2024S1 PH1012 Physics A - Electromagnetism and Circuits - Electric Fields PDF

2024S1 PH1012: Physics A Electromagnetism and Circuits Electric Fields Dr Ho Shen Yong Lecturer, School of Physical and Mathematical Scie...

2024S1 PH1012: Physics A Electromagnetism and Circuits Electric Fields Dr Ho Shen Yong Lecturer, School of Physical and Mathematical Sciences Nanyang Technological University Week 8 "You can't depend on your judgment when your imagination is out of focus." - Mark Twain Itinerary 1. Electric fields and potentials (for point charges and parallel plates). 2. Electric fields, potentials, Gauss’s law 3. Electric potential and capacitance 4. Capacitors in circuits. 5. Electric fields in circuits; resistors and D.C circuits; 6. Kirchhoff's laws; RC circuits. 6. Magnetic Fields and Forces 7. Sources of Magnetic Fields 8. Electromagnetic Induction Self-check – after completing the lessons (lecture and tutorial) for this topic, are you able to 1. cite examples of electrostatics and the effects of charge transfer 2. state Coulomb’s law of electric forces and apply it solve problems in Physics 3. calculate the net force / field acting on a charge placed in a system of fixed charges or an external electric field 4. calculate electric field in problems and demonstrate understanding of the concept of electric force, field and flux; the relationship electric potential energy and electric firce 5. apply Gauss’s law to analyze the relationship between charge and electric fields 6. calculate electric potential, electric potential energy 1 and solve problems using energy conservation  Introduction In mechanics (dynamics), calculations often involve the physical property - mass of objects. In electricity and magnetism, the key physical property of objects that is of interest is electric charge. The subject of electricity and magnetism can be divided into three realms: 1) the charges that we study are at rest with respect to each other and to observers like us; 2) the charges are moving with a relatively constant average speed in the circuitry – “electricity”. Another phenomena that involves charges moving at constant velocity is the generation of magnetic fields. 3) the charges are accelerating – fields can be generated and energy radiated into space. Gravitational force is important on scale of planets and our everyday life. However, electromagnetic interaction is the dominant force at the microscopic level down to the atomic scale. It is the force responsible for holding the electrons and the nuclei together in atoms and all forms of molecular bonding – thus responsible for all of chemical and biological phenomena. Looking at our present reliance on electrical appliances, we can also safely say that the understanding and application of electricity and magnetism is one of the cornerstone in the modernization of human civilization. [email protected] The unit of charge: Coulomb (C) The SI unit of charge is Coulomb (C) which is defined as the amount of charge that flows in 1 second when there is a steady current of 1 