Download CHAPTER (3) ELECTRIC FLUX DENSITY

CHAPTER (3) ELECTRIC FLUX DENSITY Electric flux (𝝍𝒆 ): The electric flux concept is based on the following rules: 1- Electric flux begins from (+ ve) charge and ends to (-ve) charge 2- Electric field at a point is tangent to the electric flux line passing with this point and out wide. 3- In the absence of (-ve) charge the electric flux terminates at infinity. 4- The magnitude of the electric field at a point is proportional to the magnitude of the electric flux density at this point. 5- The number of electric flux lines from a (+ ve) charge Q is equal to Q in SI unit 𝝍𝒆 = 𝑸 Electric flux density��⃗ 𝑫 displacement vector): ⃗ is defined In free space, the electric flux density vector �𝐃 as ∆𝝍 �𝐃 �⃗ = 𝒂 � 𝒏 𝒍𝒊𝒎∆𝒔→𝟎 𝒆 𝑪 𝒎−𝟐 , ∆𝒔 Where: ∆𝝍𝒆 equals the number of electric lines that are normal to the surface ∆S �⃗ . ��⃗ 𝝍𝒆 = � �𝐃 𝒅𝒔 �⃗ due to Point Charge Relation Between �𝐃⃗ and 𝐄 If we locate a point charge Q at the origin, the electric flux �⃗ can be evaluated by dividing 𝝍𝒆 by the surface area of density �𝐃 the sphere, thus ��⃗ = 𝒂 � 𝒓𝒔 𝐃 �𝐃⃗ = 𝒂 � 𝒓𝒔 𝝍𝒆 𝟒𝝅𝒓𝟐𝒔 𝑸 𝑪 𝒎−𝟐 𝟐 𝟒𝝅𝒓𝒔 �⃗ on the surface at 𝒓𝒔 due to Q, is The expression for 𝐄 �⃗ = 𝒂 � 𝒓𝒔 𝐄 𝑸 −𝟏 𝑵 𝑪 𝟒𝝅𝜺𝒐 𝒓𝟐𝒔 ��⃗ and 𝐄 �⃗, it can be seen that From the expressions for 𝐃 �𝐃 �⃗ = 𝜺𝒐 𝐄 �⃗ �⃗ and 𝐄 �⃗ was derived using a Point charge Q, The relation between �𝐃 but also it is valid for general charge distribution, �𝑬⃗ = ∭ 𝝆𝒗 𝒅𝒗𝟐 𝒂 �𝑹 𝟒𝝅𝜺𝒐 𝑹 𝝆𝒗 𝒅𝒗 �𝑫 �⃗ = � � 𝒂 𝟒𝝅𝑹𝟐 𝑹 �⃗ From Faraday’s experiment, it is found that, 𝝍𝒆 and thus 𝐃 are independent of the dielectric media in which Q is embedded. Example: Find the electric flux 𝝍𝒆 that passes through the surface shown in the figure. Where: �𝐃⃗ = �𝒚 𝒂 �𝒙 + 𝒙 𝒂 �𝒚 �𝒙 𝟏𝟎−𝟐 𝑪 𝒎−𝟐 Solution �⃗ . ��⃗ 𝝍𝒆 = � �𝐃 𝒅𝒔 𝟐 𝟑 �𝒙 + 𝒙 𝒂 �𝒚 �𝒙 𝟏𝟎−𝟐 . 