General physics book code IPHS 115-1 PDF

Zagazig University Faculty of Science Physics department General Physics Prepared by Physics department Course Code: IPHS 115 Chapter (1) Electrostatics...

Zagazig University Faculty of Science Physics department General Physics Prepared by Physics department Course Code: IPHS 115 Chapter (1) Electrostatics CHAPTER (1) ELECTROSTATICS 1.1 Introduction Electrostatics is the study of electric charge at rest. The charges at rest are generated due to the friction between two insulating bodies, which are rubbed against each other. When a glass rod is rubbed with a piece of silk, the glass rod acquires the property of attracting small pieces of papers towards its. Then, glass rod is said to be charged with electricity. The Phenomenon of electricity was discovered in the year 600 B.C. by a Greek Philosopher "Thales of Miletus". 1.2 Electrical charges and their kinds It is well known that, any material was created from atoms which contain equal amount of positive (protons) and negative (electrons) charges, so the atom is named a neutral charge. If the material has lost or gained an electron or more, hence, it can be named an electric charged object. With the help of several experiments, it can be concluded that static electricity has two types of electric charges. When a glass rod is rubbed with silk, glass rod become a positive charge (+𝑄) and the silk carry a negative charge (−𝑄). On the other hand, when ebonite rod is rubbed with wool, the charge on ebonite is found to be (−𝑄) and wool becomes positively charged (+𝑄). From figure 1.1a the two glass rods (+𝑄) repel each other, similarly two ebonite rods (−𝑄) repel each other as shown in fig 1.1b. But when a glass rod rubbed with silk (+𝑄) it attracts an ebonite rod rubbed with wool (−𝑄). figure 1.1c. Thus, we say that bodies having same kind of charge repel each other, while those with an opposite kind of charge attract each other. Characteristics of electric charge (i) Electric charge is quantized; the materials cannot loss or gain a fraction of an electron. Therefore, the electric charge either positive or negative is exist only as an integer numbers of electron charge (𝑒) i.e. any positive or negative charge (𝑄) can be written as: 𝑸 = ±𝒏𝒆 where (𝑛 = 1, 2, 3, … ) and (𝑒) is a charge of an electron (𝑒 = 6 × 10−19 𝐶𝑜𝑢𝑙𝑜𝑚𝑏 "C" ) Figure 1.1: Attraction and repulsion between electric charges 1 Chapter (1) Electrostatics (ii) Electric charge is additive in nature. It implies that the total charge on an extended body is the algebraic sum of the positive and negative charges located at different points of the body. Therefore, the electric charge is a scalar quantity and The S.I. unit of charge is coulomb and it is represented by (𝐶). (iii) Electric charge is always conserved, means charge can neither be created nor be destroyed in isolation i.e. charge can be created or destroyed only in equal and opposite pairs. Moreover, the charge can be transferred from substance to other as occurring in radioactivity examples. 92U −−−−→ 90Th 238 234 + 2He4 Example (1.1): 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 has equal positive (Protons) and negative (electrons) charge (= ±𝑍𝑒). For copper (𝑍 = 29), which means that copper has (29) protons, so each atom has charge (𝑞). ⟹ 𝑞 = 𝑍𝑒 Then the total positive charge in penny (𝑞 = 𝑁𝑍𝑒), 𝑁 is number of Cu atom. To find (𝑁), we multiply the number of moles of copper in the penny by the number of atoms in a mole (Avogadro's number, 𝑁𝐴 = 6.20 × 1023 𝑎𝑡𝑜𝑚𝑠/𝑚𝑜𝑙). The number of moles of copper in the penny is (𝑚 / 𝑀), where (𝑀) is the molar mass of copper, 63.5 𝑔/𝑚𝑜𝑙. Thus, we have 𝑚 𝑎𝑡𝑜𝑚𝑠 3.11 𝑔 𝑁 = 𝑁𝐴 = [6.02 × 1023 ][ 𝑔 ] 𝑀 𝑚𝑜𝑙 63.5 𝑚𝑜𝑙 𝑁 = 2.95 × 1022 𝑎𝑡𝑜𝑚𝑠 We then find the magnitude of the total positive or negative charge in the penny to be 𝑞 = 𝑁 𝑍 𝑒 = (2.95 × 1022 )(29)(1.6 × 10−19 𝐶) 𝑞 = 137 × 103 𝐶 Example (1.2): 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. The total number (𝑁) of electrons in one coulomb is 𝑞 (1 𝐶) 𝑁= = = 0.625 × 1019 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 𝑒 (1.6 × 10−19 𝐶) 2 Chapter (1) Electrostatics Example (1.3): 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: (a) The charge on each ball 𝑞1 + 𝑞2 + 𝑞3 [(−3 ) + (8) + (4)] × 10−12 𝐶 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 𝐶) 𝑁= = = 1.875 × 107 𝑒 (1.6 × 10−19 𝐶) Then, the number of electrons = 1.875 × 107 electrons. Example (1.4): A polythene (Plastic) piece rubbed with wool is found to have a negative charge of −𝟑. 𝟐 × 𝟏𝟎−𝟕 𝑪. Estimate the number of electrons transferred from which to which? Is there a transfer of mass from wool to polythene? Solution: The number (𝑁) of electrons transferred to plastic piece from wool 𝑞 (−3.2 × 10−7 𝐶) 𝑁= = = 2 × 1012 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 𝑒 (1.6 × 10−19 𝐶) Yes, there is a transfer of mass, whereas the mass of each electron (𝑚𝑒 = 9.11 × 10−31 𝑘𝑔). Then, the total mass (𝑀) that transferred to plastic piece is 𝑀 = 𝑛 𝑚𝑒 = (2 × 1012 )(9.11 × 10−31 𝑘𝑔) = 1.8 × 10−18 𝑘𝑔 1.3 Coulomb’s law of electrostatic In 1785, Coulomb measured the force of attraction or repulsion between two electric charges by using a torsion balance. His observation is known as the Coulomb's Law of electrostatics. It states that; “The force of attraction or repulsion between the two stationary electric charges is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between them” Consider that two charges (𝑞1 ) and (𝑞2 ) at a distance (𝑟) apart as shown in figure 1.2, Then force of attraction or repulsion between the two charges is given by 3 Chapter (1) Electrostatics Figure 1.2: The force between two charges 𝑞1 𝑞2 𝐹∝ 𝑟2 𝑞1 𝑞2 ⟹𝐹=𝐾 2 𝑟 where, (𝐾) is constant of proportionality, its value depends upon the nature of the medium in which two charges are located and also the system of units adopted to measure (𝐹), (𝑞1 ), (𝑞2 ) and (𝑟). In S.I., charge is measured in coulomb (𝐶), Force in Newton (𝑁), and distance in meter (𝑚). So that 1 𝐾= = 8.997 × 109 𝑁. 