ampere. The electron is an elementary charge carrier and has a charge of −𝑒 where 𝑒 = 1.60217733 × 10−19 𝐶. The charge of a proton is +𝑒. Thus, we see that one C is made up of many elementary charges. Giancoli pg 561 Fig 21.3: Simple model of atom 2  Examples of Daily Occurrences and Applications 1. Lightning 2. Electrical igniters at gas stove 3. Charging of cellphones, camera, flash etc 4. Computer Keyboards 5. Capacitive Touchscreens Aspects of Electrostatics https://www.youtube.com/ https://www.youtube.com/w watch?v=dwJ- atch?v=ViZNgU-Yt-Y&t=115s MM7yu4E&t=3s 3  Visualizing Charges (Knight) Pause to Ponder: A rod attracts a positively charged hanging ball. The rod is A. Positive. B. Negative. C. Neutral. D. Either A or C. E. Either B or C. 4  Charge Polarization (Knight) Charge polarization is the slight separation of the positive and negative charges in a neutral object when a charged object is brought near. The polarization force arises because the charges in a neutral object are slightly separated, not because the objects are oppositely charged. The polarization force between a charged object and a neutral one is always attractive. 5 Pause to Ponder: Metal spheres 1 and 2 are touching. Both are initially neutral. a. The charged rod is brought near. b. The charged rod is then removed. c. The spheres are separated. Afterward, the charges on the sphere are: A. Q1 is + and Q2 is + B. Q1 is + and Q2 is – C. Q1 is – and Q2 is + D. Q1 is – and Q2 is – E. Q1 is 0 and Q2 is 0 Pause to Ponder: Metal spheres 1 and 2 are touching. Both are initially neutral. a. The charged rod is brought near. b. The spheres are separated. c. The charged rod is then removed. Afterward, the charges on the sphere are: A. Q1 is + and Q2 is + B. Q1 is + and Q2 is – C. Q1 is – and Q2 is + D. Q1 is – and Q2 is – E. Q1 is 0 and Q2 is 0 6 Atomic view of charging (Knight) Molecular ions can be created when one of the bonds in a large molecule is broken. 7  Wimshurst Machine and Van der Graff generator https://commons.wikimedia.org/wiki/File:Wimshurst.jpg Hand-Cranked Laser – How it works: See Youtube TEA Laser Powered by Wimshurst Machine MIT Physics Demo – The Wimshurst Machine 8 https://en.wikipedia.org/wiki/Van_de_Graaff_generator  Newton’s Law of Gravitation and Coulomb’s Law Newton’s Law of Gravitation Coulomb’s Law on Electrostatic forces Coulomb’s Law: Gauss’s Law which was discussed for gravitational fields is also similarly applicable to electric fields. 9  Coulomb's law (discovered in 1785) The magnitude of the force (F) between two electrically charged bodies, which are small compared with their separation (r), is inversely proportional to r2 and proportional to the product of their charges (Q1 and Q2): 𝑄𝑄 𝐹 = 𝑘 122 𝑟 where the constant of proportionality k, in vacuum, expressed in SI units is given by 1 𝑘= 4𝜋𝜀 where 𝜖 is the permittivity. It is dependent on the medium in which the charges are in. For example, the permittivity of water at room temperature is about 80 times the permittivity of free space (vacuum) and thus the attractive forces between Na+ and Cl- is much weaker in water. The permittivity of free space (vacuum) −12 𝐶2 𝜀𝑜 = 8.85418781762 × 10. 