𝒂 � 𝒚 𝒅𝒙𝒅𝒛 𝝍𝒆 = � � �𝒚 𝒂 𝟎 𝟎 𝒙𝟐 𝟑 𝟗 𝝍𝒆 = � 𝟐 � [𝒛]𝟐𝟎 𝒙𝟏𝟎−𝟐 = �𝟐 𝒙 𝟐� 𝒙𝟏𝟎−𝟐 = 𝟗𝒙𝟏𝟎−𝟐 𝑪 Gauss’s law 𝟎 As it is stated before, the total electric flux emanating from a charge + Q [C] is equal to Q [C] in the SI units. The previous statement can be restated by saying that the total electric flux passing through any closed imaginary surface, enclosing the charge Q [C], is equal to Q [C] in the SI units. Since the charge Q is enclosed by the closed surface, so the charge Q will be named as 𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 . Gauss’s law states that: the total flux out of a closed surface is equal to the net charges within the surface. This can be written in integral form as: ��⃗ . ��⃗ 𝝍𝒆 = � 𝒅 𝝍 𝒆 = � 𝐃 𝒅𝒔 = 𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 ��⃗ and then 𝐄 �⃗ by Gauss’s law is used in order to determine 𝐃 �⃗ outside the closed surface integral. This can be getting 𝐃 executed by choosing Gaussian surface that satisfies the ��⃗ be independent of ds following conditions, such that 𝐃 variables. Conditions for Gauss’s law: 1- The surface or volume contained charges must has degree of symmetry. �⃗ must be defined in the surface (𝐃 ��⃗ ≠ ∞). 2- �𝐃 �⃗ must be uniform on the Gaussian surface 3- �𝐃 4- The Gaussian surface must be identical to the body contained the charge. Note: 1- Gauss’s law is not used for all cases of charges, but it can be used only for the cases where the chosen Gaussian surface satisfy the previous conditions. 