when is a function of x only in coordinates. The Laplace’s equation then become Integral eq.(25) we get, where a and b are constant. 1-b:when is a function of r only in coordinates the Laplace’s equation in this case becomes, where a and b are constant. 21

22. 1-c: when is a function of r only in coordinates the Laplace’s equation, therefore becomes,, where a and b are constant. Ex.1: find out the capacity of a parallel plate condenser using Laplace’s equation : Let A and B be two plates of the condenser and let their distance be d and area of one plate be A as shown in the diagram. from Laplace’s equation we have in this have u=ax+b in this case when x=0, u=u 1 and when x=d, u=u 2. 22

23. From Laplace’s equation we have u=ax+b (1) in this case when x=0, u=u 1 and when x=d and u=u 2 u 1 =a*0+b or u 1 =b u 1 =a*d+b u 1 -u 2 = -a.d (2) From eq. (1) Now we know that E between the two plates is given by Where is charge per unit area therefore from eq. (3) and (4), Hence 23

24. Ex. 2: find out the capacity of spherical condenser using Laplace’s equation From eq.(5) and eq.(6) 24

25. Ex. 3 find out the capacity of cylindrical condenser using Laplace’s equation From Laplace’s equation we have Let, 25

26. 2- solution of Laplace’s equation in spherical coordinate In coordinates Laplace’s equation is: To make it simple let us assume that u is a function of r and θ only and is independent of ф. Therefore Laplace’s equation become: Let u = z(r) p(θ) where Z is pure function of r and p is a pure function of θ. Form eq. (3) 26

27. Multiplying eq.(6) by we get If eq. (8) the left hand side is purely dependent on r and the right hand side is pure function of, therefore both side be equal to a constant Eq. (10) can be written as Eq. (11) is the equation known as Legendre equation and it acceptable solution are obtain when k=n(n+1) where n= 0,1,2,3,….the different value of gives different solutions and these different solutions are known as Legendre polynomial. 27

28. these polynomials are as below: n P n (θ) 01*constant 1cosθ 21/2(3cos 2 θ-1) 31/2(5cos 3 θ-3cosθ) 41/8(35cos 4 θ-30cos 2 θ+3) 51/8(63cos 5 θ-70cos 3 θ+15cosθ) Solution of eq.(13) is z = r n or z= r -(n+1) Prove,,,,, Therefore 28

29. eq. (14) will give different solution for different values of n. these solution are called (Zonal Harmonics) or spherical harmonics. Now if we give different value of n we get when n=0 when n=1 when n=2 from theorem 1, know the complete solution is given by: the complete solution in this case is In eq. (15), the term is similar to the potential due to a point charge and the term is similar to potential due to an electric dipole. Therefore eq. (15) can be used to solving different electric problems. 29

30. Ex.1: To study the behavior of a spherical conductor place in uniform electric field. Sol. Let us suppose that there is a uniform electric field E 0 as shown in fig., let a spherical conductor of radius ‘a’ be placed in this field. According to Laplace's equation, the potential is given by In this problem, the conductor has no charge and therefore in eq. (1) the term c 1 r -1 must be zero c 1 = 0 with this condition eq. (1) becomes At infinity, But 30

31. The polar direction (z- direction), and if we make the origin coincide with center of sphere. Then we apply this condition to eq.(2), we find that A 2 = - E 0 and A 3, A 4, A 5,……. should become zero. Therefore eq. (2) becomes Now, at r=a, u is constant = u 0 in eq. (3) when r=a (at the surface of sphere the potential must become independent of a) then we apply this condition, we get A 1 =u 0 and at the surface of sphere the potential must become independent of angle θ, the two term involving cosθ must be cancel each other, but the term with higher inverse power of r cannot be cancelled one against the other because the contain different Legendre function, the only possibility is to set all the c i s with i ≥ 3 equal to zero. equal zero.c 3,c 4,c 5,.. equal zero. 31