𝑁 ∙ 𝑚2 For the purpose of simple computation in vacuum, we just use 1 𝑘= = 8.99 × 109 𝑁 ∙ 𝑚2/𝐶2 4𝜋𝜀𝑜 Thus, Coulomb’s law can be written as 𝑄𝑄 𝐹 = 1 2 2. 4𝜋𝜀𝑜𝑟 Taken from Knight, pg 801, Fig 26.17 10  Comparison with Gravitational force [Serway Ex. 23.1, pg 663] The electron and proton of a hydrogen atom are separated (on average) by a distance of approximately 5.3 × 10−11 𝑚. Compare the magnitudes of the electric force and gravitational force between the two particles. 𝑒 |−𝑒| (1.60×10−19 )2 𝐹𝑒 = 𝑘 = 8.99 × 109 = 8.2 × 10−8 𝑁. 𝑟2 5.3×10−11 2 𝑚𝑒𝑚𝑝 (9.11×10−31 )(1.67×10−27 ) 𝐹𝑔 = 𝐺 = 6.67 × 10−11 = 3.6 × 10−47 𝑁. 𝑟 2 5.3×10−11 2 𝐹𝑒Τ ≈ 2 × 1039. Thus, the gravitational attraction is negligible when 𝐹𝑔 considering electric forces. Example Giancoli page 565, Conceptual Example 21-1: (modified) Which charge exerts the greater force? Two positive point charges, Q1 = 50 μC and Q2 = 1 μC, are separated by a distance 𝑙 = 0.50 m. Calculate the force that Q1 exerts on Q2. What about the force that Q2 exerts on Q1, will it be larger?. Point Charges 11  Giancoli pg 566 Example 21-2: Three charges in a line Three charged particles are arranged in a line, as shown. Calculate the net electrostatic force on particle 3 (the -4.0 μC on the right) due to the other two charges. 𝐹Ԧ32 = 𝐹Ԧ31 = 𝐹Ԧ = 12 Point Charges  Giancoli pg 566 Example 21-3: Electric force using vector components. Calculate the net electrostatic force on charge 𝑄3 shown in the figure due to the charges 𝑄1 and 𝑄2. |𝐹Ԧ32 | = |𝐹Ԧ31 | = 𝐹𝑥 = 𝐹𝑦 = Ԧ = |𝐹| 𝜃= Point Charges 13  Electric field of a point charge The electric field exist in the region of space around a charged object, the source charge Q. To study the properties of the electric field, a very small positive test charge qo is placed in this region to probe the electric field without distorting it. The test charge qo will experience a force 𝐹റ 𝑒 due to the source charge. The electric field 𝐸 of the charge Q at a point is defined as 𝐹റ 𝑒 𝐸 ≡ lim 𝑞𝑜→0 𝑞𝑜 Remarks: General 1. 𝐸 has SI units N/C. 2. The direction of 𝐸 is the direction of force the positive test charge experiences. 3. The electric field is a property of the charge Q, independent of the test charge. 4. We can write the force 𝐹റ 𝑒 acting on any charge q in an electric field 𝐸 as 𝐹റ 𝑒 = 𝑞𝐸. This is true for any electric field. 5. For a point charge Q, 𝐹റ 𝑒 𝑄 𝐸= = 𝑟Ƹ 𝑞𝑜 4𝜋𝜀𝑜𝑟2 [email protected] The electric field can be represented by field lines. These lines start on a positive charge and end on a negative charge. Point Charges 14 Giancoli page 569, Example 21-6: Electric field of a single point charge. Calculate the magnitude and direction of the electric field at a point P which is 30 cm to the right of a point charge Q = -3.0 x 10-6 C. What will be the electric field strength at another point a) 15 cm away from Q (+ve)? b) 60 cm away from Q (+ve)? Point Charges Resultant electric fields of multiple charges in general General To find 𝐸 for a group of N point charges, the procedure is as follows: 1) Calculate 𝐸 𝑖 due to charge i at the given point as if it were the only charge present. 