2- Gauss’s law is used for the following cases: • Infinite line charges and coaxial charged cylinders • Infinite charged sheet • Concentric charged spheres Example: Find the electric flux density at a point p(𝒓𝒄 , 𝝋, 𝒛) due to an infinite charged line of 𝝆𝒍 at z-axis. Solution: ��⃗ . ��⃗ (1) ∯ 𝐃 𝒅𝒔 = 𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 (2) Choice of Gaussian surface (3) 𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 = 𝝆𝒍 𝑳 𝟐𝝅 𝑳 �𝒓𝒄 (𝒓𝒄 𝒅𝝋 𝒅𝒛) = 𝟐𝝅𝒓𝒄 𝑳𝐃 (4) ∯ �𝐃⃗ . ��⃗ 𝒅𝒔 = �𝐃⃗ . ∫𝟎 ∫𝟎 𝒂 z (5) 𝟐𝝅𝒓𝒄 𝑳𝐃 = 𝝆𝒍 𝑳 �⃗ = 𝒂 � 𝒓𝒄 (6) 𝐃 𝝆𝒍 𝟐𝝅𝒓𝒄 𝑪 𝒎−𝟐 �⃗ = �𝐃⃗�𝜺 = 𝒂 � 𝒓𝒄 (7) 𝐄 𝒐 𝝆𝒍 𝟐𝝅𝒓𝒄 𝜺𝒐 𝐍� 𝑪 + + + + + + + + + + + + + + + r (rc, L y Example: �⃗ and 𝐄 �⃗ inside and outside a sphere of radius (a) and Find �𝐃 surface charge density𝝆𝒔 . Solution: Region 1 𝐫𝐬 < 𝐚 �⃗ . ��⃗ (1) ∯ �𝐃 𝒅𝒔 = 𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 (2) Choice of Gaussian surface (3) 𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 = 𝟎 �⃗ . ��⃗ �⃗ . ∫𝟐𝝅 ∫𝝅 𝒂 �𝒓𝒔 �𝒓𝟐𝒔 𝒔𝒊𝒏 𝜽𝒅𝜽𝒅𝝋 � = 𝟒𝝅 𝒓𝟐𝒔 𝐃 (4) ∯ �𝐃 𝒅𝒔 = �𝐃 𝟎 𝟎 (5) 𝟒𝝅𝒓𝟐𝒔 𝐃 = 𝟎 �⃗ = 𝒂 �𝒓𝒔 𝟎 𝑪 𝒎−𝟐 (6) �𝐃 �⃗� �⃗ = �𝐃 �𝒓𝒔 𝟎 𝑵 𝑪−𝟏 (7) 𝐄 𝜺𝒐 = 𝒂 Region 2 𝐫𝐬 > 𝑎 (1) ∯ �𝐃⃗ . ��⃗ 𝒅𝒔 = 𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 (2) Choice of Gaussian surface (3) 𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 = 𝟒𝝅 𝒂𝟐 𝝆𝒔 𝟐𝝅 𝝅 �𝒓𝒔 �𝒓𝟐𝒔 𝒔𝒊𝒏 𝜽𝒅𝜽𝒅𝝋 � = 𝟒𝝅 𝒓𝟐𝒔 𝐃 (4) ∯ �𝐃⃗ . ��⃗ 𝒅𝒔 = �𝐃⃗ . ∫𝟎 ∫𝟎 𝒂 (5) 𝟒𝝅𝒓𝟐𝒔 𝐃 = 𝟒𝝅 𝒂𝟐 𝝆𝒔 𝒂𝟐 � ⃗ �𝒓 𝟐 𝝆𝒔 𝑪 𝒎−𝟐 (6) 𝐃 = 𝒂 𝒔 𝒓𝒔 �⃗ = �𝐃⃗�𝜺 = 𝒂 � 𝒓𝒔 (7) 𝐄 𝒐 𝒂𝟐 𝜺𝒐 𝒓𝟐𝒔 𝝆𝒔 𝑵 𝑪−𝟏 a Rs E εΟ rs Example: ��⃗ and 𝐄 �⃗ in all regions for a spherical shell of Find 𝐃 radii a, b and volume charge density 𝝆𝒗 b a rs ρv Solution: Region 1 𝐫𝐬 < 𝐚 ��⃗ . ��⃗ (1) ∯ 𝐃 𝒅𝒔 = 𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 (2) Choice of Gaussian surface (3) 𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 = 𝟎 𝟐 ��⃗ . ��⃗ �⃗ . ∫𝟐𝝅 ∫𝝅 𝒂 � (4) ∯ 𝐃 𝒅𝒔 = �𝐃 �𝒓 𝒔𝒊𝒏 𝜽𝒅𝜽𝒅𝝋 � = 𝟒𝝅 𝒓𝟐 𝒓 𝒔 𝒔𝐃 𝒔 𝟎 𝟎 (5) 𝟒𝝅𝒓𝟐𝒔 𝐃 = 𝟎 �⃗ = 𝒂 �𝒓𝒔 