𝑚2 /𝐶 2 4𝜋𝜀𝑜 where (𝜀𝑜 ) is called absolute electrical permittivity of free space and its equal 8.854 × 10−12 𝐶 2 𝑁 −1 𝑚−2. It is worth to known the physical meaning of (𝜀𝑜 ) , it equal numerically the amount of interaction between the molecules of this medium with electric field that pass through it. Moreover, the modified Coulomb’s law for a medium (e.g. plastic or ceramic) could be written as; 1 𝑞1 𝑞2 𝐹= 4𝜋𝜀𝑟 𝜀𝑜 𝑟 2 The ratio (𝜅 = 𝜀 ⁄𝜀𝑜 ) is called the relative permittivity or dielectric constant of the medium. The value of κ for air or vacuum is unity. Example (1.5): 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, 𝑞1 = 𝑞2 = 𝑞 Then, 𝑞1 𝑞2 𝑞2 𝐹 = 𝑘 2 = 𝑘 2 = mg 𝑟 𝑟 𝑚𝑔𝑟 2 (4.5 𝑘𝑔)(9.8 𝑚/𝑠 2 )(0.03 m)2 2 ⟹𝑞 = = 𝑘 (9 × 109 𝑁𝑚2 /𝐶 2 ) ⟹ 𝑞 = ±2.1 × 10−14 𝐶 Example (1.6): An electron and a proton are at a distance of 𝟏𝟎−𝟗 𝒎 from each other in a free space. Compute the force between them. 4 Chapter (1) Electrostatics Solution: 𝑞1 𝑞2 𝑞2 9 (1.6 × 10−19 𝐶)2 𝐹=𝑘 = 𝑘 = (9 × 10 𝑁𝑚 2 /𝐶 2 ) 23.04 × 10−11 𝑁 𝑟2 𝑟2 (10−9 𝑚)2 Example (1.7): 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 (𝒌 = 80). Solution: 1 𝑞1 𝑞2 (9 × 109 𝑁𝑚2 /𝐶 2 )(6.5 × 10−7 𝐶)2 𝐹= = = 1.52 × 10−2 𝑁 4𝜋𝜀𝑜 𝑟 2 (0.5 𝑚)2 (i) If the charge is doubled and separation between them is halved then, 1 (2𝑞1 )(2𝑞2 ) 1 𝑞1 𝑞2 𝐹1 = 2 = 16 [ ] = 0.24 𝑁 4𝜋𝜀𝑜 𝑟 4𝜋𝜀𝑜 𝑟 2 (2) (ii) When placed in water of (𝜅 = 80) 𝐹 (1.52 × 10−2 𝑁) 𝐹2 = = = 1.9 × 10−4 𝑁 𝜅 80 Example (1.8): 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 (𝐹𝑔 ) between the electron and the proton is found from Newton’s law of universal gravitation, as 𝑚𝑃 𝑚𝑒 𝐹𝑔 = 𝐺 𝑟2 (1.67 × 10−27 𝑘𝑔)(9.11 × 10−31 𝑘𝑔) 𝐹𝑔 = (6.67 × 10−11 𝑁m2 kg −2 ) = (3.63 × 10−47 ) N (5.29 × 10−11 𝑚)2 The electric force (𝐹𝑒 ) between the electron and the proton is found from Coulomb’s law of electrostatics, as 𝑞𝑃 𝑞𝑒 9 2 2) (1.6 × 10−19 𝐶)(1.60 × 10−19 𝐶) 𝐹𝑒 = 𝑘 = (9 × 10 𝑁𝑚 /𝐶 𝑟2 (5.29 × 10−11 𝑚)2 𝐹𝑒 = (8.23 × 10−8 ) N 5 Chapter (1) Electrostatics Although both forces seem quite small, let us compare the relative magnitude of these forces by taking the ratio of the electric force to the gravitational force, that is 𝐹𝑒 (8.23 × 10−8 ) 𝑁 = = (2.27 × 1039 ) 𝐹𝑔 (3.63 × 10−47 ) 𝑁 This shows that the electrostatic force is (2.27 × 1039 ) times stronger than gravitational force. Example (1.9): 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: Let a third charge (𝑞) be kept at a distance (𝑥) from (+ 9𝑒) and (𝑟 – 𝑥) from (+ 𝑒) 1 𝑞1 𝑞2 𝐹= 4𝜋𝜀𝑜 𝑟 2 𝐹+9𝑒,𝑞 = 𝐹𝑞,𝑒 1 (9𝑒)(𝑞) 4𝜋𝜀𝑜 𝑥 2 1 (𝑞)(𝑒) = 4𝜋𝜀 (𝑟 − 𝑥)2 9 1 2 = 𝑥 (𝑟 − 𝑥)2 𝑥2 =9 (𝑟 − 𝑥)2 Take the root of both sides 𝑥 =3 (𝑟 − 𝑥) 𝑥 =3 (16 − 𝑥) 𝑥 = 12 𝑐𝑚 The third charge should be placed at a distance of 0.12 𝑚 from charge (9𝑒). 6 Chapter (1) Electrostatics 1.4 Coulomb’s law – vector form ⃗⃗⃗⃗⃗⃗ (i) If (𝐹 21 ) is the force exerted on charge (𝑞2 ) by charge (𝑞1 ) 𝑞1 𝑞2 ⃗⃗⃗⃗⃗⃗ 𝐹21 = 𝑘 2 𝑟̂12 𝑟12 where (𝑟̂12 ) is the unit vector from (𝑞1 ) to (𝑞2 ) ⃗⃗⃗⃗⃗ (ii) If (𝐹 12 ) is the force exerted on (𝑞1 ) due to (𝑞2 ); 𝑞1 𝑞2 ⃗⃗⃗⃗⃗ 𝐹12 = 𝑘 2 𝑟̂21 𝑟21 where (𝑟̂21 ) is the unit vector from (𝑞2 ) to (𝑞1 ), notice, both (𝑟̂21 ) and (𝑟̂12 ) has the same magnitude but oppositely directed. q1 q 2 q1 q 2 ⃗⃗⃗⃗⃗ 𝐹12 = k ̂12 ) or ⃗⃗⃗⃗⃗ 2 (−r 𝐹12 = −k 2 r̂12 r12 r12 ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ 𝐹12 = −𝐹 21 So, the forces exerted by charges on each other are equal in magnitude and opposite in direction. Principle of superposition The principle of superposition is to calculate the total electric force experienced by a charge say (𝑞1 ) due to other charges 𝑞2 , 𝑞3 … …. 𝑞𝑛. Therefore, ⃗⃗⃗ 𝐹1 = ⃗⃗⃗⃗⃗ 𝐹12 + ⃗⃗⃗⃗⃗ 𝐹13 + ⃗⃗⃗⃗⃗ 𝐹14 + ⋯.. ⃗⃗⃗⃗⃗⃗ 𝐹1𝑛 = or 1 𝑞1 𝑞2 𝑞1 𝑞3 𝑞1 𝑞4 𝑞1 𝑞𝑛 ⃗⃗⃗ 𝐹1 = [ 2 r̂12 + 2 r̂13 + 2 r̂14 + ⋯ + 2 r̂1n ] 4𝜋𝜀𝑜 𝑟12 𝑟13 𝑟14 𝑟1𝑛 Steps to solve the problems of multi-charges exist at one plane 1. 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 2. Plot the origin point of (𝑥 − 𝑦)-axis at its position. ‫نرسم محوري س و ص عند هذه الشحنة‬ 3. 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 األخرى وهذه الشحنة لنحدد المركبات األفقية والرأسية لمحصلة القوة الكهربية‬ 4. Calculate the horizontal (𝐹𝑥 ) and vertical (𝐹𝑦 )component of electric forces values..‫نحسب عدد يا المركبات األفقية والرأسية لمحصلة القوة الكهربية باستخدام قانون كولوم‬ 5. Calculate the magnitude of resultant force by using this relation 𝐹 = √𝐹𝑥2 + 𝐹𝑦2 7 Chapter (1) Electrostatics Example (1.10): The figure shows an arrangement of six fixed charged particles, where 𝒂 = 𝟐. 𝟎 𝒄𝒎 and 𝜽 = 𝟑𝟎°. All six particles have the same magnitude of charge, 𝒒 = 𝟑. 𝟎 × 𝟏𝟎−𝟔 𝑪: Their electrical signs are as indicated. What is the net electrostatic force acting on (𝒒𝟏 ) due to the other charges? Solution: 1 𝑞1 𝑞2 𝐹12 = 𝐹14 = 4𝜋𝜀𝑜 (2𝑎)2 1 𝑞1 𝑞3 𝐹13 = 𝐹15 = 𝐹16 = 4𝜋𝜀𝑜 𝑎2 The above figure is a free body diagram for (𝑞1 ). It is cleared that 𝐹12 and (𝐹14 ) are equal in magnitude but opposite in direction: thus, those forces cancel. Inspection of above figure. reveals that the 𝑦-components of (𝐹13 ) and (𝐹15 ) are equal and opposite direction therefore they cancel each other. 𝐹13 cos 30𝑜 − 𝐹15 cos 30𝑜 = 0 And that their 𝑥 components are identical in magnitude and both points in the direction of decreasing (𝑥). The figure shows also shows that (𝐹16 ) points in the direction of increasing (𝑥). thus (𝐹1 ) must be parallel to the 𝑥 − axis, its magnitudes are the difference between (𝐹16 ) and twice the 𝑥-component of (𝐹13 ): 8 Chapter (1) Electrostatics 𝐹1 = 𝐹16 − 2𝐹13 𝑠𝑖𝑛 𝜃 1 𝑞1 𝑞6 2 𝑞1 𝑞3 = − 𝑠𝑖𝑛 𝜃 4𝜋𝜀𝑜 𝑎2 4𝜋𝜀𝑜 𝑎2 Setting (𝑞1 = 𝑞6 ) and (𝜃 = 30°), we find 1 𝑞1 𝑞6 2 𝑞1 𝑞2 𝐹1 = 2 − 𝑠𝑖𝑛 30𝑜 = 0 4𝜋𝜀𝑜 𝑎 4𝜋𝜀𝑜 𝑎2 Note that the presence of (𝑞6 ) along the line between (𝑞1 ) and (𝑞4 ) does not in any way alter the electrostatic force exerted by (𝑞4 ) on (𝑞1 ). Example (1.11): 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. Solution: the repulsion force between charge (2𝑞) and (𝑞) at the corner of square along 𝑦 − 𝑎𝑥𝑖𝑠 (2𝑞)(𝑞) 𝐹1 = 𝑘 𝑎2 2(0.1 × 10−6 𝐶)2 𝐹1 = (9 × 109 𝑁𝑚2 /𝐶 2 ) ( ) = 0.072 𝑁 (0.05 𝑚)2 the attraction force between charge (2𝑞) and (𝑞) in the corner on the diagonal of the square. (2𝑞 )(𝑞 ) 𝐹2 = 𝑘 2 (𝑎√2) 2(0.1 × 10−6 𝐶)2 𝐹2 = (9 × 109 𝑁𝑚2 /𝐶 2 ) ( 2 ) = 0.036 𝑁 (0.05√2 𝑚) the attraction force between charge (2𝑞) and (−2𝑞) in the corner on horizontal axis of the square. 