Chapter 1 Electrostatics PDF

Chapter 1 Electrostatics Dr/ Ashraf selim  What is the electrostatic? Electrostatics is the study of electric charges at rest (not move) under the action of electric forces. Methods of charging i. Friction or rubbing between dielectric materials. ii. Induction method apply only for conducto...

Chapter 1 Electrostatics Dr/ Ashraf selim  What is the electrostatic? Electrostatics is the study of electric charges at rest (not move) under the action of electric forces. Methods of charging i. Friction or rubbing between dielectric materials. ii. Induction method apply only for conductors. iii.Conduction method apply only for conduction. Grounding (the removal of a charge) - the process of removing the excess charge on an object by another object of substantial size. Types of electric charges 1-Positive charge (+Q); The substances that loss electrons. [e.g. a glass rod is rubbed with a silk cloth] 2-Negative charge (-Q); The substances that gain electrons. [e.g. an ebonite rod is rubbed with wool] Charges law “Similar charges repel, and different charges attract” Properties of Electric Charge Quantization of electric charge The substance cannot loss or gain fraction of an electron or more. An electric charge has an integer number(n) of electron charge (e). ∵ 𝐐 = ∓𝒏𝒆 ∵ 𝐞 = 𝟏. 𝟔 × 𝟏𝟎−𝟏𝟗 ∵ 𝐧 = 𝟏, 𝟐, 𝟑, 𝟒 ….. Conservation of Electric Charge ▪ The Electric charges cannot create or destroy, but they may be transferred from substance to another. 92U 238 −−−−→ 90Th234 + 2He4 Additive of Electric Charge The total charge on a body is the algebraic sum of the positive and negative charges. + - + ∵ 𝐐 = 𝒒𝟏 + 𝒒𝟐 + ⋯ - + - + -  Example (1) An electrically neutral penny, of mass 𝒎 = 𝟑. 𝟏𝟏 𝒈, contains equal amounts of positive and negative charge. Assuming the penny is made entirely of copper, what is the magnitude of 𝒒 of the total positive (or negative) charge in the penny? Solution Copper ( 63.5 29𝐶𝑢 ) atom which means that copper has (29) protons, so each atom has charge (𝑞). [ Z = 29]. 𝑸 = ±𝒁𝒆 Then the total positive charge in penny [𝑁 is number of Cu atom]. 