𝟎 𝑪 𝒎−𝟐 (6) �𝐃 �⃗ = �𝐃⃗�𝜺 = 𝒂 �𝒓𝒔 𝟎 𝑵 𝑪−𝟏 (7) 𝐄 𝒐 Region 2 𝒂 < 𝐫𝐬 < 𝑏 �⃗ . ��⃗ (1) ∯ �𝐃 𝒅𝒔 = 𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 (2) Choice of Gaussian surface 𝟒𝝅 (3) 𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 = 𝟑 (𝒓𝟑𝒔 − 𝒂𝟑 ) 𝝆𝒗 ��⃗ . ��⃗ �⃗ . ∫𝟐𝝅 ∫𝝅 𝒂 � �𝒓𝟐𝒔 𝒔𝒊𝒏 𝜽𝒅𝜽𝒅𝝋 � = 𝟒𝝅 𝒓𝟐𝒔 𝐃 (4) ∯ 𝐃 𝒅𝒔 = �𝐃 𝟎 𝟎 𝒓𝒔 (5) (6) 𝟒𝝅𝒓𝟐𝒔 𝐃 = �𝐃⃗ = 𝒂 � 𝟒𝝅 𝟑 (𝒓𝟑𝒔 − 𝒂𝟑 ) 𝝆𝒗 (𝒓𝟑𝒔 −𝒂𝟑 ) 𝒓𝒔 𝟑𝒓𝟐𝒔 �⃗ = �𝐃⃗�𝜺 = 𝒂 �𝒓𝒔 (7) 𝐄 𝒐 Region 3 𝐫𝐬 > 𝑏 𝝆𝒗 𝑪 𝒎−𝟐 (𝒓𝟑𝒔 −𝒂𝟑 ) 𝟑𝜺𝒐 𝒓𝟐𝒔 𝝆𝒗 𝑵 𝑪−𝟏 �⃗ . ��⃗ (1) ∯ �𝐃 𝒅𝒔 = 𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 (2) Choice of Gaussian surface (3) 𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 = 𝟒𝝅 𝟑 (𝒃𝟑 − 𝒂𝟑 ) 𝝆𝒗 ��⃗ . ��⃗ �⃗ . ∫𝟐𝝅 ∫𝝅 𝒂 � �𝒓𝟐𝒔 𝒔𝒊𝒏 𝜽𝒅𝜽𝒅𝝋 � = 𝟒𝝅 𝒓𝟐𝒔 𝐃 (4) ∯ 𝐃 𝒅𝒔 = �𝐃 𝟎 𝟎 𝒓𝒔 𝟒𝝅 (5) 𝟒𝝅𝒓𝟐𝒔 𝐃 = (𝒃𝟑 − 𝒂𝟑 ) 𝝆𝒗 𝟑 ��⃗ = (6) 𝐃 𝟑 𝟑 (𝒃 −𝒂 ) � 𝒓𝒔 𝒂 𝟑𝒓𝟐𝒔 �⃗� �⃗ = �𝐃 (7) 𝐄 𝜺𝒐 = 𝝆𝒗 𝑪 𝒎−𝟐 𝟑 𝟑 (𝒃 −𝒂 ) � 𝒓𝒔 𝒂 𝟑𝜺𝒐 𝒓𝟐𝒔 𝝆𝒗 𝑵 𝑪−𝟏 Example: In the figure shown, find the electric field intensity in all regions. ρV b a (I) (II) (III) L Solution Region 1 𝐫𝐜 < 𝐚 ��⃗ . ��⃗ (1) ∯ 𝐃 𝒅𝒔 = 𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 (2) Choice of Gaussian surface (3) 𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 = 𝟎 ��⃗ . ��⃗ �⃗ . ∫𝟐𝝅 ∫𝑳 𝒂 �𝒓𝒄 (𝒓𝒄 𝒅𝝋 𝒅𝒛) = 𝟐𝝅𝒓𝒄 𝑳𝐃 (4) ∯ 𝐃 𝒅𝒔 = �𝐃 𝟎 𝟎 (5) 𝟐𝝅𝒓𝒄 𝑳𝐃 = 𝟎 �⃗ = 𝒂 �𝒓𝒄 𝟎 𝑪 𝒎−𝟐 (6) �𝐃 �⃗� �⃗ = �𝐃 � 𝒓𝒄 𝟎 (7) 𝐄 𝜺𝒐 = 𝒂 𝑵 𝑪−𝟏 Region 2 𝒂 < 𝐫𝐜 < 𝑏 ��⃗ . ��⃗ (1) ∯ 𝐃 𝒅𝒔 = 𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 (2) Choice of Gaussian surface (3) 𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 = �𝝅𝒓𝟐𝒄 − 𝝅𝒂𝟐 � 𝑳 𝝆𝒗 ��⃗ . ��⃗ �⃗ . ∫𝟐𝝅 ∫𝑳 𝒂 �𝒓𝒄 (𝒓𝒄 𝒅𝝋 𝒅𝒛) = 𝟐𝝅𝒓𝒄 𝑳𝐃 (4) ∯ 𝐃 𝒅𝒔 = �𝐃 𝟎 𝟎 (5) 𝟐𝝅𝒓𝒄 𝑳𝐃 = �𝝅𝒓𝟐𝒄 − 𝝅𝒂𝟐 � 𝑳 𝝆𝒗 �⃗ = 𝒂 � 𝒓𝒄 (6) �𝐃 �𝒓𝟐𝒄 − 𝒂𝟐 � 𝟐𝒓𝒄 𝟐 �𝒓𝒄 − 𝒂𝟐 � 𝝆𝒗 𝑪 𝒎−𝟐 �⃗� −𝟏 �⃗ = �𝐃 � 𝒓𝒄 (7) 𝐄 𝝆 