9 Chapter (1) Electrostatics (2𝑞)(2𝑞) 𝐹3 = 𝑘 (𝑎)2 9 2 4(0.1 × 10−6 𝐶)2 2 𝐹3 = (9 × 10 𝑁𝑚 /𝐶 ) ( ) = 0.144 𝑁 (0.05 𝑚)2 The horizontal component of electric force act on the lower left charge. Force component ∑ 𝐹𝑥 = 𝐹3 + 𝐹2 𝑐𝑜𝑠 4 5° = (0.144 𝑁) + (0.036 cos 45𝑜 𝑁) = 0.169 𝑁 The vertical component of electric force act on the lower left charge. ∑ 𝐹𝑦 = 𝐹2 𝑠𝑖𝑛 4 5° − 𝐹1 = (0.036 𝑠𝑖𝑛 45° 𝑁) − (0.072 𝑁) = 0.047 𝑁 Example (1.12): 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 (𝜃) with vertical plane, (ii) weight of ball and its direction downward and (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 𝑘𝑞 2 𝑞2 tan 𝜃 = = 𝑚𝑔𝑥 2 4𝜋𝜀0 𝑚𝑔𝑥 2 From statement of problem, we get 𝑥 𝑞2 𝑥3 𝑞2 sin 𝜃 ≅ tan 𝜃 = = ⟹ = 2𝐿 4𝜋𝜀0 𝑚𝑔𝑥 2 2𝐿 4𝜋𝜀0 𝑚𝑔 1 𝑞2𝐿 3 𝑥=( ) 2𝜋𝜀𝑜 𝑚𝑔 10 Chapter (1) Electrostatics 1.5 Electric field (𝑬) The space around an electrical charge in which its effect can be displayed in this region is known as electric field region. Consider an electric charge (𝑸) located in space. If we bring another charge (𝒒𝒐 ) near the charge (𝑸), it will acquire a force of attraction or repulsion due to presence of charge (𝒒𝒐 ). The force exerted on charge (𝒒𝒐 ) is named electric field intensity or strengthen that done on charge (𝒒𝒐 ). Electric field intensity The electric field intensity or strength of electric field at a point may be defined as “It is electric force act on unit positive test charge +𝒒𝒐 placed at that point”. If (𝐹) is the force acting on test charge (+𝑞𝑜 ) at any point a, then electric field intensity may be given by. 𝐹 𝐸⃗ = 𝑞𝑜 (𝐸) is a vector quantity which has a magnitude and direction, and the S.I unit of (𝐸) is Newton per Coulomb (𝑁/𝐶). Then, the electric force done by an electric field (𝐸) on a charge (𝑞𝑜 ) is 𝐹 = 𝑞𝑜 𝐸⃗ 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 𝑞𝑞𝑜 𝐹=𝑘 2 𝑟 The electric field at a point (𝑃) is, by definition, the force per unit test charge. 𝐹 1 𝑞 𝐸= = 𝑞𝑜 4𝜋𝜀𝑜 𝑟 2 Lines of electric field: Electric field lines are an imaginary straight or curved path along which a unit positive charge tends to move in an electric field. Hence, the electric fields due to simple arrangements of Figure 1.3: The electric field lines for a positive point point charges are shown in figure 1.3. charge, the lines are directed radially outward and are directed radially inward for a negative point charge 11 Chapter (1) Electrostatics Properties of electric field lines: 1. Electric field lines get out from (+𝑄) but get into through (−𝑄) 2. Lines of force start from (+𝑄) and terminate at (−𝑄). 3. Electric Lines of forces are imaginary. 4. The tangent to a line of force at any point gives the direction of the electric field (𝐸) at that point. 4. These lines cannot intersect, and crowded near to charges but divergent away of them. Electric dipole  It is a molecule (figure 1.4) has two charges have same value but one of them is positive (+𝑞) and other is negative (−𝑞) as well as the distance between them is (𝑑 = 2𝑎) e.g. salt (𝑁𝑎𝐶𝑙) molecule and water vapor 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 = 𝒒𝒅 Figure 1.4: An electric dipole Electric field due to an electric dipole at a point on its axis As shown in figure 1.5 the total electric field is: 𝑞 𝑞 𝐸 = 𝐸+ − 𝐸− = 𝑘 2 − 𝑘 2 𝑟+ 𝑟− 𝑞 𝑞 =𝑘 − 𝑘 (𝑧 − 𝑑/2)2 (𝑧 + 𝑑/2)2 𝑞 𝑑 −2 𝑑 −2 = 𝑘 2 [(1 − ) − (1 + ) ] 𝑧 2𝑧 2𝑧 We can use this approximation and use the binomial equation of negative power 𝑑 −2 𝑑 −2 𝑑 𝑑 2𝑑 (1 − ) − (1 + ) = [(1 + − ⋯ ) − (1 − + ⋯ )] ≈ 2𝑧 2𝑧 𝑧 𝑧 𝑧 Figure 1.5: Electric field due to an electric dipole at a point on its Therefore, axis 𝑞 𝑑 𝑞𝑑 𝑷 𝐸 = (2𝑘 ) ( ) = 2𝑘 = 𝟐𝒌 𝑧2 𝑧 𝑧3 𝒛𝟑 12 Chapter (1) Electrostatics Electric field due to an electric dipole at a point on the equatorial line. For the dipole shown in figure 1.6.  The electric field (𝐸1 ) at a point (𝑃) due to the charge (+𝑞) of the dipole, 1 𝑞 𝐸1 = 4𝜋𝜀𝑜 (𝑎2 + 𝑦 2 ) Notice (𝑟 = (a2 + y 2 )1/2 )  The electric field (𝐸2 ) at a point (𝑃) due to the charge (−𝑞) of the dipole 1 𝑞 𝐸2 = 4𝜋𝜀𝑜 (𝑎2 + 𝑦 2 ) Figure 1.6: Electric field due to  The magnitudes of (𝐸1 ) and (𝐸2 ) are equal. an electric dipole at a point on the  Resolving (𝐸1 ) and (𝐸2 ) into their horizontal and vertical equatorial line. components as in figure1.6.  the resultant electric field at the point (𝑃) due to the dipole in vertical axis is zero because (𝐸1 = 𝐸2 = 𝐸) ⟹ 𝐸 = 𝐸1 𝑠𝑖𝑛 𝜃 − 𝐸2 𝑠𝑖𝑛 𝜃 = 0  the resultant electric field at the point (𝑃) due to the dipole in horizontal axis is ⟹ 𝐸 = 𝐸1 𝑐𝑜𝑠 𝜃 + 𝐸2 𝑐𝑜𝑠 𝜃 (𝐸1 = 𝐸2 = 𝐸) 𝐸 = 2𝐸1 𝑐𝑜𝑠 𝜃 1 𝑞 𝐸=2 𝑐𝑜𝑠 𝜃 4𝜋𝜀𝑜 (𝑎 + 𝑦 2 ) 2 𝑎 𝐵𝑢𝑡, 𝑐𝑜𝑠 𝜃 = √(𝑎2 + 𝑦 2 ) 1 𝑞 𝑎 𝐸=2 4𝜋𝜀𝑜 (𝑎2 + 𝑦 2 ) √(𝑎2 + 𝑦 2 ) 1 2𝑞𝑎 𝐸= 3 4𝜋𝜀𝑜 (𝑎2 + 𝑦 2 )2 1 𝑃 𝐸= 3 4𝜋𝜀𝑜 (𝑎2 + 𝑦 2 )2  where (𝑃) is the electric dipole moment (𝑃 = 2𝑞𝑎). 13 Chapter (1) Electrostatics  For a dipole, (𝑎2 ) is very small when compared to y i.e. (𝑎 ≪ 𝑦) then, you can neglect (𝑎2 ) value from above equation 1 P P E= 𝑜𝑟 𝐸=𝑘 4πεo y 3 y3  The direction of (𝐸) is parallel to the axis of the dipole and directed opposite to the direction of dipole moment (𝑃). Electric charge configuration and distribution (i) Point charge that represented by (𝑸) (ii) Linear charge distribution When the distribution of charge is uniformly along a line, Q λ= L where (𝑄) is total charge distributed over a long conductor of length (𝐿) and (𝜆)is leaner charge density, so the S.I unit of (𝜆)is 𝐶/𝑚. Therefore, the small amount of charge (𝑑𝑞) on this element (𝑑𝐿) is 𝑑𝑞 = 𝜆 𝑑𝐿 (iii) Surface charge distribution When the distribution of charge is uniformly over a particular area, Q σ= A where (𝑄) is total charge distributed over a particular area (𝐴) and is (𝜎) surface charge density, so the S.I unit of (𝜎) is (𝐶/𝑚2 ). (iv) Volume charge distribution When the distribution of charge is uniformly through a certain volume,  𝑄 𝜌= 𝑉 where (𝑄) is total charge distributed through a particular volume (𝑉) and (𝜌) is volume charge density, so the S.I unit of (𝜌) is (𝐶/𝑚3 ). 14 Chapter (1) Electrostatics The electric field of a uniform charged ring along its central axis Assume a circular ring of radius (𝑎) carries (+𝑄) charge that uniformly distributed along its circumference as in figure 1.7. 𝑄 = 𝜆(2𝜋𝑟) The electric field at (𝑃) due to segment of charge 𝑑𝑞 = 𝜆𝑑𝑙 is 𝑑𝑞 1 𝜆 × 𝑑𝑙 𝑑𝐸 = 2 = Figure 1.7: The electric field of a uniform charged ring 4𝜋𝜖𝑜 𝑟 4𝜋𝜖𝑜 𝑟 2 along its central axis. The electric field component in 𝑥 − 𝑎𝑥𝑖𝑠 is 𝐸𝑥 = 𝑑𝐸 cos 𝜃 The electric field component in 𝑦 − 𝑎𝑥𝑖𝑠 is 𝐸𝑦 = 𝑑𝐸 𝑠𝑖𝑛𝜃 Due to symmetry of circular ring, the components of y-axis are Zero. (The perpendicular components of the field created by any charge element are canceled by the perpendicular component created by an element on the opposite side of the ring as in figure). 