2) Add these separately calculated fields vectorially to find the resultant field 𝐸 at the point. In equation form, 𝐸 = 𝐸1 + 𝐸2 + 𝐸3 + ⋯ 3) The equation is an example of the application of superposition. However, this approach may fail when the magnitudes of the fields are extremely large (perhaps causing polarization of non-elementary source charges) but in the present course, it is always valid. 15 Gianocoli pg 570 Example 21-7: E at a point between two charges. Two point charges are separated by a distance of 10.0 cm. One has a charge of -25 μC and the other +50 μC. (a) Determine the direction and magnitude of the electric field at a point P between the two charges that is 2.0 cm from the negative charge. (b) If an electron (mass = 9.11 x 10-31 kg) is placed at rest at P and then released, what will be its initial acceleration (direction and magnitude)? Point Charges https://www.youtube.com/watch?v=p FM-a7_qjFM&t=3s Problem solving in electrostatics: electric forces and electric fields 1. Draw a diagram; show all charges, with signs, and electric fields and forces with directions. 2. Calculate forces using Coulomb’s law. 3. Add forces vectorially to get result. 4. Check your answer! General 16  Giancoli page 570, Example 21-8: Electric field of two point charges. Calculate the total electric field (a) at point A and (b) at point B in the figure due to both charges, Q1 and Q2. 𝑚2 9.0 × 109 𝑁 50 × 10−6 𝐶 𝑘𝑄2 𝐶2 𝐸𝐴2 = = = 𝑚2 9.0 × 109 𝑁 −50 × 10−6 𝐶 𝑘𝑄1 𝐶2 𝐸𝐴1 = = = 𝐸𝐴𝑥 = 𝐸𝐴𝑦 = 𝐸𝐴 = 𝜃= Solve for 𝐸𝐵 yourself 17 Point Charges Electric fields of various systems of charges Parallel Plates https://www.youtube.com/w atch?v=pOPQBlknBcM&t=3s https://phet.colorado.edu/ en/simulati on/charges- and-fields Point Charges 18 Giancoli pg 572 Example 21-9: A ring of charge. [Non-Examinable] A thin, ring-shaped object of radius a holds a total charge +Q distributed uniformly around it. Determine the electric field at a point P on its axis, a distance x from the center. Let λ be the charge per unit length (C/m). 𝑟= 𝑎2 + 𝑥 2 cos 𝜃 = sin 𝜃 = 𝑄 A small segment of length 𝑑𝑙 has charge 𝑑𝑞 = 𝜆𝑑𝑙 where 𝜆 = 2𝜋𝑎 The contribution of charge 𝑑𝑞 to the magnitude of electric field at 𝑃 is 𝑘𝑑𝑞 𝑘𝜆𝑑𝑙 𝑑𝐸 = = 2 𝑟2 𝑟 𝑑𝐸 has two components - 𝑑𝐸⊥ and 𝑑𝐸𝑥 Summing up all the components perpendicular (⊥) to 𝑥 direction, we get 𝐸⊥ = 0 Summing up all the components along the 𝑥 direction, we get 𝑑𝑙 𝐸𝑥 = ∫ 𝑑𝐸𝑥 = න𝑑𝐸 cos 𝜃 = 𝑘𝜆 න cos 𝜃 𝑟2 2𝜋𝑎 𝑘𝜆 = 2 cos 𝜃 න 𝑑𝑙 = 𝑟 0 Check for reasonableness – when 𝑥 ≫ 𝑎, 𝐸 = Point Charges -> Ring of charges 19  2024S1 PH1012: Physics A Electric Potential Dr Ho Shen Yong Lecturer, School of Physical and Mathematical Sciences Nanyang Technological University Week 8 Today is cruel and tomorrow is crueller. The day after tomorrow will be wonderful; however, most people die on tomorrow night and won’t see the sunshine the day after tomorrow. — From Ma Yun, CEO of Alibaba Point Charges Parallel Plates 2 General Δ𝑈 𝑥 = 𝑈2 − 𝑈1 = − න 𝐹𝑑𝑥 1 Point Charges 𝑄1𝑄2 𝑄1 𝑄2 1 𝐹റ 𝑒 = 𝑄2 𝐸 𝐹= 𝑈=. 