32. So eq.(2) becomes Where r=a, the field is E r the charge density 32

33. The total charge on the conductor x=sinθ Q:Show that the charge on spherical conductor in uniform field is zero. 33

34. 2- Solution of Laplace’s equation in cylindrical coordinates: In this coordinates Laplace’s equation is For problems in which is independent of eq.(1) becomes Therefore where k is the separation constant And Let where Y is pure function of r and S is pure function of θ OR 34

35. We know that solution of eq. (6) is If this solution is correct then Placing this value of k in eq. (4) we get, The solution of eq. (9) is solution eq. (2) And When n=0 The solution are 1 and lnr When n=1,2,3,4 The solution are These solutions are known as cylindrical harmonics. 35

36. Ex. 2 A long cylindrical conductor of radius ‘a’ bearing no net charge is placed in an infinity uniform electric field E 0.The direction of E 0 is perpendicular to the cylinder axis. Find the potential and the electric field at point exterior to the cylinder. Sol. Let the a be small, so the cylinder can only change the potential nearby at large distance, the potential will remain unchanged. This gives us the boundary condition Another boundary condition is that the electric field at the cylindrical must be normal to the surface According to Laplace’s equation, the potential given by According to the boundary condition any where in space consist of -E 0 ρcosθ plus extra terms that vanish as 36

37. 1- at ρ→∞ 2- at Therefore 37

38. 3- solution of Laplace’s equation in(x,y,z) coordinates: The Laplace’s equation in(x,y,z) coordinates is written as Let Where and are independent function of x,y,z respectively. Finding out and from eq. (2) and substituting in eq. (1) we get, dividing by, we get Let Where k is the separation constant. 38

39. ,, The solution of eq. (5) is From 2 nd part of eq. (4) we have, Where m is the second separation constant. The solution of eq. (7) is Therefore From (2) or eq. (8) give different solution. 39

40. A very interesting case is when k=0 and m=0 we get Where are constant 40

41. Ex.1 If the electric potential in the y=0 plane is given by find the potential and the electric field component at any point like, p Sol. The Laplace’s equation in coordinate is written as We will look for special solution of the type. سنحاول حل هذه المعادلة التفاضلية بدلالة فصل المتغيرات المطلوب هو حل المعادلات التفاضلية الثلاث و إيجاد قيم الدوال و من ثم الجهد من الشروط الحدودية للمسألة عندما y=0 عندما 41

42. من الشرط الحدودي الأول يتضح إن u ليست دالة الى z و عليه فان z كمية ثابتة و بإمكاننا إن نختارها تساوي واحد و هكذا نستنتج بأن و عليه فان من الشرط الأول أيضا يتبين إن كمية خيالية نفترض تساوي و هكذا فان و إن الحل العام للدالة f 1 هو و إن الحل العام للدالة f 2 هو بينما الحل العام للدالة f 3 هو و هكذا نجد الحل العام للدالة يأخذ الشكل التالي : 42

43. الشرط الحدودي الأول يبين أن A=0 الشرط الحدودي الثاني يتطلب أن C=0 for الشرط الحدودي الثالث يتطلب أن D=0 for عندما y=0 فان عندما فان 43

44. Ex. 2 Find the electric potential distribution functions inside rectangular as shown in diagram. Note the potential is constant in the z – direction. In this problem is independent of z, Laplace’s equation becomes و هكذا يتضح أن كمية حقيقية عندما كمية خيالية و بالعكس. أن اختيار كمية حقيقية آو خيالية يعتمد على الشروط الحدودية التي تمليها المسألة التي نحن بصددها، حيث سنجد أن حقيقية و خيالية كما سنرى و عليه فان 44

45. و من الشروط الحدودية للمسألة و هي و هكذا فان when m=0,1,2,3,… 45

46. و لإيجاد قيم دعنا نستعين بخاصية التعامد لمتسلسلة فورير فعند ضرب طرفي المعادلة الأخيرة (*) بـ ونجري التكامل نحصل على 46