𝑸 = 𝑵𝒁𝒆 𝑵 = the number of atoms in a mole (Avogadro′s number) x number of moles of copper in the penny 𝑵𝑨 = 6.2 x 𝟏𝟎𝟐𝟑 (Avogadro′s number) 𝒎 number of moles of copper in the penny = 𝑴 m is mass of penny (𝑀) is the molar mass of copper, 63.5 𝑔/𝑚𝑜𝑙. 𝒎 𝟐𝟑 𝟑. 𝟏𝟏 𝑵 = 𝑵𝑨 = 𝟔. 𝟐𝒙 𝟏𝟎 𝒙 = 𝟐. 𝟗𝟓 𝐱𝟏𝟎𝟐𝟐 𝒂𝒕𝒐𝒎 𝑴 𝟔𝟑. 𝟓 𝑸 = (𝟐. 𝟗𝟓 𝐱𝟏𝟎𝟐𝟐 )( 𝟐𝟗) ( 𝟏. 𝟔𝒙 𝟏𝟎−𝟏𝟗 ) 𝑸 = 𝟏𝟑𝟕𝐱𝟏𝟎𝟑 𝑪 Example (2) Which is bigger a Coulomb or a charge on an electron? How many electrons in one Coulomb of charge? Solution A Coulomb of charge is bigger than the charge of an electron. Magnitude of electron charge (e) = 1.6×l0-19 C. Then the number of electrons in one coulomb. 𝑸 𝟏 𝟏𝟖 𝒆𝒍𝒆𝒄𝒕𝒓𝒐𝒏𝒔 𝒏= = = 𝟔. 𝟐𝟓 × 𝟏𝟎 𝒆 𝟏. 𝟔 × 𝟏𝟎−𝟏𝟗  Example (3): Three small identical balls have charges –3 × 10−12 C, 8 × 10−12 C and 4 × 10−12 C respectively. They are brought in contact and then separated. Calculate (a) charge on each ball (b) number of electrons in excess or deficit on each ball after contact. Solution [q1 = −3 × 10−12 C, q2 = 8 × 10−12 C, q3 = 4 × 10−12 C] They are brought in contact They are brought in contact then separated (a) The charge on each ball q1 + q 2 + q 3 (−3) + (8) + (4) × 10−12 q= = = 3 × 10−12 C 3 3 (b) Since the charge is positive, there is a shortage of electrons on each ball. q 3 × 10−12 7 N= = = 1.875 × 10 e 1.6 × 10−19 Then, the number of electrons = 1.875 × 107 electrons. Example (4) A polythene (Plastic) piece rubbed with wool is found to have a negative charge of 3.2×l0-7 C. Estimate the number of electrons transferred from which to which? is there a transfer of mass from wool to polythene? 𝒎𝒆 = 𝟗. 𝟏𝟏 × 𝟏𝟎−𝟑𝟏 kg Solution; Then, Q = -3.2×l0-7 C, charge on electron (e) = -1.6×l0-19C, number of electrons transferred to plastic piece from wool. 𝑸 −𝟑. 𝟐 × 𝟏𝟎−𝟕 𝟏𝟐 𝒆𝒍𝒆𝒄𝒕𝒓𝒐𝒏𝒔 𝒏= = = 𝟐 × 𝟏𝟎 𝒆 𝟏. 𝟔 × 𝟏𝟎−𝟏𝟗 Yes, there is a transfer of mass, whereas the mass of each electron = 9.11×10-31 kg. Then, total mass that transferred to plastic piece (M) = number of electrons × mass of an electron. 𝑴 = 𝒏 × 𝒎𝒆 = 𝟐 × 𝟏𝟎𝟏𝟐 × 𝟗. 𝟏𝟏 × 𝟏𝟎−𝟑𝟏 = 𝟏. 