𝑈 = 𝑄2 𝑉 4𝜋𝜀𝑜𝑟2 4𝜋𝜖𝑜 𝑟 𝑄1 1𝑄 1 𝐸= 𝑉 = 4𝜋𝜖. 4𝜋𝜀𝑜𝑟2 𝑜 𝑟 General 𝜕𝑉 General 𝐸𝑟 = − 𝐸 = 𝐸1 + 𝐸2 + 𝐸3 + ⋯ 𝜕𝑟 𝑉 = 𝑉1 + 𝑉2 + ⋯ https://www.youtube. https://www.youtu com/watch?v=9 be.com/watch?v=1 kNkpQTwPc0&t=2s XI4D4SgHTw&t=2s Electric force and electric potential energy Earlier we learnt that the change in potential energy associated with a particular conservative force 𝐹Ԧ as the negative of the work done by that force: 2 Δ𝑈 = 𝑈2 − 𝑈1 = − න 𝑭. 𝒅𝒍Ԧ = −𝑊 1 Remembering that integration and differentiation are inverse operations” 𝑑 න𝐹(𝑥) 𝑑𝑥 = 𝐹(𝑥) 𝑑𝑥 Thus, 𝑑𝑈 𝑥 𝐹 𝑥 =− General 𝑑𝑥 The same is applicable to the charges and electric field. Previously, it was mass and the gravitational field. Thus, electric potential energy can be written as 2 Δ𝑈𝐸 = 𝑈2 − 𝑈1 = − න 𝑭𝑬. 𝒅𝒍Ԧ General 1 We will consider the simplest electric field, the uniform field between two parallel plates where the electric field is a constant value 𝐸 anywhere in between the plates. 𝑏 Δ𝑈𝐸 = 𝑈𝑏 − 𝑈𝑎 = − න 𝑭𝑬. 𝒅𝒍Ԧ 𝑎 𝑏 𝟎 − 𝑼𝒂 = - ∫𝑎 𝒒𝑬𝒅𝒙 𝒄𝒐𝒔 𝟎∘ Parallel Plates As in gravitational PE, we need to define a point as zero PE. If the charge is –ve instead, the change in electrical potential energy moving from a to b will be Parallel Plates 21 Electric potential Electric potential is defined as electrical potential energy per unit charge: 𝑈𝐸 𝑉= General 𝑞 Unit of electric potential: the volt (V): 1 V = 1 J/C. General A simple numerical example: If the separation of the two parallel plate is 𝑑 = 10 cm, what is the electric potential at a point 3 cm from the positive plate? Parallel Plates What is the electric potential energy of an object with charge 𝑞 = 2 nC at this point? If the mass of the object is 0.01 kg, what is its velocity just before it hits the –ve plate? 22 Electric potential energy 2 Consider two charges +𝑄1 and +𝑄2 at a distance 𝑟𝑖 apart. The position of one of the charge is fixed while the other is free to move starting from rest at point i. A repulsive force of magnitude 𝑄 𝑄 𝐹Ԧ = 1 2 2 𝑟Ƹ Point Charges 4𝜋𝜀 𝑟𝑜 acts between the protons while the free proton moves to a new position 𝑟𝑓. 𝑟𝑖 𝑟𝑓 𝑄1 𝑄2 + + + Fixed i f The change in potential energy of charge 𝑄2 in the electric field of the fixed charge 𝑄1 is given by 𝑟𝑓 𝑟𝑓 Δ𝑈𝐸 = − න 𝐹Ԧ. 𝑑 𝑟Ԧ = − න 𝐹Ԧ 𝑑 𝑟Ԧ cos 0∘ General 𝑟𝑖 𝑟𝑖 𝑟𝑓 𝑄1 𝑄2 = −න 𝑑𝑟 𝑟𝑖 4𝜋𝜀𝑜𝑟2 𝑄1 𝑄2 𝑟𝑓 1 𝑄1 𝑄2 1 1 =− න 2 𝑑𝑟 = − − − − 4𝜋𝜀𝑜 𝑟𝑖 𝑟 4𝜋𝜀𝑜 𝑟𝑓 𝑟𝑖 𝑄1 𝑄2 1 1 = − = 4𝜋𝜀𝑜 𝑟𝑓 𝑟𝑖 Point Charges As we would expect because the loss in electrical potential energy is transformed to kinetic energy. If we fix the electrical potential energy at 𝑟𝑓 = ∞ as 𝑈𝐸 = 0, then 𝑄1 𝑄2 1 1 𝑈𝑓 − 𝑈𝑖 = − Point Charges 4𝜋𝜀𝑜 𝑟𝑓 𝑟𝑖 𝑄1 𝑄2 1 ⇒ for a general 𝑟 for Point Charges, 𝑈 =. As before, we can define 4𝜋𝜀𝑜 𝑟 the electric potential at a distance 𝑟 from of a point charge 𝑄 as 𝑈 𝑄 𝑉= = Point Charges 𝑞 4𝜋𝜀𝑜 𝑟 23  Electric Field of Point charges Point Charges Giancoli pg 616, fig 23.17 The diagram shows the electric field lines and the equipotential lines of a positive