𝐸𝑦 = 𝑑𝐸 𝑠𝑖𝑛𝜃 − 𝑑𝐸 𝑠𝑖𝑛𝜃 = 0 Then, the resultant field at (𝑃) must lie along the 𝑥 − 𝑎𝑥𝑖𝑠. 1 Notice that, from figure, 𝑟 = (𝑥 2 + 𝑎2 )2 and 𝑐𝑜𝑠 𝜃 = 𝑥/𝑟, we find that 𝑑𝑞 𝑥 𝑥 𝑑𝑞 𝑑𝐸𝑥 = 𝑑𝐸 𝑐𝑜𝑠 𝜃 = 𝑘. ( ) = 𝑘 𝑟2 𝑟 𝑟3 All segments of the ring make the same contribution to the field at (𝑃) because they are all equidistant from this point. Thus, we can integrate to obtain the total field at (𝑃): 𝑥 𝑑𝑞 𝑥 𝐸 = ∫ 𝑑𝐸𝑥 = ∫ 𝑘 3 =𝑘 3 ∫ 𝑑𝑞 (𝑥 2 + 𝑎 2 )2 (𝑥 2 + 𝑎2 )2 𝑸𝒙 =𝒌 𝟑 (𝒙𝟐 + 𝒂𝟐 )𝟐 15 Chapter (1) Electrostatics PROBLEMS 1. The sum of two-point charges is 𝟔 𝝁𝑪. They attract each other with a force of 𝟎. 𝟗 𝑵, when kept 𝟒𝟎 𝒄𝒎 apart in vacuum. Calculate the charges. 2. Two small charged spheres repel each other with a force of 𝟐 × 𝟏𝟎−𝟑 𝑵. The charge on one sphere is twice that on the other. When one of the charges is moved 𝟏𝟎 𝒄𝒎 away from the other, the force is 𝟓 × 𝟏𝟎−𝟒 𝑵. Calculate the charges and the initial distance between them. 3. Two identical metal spheres are placed 𝟎. 𝟐 𝒎 apart. A charge (𝒒𝟏 ) of 𝟗 𝝁𝑪 is placed on one sphere while a charge (𝒒𝟐 ) of −𝟑 𝝁𝑪 is placed upon the other. (a) What is the force on each of the spheres? (b) If the two spheres are brought together and touched and then returned to their original positions, what will be the force on each sphere? 4. Two charges lie along the 𝒙 − 𝒂𝒙𝒊𝒔 as in figure. The positive charge 𝒒𝟏 = 𝟏𝟓 µ𝐂 is at 𝒙 = 𝟐 𝒎, and the positive charge 𝒒𝟐 = 𝟔 µ𝐂 is at the origin. Where a negative charge (𝒒𝟑 ) must be placed on the 𝒙 − 𝒂𝒙𝒊𝒔 so that the resultant electric force on it is 𝒛𝒆𝒓𝒐? 5. Charges (𝒒𝟏 ) and (𝒒𝟐 ) lie on the 𝒙 − 𝒂𝒙𝒊𝒔 at points (𝒙 = − 𝒂), and (𝒙 = 𝒂), respectively, (a) How must (𝒒𝟏 ) and (𝒒𝟐 ) be related for the net electrostatic force on charge (𝑸) placed at (𝒙 = 𝒂/ 𝟐), to be zero? 6. Three-point charges, 𝒒𝟏 = 𝟑 𝝁𝑪, 𝒒𝟐 = 𝟓 𝝁𝑪, 𝒒𝟑 = 𝟒 𝝁𝑪, are fixed at the corners of a right triangle, as shown in figure. (a) Find the resultant force on charge (𝒒𝟐 ) and (b) Find the resultant force on charge (𝒒𝟑 ). (𝒓𝟏𝟐 = 𝟎. 𝟒 𝒎 𝒂𝒏𝒅 𝒓𝟐𝟑 = 𝟎. 𝟑 𝒎). 7. Four equal point charges each of 𝟑 𝒎𝑪 are placed at the corners of a square of side length 𝟏 𝒎. Calculate the electric field at the intersection of the diagonals of the square. 8. Two equal and opposite charges of magnitude 𝟎. 𝟐 µ𝑪 are held 𝟏𝟓 𝒄𝒎 apart (i) what are the magnitude and direction of (𝑬) at the point midway between the charges? (ii) What is the magnitude and the direction of the force that would act on an electron placed there? 16 Chapter (1) Electrostatics 9. Figure shows three particles with charges 𝒒𝟏 = +𝟐𝑸, 𝒒𝟐 = −𝟐𝑸, and 𝒒𝟑 = −𝟒𝑸, each a distance (𝒅) from the origin. What net electric field is produced at the origin? 10. In the figure, four charges form the corners of a square and four more charges lie at the midpoints of the sides of the square. The distance between adjacent charges on the perimeter of the square is (𝒅). What are the magnitude and direction of the electric field at the center of the square? 11. In the figure, charges 𝑸𝟏 = 𝒒 and 𝑸𝟐 = −𝟐𝒒 are fixed a distance “𝒅” apart (i) find (𝑬) at points (𝑨), (𝑩), and (𝑪) (b) sketch the electric field lines. 𝟏𝟏. In figure, the four particles are fixed in place and have charges 𝒒𝟏 = 𝒒𝟐 = +𝟓𝒆, 𝒒𝟑 = +𝟑𝒆, and 𝒒𝟒 = −𝟏𝟐𝒆. Distance 𝒅 = 𝟓 𝒎𝒎. What is the magnitude of the net electric field at point 𝑷 due to the particles? 17 Chapter (1) Electrostatics 13. At which points is the net electric field due to these two charges is zero? 14. An electron is accelerated eastward at 𝟏. 𝟖 × 𝟏𝟎𝟗 𝒎/𝒔𝟐 by an electric field. Determine the magnitude and direction of the electric field. 15. An electric dipole of charges 𝟐 × 𝟏𝟎−𝟔 𝑪, −𝟐 × 𝟏𝟎−𝟔 𝑪 are separated by a distance 𝟏 𝒄𝒎. Calculate the electric field due to dipole at a point on its. (i) Axial line 𝟏 𝒎 from its center (ii) equatorial line 𝟏 𝒎 from its center. 16. A neutral water molecule (𝑯𝟐 𝑶) in its vapor state has an electric dipole moment of magnitude 𝟔. 𝟐 × 𝟏𝟎−𝟑𝟎 𝑪/𝒎.(i) How far apart are the molecule’s centers of positive and negative charge? (ii) If the molecule is placed in an electric field of 𝟏. 𝟓 × 𝟏𝟎𝟒 𝑵/𝑪, what maximum force act on each charge? 17. In figure shows charged particles on an 𝑥 − 𝑎𝑥𝑖𝑠: 𝒒 = −𝟑. 𝟐 × 𝟏𝟎−𝟏𝟗 𝑪 𝒂𝒕 𝒙 = −𝟑 𝒎 𝒂𝒏𝒅 𝒒 = 𝟑. 𝟐 × 𝟏𝟎−𝟏𝟗 𝑪 𝒂𝒕 𝒙 = +𝟑 𝒎. What are the (a) magnitude and direction of the net electric field produced at point 𝑷 at 𝒚 = 𝟒. 𝟎𝟎 𝒎? 18. The Figure shows two parallel non conducting rings with their central axes along a common line. Ring 1 has uniform charge (𝑸𝟏 ) and radius (𝑹); ring 𝟐 has uniform charge (𝑸𝟐 ) and the same radius (𝑹). The rings are separated by distance 𝒅 = 𝟑𝑹.The net electric field at point P on the common line, at distance (𝑹) from ring1, is zero. What is the ratio (𝑸𝟏 / 𝑸𝟐 ). 18 Chapter (2) Gauss’ Law CHAPTER (2) GAUSS’ LAW 2.1 Introduction In physics, Gauss's law, also known as Gauss's flux theorem, is a law relating the distribution of electric charge to the resulting electric field. The surface under consideration may be a closed one enclosing a volume such as a spherical surface. The law was first formulated by Joseph-Louis Lagrange in 1773, followed by Carl Friedrich Gauss in 1813, both in the context of the attraction of ellipsoids. It is one of Maxwell's four equations, which form the basis of classical electrodynamics. Gauss's law can be used to derive Coulomb's law and vice versa. and is used to calculate the electric field results for many shapes of electric charge distribution. 2.2 Electric flux ሺ𝚽ሻ "Electric flux may be defined as the total number of electric filed lines passing through the normal surface area placed at this point" Consider a small area element ሺ𝑑𝐴ሻ of a closed surface area ሺ𝐴ሻ (figure 2.1). ሺ𝐸ሻ is the electric field at the area element ሺ𝑑𝐴ሻ. Let ሺ𝜃ሻ is the angle between area ⃗⃗⃗⃗⃗ ) and electric field vector (𝐸⃗ ), then electric flux ሺ∅ሻ through the area vector (𝑑𝐴 element ሺ𝑑𝐴ሻ is given by Figure 2.1: A small element of surface Φ = ∮ 𝐸⃗. ⃗⃗⃗⃗⃗ 𝑑𝐴 = ∮ 𝐸𝑑𝐴 𝐶𝑜𝑠𝜃 area ሺ∆𝐴ሻ The above equation represents the electric flux for a small area element ሺ𝑑𝐴ሻ, and the normal component of the electric field. The electric flux through the whole surface ሺ𝐴ሻ, may be calculate by applying the closed integration in the right side. Over the surface ሺ𝐴ሻ, 𝛷 = 𝐸. 