𝟖 × 𝟏𝟎−𝟏𝟖 𝒌𝒈  Electric Charge Configuration and distribution (i) linear charge distribution; The charge (Q) is uniformly distribute along a line of length (L) 𝑸 𝝀= 𝑳  is maned leaner charge density and its unit is C/m. (ii) Surface Charge distribution; The charge (Q) is uniformly distribute through an area (A) 𝑸 𝝈= 𝑨 () is named surface charge density and its unit is (C/m2). iii) Volume Charge distribution The charge (Q) is uniformly distribute through a volume (V) 𝐐 𝛒= 𝐕  is named volume charge density and its unit is (C/m3). Electrostatic Force (F) [Coulomb’s Law] “The force of attraction or repulsion between two-point charges is directly proportional to their product of them and inversely proportional to the square distance between them”. 𝑸𝟏 𝑸𝟐 𝑭∝ 𝒓𝟐 𝑸 𝟏 𝑸𝟐 ∴𝑭=𝒌 𝒓𝟐 Electrostatic Force (F) [Coulomb’s Law] 𝑸𝟏 𝑸𝟐 𝑸𝟏 𝑸𝟐 𝑭∝ ∴𝑭=𝒌 𝒓𝟐 𝒓𝟐 where k is a dielectric constant. 𝟏 ∴ 𝒌𝒂𝒊𝒓 = = 𝟗 × 𝟏𝟎𝟗 𝑵. 𝒎𝟐 /𝑪𝟐 𝟒𝝅𝜺𝒐 o= 8.854×10−12 C2/N m2. o is an electrical permittivity of free space “The force of attraction or repulsion between two-point charges is directly proportional to their product of them and inversely proportional to the square distance between them”. Example (5) Two equal and similar charges kept 3 cm apart in air repel each other with a force equivalent to weight of mass 4.5 kg. Find charges in coulomb Solution Let each charge be q, then q1=q2=q and r = 3 cm = 0.03 m, 𝒒𝟏 𝒒𝟐 𝒒𝟐 ∴𝑭=𝒌 𝟐 =𝒌 𝟐 =𝐦×𝐠 F = m × g = 4.5×9.8 N 𝒓 𝒓 𝑭𝒓𝟐 𝑭𝒓𝟐 ∴ 𝒒𝟐 = ∴𝒒= 𝒌 𝒌 𝟒. 𝟓 × 𝟗. 𝟖 × 𝟎. 𝟎𝟑 𝟐 ∴ 𝒒𝟐 = ∴ 𝐪 = ±𝟐. 𝟏 × 𝟏𝟎−𝟏𝟒 𝐂 𝟗 × 𝟏𝟎𝟗 Example (6) An electron and a proton are at a distance of 10-9m from each other in a free space. Compute the force between them. Solution Given, q1 = charge of an electron = -1.6×10 -19 C, q2 = charge of proton =1.6×10-19C and r =10-9m 𝒒𝟏 𝒒𝟐 𝒒𝟐 ∵𝑭=𝒌 𝟐 =𝒌 𝟐 𝒓 𝒓 −𝟏𝟗 𝟐 𝟏.𝟔×𝟏𝟎 F= 𝟗 × 𝟏𝟎𝟗 𝟐 𝟏𝟎−𝟗 𝐅 = 𝟐𝟑. 𝟎𝟒 × 𝟏𝟎−𝟏𝟏 𝑵N Example (7): Two insulated charged spheres of charges 6.5×10−7 C each are separated by a distance of 0.5 m. Calculate the electrostatic force between them. Also calculate the force (i) when the charges are doubled and the distance of separation is halved. (ii) when the charges are placed in a dielectric medium water ( = 80) Solution Given, q1 = q2 = 6.5 × 10−7C, r = 0.5 m 𝟐 𝟗 −𝟕 𝟐 𝒒 𝟗 × 𝟏𝟎 𝟔. 𝟓 × 𝟏𝟎 −𝟐 𝐅=𝑲 𝟐 = 𝟐 = 𝟏. 𝟓𝟐 × 𝟏𝟎 𝐍 𝐫 𝟎. 