point charge Giancoli pg 613, fig 23.10/11 The electric potential of a charge 𝑄 as a function of distance 𝑟. Forces, Fields are vectors / scalars General Potential energy, potential are vectors / scalars Uniform Electric Field 24 Parallel Plates  Electric Field and Equipotential Lines Parallel-plate Point charge Dipole Electric Field Electric Potential 25 Giancoli page 610, Example 23-2: Electron in CRT. Suppose an electron in a cathode ray tube is accelerated from rest through a potential difference Vb – Va = Vba = +5000 V. (a) What is the change in electric potential energy of the electron? (b) What is the speed of the electron (m = 9.1 × 10-31 kg) as a result of this acceleration? 26 Modified Giancoli pg 566 Example 21-2: Two charges in a line Two charged particles are arranged in a line, as shown. Calculate the electric potential at the point mark “x". x If a charge 𝑄3 (= +4.0 𝜇𝐶) is placed at “x”, calculate the electric potential energy of 𝑄3. The particle with 𝑄3 , mass 3.2 × 10−14 kg, is released from rest. What is its velocity when it is 0.50 𝑚 from 𝑄2 ? 2.6 × 106 𝑚/𝑠 27 Modified Giancoli pg 566 Example 21-3: Calculate the electric potential energy of charge 𝑄3 shown in the figure due to the charges 𝑄1 and 𝑄2. Electric potential V We can also write the vectorial form of the definition for two points in 3-D space 𝑏 𝑏 𝑉𝑏 − 𝑉𝑎 = − න 𝐸. 𝑑 𝑠റ = − න(𝐸𝑥 𝑑𝑥 + 𝐸𝑦𝑑𝑦 + 𝐸𝑧𝑑𝑧) 𝑎 𝑎 Conversely, given the potential V(x, y, z), we can obtain the components of the electric field 𝜕𝑉 𝜕𝑉 𝜕𝑉 𝐸𝑥 = − ; 𝐸𝑦 = − ; 𝐸𝑧 = − 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑉 Here, 𝜕𝑥 denotes a partial derivative – differentiating w.r.t. x and treating variables 𝜕𝑉 𝜕𝑉 y and z in V as constants. It applies similarly to 𝜕𝑦 and 𝜕𝑧. If the electric field only varies in the radial direction with r, then 𝜕𝑉 𝐸𝑟 = −. 𝜕𝑟 The negative sign is associated with the fact that the electric field is always pointing from higher potential to lower potential. We note that another unit for the electric field is volts per metre (V/m). 28 Electric Field and Electric Potential http://dev.physicslab.org/Document.aspx?doctype=3&filename=Electrostatics_PointChargesEPE.xml i. Where is the electric field strongest? L, M, N, R, S, T, U ii. Where is the electric field weakest? L, M, N, R, S, T, U iii. What is the direction of the electric field at R? iv. How much work would it take an external agent to move a charge 2 𝜇C from R to N? v. Does it take more work to move a 2 µC charge from R to L and then to T compared to going directly to T? vi. How much EPE did the 2 µC charge have while it was at rest at position R? vii. What is the 2 µC charge’s EPE at point T? viii. How much work was required to move the 2 µC charge from R to T? 29  Force, Field and Flux 𝑛ො Area A Force, 𝐹Ԧ Field Lines, 𝐸 Flux Φ𝐸 𝐹Ԧ = 𝑞𝐸 Φ𝐸 = න𝐸. 