𝑵 𝑪 𝒗 𝜺𝒐 = 𝒂 𝟐𝜺𝒐 𝒓𝒄 Region 3 𝐫𝐜 > 𝑏 ��⃗ . ��⃗ (1) ∯ 𝐃 𝒅𝒔 = 𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 (2) Choice of Gaussian surface (3) 𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 = �𝝅𝒃𝟐 − 𝝅𝒂𝟐 � 𝑳 𝝆𝒗 ��⃗ . ��⃗ �⃗ . ∫𝟐𝝅 ∫𝑳 𝒂 �𝒓𝒄 (𝒓𝒄 𝒅𝝋 𝒅𝒛) = 𝟐𝝅𝒓𝒄 𝑳𝐃 (4) ∯ 𝐃 𝒅𝒔 = �𝐃 𝟎 𝟎 (5) 𝟐𝝅𝒓𝒄 𝑳𝐃 = �𝝅𝒃𝟐 − 𝝅𝒂𝟐 � 𝑳 𝝆𝒗 ⃗=𝒂 �𝒓𝒄 (6) �𝐃 �𝝅𝒃𝟐 − 𝝅𝒂𝟐 � 𝟐𝒓𝒄 �⃗ = �𝐃⃗�𝜺 = 𝒂 �𝒓𝒄 (7) 𝐄 𝒐 𝝆𝒗 𝑪 𝒎−𝟐 �𝝅𝒃𝟐 − 𝝅𝒂𝟐 � 𝟐𝜺𝒐 𝒓𝒄 𝝆𝒗 𝑵 𝑪−𝟏 Example: Find the electric field intensity in all regions for the following charge configurations: - Point charge Q is located at the center. - Conducting sphere of radius a and of charge ρs. - A volume charge of ρv in a spherical shell of radii b, c. b IV Solution: + + + + + + II+ +a D( + Q + + + + +ρ + + s C Region 1 𝐫𝐬 < 𝐚 ρv ��⃗ . ��⃗ (1) ∯ 𝐃 𝒅𝒔 = 𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 (2) Choice of Gaussian surfa (3) 𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 = 𝑸 𝟐 ��⃗ . ��⃗ �⃗ . ∫𝟐𝝅 ∫𝝅 𝒂 � (4) ∯ 𝐃 𝒅𝒔 = �𝐃 �𝒓 𝒔𝒊𝒏 𝜽𝒅𝜽𝒅𝝋 � = 𝟒𝝅 𝒓𝟐 𝒓 𝒔 𝒔𝐃 𝒔 𝟎 𝟎 (5) 𝟒𝝅𝒓𝟐𝒔 𝐃 = 𝑸 𝑸 −𝟐 �⃗ = 𝒂 � 𝒓𝒔 (6) 𝐃 𝑪 𝒎 𝟐 𝟒𝝅𝒓 𝒔 𝑸 �⃗� −𝟏 �⃗ = �𝐃 � 𝒓𝒔 (7) 𝐄 𝑵 𝑪 𝟐 𝜺𝒐 = 𝒂 𝟒𝝅𝜺𝒐 𝒓𝒔 Region 2 𝒂 < 𝐫𝐬 < 𝑏 ��⃗ . ��⃗ (1) ∯ 𝐃 𝒅𝒔 = 𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 (2) Choice of Gaussian surface (3) 𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 = 𝑸 + 𝟒𝝅𝒂𝟐 𝝆𝒔 𝟐 𝟐 ��⃗ . ��⃗ �⃗ . ∫𝟐𝝅 ∫𝝅 𝒂 � (4) ∯ 𝐃 𝒅𝒔 = �𝐃 �𝒓 𝒔𝒊𝒏 𝜽𝒅𝜽𝒅𝝋 � = 𝟒𝝅 𝒓 𝒓 𝒔 𝒔𝐃 𝒔 𝟎 𝟎 (5) 𝟒𝝅𝒓𝟐𝒔 𝐃 = 𝑸 + 𝟒𝝅𝒂𝟐 𝝆𝒔 (6) (7) �𝐃⃗ = 𝒂 � 𝑸+𝟒𝝅𝒂𝟐 𝝆𝒔 𝒓𝒔 𝟒𝝅𝒓𝟐𝒔 �⃗ = �𝐃⃗�𝜺 = 𝒂 �𝒓𝒔 𝐄 𝒐 Region 3 𝐛 < 𝐫𝐬 < 𝑐 𝑪 𝒎−𝟐 𝑸+𝟒𝝅𝒂𝟐 𝝆𝒔 𝟒𝝅𝜺𝒐 𝒓𝟐𝒔 ��⃗ . ��⃗ (1) ∯ 𝐃 𝒅𝒔 = 𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 (2) Choice of Gaussian surface (3) 𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 = 𝑸 + 𝟒𝝅𝒂𝟐 𝝆𝒔 + (4) (5) (6) (7) 𝝆𝒗 𝑵 𝑪−𝟏 𝟒𝝅 𝟑 (𝒓𝟑𝒔 − 𝒂 ) 𝝆𝒗 𝟑 𝟐𝝅 𝝅 𝟐 ��⃗ . ��⃗ �⃗ . ∫ ∫ 𝒂 � 𝒅𝒔 = �𝐃 �𝒓 ∯𝐃 𝒓 𝒔 𝒔𝒊𝒏 𝜽𝒅𝜽𝒅𝝋 � 𝒔 𝟎 𝟎 𝟒𝝅 𝟑 𝟒𝝅𝒓𝟐𝒔 𝐃 = 𝑸 + 𝟒𝝅𝒂𝟐 𝝆𝒔 + (𝒓𝟑𝒔 − 𝒂 ) 𝝆𝒗 𝟑 𝟒𝝅 𝑸+𝟒𝝅𝒂𝟐 𝝆𝒔 + (𝒓𝟑𝒔 −𝒂𝟑 ) 𝝆𝒗 𝟑 𝒓𝒔 𝟒𝝅𝒓𝟐𝒔 𝟐 𝝆 +𝟒𝝅 (𝒓𝟑 −𝒂𝟑 ) 𝝆 𝑸+𝟒𝝅𝒂 𝒔 �⃗� 𝒔 𝟑 𝒗 �⃗ = �𝐃 � = 𝒂 𝐄 𝒓 𝜺𝒐 𝒔 𝟒𝝅𝜺𝒐 𝒓𝟐𝒔 �𝐃⃗ = 𝒂 � = 𝟒𝝅 𝒓𝟐𝒔 𝐃 𝑪 𝒎−𝟐 Region 4 𝐫𝐬 > 𝑐 ��⃗ . ��⃗ (1) ∯ 𝐃 𝒅𝒔 = 𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 (2) Choice of Gaussian surface (3) 𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 = 𝑸 + 𝟒𝝅𝒂𝟐 𝝆𝒔 + (4) (5) (6) (7) 𝟒𝝅 (𝒃𝟑 − 𝒂𝟑 ) 𝝆𝒗 𝟑 𝟐𝝅 𝝅 𝟐 ��⃗ . ��⃗ �⃗ . ∫ ∫ 𝒂 � 𝒅𝒔 = �𝐃 �𝒓 ∯𝐃 𝒓 𝒔 𝒔𝒊𝒏 𝜽𝒅𝜽𝒅𝝋 � = 𝒔 𝟎 𝟎 𝟒𝝅 𝟒𝝅𝒓𝟐𝒔 𝐃 = 𝑸 + 𝟒𝝅𝒂𝟐 𝝆𝒔 + (𝒃𝟑 − 𝒂𝟑 ) 𝝆𝒗 𝟑 �𝐃⃗ = 𝒂 �𝒓𝒔 𝟒𝝅 𝑸+𝟒𝝅𝒂𝟐 𝝆𝒔 + �⃗� �⃗ = �𝐃 � 𝒓𝒔 𝐄 𝜺𝒐 = 𝒂 𝟑 𝟒𝝅𝒓𝟐𝒔 𝑸+𝟒𝝅𝒂𝟐 𝟑 (𝒃𝟑 −𝒂 ) 𝝆𝒗 𝑪 𝒎−𝟐 𝟑 𝟒𝝅 𝟑 𝝆𝒔 + (𝒃 −𝒂 ) 𝝆𝒗 𝟑 𝟒𝝅𝜺𝒐 𝒓𝟐𝒔 𝑵 𝑪−𝟏 𝟒𝝅 𝒓𝟐𝒔 𝐃 𝑵 𝑪−𝟏 Divergence ⃗ equals the net flux of the vector �𝐃 �⃗ that The divergence of �𝐃 flows outwardly through a closed surface S per unit volume (enclosed by ∯) as the volume goes to zero. Divergence Law ��⃗ 𝑫 . 𝒅𝒔 ∯ ��⃗ �⃗ = 𝛁. �𝑫 �⃗ ≜ 𝒍𝒊𝒎∆𝒗→𝟎 𝑫𝒊𝒗 �𝑫 ∆𝒗 �⃗ = 𝝆𝒗 [𝑪𝒎−𝟑 ] 𝛁. �𝑫 The general form of the divergence can be written as �⃗ = 𝛁. �𝑫 𝟏 𝝏 𝝏 � (𝒉𝟐 𝒉𝟑 𝑫𝝁𝟏 ) + (𝒉 𝒉 𝑫 ) 𝒉𝟏 𝒉𝟐 𝒉𝟑 𝝏𝝁𝟏 𝝏𝝁𝟐 𝟏 𝟑 𝝁𝟐 𝝏 + (𝒉 𝒉 𝑫 )� 𝝏𝝁𝟑 𝟏 𝟐 𝝁𝟑 Where, 𝝁𝟏 , 𝝁𝟐 , 𝒂𝒏𝒅 𝝁𝟑 are the variables of the coordinates system, and 𝒉𝟏 , 𝒉𝟐 , 𝒂𝒏𝒅 𝒉𝟑 are the factors multiplied by the differentiable of the variables. So For Cartesian coordinates �⃗ = � 𝛁. �𝑫 𝝏 𝝏 𝝏 𝑫 + 𝑫 + 𝑫 � 𝝏𝒙 𝒙 𝝏𝒚 𝒚 𝝏𝒛 𝒛 For Cylinderical coordinates 𝟏 𝝏 𝝏 𝝏 � 𝒓 𝒄 𝑫 𝒓𝒄 + 𝑫𝝋 + 𝒓 𝑫 � 𝒓𝒄 𝝏𝒓𝒄 𝝏𝝋 𝝏𝒛 𝒄 𝒛 For Spherical coordinates �⃗ = 𝛁. �𝑫 �⃗ = 𝛁. �𝑫 𝟏 𝝏 𝝏 𝝏 𝟐 ( ) � �𝒓 𝒔𝒊𝒏 𝜽 𝑫 � + 𝒓 𝒔𝒊𝒏𝜽 𝑫 + �𝒓 𝑫 �� 𝒓𝒔 𝜽 𝒓𝟐𝒔 𝒔𝒊𝒏 𝜽 𝝏𝒓𝒔 𝒔 𝝏𝜽 𝒔 𝝏𝝋 𝒔 𝝋 Proof of Divergence Law Let a cube enclosed at its center the point (xo,yo,zo) and the �⃗ crossing the cube surface at this electric field density 𝐷 point and is giving by: �⃗ = 𝒂 �𝒙 𝑫𝒙𝒐 + 𝒂 �𝒚 𝑫𝒚𝒐 + 𝒂 � 𝒛 𝑫𝒛 𝒐 𝐷 D(x+∆x) D(x) dS b ∆x ��⃗ for the cube, all six faces ��⃗ . 