𝟓 (i) If the charge is doubled and separation between them is halved then, (𝟐𝒒)𝟐 𝒒𝟐 𝑭𝟏 = 𝑲 𝟐 = 𝟏𝟔 𝒌 𝟐 = 𝟎. 𝟐𝟒 𝑵 𝒓 𝐫 𝟐 (ii) When placed in water of 𝜿 = 80 𝑭 𝟏. 𝟓𝟐 × 𝟏𝟎−𝟐 𝑭𝟐 = = = 𝟏. 𝟗 × 𝟏𝟎−𝟒 𝑵 𝜿 𝟖𝟎 Example (8): Compare the magnitude of the electrostatic and gravitational force between an electron and a proton at a distance r apart in hydrogen atom. (Given: me = 9.11 × 10−31 kg, mP = 1.67 × 10−27 kg, G = 6.67 × 10−11 Nm2 kg−2; e = 1.6 × 10−19 C) Solution: The gravitational force 𝒎𝑷 𝒎𝒆 −𝟏𝟏 𝐍𝒎 𝟐 𝟏. 𝟔𝟕 × 𝟏𝟎−𝟐𝟕 𝑲𝒈 𝟗. 𝟏𝟏 × 𝟏𝟎−𝟑𝟏 𝑲𝒈 −𝟒𝟕 𝑭𝒈 = 𝐆 = 𝟔. 𝟔𝟕 × 𝟏𝟎 = 𝟑. 𝟔𝟑 × 𝟏𝟎 𝐍 𝒓𝟐 𝑲𝒈𝟐 𝟓. 𝟐𝟗 × 𝟏𝟎−𝟏𝟏 𝒎 𝟐 The electrostatic force 𝒒𝑷 𝒒𝒆 𝐍𝒎 𝟐 (𝟏. 𝟔𝟎 × 𝟏𝟎−𝟏𝟗 𝑪)𝟐 𝑭𝒆 = 𝐤 𝟐 = 𝟗 × 𝟏𝟎𝟗 𝟐 = 𝟖. 𝟐𝟑 × 𝟏𝟎 −𝟖 𝐍 𝒓 𝑪 𝟓. 𝟐𝟗 × 𝟏𝟎−𝟏𝟏 𝒎 𝟐 𝑭𝒆 𝟖. 𝟐𝟑 × 𝟏𝟎−𝟖 𝑵 𝟑𝟗 = = 𝟐. 𝟐𝟕 × 𝟏𝟎 𝑭𝒈 𝟑. 𝟔𝟑 × 𝟏𝟎−𝟒𝟕 𝑵 This shows that the electrostatic force is 2.27 × 1039 times stronger than gravitational force.  ‫?‪Where the net electric force is equal zero‬‬ ‫‪When you have two charges and you want to know where you can put a third charge in order‬‬ ‫‪that the net force on this third charge equal zero‬‬ ‫ اذا كانت اشارة الشحنتين متشابهة يتم وضع الشحنة الثالثة بينهما على الخط الواصل بين الشحنتين‬ ‫ اذا كانت اشارة الشحنتين مختلفة يتم وضع الشحنة الثالثة خارجهما على الخط الواصل بين الشحنتين واقرب للصغرى فى‬ ‫المقدار بدون اشارة‬ (i) If (𝐹21) is the force exerted on charge (𝑞2 ) by charge (𝑞1 ) (ii) If (𝐹12) is the force exerted on (𝑞1 ) due to (𝑞2 ); 𝐹21 and 𝐹12 has the same magnitude but oppositely directed. Example (9): Two-point charges +9e and +1e are kept at a distance of 16 cm from each other. At what point between these charges, should a third charge q to be placed so that it remains in equilibrium? Solution r = 16 cm, q1 = 9e and q2 = e Let a third charge q be kept at a distance x from + 9e and (r – x) from + e 𝒒𝟏 𝒒𝟐 𝑭=𝒌 𝟐 𝒓 𝑭+𝟗𝒆,𝒒 = 𝑭𝒒,𝒆 𝟗𝒆 𝒒 𝒒 𝒆 𝒌 =𝒌 𝒙𝟐 (𝒓 − 𝒙)𝟐 𝟗 𝟏 𝒙𝟐 = ∴ =𝟗 Take the root of both sides 𝒙𝟐 (𝒓 − 𝐱)𝟐 𝒓−𝒙 𝟐 𝒙 𝒙 =𝟑 ∴ =𝟑 x = 12 cm (𝒓 − 𝐱) (𝟏𝟔 − 𝐱) The third charge should be placed at a distance of 0.12 m from charge 9e.  