𝐴𝑛𝑜𝑟𝑚𝑎𝑙 It is the number of electric filed lines pass through a normal unit area placed at certain point (figure 2.2). The number of electric field lines and/or electric flux is directly Figure 2.2: Field lines proportional to the magnitude of the charge. representing a uniform electric field penetrating a 𝑄 ሺ𝐴ሻ 𝛷∝𝑄 ⟹ 𝛷= plane of area 𝜀𝑜 perpendicular to the field. 19 Chapter (2) Gauss’ Law 2.3 Some physical concepts 1. Gaussian surface (G.S.):  It is an imaginary closed surface, with high symmetry and surrounded the charge from all direction.  It is suggested according the shape of electric charge distribution. 2. Closed integrationሺ∮ 𝒅𝑨ሻ:  It is one kind of integration and applying on closed surface (i.e G. S)  Its value equals the area of Gaussian surface. 3. Vector area: Consider a small area element ሺ𝑑𝐴ሻ (figure 2.3) of a closed surface as shown in figure. The arrow representing the area vector is drawn perpendicular to the area of the element. If ሺ𝑛̂ሻ represents unit vector along the outdrawn normal to the area element, then, ⃗⃗⃗⃗⃗ = 𝑛̂𝑑𝐴 𝑑𝐴 Figure 2.3: A small element of surface area Therefore, it is a vector its value is one and its direction outward and ሺ∆𝐴ሻ of a closed normal on Gaussian surface. surface 4. Scalar product It is the kind of vectors product which obtains a numerical value positive or negative. The scalar product depends on the angle between two vectors (figure 2.4) and general formula can be written as ⃗𝑨 ⃗ ⊙ ⃗𝑩 ⃗ = |𝑨 ⃗ | × |𝑩 ⃗⃗ | 𝒄𝒐𝒔𝜽 Therefore; Figure 2.4: A and B ⃗ = |𝐴|. |𝐵 ⃗ | 𝑐𝑜𝑠 0° = 𝐴𝐵 vectors 𝜃 =0 ⟹𝐴⊙𝐵 ⃗ = |𝐴|. |𝐵 𝜃 =𝜋 ⟹ 𝐴⊙𝐵 ⃗ | 𝑐𝑜𝑠 𝜋 𝑜 = −𝐴𝐵 ⃗ = |𝐴|. |𝐵 𝜃 = 90𝑜 ⟹ 𝐴 ⊙ 𝐵 ⃗ | 𝑐𝑜𝑠 90𝑜 = 0 5. The two-shell theorem of spherical conductor. (i) The charges on the surface of metallic spherical shell have attracted or repelled until they uniformly distributed on its surface. Moreover, these charges are accumulated outside the surface of metallic shell and can be imagined them as if were concentrated at its center. (ii) A charged metallic spherical shell does not generate an electrostatic force on any charged particle that is located inside this metallic shell. So, the electric field ሺ𝐸ሻ inside the metallic shell is equal 20 Chapter (2) Gauss’ Law zero this because the charge is equal zero ሺ𝑄 = 0ሻ inside it. Furthermore, if excess charge is placed on a spherical shell that is made of conducting material, this excess charge spreads uniformly outside the (external) surface. For example, if we place excess electrons on a spherical metal shell, those electrons repel one another and tend to move apart, spreading over the available surface until they are uniformly distributed. 2.4 Formula of Gauss’ law 𝑸𝒊𝒏 𝑮.𝑺 𝛷= = ∮ 𝐸⃗. ⃗⃗⃗⃗⃗ 𝑑𝐴 𝜺𝟎 where, ሺ𝛷ሻ is electric flux, ሺ𝑄𝑖𝑛 ሻ is sum of electric charge within Gaussian surface, 𝜀𝑜 = 8.85 × ⃗⃗⃗⃗⃗ ) is vector area and ∮ 𝑖𝑠 10−12 𝐶 2 /𝑁𝑚2 is electric permittivity of air, (𝐸⃗ )is electric field vector, (𝑑𝐴 closed integration that apply on Gaussian closed surface. Example (2.1): A point charge of 𝟏. 𝟖 µ𝒄 is at the center of a cubical Gaussian surface 𝟓𝟓 𝒄𝒎 on edge. What is the net electric flux through the surface? Solution: 𝑞 ሺ1.8 × 10−6 𝐶ሻ 𝛷= = = 2 × 105 𝑁. 𝑚2 /𝐶 𝜀0 ሺ8.854 × 10−12 𝐶 2 /𝑁𝑚2 ሻ Procedure for deriving or solving the Gauss’ law problem; ‫لحل مسائل الشحنات الموزعة باستخدام قانون جاوس يتم االتى‬ 1. Detriment the shape of electric charge (‫(تحديد شكل الشحنة‬  Point charge or charged sphere ሺ𝑸ሻ  Charged line or charged cylinder 𝑸 = 𝝀𝒍  Charged surface (plate, plane, sheet) 𝑸 = 𝝈𝑨 2. Assume Gaussian surface according to the shape of charge ‫إختيار سطح مغلق(جاوسى) الحتواء الشحنة الموزعة‬ 3. Apply Gauss law ‫تطبيق قانون جاوس‬ 𝑸𝒊𝒏 𝑮.𝑺 Φ= = ∮ 𝐸⃗. ⃗⃗⃗⃗⃗ 𝑑𝐴 𝚽 𝜺𝒐 𝑸𝒊𝒏 𝑮.𝑺 = ∮ 𝑬 𝒅𝑨 𝒄𝒐𝒔 𝜽 = 𝑬 ∮ 𝒅𝑨 𝜺𝒐 𝑸𝒊𝒏 𝑮.𝑺 = 𝑬𝑨 𝑮.𝑺. 𝜺𝒐 4. Find the formula of electric field ሺ𝐸ሻ. 21 Chapter (2) Gauss’ Law 2.5 Derivation of Coulomb’ s law by applying Gauss’ law: To calculate the electric field at a distance ሺ𝑟ሻ from a point charge ሺ𝑞ሻ using Gauss’ law consider a spherical Gaussian surface has a radius ሺ𝑟ሻ, ⃗⃗⃗⃗⃗ ) is Zero and the angle ሺ𝜃ሻ between (𝐸⃗ ) and (𝑑𝐴 𝑞 ⃗⃗⃗⃗⃗ = ∮ 𝐸⃗. 𝑑𝐴 𝜀0 Figure: 2.5: Gauss’ 𝑞 law for sphere = ∮ 𝐸 𝑑𝐴 𝑐𝑜𝑠 𝜃 = 𝐸 ∮ 𝑑𝐴 = 𝐸𝐴 𝜀0 𝑞 = 𝐸ሺ4𝜋𝑟 2 ሻ 𝜀𝑜 𝟏 𝒒 𝒒 ⟹𝑬= 𝟐 ⟹𝑬=𝒌 𝟐 𝟒𝝅𝜺𝒐 𝒓 𝒓 2.6 Electric field due to charged line or cylinder (Cylindrical symmetry)  The electric field at a distance ሺ𝑟ሻ from an infinite line of charges of constant charge per unit length ሺ𝜆ሻ ⟹ 𝑞 = 𝜆𝑙  Gaussian surface is a cylinder of radius ሺ𝑟ሻ and its length is ሺ𝐿ሻ. the total flux from both circular bases is zero because the angle ሺ𝜃ሻ between (𝐸⃗ ) and (𝑑𝐴 ⃗⃗⃗⃗⃗ ) is ሺ90°ሻ Φ = ∮ 𝐸⃗. ⃗⃗⃗⃗⃗ 𝑑𝐴 = ∮ 𝐸 𝑑𝐴 𝑐𝑜𝑠 90𝑜 = 0 Figure 2.6:  Then, all electric flux gets from the side area of cylinder because the angle ሺ𝜃ሻ Cylindrical ⃗⃗⃗⃗⃗ ) is Zero in all position. between (𝐸⃗ ) and (𝑑𝐴 symmetry 𝑞 ⃗⃗⃗⃗⃗ = ∮ 𝐸⃗. 𝑑𝐴 𝜀𝑜 𝜆𝑙 = ∮ 𝐸 𝑑𝐴 𝑐𝑜𝑠 𝜃 = 𝐸 ∮ 𝑑𝐴 𝜀𝑜 𝜆𝑙 = 𝐸ሺ4𝜋𝑟𝑙ሻ 𝜀0 𝟏 𝝀 𝝀 ⟹𝑬= 𝒐𝒓 𝑬 = 𝟐𝒌 𝟐𝝅𝜺𝒐 𝒓 𝒓 Example (2.2): An infinite line of charge produces a field of 𝟒. 𝟓 𝒙 𝟏𝟎𝟒 𝑵/𝑪 at a distance of 𝟐 𝒎. Calculate the linear charge density. Solution: 𝜆 𝐸= ⟹ 𝜆 = 2𝜋𝜀𝑜 𝑟𝐸 = ሺ2𝜋ሻሺ8.854 × 10−12 𝐶 2 /𝑁𝑚2 ሻሺ2 𝑚ሻሺ4.5 × 104 𝑁/𝐶ሻ = 5𝜇𝐶/𝑚 2𝜋𝜀𝑜 𝑟 22 Chapter (2) Gauss’ Law 2.7 Electric field due to charged planer symmetry (i) For conducting sheet (conducting material)  This material has a free charge so it concentrated on one face of sheet.  To find the electric field due to an infinite-conducting sheet that has a uniform surface charge density ሺ𝜎ሻ. 𝑞 = 𝜎𝐴 Figure 2.7: Electric field for  Gaussian surface is a cylinder one of its base on the conducting sheet charged sheet as in figure, and the angle ሺ𝜃ሻ between (𝐸⃗ ) ⃗⃗⃗⃗⃗ ) is Zero and (𝑑𝐴 𝑞 = ∮ 𝐸⃗. ⃗⃗⃗⃗⃗ 𝑑𝐴 𝜀0 𝜎𝐴 = ∮ 𝐸 𝑑𝐴 𝑐𝑜𝑠 𝜃 = 𝐸 ∮ 𝑑𝐴 𝜀0 𝜎𝐴 = 𝐸ሺ𝐴ሻ 𝜀0 𝜎 ⟹𝐸= 𝜀𝑜 (ii) For non-conducting Sheet (Insulating material)  This material hasn’t free charge so it can be placed on both faces of sheet.  To find the electric field due to an infinite non-conducting sheet that has a uniform surface charge density ሺ𝜎ሻ on each face of sheet Figure 2.8: Electric field for non 𝑞 = 𝜎𝐴 conducting sheet  Gaussian surface is two cylinder their one of base on both ⃗⃗⃗⃗⃗⃗ and (𝑑𝐴 sides of the sheet as in figure, and the angle ሺ𝜃ሻ between ሺ𝐸ሻ ⃗⃗⃗⃗⃗ ) is Zero 𝑞 = ∮ 𝐸⃗. ⃗⃗⃗⃗⃗ 𝑑𝐴 𝜀0 𝜎𝐴 = ∮ 𝐸 𝑑𝐴 𝑐𝑜𝑠 𝜃 + ∮ 𝐸 𝑑𝐴 𝑐𝑜𝑠 𝜃 = 2𝐸 ∮ 𝑑𝐴 𝜀0 𝜎𝐴 = 2𝐸ሺ𝐴ሻ 𝜀0 𝝈 ⟹𝑬= 𝟐𝜺𝒐 23 Chapter (2) Gauss’ Law (iii) Electric field due to two parallel conducting charged sheets  Consider two plane parallel infinite non-conducting sheets with equal and opposite charge densities ሺ+𝜎ሻ and ሺ+𝜎ሻ as shown in figure.  