𝑑𝑠 In order to express ∯ 𝐷 ��⃗ is outward since the must be taken, the direction of 𝑑𝑠 faces are normal to the three axes. Only one component of ��⃗ 𝐷 will cross any two surface. Thus, It’s required to ��⃗. We take at the first the surface in + x ��⃗ . 𝑑𝑠 find∯ 𝐷 direction and in – x direction. ∫ D ⋅ d S = (D Xo ( ) xˆ + DYo yˆ + D Zo zˆ ) ⋅ ∆y∆z Xˆ ↓ left + D(x + ∆x ) xˆ ⋅ ∆y∆z xˆ ↓ right ∂D ∆x +  ∂x ∂D   ⋅ = − ∆ ∆ + ∆ ∗ D d S D z x Xo Y ∫  o ∂x ∆x  ∆y∆z   ∂D ∫ D ⋅ d S = ∂x Dx∆y∆z ∂D y ⋅ ↓ = D d S Dx∆y∆z backed , front ∫ ∂y ∂D z ⋅ D d S D x ∆y∆z ↓ top ,bottom = ∫ ∂z ∂D   ∫ D ⋅ d S = − D Xo ∆ Y ∆z + ∆xo ∗ ∂x ∆x ∆y∆z ∂D D ⋅ d S = Dx∆y∆z ∫ ∂x ∂Dy D ⋅ d S ↓ = Dx∆y∆z backed , front ∫ ∂y  D(x + ∆x ) = D(x O ) + ∫ D ⋅ d S ↓top ,bottom =  ∂D X  ⋅ = + D d S ∫ ∂ x   ∂D X ∴ ∫ D ⋅ d S =   ∂x ∴ ρυ = ∂Dz Dx∆y∆z ∂z ∂DY ∂D Z + ∂y ∂z  ∆x∆y∆z  ∂DY ∂D Z  ∆v = Q + + ∂y ∂z  ∂DY ∂D Z Q ∂D X = + + =∇⋅D ∆v ∂x ∂y ∂z Example: A charged sphere of ρv and radius a, the electric flux density D for rs < a is given by: and for rs > a is given by: D = 10 −5 rs D= rˆs , 3 10 −5 a 3 3rs 2 . Find ρv in the previous two regions. Solution: ∇⋅D= 1 h1 h2 h3 ∂  ∂ ∂ ( ) ( ) ( ) h h D h h D h h D + + 2 3 rs 1 3 θ 1 2 ϕ   ∂θ ∂φ  ∂rs  where : h1 = 1, h2 = rs , h3 = rs sin θ for rs < a : 1 ρv = ∇ ⋅ D = 2 rs sin θ  ∂   ∂rs   rs 2 sin θ ⋅ 10 −5 rs    + 0 + 0   3    10 −5 ρ v = ∇ ⋅ D = 2 ∗ 3rs 2 = 10 −5 C / m 3 3rs for rs > a : 2 1  ∂  rs sin θ ⋅10 −5 a 3  ρv = ∇ ⋅ D = 2   +0+0 = 0 2  ∂ r 3rs rs sin θ  s   ρv = ? ρv = ? a Divergence Theorem: ��⃗ . ��⃗ �𝐃 𝒅𝒔 = 𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 = � 𝝆𝒗 𝒅𝒗 From divergence law, So �⃗ = 𝝆𝒗 [𝑪𝒎−𝟑 ] 𝛁. �𝑫 ��⃗ . ��⃗ �𝐃 𝒅𝒔 = 𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 = � 𝝆𝒗 𝒅𝒗 = � ∇. ��⃗ 𝐷 𝒅𝒗 We can transfer the surface integral into a volume integral. For the left-hand side to be equal the right hand side of divergence theorem, the following conditions must be fulfilled: “��⃗ 𝐷 Must be well behaved within the volume v and on the surface” Note: �⃗ and ∇. 𝐷 �⃗ are Well behaved means that 𝐷 continuous and defined (not infinite). Example: 10 x 3 Given D = 3 xˆ evaluate both sides of the divergence theorem for the volume of cube 2m on edge centered at the origin and with edges parallel to the axis. Solution: z z x ∫ D ⋅ d S = ∫ ∇ ⋅ D dv L.H .S = ∫ D ⋅ dS 1 1 1 1 10 x 3 ˆ 10 x 3 ˆ ˆ =∫∫ X ⋅ dydzX ↓ X =! + ∫ ∫ X ⋅ dydzXˆ ↓ X = −! +0 + 0 3 3 −1 −1 −1 −1 10(1) 40 ⋅2⋅2= C 3 3 40 40 80 = + = C 3 3 3 3 = R.H .S = ∫∇⋅ D dv ∂ 10 x 3  2 ∴∇ ⋅ D =   = 10 x ∂x  3  1  x3  2 ∴ ∫ ∇ ⋅ D dv = ∫ ∫ ∫ 10 x dxdydz = 10   ⋅ 2 ⋅ 2  3  −1 −1 −1 −1 10 [1 + 1]⋅ 2 ⋅ 2 = 80 C = 3 3 1 1 1 Example: 2 5r D = s rˆs 4 Given evaluate both sides of divergence π theorem for volume: r = 4m , θ = 4 z Solution: r = 4m θ θ= 45o y x ∫ D ⋅ d S = ∫ ∇ ⋅ D dv L.H .S = ∫ D ⋅ d S π 2π 4 2 5r 2 = ∫ ∫ s rˆs ⋅ rˆs rs sin θdθdφ ↓ rs = 4 4 0 0 π π 5(4 ) 4 5(4 ) (− cos θ )04 = ⋅ ∫ sin θdθ ⋅ ∫ dφ = 4 0 4 0 4 = 589.1 C 2π 4 ∇⋅D = = rs 2 2 4rs  ∂  2 1  rs sin θ  ∂ r 4  sin θ s  5 2 ⋅ 4rs = 5rs 2 4rs π RHS = ∫ ∇ ⋅ D dv = 5rs = 4 4 2π 4 4 ∫ ∫ ∫ 5rs ⋅ r 2 sin θdrs dθ dφ 0 0 0 π (− cos θ )04 ⋅ 2π = 589.1C

Top subcategories

Top subcategories

Top subcategories

Top subcategories

Top subcategories

Top subcategories

Top subcategories

Top subcategories

Download CHAPTER (3) ELECTRIC FLUX DENSITY