Principle of Superposition The principle of superposition is to calculate the electric force experienced by a charge q1 due to other charges q2, q3 ……. qn. The total force on a given charge is the vector sum of the forces exerted on it due to all other charges. The force on q1 due to q2 𝒒 𝟏 𝒒𝟐 𝒒𝟏 𝒒𝟑 𝒒𝟏 𝒒𝟒 ∴𝑭=𝒌 + + +⋯ 𝒓𝟐𝟐𝟏 𝒓𝟐𝟑𝟏 𝒓𝟐𝟒𝟏  Steps to solve the problems of multi-charges exist at one plane Detriment the charge that calculating the resultant force act on it (from text of problem) ‫نحدد الشحنة المراد حساب محصلة القوي الكهربية عندها من نص المسألة‬ Plot the origin point of X-Y axis at its position. ‫نرسم محوري س وص عند هذه الشحنة‬ Apply the charge law (like charge repel and unlike charge attract) between other charge and the desired charge. (i.e find the direction of force component Fx and Fy. ‫نطبق قانون التنافر والتجاذب بين الشحنات الكهربية األخري و هذه الشحنة لتحدد المركبات األفقية والرأسية للمحصلة القوة الكهربية‬ Calculate the horizontal (Fx) and vertical (Fy) component of electric forces values. ‫نحسب عدديا المركبات األفقية والرأسية للمحصلة القوة الكهربية بإستخدام قانون كولوم‬ Calculate the magnitude of resultant force by using this relation 𝑭= 𝑭𝟐𝒙 + 𝑭𝟐𝒚 Example (9) Figure 5 shows an arrangement of six fixed charged particles, where a = 2.0 cm and θ = 30o. All six particles have the same magnitude of charge, q = 3.0 x 10-6 C: Their electrical signs are as indicated. What is the net electrostatic force acting on q1 due to the other charges? Solution  Y-components of F13 and F15 are equal and opposite direction. ∴ 𝑭𝒀 = 𝑭𝟏𝟑 𝐜𝐨𝐬 𝜽𝒐 − 𝑭𝟏𝟓 𝐜𝐨𝐬 𝜽𝒐 X-components are identical in magnitude and both point in the direction of decreasing x. 𝑭𝒙 = 𝑭𝟏𝟐 − 𝑭𝟏𝟒 + 𝑭𝟏𝟔 − 𝑭𝟏𝟑 𝒔𝒊𝒏 𝜽 + 𝑭𝟏𝟓 𝒔𝒊𝒏 𝜽 q1 = q6 and θ = 30o, we find 𝒒𝟏 𝒒𝟐 𝒒𝟐 𝟏 𝒒𝟐 𝑭 𝑭𝟏𝟐 = 𝑭𝟏𝟒 =𝒌 𝟐 =𝒌 𝟐= 𝒌 𝟐= 𝟐𝒂 𝟒𝒂 𝟒 𝒂 𝟒 𝒒𝟏 𝒒𝟑 𝒒𝟐 𝑭𝟏𝟑 = 𝑭𝟏𝟓 = 𝑭𝟏𝟔 =𝒌 𝟐 =𝒌 𝟐=𝑭 𝒂 𝒂  Y-components of F13 and F15 are equal and opposite direction. 𝑭𝒚 = 𝑭𝐜𝐨𝐬 𝟑𝟎 − 𝑭 𝐜𝐨𝐬 𝟑𝟎 = 𝒛𝒆𝒓𝒐 X-components are identical in magnitude, and both point in the direction of decreasing x. 