𝑑 𝐴መ 𝑑𝑉 𝐸=− 𝑑𝑟 Force + Field 𝑑𝑈 𝐹Ԧ = − 𝑑𝑟 Equipotential Lines 𝑉 𝑈 = 𝑞𝑉 Using dipole as example Potential Energy [email protected]  2024S1 PH1012: Physics A Gauss’s Law Dr Ho Shen Yong Lecturer, School of Physical and Mathematical Sciences Nanyang Technological University Week 8/9 "Genius is one percent inspiration, ninety-nine percent perspiration." - Thomas Alva Edison (1847-1931) E-field / potential of Gauss’s Law system of charges Electrostatic https://www.youtub equilibrium e.com/watch?v=p qTfk9HMLj4&t=2s Three key charge distributions related charge storing configurations Giancoli pg 573, Example 21.11: Electric field due to a long line of charge Determine the magnitude of the electric field at any point P a distance x from a very long line (a wire, say) of uniformly distributed charge. Assume x is much smaller than the length of the wire, and let λ be the charge per unit length (C/m). 1 𝑑𝑄 1 𝜆𝑑𝑦 𝑑𝐸 = = 4𝜋𝜀𝑜 𝑟 2 4𝜋𝜀𝑜 (𝑥 2 + 𝑦 2 ) 𝑃 is along bisector of rod so 𝑦 component of 𝐸 is zero. 𝐸𝑦 = ∫ 𝑑𝐸 sin 𝜃 = 0 Along the 𝑥 direction, 𝜆 cos 𝜃 𝑑𝑦 𝐸𝑥 = ∫ 𝑑𝐸 cos 𝜃 = න 2 =… 4𝜋𝜀𝑜 𝑥 + 𝑦2 For an infinitely long wire, 𝜆 𝐸= 2𝜋𝜀𝑜 𝑥 Is there an easier way to work out the electric field of an infinitely long line of charge? 32 Giancoli pg 592 Example 22-1 Calculate the electric flux through the rectangle shown. The rectangle is 10 cm by 20 cm, the electric field is uniform at 200 N/C, and the angle θ is 30°. Non-uniform field and/or flux through curved surfaces In general, when the field is not uniform and/or if the surface is not flat, we have to divide up the chosen surface into 𝑛 smaller patches of area Δ𝐴1 , Δ𝐴2 , … small enough that 1) it can be considered (nearly) flat 2) the variation of the field over the selected area is (nearly) constant. Thus, the entire (electric) flux ΦE over the surface is 𝑛 Φ𝐸 ≈ ෍ 𝐸𝑖. Δ𝐴Ԧ𝑖 𝑖=1 where 𝐸 is the field passing through area Δ𝐴Ԧ𝑖. In the limit, where the elemental area is infinitesimally small Δ𝐴Ԧ𝑖 → 0, the 𝐸 is constant and we have to move from a discrete sum to continuous sum Φ𝐸 = න𝐸. 𝑑 𝐴Ԧ. Later, we will be dealing with total flux through closed surface – a surface of any shape that completely encloses a volume of space, we write Φ𝐸 = ර𝐸. 𝑑 𝐴Ԧ Here, the symbol ‫ׯ‬ denotes an integral over a closed surface. 33  Revising the Concept of Flux Heavy Rain – “high rain flux per unit area” Light Rain – “low rain flux per unit area” (high flux density) (low flux density) http://99fm.com.na/unseasonably-heavy-rain-expected/ http://www.chachacharming.com/music/i-like-london-in-the-rain-mix/ 𝑓𝑙𝑢𝑥 𝑑𝑒𝑛𝑠𝑖𝑡𝑦, 𝑓Ԧ 𝑛ො Area A 𝜃 𝜃′ Vector area 𝐴𝑛ො Projected area “shadow” [email protected] Total flux “captured” by area Φ = “flux density”× Projected Area = “flux density”× Area 𝑐𝑜𝑠 𝜃 = 𝐴 𝑓Ԧ 𝑛ො cos 𝜃 = 𝑓.Ԧ 𝐴𝑛ො Earlier discussion on gravitational fields: Ԧ For curved surface, 𝑑 𝐴 = 𝐴𝑑𝑛 Ԧ 𝑑 𝐴Ԧ 𝑇𝑜𝑡𝑎𝑙 𝑓𝑙𝑢𝑥, Φ = න𝑓. For closed curved surface, (convention, outwards +ve) Ԧ 𝑑 𝐴Ԧ 𝑇𝑜𝑡𝑎𝑙 𝑓𝑙𝑢𝑥, Φ = ර𝑓. Thus, for electric fields, the total electric flux Φ𝐸 http://commons.wikimedia.or passing through an imaginary surface g/wiki/File:Gravitational_field_ Earth_lines_equipotentials.svg (“Gaussian surface”) is given by Φ𝐸 = ර𝐸. 