The electric field on either side of a plane is ሺ𝐸 = 𝜎⁄2εo ሻ , directed outward (if the charge is positive) or inward (if the charge is negative).  At a point ሺ𝑃1 ሻ between the two sheets, Figure 2.9: Two parallel infinite non-conducting 𝜎 𝜎 𝜎 𝐸 = 𝐸1 + 𝐸2 = + = ሺ𝑡𝑤𝑜𝑟𝑑𝑠 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡ሻ sheets 2𝜀𝑜 2𝜀𝑜 𝜀𝑜  At a point ሺ𝑃2 ሻ outside the two sheets, 𝜎 𝜎 𝐸 = 𝐸1 − 𝐸2 = − =0 2𝜀𝑜 2𝜀𝑜 Notice; If the two plate are conducting sheets  At a point ሺ𝑃1 ሻ between the two sheets, 𝜎 𝜎 𝐸 = 𝐸1 − 𝐸2 = − =0 𝜀𝑜 𝜀𝑜  At a point ሺ𝑃2 ሻoutside the two sheets, 𝜎 𝜎 2𝜎 𝐸 = 𝐸1 + 𝐸2 = + = ሺ𝑡𝑤𝑜𝑟𝑑𝑠 𝑡ℎ𝑒 𝑙𝑒𝑓𝑡 𝑤𝑎𝑟𝑑ሻ 𝜀𝑜 𝜀𝑜 𝜀𝑜 Example (2.3): The figure shows portions of two large non-conducting sheets, each with a fixed uniform charge on one side. The magnitude of the surface charge densities is 𝝈+ = 𝟔. 𝟖 𝝁𝑪/𝒎𝟐 and 𝝈+ = 𝟒. 𝟑 𝝁𝑪/𝒎𝟐. Find the electric field, (i) between the Sheets, (ii) to the right of the sheets and (iii) to the left of the sheets. Solution: (i) Between two sheets (Towards +𝑣𝑒 𝑥 − 𝑎𝑥𝑖𝑠) 𝜎ሺ−ሻ 𝜎ሺ+ሻ ሺ4.3 + 6.8ሻ × 10−6 𝐶/𝑚2 𝐸𝑏𝑒𝑡 = 𝐸ሺ−ሻ + 𝐸ሺ+ሻ = + = 6.2 × 105 𝑁/𝐶 2𝜀𝑜 2𝜀𝑜 2ሺ8.854 × 10−12 𝐶 2 /𝑁𝑚2 ሻ (ii) At the right from two sheet (Towards +𝑣𝑒 𝑥 − 𝑎𝑥𝑖𝑠) 𝜎ሺ+ሻ 𝜎ሺ−ሻ ሺ6.8 − 4.3ሻ × 10−6 𝐶/𝑚2 𝐸𝑟𝑖𝑔ℎ𝑡 = 𝐸ሺ+ሻ − 𝐸ሺ−ሻ = − = = 1.4 × 105 𝑁/𝐶 2𝜀𝑜 2𝜀𝑜 2ሺ8.854 × 10−12 𝐶 2 /𝑁𝑚2 ሻ 24 Chapter (2) Gauss’ Law (iii) At the left from two sheet (Towards −𝑣𝑒 𝑥 − 𝑎𝑥𝑖𝑠) 𝜎ሺ−ሻ 𝜎ሺ+ሻ ሺ4.3 − 6.8ሻ × 10−6 𝐶/𝑚2 𝐸𝑟𝑖𝑔ℎ𝑡 = 𝐸ሺ−ሻ − 𝐸ሺ+ሻ = − −12 2 2 = −1.4 × 105 𝑁/𝐶 2𝜀𝑜 2𝜀𝑜 2ሺ8.854 × 10 𝐶 /𝑁𝑚 ሻ Example (2.4): The magnitude of the average electric field normally present in the earth’s atmosphere just above the surface of the earth is about 𝟏𝟓𝟎 𝑵/𝑪, directed downward. What is the total net surface charge of the earth? Assume the earth to be a conductor with a uniform surface charge density Solution: 𝐸 = 𝜎⁄𝜀0 −12 𝐶2 𝑁 ⟹ 𝜎 = 𝜀𝑜 𝐸 = (8.854 × 10 ) (150 ) = 1.33 × 10−9 𝐶/𝑚2 𝑁𝑚2 𝐶 𝑞 = 𝜎𝐴 = 𝜎ሺ4𝜋𝑟 2 ሻ = ሺ1.33 × 10−9 𝐶/𝑚2 ሻሺ4𝜋ሻሺ6.37 × 106 𝑚ሻ2 = −6.8 × 105 𝐶 Example (2.5): In the figure, a small, non-conducting ball of mass 𝒎 = 𝟏 𝒎𝒈 and uniformly distributed charge 𝒒 = 𝟐 × 𝟏𝟎−𝟖 𝐂 hangs from an insulating thread that makes an angle 𝜽 = 𝟑𝟎𝒐 with a vertical, uniformly charged non-conducting sheet. Considering the ball’ s weight and assuming that the sheet extends far in all directions, calculate the surface charge density σ of the sheet. Solution: The sheet is non-conducting, so 𝑞𝜎 𝐹 = 𝑞𝐸 = 2𝜀𝑜 𝑞𝜎 ∑ 𝐹𝑥 = 0 ⇒ 𝑇 sin 𝜃 = 2𝜀𝑜 ∑ 𝐹𝑦 = 0 ⇒ 𝑇 cos 𝜃 = 𝑚𝑔 𝑞𝜎 tan 𝜃 = 2𝑚𝑔𝜀𝑜 2𝑚𝑔𝜀𝑜 tan 𝜃 2ሺ10−6 𝑘𝑔ሻሺ9.8 𝑚/𝑠 2 ሻሺ8.854 × 10−12 𝐶 2 /𝑁𝑚2 ሻ tan 30° ⇒𝜎= = = 5 𝑛𝐶/𝑚2 𝑞 ሺ2 × 10−8 𝐶ሻ 25 Chapter (2) Gauss’ Law 2.8 The Electric field due to charged sphere (Spherical symmetry) To find the electric field inside, on the surface and outside an insulating solid sphere of radius ሺ𝑎ሻ has a uniform volume charge density ρ and carries a total charge ሺ𝑄ሻ.  First inside the sphere (i) The Gaussian surface is a sphere of radius 𝑟 ˂ 𝑎 (ii) The charge within the sphere is direct proportional with its volume 4 4 Figure 2.10: Spherical 𝑄 ∝ 𝜋𝑎3 𝑎𝑛𝑑 𝑞 ∝ 𝜋𝑟 3 symmetry 3 3 4 3 𝑞 3 𝜋𝑟 𝑟3 ⟹ = = 3 𝑄 4 𝜋𝑎3 𝑎 3 𝑟3 ⟹𝑞= 𝑄 𝑎3 Using above equation in electric field formula for point charge 𝑞 𝑟 3𝑄 𝑄 ⟹𝐸=𝑘 2 = 𝑘 2 3 =𝑘 3 𝑟 𝑟 𝑟 𝑎 𝑎 𝑄 Figure 2.11: Relation ⟹𝐸=𝑘 3 𝑟 𝑎 between E&r Notice: At a point inside the conducting spherical shell (metal sphere): q in EA= εo 𝑞𝑖𝑛 = 0 ⟹𝐸 =0 (i.e) the field inside the conducting sphere is zero because the charge inside conductor is zero.  At a point outside the charged sphere of radius ሺ𝑟 > 𝑅 ሻ For conducting (metal) or nonconducting (plastic) sphere. 𝑞 ⟹𝐸 =𝑘 2 𝑟  At a point on the surface ሺ𝑟 = 𝑅 ሻ For conducting (metal) or nonconducting (plastic) sphere. 𝑞 ⟹𝐸 =𝑘 2 𝑅 26 Chapter (2) Gauss’ Law Example (2.6): The nucleus of an atom of gold has a radius 𝟔. 𝟐 𝒙𝟏𝟎−𝟏𝟓 𝒎 and a positive charge 𝒒 = 𝒁𝒆, where the atomic number 𝒁 of gold is 𝟕𝟗. Plot the magnitude of the electric field from the center of the gold nucleus outward to a distance of about twice its radius. (Assume that the nucleus is spherical and that the charge is distributed uniformly throughout its volume) Solution: (i) First inside the nucleus 𝑞 = 𝑍𝑒 = ሺ79ሻሺ1.6 × 10−19 𝐶ሻ = 1.264 × 10−17 𝐶 𝑞𝑟 9 2 2ሻ ሺ1.264 × 10−17 𝐶ሻ𝑟 𝐸=𝑘 = ሺ9 × 10 𝑁𝑚 /𝐶 𝑎3 ሺ6.2 × 10−15 𝑚ሻ3 𝐸𝑖𝑛𝑠𝑖𝑑𝑒 = 4.8 × 1035 𝑟 𝑁/𝐶 (ii) Second on the surface of the nucleus 𝑞 9 2 2 ሺ1.264 × 10−17 𝐶ሻ 𝐸=𝑘 = ሺ9 × 10 𝑁𝑚 /𝐶 ሻ = 3 × 1021 𝑁/𝐶 𝑎2 ሺ6.2 × 10−15 𝑚ሻ2 (iii) Third outside the sphere 𝑞 9 𝑁𝑚2 ሺ1.264 × 10−17 𝐶ሻ 1.1 × 107 𝐸=𝑘 = (9 × 10 ) = 𝑁/𝐶 𝑟2 𝐶2 𝑟2 𝑟2 Example (2.7): Consider an atom containing a point positive charge 𝒁𝒆 at its center and surrounded by a distribution of negative electricity, − 𝒁𝒆 uniformly distributed within a sphere of radius 𝑹. Prove that the electric field 𝑬 at a distance 𝒓 from the center for a point inside the atom is: 𝑍𝑒 1 𝑟 𝐸= ( 2 − 3) 4𝜋𝜀𝑜 𝑟 𝑅 Solution: There are two electric fields at this point inside the sphere 𝐸 = 𝐸𝑜𝑢𝑡 − 𝐸𝑖𝑛 1. 𝐸 get out due to the +𝑣𝑒 point charge (Ze) at the center 𝑘𝑞 𝑍𝑒 𝐸𝑜𝑢𝑡 = 2 = 𝑟 4𝜋𝜀𝑜 𝑟 2 27 Chapter (2) Gauss’ Law 2. 𝐸 get into due to the – 𝑣𝑒 charge is uniform distributed in volume of atom 𝑘𝑞𝑟 𝑍𝑒𝑟 𝐸𝑖𝑛 = = 𝑅3 4𝜋𝜀𝑜 𝑅3 𝑍𝑒 𝑍𝑒𝑟 𝐸= − 4𝜋𝜀𝑜 𝑟 2 4𝜋𝜀𝑜 𝑅 3 𝑍𝑒 1 𝑟 ∴𝐸 = ( 2 − 3) 4𝜋𝜀𝑜 𝑟 𝑅 Example (2.8): It is found experimentally that the electric field in a certain region of the earth’s atmosphere is directed vertically down. At an altitude of 𝟑𝟎𝟎 𝒎 the field has magnitude 𝟔𝟎 𝑵/𝑪, and at an altitude of 𝟐𝟎𝟎 𝒎, 𝟏𝟎𝟎 𝑵/𝑪. Find the net amount of charge contained in a cube 𝟏𝟎𝟎 𝒎 on edge, with horizontal faces at altitudes of 𝟐𝟎𝟎 and 𝟑𝟎𝟎 𝒎. Neglect the curvature of the earth. Solution: Apply the Gauss law on upper and lower surface of cube. ∵ Φ𝑢𝑝𝑝𝑒𝑟 = ∮ 𝐸 𝑑𝐴 𝑐𝑜𝑠 𝜃 = 𝐸 ∮ 𝑑𝐴 𝑐𝑜𝑠𝜋 Φ𝑢𝑝𝑝𝑒𝑟 = −𝐸𝐴 = −ሺ100 𝑁/𝐶ሻሺ100 𝑚ሻ2 Φ𝑙𝑜𝑤𝑒𝑟 = ∮ 𝐸 𝑑𝐴 𝑐𝑜𝑠 𝜃 = 𝐸 ∮ 𝑑𝐴 cos 0 Φ𝑙𝑜𝑤𝑒𝑟 = 𝐸𝐴 = ሺ60 𝑁/𝐶ሻሺ100 𝑚ሻ2 𝑄𝑖𝑛 ∆Φ = 𝜀𝑜 ⟹ 𝑄𝑖𝑛 = 𝜀𝑜 (Φ𝑢𝑝𝑝𝑒𝑟 − Φ𝑙𝑜𝑤𝑒𝑟 ) = ሺ8.854 × 10−12 𝐶 2 /𝑁𝑚2 ሻሺ100 𝑚ሻ2 ሺ60 − 100ሻ𝑁/𝐶 = −3.54 𝜇𝐶 28 Chapter (2) Gauss’ Law PROBLEMS 1. The figure shows a section of a long, thin-walled metal tube of radius 𝑹 = 𝟑 𝒄𝒎, with a charge per unit length of 𝝀 = 𝟐 × 𝟏𝟎 −𝟖 𝑪/𝒎. What is the magnitude ሺ𝑬ሻ of the electric field at radial distance (a) 𝒓 = 𝑹/𝟐 and (b) 𝒓 = 𝟐 𝑹? 2. An electron is released 𝟗 𝒄𝒎 from a very long non-conducting rod with a uniform 𝟔 𝒎𝑪/𝒎. What is the magnitude of the electron’s initial acceleration? 