𝑭 𝑭 𝟏 𝑭𝒙 = − + 𝑭 − 𝑭 𝒔𝒊𝒏 𝟑𝟎 + 𝑭 𝒔𝒊𝒏 𝟑𝟎 = 𝑭 − 𝟐 𝒔𝒊𝒏 𝟑𝟎 = 𝑭 − 𝟐𝑭𝒙 = 𝑭 − 𝑭 = 𝒛𝒆𝒓𝒐 𝟒 𝟒 𝟐 𝑭𝒕 = 𝑭𝒙 + 𝑭𝒚 = 𝒁𝒆𝒓𝒐  Example (10) What are the horizontal and vertical components of the resultant electrostatic force on the charge in the lower left corner of the square if q = 0.1 µC and a = 5 cm. Calculate electric force between different charge and charge 2q that lies in lower left corner of square using coulomb’s law Solution The horizontal component of electric force act on the lower left charge. ෍ 𝑭𝒙 = 𝑭𝟏𝟐 + 𝑭𝟏𝟑 𝒄𝒐𝒔 𝟒 𝟓° = The vertical component of electric force act on the lower left charge. ෍ 𝑭𝒚 = 𝑭𝟏𝟑 𝒔𝒊𝒏 𝟒 𝟓° − 𝑭𝟏𝟒 the repulsion force between charge 2q and q at the corner of square along Y-axis −𝟔 )𝟐 (𝟐𝒒)(𝒒) (𝟐)(𝟎. 𝟏 × 𝟏𝟎 𝑭𝟏𝟐 =𝒌 𝟐 = 𝟗 × 𝟏𝟎𝟗 𝟐 = 𝟎. 𝟎𝟕𝟐 𝑵 𝒂 (𝟎. 𝟎𝟓) The attraction force between charge 2q and q in the corner on the diagonal of the square −𝟔 )𝟐 (𝟐𝒒)(𝒒) 𝟗 (𝟐)(𝟎. 𝟏 × 𝟏𝟎 𝑭𝟏𝟐 = 𝒌 𝟐 = 𝟗 × 𝟏𝟎 = 𝟎. 𝟎𝟑𝟔 𝑵 𝒂 𝒂 (𝟎. 𝟎𝟓 𝟐)𝟐 The attraction force between charge 2q and -2q in the corner on horizontal axis of the square. −𝟔 )𝟐 (𝟐𝒒)(𝟐𝒒) 𝟗 (𝟒)(𝟎. 𝟏 × 𝟏𝟎 𝑭𝟑 = 𝒌 𝟐 = 𝟗 × 𝟏𝟎 𝟐 = 𝟎. 𝟏𝟒𝟒 𝑵 𝒂 (𝟎. 𝟎𝟓) The horizontal component of electric force act on the lower left charge. ෍ 𝑭𝒙 = 𝑭𝟏𝟐 + 𝑭𝟏𝟑 𝒄𝒐𝒔 𝟒 𝟓° = 𝟎. 𝟏𝟒𝟒 + 𝟎. 𝟎𝟑𝟔 × 𝐜𝐨𝐬 𝟒𝟓𝒐 = 𝟎. 𝟏𝟔𝟗 𝑵 The vertical component of electric force act on the lower left charge. ෍ 𝑭𝒚 = 𝑭𝟏𝟑 𝒔𝒊𝒏 𝟒 𝟓° − 𝑭𝟏𝟒 = 𝟎. 𝟎𝟑𝟔 𝒔𝒊𝒏 𝟒 𝟓° − 𝟎. 𝟎𝟕𝟐 = 𝟎. 𝟎𝟒𝟕 𝑵 The resultant electrostatic force on the charge in the lower left corner of the square 𝑭= 𝑭𝒙 + 𝑭𝒚 = 𝟎. 𝟏𝟔𝟗 + 𝟎. 𝟎𝟒𝟕 = The resultant electrostatic force on the charge in the lower left corner of the square 𝑭𝒚 𝟎. 𝟎𝟒𝟕 𝐹𝑡 𝜽= tan−𝟏 = tan −𝟏 𝐹𝑦 𝑭𝒙 𝟎. 𝟏𝟔𝟗 𝛉 𝐹𝑥  Example (11) Two tiny conducting balls of identical mass m and identical charge q hang from non-conducting threads of length L. Assume that θ is so small that tan θ can be replaced by sin θ; show that, for equilibrium, 𝐪𝟐 𝐋 𝟏 𝐗=( )𝟑 𝟐𝛑𝛆𝐨 𝐦𝐠 Solution The forces effected on each ball (i) tension force (T) in non-conducting thread that make an angle q with vertical plane, (ii) (ii) weight of ball and its direction downward and (iii) (iii) repulsion electric force between the charged ball. at equilibrium the net forces act on each ball equal zero, therefore the horizontal component of resultant fore is zero 𝒒𝟐 ෍ 𝑭𝒙 = 𝟎 ⇒ 𝑻 𝐬𝐢𝐧 𝜽 = 𝒌 𝟐 𝒙 Also, the vertical component of resultant force is zero, ෍ 𝑭𝒚 = 𝟎 ⇒ 𝑻 𝐜𝐨𝐬 𝜽 = 𝒎𝒈 Divided the both equations, we get 𝒌𝒒𝟐 𝒒𝟐 𝐭𝐚𝐧 𝜽 = = 𝒎𝒈𝒙𝟐 𝟒𝝅𝜺𝟎 𝒎𝒈𝒙𝟐 From statement of problem, we get 𝒙 𝒒𝟐 𝐬𝐢𝐧 𝜽 ≅ 𝐭𝐚𝐧 𝜽 = = 𝟐𝑳 𝟒𝝅𝜺𝟎 𝒎𝒈𝒙𝟐 𝟏 𝒒𝟐 𝑳 𝟑 𝒙= 𝟐𝝅𝜺𝟎 𝒎𝒈  The Electric Field Intensity (E) Electric field : The region around the electric charge in which the electric force act The Electric Field Intensity (E) Electric field at a point is measured in terms of electric force (F) and the unit of charge(qo) at certain point. The Electric Field Intensity (E) It is electric force act on unit positive test charge +𝒒𝒐 placed at that point  E is a vector quantity which has magnitude and direction. 𝑭 𝑬= 𝒒𝒐 The S.I unit of E is Newton per Coulomb (N/C). The electric force done by an electric field (E) on a charge (qo) is 𝑭 = 𝐪𝐨 𝑬  Electric field due to a point charge Let (Q) is a point charge and (qo) is a test charge is placed at point P at a distance r from Q. According to Coulomb’s law, the force acting on qo due to Q is 𝑸 × 𝒒𝒐 𝑭=𝒌 𝒓𝟐 The electric field at a point P is, by definition, the force per unit test charge. 𝑸 × 𝒒𝒐 𝑭 𝒌 𝟐 𝑸 𝑬= = 𝒓 =𝒌 𝟐 𝒒𝒐 𝒒𝒐 𝒓 𝑸 𝑬=𝒌 𝟐 𝒓  Electric Lines of Force: Electric field lines are an imaginary straight or curved path along which a unit positive charge tends to move in an electric field. Properties of Electric field Lines : 1. The (E) lines get out from +Q but get into through -Q 2. Lines of E start from +Q and terminate at -Q. 3. The tangent to a line at any point gives the direction of field (E) at a point. 4. these lines cannot intersect. and Crowded near to charges but divergent away of them. Calculation the net field (E) of two charges The field at pint (B) Between two charges 𝑬𝑩 = 𝑬 + + 𝑬− The field at pint (L) left two charges 𝑬𝑳 = 𝑬− − 𝑬+ The field at pint (R) right two charges 𝑬𝑹 = 𝑬+ − 𝑬− Calculation the net field (E) of two charges The field at pint (B) Between two charges 𝑬𝑩 = 𝑬𝟏 − 𝑬𝟐 The field at pint (L) left two charges 𝑬𝑳 = 𝑬𝟏 + 𝑬𝟐 The field at pint (R) right two charges 𝑬𝑹 = 𝑬𝟏 + 𝑬𝟐 𝑬𝟏 𝑬𝟏 𝑬𝟏 𝑬𝟏 𝑬𝟐 𝑬𝟐 𝑬𝟐 𝑬𝟐 𝑬𝟑 𝑬𝟑 𝑬𝟑 𝑬𝟑 - + - 𝒒𝟏 𝒒𝟐 𝒒𝟑  Electrical Dipole Electrical Dipole two equal and opposite charges separate by distance between them e.g. salt (NaCl), H2O molecule. The axis of dipole; it is straight line pass through the two charges of dipole. Dipole moment (p); it is product of one of its charge and the distance between them 𝑷=𝒒×𝒅 Derivation of Electric Field Due to an Electric Dipole at a Point on its central axis. From figure; 𝒒 𝒒 ∴ 𝑬 = 𝑬+ − 𝑬− = 𝒌 𝟐 − 𝒌 𝟐 𝒓+ 𝒓− 𝒒 𝒒 ∴𝑬=𝒌 𝟐 −𝒌 𝟐 𝒅 𝒅 𝒁− 𝒁+ 𝟐 𝟐 𝒌𝒒 𝟏 𝟏 ∴𝑬 = 𝟐 𝟐 − 𝟐 𝒁 𝒅 𝒅 𝟏− 𝟏+ 𝟐𝒁 𝟐𝒁 𝒅 𝟐 𝒅 𝟐 𝒌𝒒 𝟏+ − 𝟏− 𝟐𝒁 𝟐𝒁 ∴𝑬 = 𝟐 𝒓+ = 𝒛 − 𝒅/𝟐 𝒁 𝒅 𝟐 𝒅 𝟐 𝟏− 𝟏+ 𝒓− = 𝒛 + 𝒅/𝟐 𝟐𝒁 𝟐𝒁  𝟐 𝟐 𝒅 𝒅 𝒅 𝒅 𝒅 𝟏+ + − 𝟏− + 𝟐 𝒒 𝒁 𝒁 𝒁 𝒁 𝒒 𝒁 ∴𝑬 =𝒌 𝟐 =𝒌 𝟐 𝒁 𝟐 𝟐 𝒁 𝟐 𝟐 𝒅 𝒅 𝟏− 𝟏− 𝟐𝒁 𝟐𝒁 ∵ 𝒅 = 𝟏𝟎−𝟔 𝒎 ∴ 𝒁 ⋙ 𝒅 𝟐 𝒅 ≈ 𝒛𝒆𝒓𝒐 𝒂𝒏𝒅 𝒄𝒂𝒏 𝒃𝒆 𝒏𝒆𝒈𝒍𝒆𝒄𝒕 𝒕𝒉𝒊𝒔 𝒕𝒆𝒓𝒎 𝟐𝒁 𝒒𝒅 ∴ 𝑬 = 𝟐𝒌 𝟑 𝒛 ∵ 𝒑 = 𝒒𝒅 𝑷 ∴ 𝑬 = 𝟐𝒌 𝒁𝟑 Derivation the Electric Field Due to an Electric Dipole at a Point on the Equatorial Line. From Figure. 𝒒 𝒒 ∵ 𝑬𝟏 = 𝑬𝟐 = 𝒌 𝟐 ∵ 𝒓𝟐 = 𝒙𝟐 + 𝒂𝟐 ∴ 𝑬𝟐 = 𝑬𝟏 = 𝒌 𝟐 𝒓 (𝒂 + 𝒚𝟐 ) Resolving E1 and E2 the resultant of E at P in vertical axis is zero because (𝑬𝟏 = 𝑬𝟐 ) ∴ 𝐄 = 𝑬𝟏 𝒔𝒊𝒏 𝜽 − 𝑬𝟐 𝒔𝒊𝒏 𝛉 = 𝟎 the resultant E at P in horizontal axis is 𝒂 ∵ 𝒄𝒐𝒔 𝜽 = 𝒓 ∴ 𝑬 = 𝑬𝟏 𝒄𝒐𝒔 𝜽 + 𝑬𝟐 𝒄𝒐𝒔 𝜽 ∵ (𝑬𝟏 = 𝑬𝟐 = 𝐄) 𝒒 𝒂 𝟐𝒂𝒒 ∴ 𝑬 = 𝟐𝑬𝟏 𝒄𝒐𝒔 𝜽 = 𝒌 𝟐 × 𝟐 =𝒌 𝟐 𝒓 𝒓 𝒓 𝟑Τ𝟐 ∵ 𝒓𝟑 = 𝒙𝟐 + 𝒂 𝟐  𝟐𝒂 𝒒 ∴𝑬=𝒌 𝟐 (𝒂 + 𝒚𝟐 )𝟑/𝟐 𝑷 = 𝟐𝒒𝒂 𝑷 𝑬=𝒌 𝟐 (𝒂 + 𝒚𝟐 )𝟑/𝟐 For a dipole, (a) is very small when compared to y i.e. 𝒂 ≪ 𝒚 then, you can neglect (a2) value from above equation 𝑷 𝐄=𝒌 𝟑 𝒚 Notice ; The direction of E is parallel to the axis of the dipole and directed opposite to the direction of dipole moment P.  The Electric Field of a Uniform charged Ring At a point on its central axis Assume a circular ring of radius (a) carries +Q charge that uniformly distributed along its circumference as in Fig. ∴ 𝑸 = 𝝀(𝟐𝝅𝒓)  The electric field at P due to segment of charge 𝒅𝒒 is 𝒅𝒒 𝒅𝑬 = 𝒌 𝟐 𝒓 The electric field component in X-axis is 𝒅𝒒 𝒙 𝒅𝑬𝒙 = 𝒅𝑬. 𝐜𝐨𝐬 𝜽 = 𝒌 𝟐 ∙ 𝒓 𝒓 The electric field component in Y -axis is 𝒙 𝒂 𝑬𝒀 = න 𝒅𝑬𝒚 = 𝒅𝑬. 𝐬𝐢𝐧 𝜽 − 𝒅𝑬. 𝐬𝐢𝐧 𝜽 = 𝟎 𝐜𝐨𝐬 𝜽 = sin 𝜽 = 𝒓 𝒓 Due to symmetry of circular ring, the components of Y-axis are Zero. 𝟏/𝟐 𝟑/𝟐 ∵𝒓= 𝒙𝟐 + 𝒂𝟐 = 𝒙𝟐 + 𝒂𝟐 ∵ 𝒓𝟑 = 𝒙𝟐 + 𝒂 𝟐 Then, the resultant field at P must lie along the X-axis. 𝑸 𝒙 𝑸 𝐱 𝑸 𝑸𝒙 ∴ 𝐄 = න 𝒅 𝑬𝒙 = 𝒌 𝟑 න 𝒅𝒒 = 𝒌 𝑬= 𝒌 𝟐 𝒓 𝟎 𝟑න 𝒅𝒒 𝒙 + 𝒂𝟐 𝟑/𝟐 𝟎 𝒙𝟐 + 𝒂𝟐 𝟐 𝟎  Notice ; The electric field at the center of the ring is zero The direction of E is parallel to the horizontal plane of the ring