3. In figure is a section of a conducting rod of radius 𝑹𝟏 = 𝟎. 𝟑 mm and length 𝑳 = 𝟏𝟏 𝒎 inside a thin-walled coaxial conducting cylindrical shell of radius 𝑹𝟐 = 𝟏𝟎 𝑹𝟏 and the same length ሺ𝑳ሻ. The net charge on the rod is 𝑸𝟏 = +𝟑. 𝟒 × 𝟏𝟎−𝟏𝟐 𝑪; that on the shell is 𝑸𝟐 = −𝟐 𝑸𝟏. What are the magnitude and direction at radial distance 𝒓 = 𝟐 𝑹𝟏 and 𝒓 = 𝟓 𝑹𝟏 ? 4. Two charged concentric spherical shells have radii 𝟏𝟎 𝒄𝒎 and 𝟏𝟓 𝒄𝒎. The charge on the inner shell is 𝟏𝟎−𝟖 𝑪, and that on the outer shell is 𝟐 × 𝟏𝟎−𝟖 𝑪. Find the electric field (i) at 𝒓 = 𝟏𝟐 𝒄𝒎 and (ii) at 𝒓 = 𝟐𝟎 𝒄𝒎. 5. The figure shows a very large non-conducting sheet that has a uniform surface charge density of 𝝈 = 𝟐 𝒎𝑪/𝒎𝟐 ; it also shows a particle of charge 𝑸 = 𝟔 𝝁𝑪, at distance ሺ𝒅ሻ from the sheet. If 𝒅 = 𝟎. 𝟐 𝒎, at what (i) positive and (ii) negative coordinate on the 𝒙 − 𝒂𝒙𝒊𝒔 (other than infinity) is the net electric field ሺ𝑬𝒏𝒆𝒕 ሻ of the sheet and charged particle is zero? 6. The figure shows a closed Gaussian surface in the shape of a cube of edge length 𝟐 𝒎, with one corner at 𝒙𝟏 = 𝟓 𝒎, 𝒚 = 𝟒 𝒎. The cube lies in a region where the electric field vector is given by ̂ , with y in meters. What is the net charge ⃗⃗ = −𝟑𝒊̂ − 𝟒𝒚𝟐 𝒋̂ + 𝟑𝒌 𝑬 contained by the cube? 7. Charge is distributed uniformly throughout the volume of an infinitely long solid cylinder of radius ሺ𝑹ሻ. 𝝆𝒓 (a) Show that, at a distance ሺ𝒓 < 𝑹ሻ from the cylinder axis, (𝑬 = 𝟐𝜺 ) where ሺ𝝆ሻ is the volume charge 𝒐 density. (b) Write an expression for ሺ𝑬ሻ when ሺ𝐫 > 𝐑ሻ. 29 Chapter (2) Gauss’ Law 8. The drum of a photocopying machine has a length of 𝟒𝟐 𝒄𝒎 and a diameter of 𝟏𝟐 𝒄𝒎. The electric field just above the drum’s surface is 𝟐. 𝟑 × 𝟏𝟎𝟓 𝑵/𝑪. What is the total charge on the drum? 9. In figure, a plastic solid sphere of radius 𝒂 = 𝟐 𝒄𝒎 is concentric with a spherical conducting shell of inner radius 𝒃 = 𝟐𝒂 and outer radius 𝒄 = 𝟐. 𝟒𝒂. The sphere has a net uniform charge 𝒒𝟏 = 𝟓 𝒇 𝑪; the shell has a net charge 𝒒𝟐 = −𝒒𝟏.What is the magnitude of the electric field at radial distances (a) 𝒓 = 𝟎, (b) 𝒓 = 𝒂/𝟐, (c) 𝒓 = 𝒂, (d) 𝒓 = 𝟏. 𝟓𝒂, (e) 𝒓 = 𝟐. 𝟑 𝒂, and (f) 𝒓 = 𝟑. 𝟓 𝒂 ? 10. The figure gives the magnitude of the electric field inside and outside a sphere with a positive charge distributed uniformly throughout its volume. The scale of the vertical axis is set by 𝑬𝒔 = 𝟓 × 𝟏𝟎𝟕 𝑵/𝑪. What is the charge on the sphere? 30 Chapter (3) Electrostatic Potential Energy CHAPTER (3) ELECTROSTATIC POTENTIAL ENERGY 3.1 Introduction The concept of potential energy has a great value in the study of electricity. Moreover, most of electrostatic phenomena can be described in terms of an electric potential energy. This idea enables us to define a scalar quantity named as “electric potential”. The electric potential at any point in an electric field region is a scalar quantity. Therefore, it helps us to describe the electrostatic phenomena more simply than the vector quantities (e.g. the electric field and electrostatic forces). Furthermore, the concept of electric potential has effective value in the operation of electric circuits and devices that we will study in later chapters. When an external force (𝐹) does work to rise up a body from a point to another against gravitational force, this work is stored as potential energy for body in the field of gravity. When the external force is removed, the body moves down toward the earth surface and its gravitational potential energy change into a kinetic based on the conservation energy law. The sum of kinetic and potential energies is thus conserved. So, forces of this kind are called conservative forces. 𝑞1 𝑞2 It is known that, Coulomb force (𝐹 = 𝑘 ) between two (stationary) charges, like the 𝑟2 𝑚1 𝑚2 gravitational force (𝐹 = 𝐺 ), is also a conservative force. This is not surprising, since both have 𝑟2 inverse-square dependence on distance and differ mainly in the proportionality constants. The masses in the gravitational law are replaced by charges in Coulomb’s law. Thus, like the potential energy of a mass in a gravitational field, we can define electrostatic potential energy of a charge in an electrostatic field. Consider an electrostatic field (𝐸) due to a charge (+𝑄) placed at the origin. Now, imagine that, we transport a test charge (+𝑞) from a point (𝑅) to a point (𝑃) against the repulsive force that produce between two positive charges. Figure 3.1: Electric field at test charge (+𝑞) due to a charge (+𝑄) In figure 3.1 , to bring the charge (+𝑞) from (𝑖) to (𝑓), we must apply an external force (𝐹𝑒𝑥𝑡 ). to overcome the repulsive electric force (𝐹𝑒 = 𝑞𝐸) between them, i.e. (𝐹𝑒𝑥𝑡. = −𝐹𝑒 ). In this situation, work done by the external force (𝑭𝒆𝒙𝒕 ) is the negative but the work done by the electric force (𝑭𝒆 ) is the positive. This work is fully stored in the form of electric potential energy for the charge (+𝑞). If the external force is removed after reaching to point (𝑃), 31 Chapter (3) Electrostatic Potential Energy the electric force will take the charge away from (+𝑄) and the stored energy (i.e. electric potential energy) at point (𝑓) is converted into a kinetic energy to the charge (+𝑞), So the sum of the kinetic and potential energies is conserved. 3.2 The electric potential energy Thus, work done by external forces (𝐹𝑒𝑥𝑡 ). to move a charge (+𝑞) from initial point (𝒊) to final point (𝑓) is 𝑓 𝑊𝑖𝑓 = ∫𝑖 𝐹𝑒𝑥𝑡 ∙ 𝑑𝑟 (3 − 1) Also, we can write that, 𝑓 𝑓 𝑊𝑖𝑓 = − ∫𝑖 𝐹𝑒 ∙ 𝑑𝑟 = −𝑞 ∫𝑖 𝐸 ∙ 𝑑𝑟 (3 − 2) This work done is against electrostatic repulsive force (𝐹𝑒 ) and gets stored as electric potential energy which is negative value of the work done by external force. (Note, the displacement direction of charge (+𝑞) is opposite to the electric force (𝐹𝑒 ) direction and hence work done by electric field is negative). 𝑓 𝑈𝑖𝑓 = −𝑊𝑖𝑓 = −𝑞 ∫𝑖 𝐸⃗ ∙ ⃗⃗⃗⃗ 𝑑𝑟 (3 − 3) Furthermore, at every point in electric field (𝐸), a particle with charge (+𝑞) has a certain electrostatic potential energy (𝑈), this work done increases its potential energy by an amount equal to potential energy difference between points (𝑖) and (𝑓). Thus, potential energy difference is; (Δ𝑈𝑖𝑓 = 𝑈𝑓 − 𝑈𝑖 ) Therefore, we can define electric potential energy difference between two points as “it is work done by an external force to move charge (+𝒒) between two points in an electric field” Two important comments may be made at this stage; (i) The right side of above equation depends only on the initial and final positions of the charge. It means that the work done by an electric field to move a charge from one point to another is not depend on its path of motion. (ii) We can select an infinity point (∞) at which the electric potential energy is zero (𝑈∞ = 0). Then, we take the initial point (𝑖) at infinity, so (𝑈𝑖 = 0) and have obtained the value of electric potential energy at point (𝑓). 𝑈∞𝑓 = 𝑈𝑓 − 𝑈∞ (3 − 4) Since the (𝑓) is an arbitrary point, So the equation (3-4) provides us with a definition of potential energy (𝑈) of a charge (+𝑞) at any point. “It is the work done by the external force (equal and opposite to the electric force) to transport the charge (+𝒒) from infinity to that point. 32 Chapter (3) Electrostatic Potential Energy Example (3.1) Helium balloon has a charge 𝒒 = −𝟓. 𝟓 × 𝟏𝟎−𝟖 𝑪 rises vertically into the air by a distance 𝟓𝟐𝟎 𝒎 from an initial position (𝑖) to a final position (𝑓). If the magnitude of electric field exists in the atmosphere near to the earth surface 𝟏𝟓𝟎 𝑵/𝑪 and is directed downward. What is the difference in electric potential energy of the balloon between positions (𝑖) and (𝑓)? Solution: The work W done by the electric force (𝐹𝑒 = 𝑞𝐸)on the balloon is 𝑓 𝑊 = ∫ 𝐹𝑒 ∙ 𝑑𝑟 𝑖 𝐷 𝐷 ∴ 𝑊 = 𝑞 ∫ 𝐸 ∙ 𝑑𝑟 = 𝑞 ∫ 𝐸𝑑𝑟 𝑐𝑜𝑠180𝑜 0 0 𝐷 𝑊 = 𝑞𝐸(−1) ∫ 𝑑𝑟 = (−1)𝑞𝐸𝐷 0 𝑊 = (−1)(−5.5 × 10−8 𝐶)(150 𝑁/𝐶)(520 𝑚) = 4.3 × 10−3 𝐽  The work done by the electrostatic force on the balloon is positive.  Thus, the difference in the electric potential energy of the balloon is Δ𝑈𝑖𝑓 = 𝑈𝑓 − 𝑈𝑖 = −𝑊𝑖𝑓 Δ𝑈𝑖𝑓 = −4.3 × 10−3 𝐽 = −4.3 𝑚𝐽  The minus sign indicates that, the electric potential (𝑈) decreases as the negatively charged balloon rises. Because its motion opposite the direction of electric field (downward directed). 3.3 Electrostatic potential (V) In the previous section, we define the electric potential energy (𝑈) of a test charge (+𝑞) in terms of the work done on it. This work is clearly proportional to magnitude of (+𝑞), since the electric force at any point is (𝐹𝑒 = 𝑞𝐸), where (𝐸) is the electric field at that point due to the given charge configuration. It is suitable to divide the work by the amount of charge (+𝑞), so that the resulting quantity is independent of (+𝑞). In other words, work done per unit test charge is characteristic of the electric field associated with the charge configuration. This leads to the idea of electrostatic potential (𝑉) due to a given charge configuration. From equation (3-2), we get: “Work done by external force to move a unit positive charge between two points” 𝑈𝑖𝑓 −𝑊𝑖𝑓 𝑈𝑓 𝑈𝑖 𝑉𝑖𝑓 = = = − = 𝑉𝑓 − 𝑉𝑖 (3 − 5) 𝑞 𝑞 𝑞 𝑞 where, (𝑉𝑓 ) and (𝑉𝑖 ) are the electrostatic potentials at points (𝑓) and (𝑖), respectively. Note, as before, that it is not the actual value of potential but the potential difference that is physically significant. 33 Chapter (3) Electrostatic Potential Energy If, we take the electric potential at initial point (𝑖) which was placed at infinity is equal zero (𝑉∞ = 0) in equation (3-5). Then, the work done by an external force to transport a unit of positive charge from infinity to a point is equal the electrostatic potential (𝑉) at this point. In other words, the electrostatic potential (V) at any point in a region with electrostatic field is the work done to transport a unit positive charge from infinity to this point. 𝑈𝑓 𝑉𝑓 = (3 − 6) 𝑞 Moreover, the S.I. unit of both electric potential and potential difference is joules per coulomb, which is defined as a volt (𝑉): 1V ≡ 1 𝐽⁄𝐶 Electron volt: The energy usually measures in the S.I. unit of joules but in atomic, solid state and nuclear physics often needs a very small unit of energy. A more suitable unit for that is the electron-volt (𝑒𝑉), which is defined as; “It is energy required to transport an elementary charge (𝒆 = 𝟏. 𝟔 × 𝟏𝟎−𝟏𝟗 𝑪) (e.g. an electron or proton) through an electric potential difference of one volt”. Furthermore, the electron volt (𝑒𝑉) is related to the joule as follows: 1 𝑒𝑉 = 1.6 × 10−19 𝐽 (3 − 7) 3.4 Calculating the potential (𝑽) from the field (𝑬) Consider an arbitrary nonuniform electric field (𝐸), represented by the field lines in figure, and a positive test charge 𝑞𝑜 that moves along the path shown from point (𝑖) to point (𝑓). At ⃗⃗⃗𝑒 = 𝑞𝑜 𝐸⃗ ) acts on the any point on the path, an electric force (𝐹 ⃗⃗⃗⃗ ). We charge as it moves through a differential displacement (𝑑𝑠 know that the differential work (𝑑𝑊) done on a particle by a force during a displacement is given by the dot product of the Figure 3.2: Moveing of test charge (𝑞0 ) in an electric field. force and the displacement: 𝑑𝑊 = ⃗⃗⃗ ⃗⃗⃗⃗ = 𝑞𝑜 𝐸⃗ ∙ 𝑑𝑠 𝐹𝑒 ∙ 𝑑𝑠 ⃗⃗⃗⃗ To find the total work (𝑊) done on the particle by the field as the particle moves from point (𝑖) to point (𝑓), we integration the differential works done on the charge as it moves through all the displacements ⃗⃗⃗⃗⃗⃗ along the path: (𝑑𝑠) 𝑓 𝑊𝑖𝑓 = 𝑞𝑜 ∫𝑖 𝐸⃗ ∙ ⃗⃗⃗⃗ 𝑑𝑠 34 Chapter (3) Electrostatic Potential Energy 𝑓 𝑊𝑖𝑓 = ∫ 𝐸⃗ ∙ ⃗⃗⃗⃗ 𝑑𝑠 𝑞𝑜 𝑖 but, 𝑈𝑖𝑓 −𝑊𝑖𝑓 𝑉𝑖𝑓 = = 𝑞𝑜 𝑞𝑜 therefore, 𝑓 𝑊𝑖𝑓 𝑉𝑖𝑓 = − ⃗⃗⃗⃗ = − ∫ 𝐸⃗ ∙ 𝑑𝑠 𝑞𝑜 𝑖 or, 𝑓 𝑓 ⃗⃗⃗⃗ = − ∫ 𝐸 𝑑𝑠 cos 𝜃 𝑉𝑓 − 𝑉𝑖 = − ∫ 𝐸⃗ ∙ 𝑑𝑠 (3 − 8) 𝑖 𝑖 Thus, the potential difference (𝑉𝑓 − 𝑉𝑖 ) between any two points (𝑖) and (𝑓) in an electric field (𝐸) is equal to the negative of the line integral (meaning the integral along a particular path) of from (𝑖) to (𝑓). However, because the electric force is conservative, all paths (whether easy or difficult to use) produce the same result. Equation (3-8) helps us to calculate (∆𝑉𝑖𝑓 = 𝑉𝑓 − 𝑉𝑖 ) between any two points in the field. If we set potential (𝑉𝑖 = 0) at infinity (∞) point, then equation (3-8) becomes 𝑓 𝑉𝑓 − 𝑉𝑖 = − ∫ 𝐸⃗ ∙ ⃗⃗⃗⃗ 𝑑𝑠 (3 − 9) i In above equation we have dropped the subscript (𝑓) on (𝑉𝑓 ) term to obtain the potential (𝑉) at any point (𝑓) in the electric field relative to the zero potential at point (𝑖) was existed at infinity.