𝑑 𝐴Ԧ Ԧ 𝑑𝐴Ԧ Φ𝐺 = ර𝑔. 34 A simple example This all can look complicated later but let us try it on something not too difficult first. Consider a charge 𝑄 = 1.2nC enclosed by an imaginary sphere of radius 10 cm with the charge in the centre of the sphere. Calculate the total electric flux that goes through the surface of the sphere. [Note here that electric field the magnitude of the electric field is constant on the surface and the elemental area 𝑑 𝐴Ԧ is parallel to the 𝐸 at all points on the sphere.] [Convention: Flux leaving enclosed volume is +ve Flux entering enclosed volume is –ve. This convention is consistent with E-field pointing away from +ve charge and towards –ve charge.] What is the total electric flux if a large sphere of 20 cm is used? What is the total electric flux if Q is replaced with a –ve charge with the same magnitude? How will the total electric flux change if the charge is placed in the centre of a cube of sides 6 cm instead? How will the total electric flux change if the charge is not in the center of sphere or the cube? 35 Giancoli pg 593, 22-2 Gauss’s Law The precise relationship between the electric flux through a closed surface (Gaussian surface) and the net charge enclosed within that surface is given by Gauss’s Law: 𝑄𝑒𝑛𝑐 Φ𝐸 = ර𝐸. 𝑑𝐴Ԧ = 𝜖 where 𝜖 is permittivity as mentioned previously. In principle, Gauss’s law can be used to find the expression for 𝐸 for a system of charge or charge distribution. However, in practice, the application of Gauss’s law is only limited to a number of highly symmetric electric fields. On the other hand, Gauss’s law gives us a simple perception so that we can deal easily with situations such as when there are charges within the interior of a hollow sphere etc. Why was that point charge / sphere example simple? 1. The magnitude of the electric field is constant on the surface and 2. the elemental area 𝑑𝐴Ԧ is parallel to the 𝐸 at all points on the surface. What are the other simple examples? 1. Point (Charge) 2. (Very long) Line (Charge per unit length, 𝜆) 3. (Very Wide) Plane (Charge per unit area, 𝜎) 4. 3D Sphere (Charge per unit volume, 𝜌) 36 Symmetry and regions of constant electric field Total flux through surface of ‘imaginary’ sphere of radius 𝑟 = 𝟒𝝅𝒓𝟐 𝑬𝒑𝒕 Total charge enclosed = 𝑸 Applying Gauss’s Law, Wire with uniform charge per unit length = 𝜆 Total flux through surface of ‘imaginary’ cylinder of radius 𝑟 and length 𝑙 = 2𝝅𝑹𝒍𝑬𝑳 Total charge enclosed = 𝝀𝒍 Applying Gauss’s Law, Conducting Plate with charge per unit area = 𝜎 Total flux through surface of ‘imaginary’ disc of cross sectional area 𝐴 = 𝟐𝑬𝒑𝒍 𝑨 Total charge enclosed = 𝝈𝑨 Applying Gauss’s Law, Solid sphere of insulator with uniform charge density = 𝜌 Total flux through surface of ‘imaginary’ sphere with radius 𝑟1 (> 𝑟𝑜 ) = 𝟒𝝅𝒓𝟐𝟏 𝑬𝒔𝒑,𝒆𝒙𝒕 𝟒 Total charge enclosed = 𝝆 𝝅𝒓𝟑𝒐 =𝑸 𝟑 Applying Gauss’s Law, 37 Spherically Symmetric Charge Distribution [Giancoli pg 596 Example 22-4] An electric charge Q is distributed uniformly throughout a nonconducting sphere of radius r0. Determine the electric field (a) outside the sphere (r > r0) and (b) inside the sphere (r < r0). Mathematical concept test: A sphere has mass M. What is the mass of a sphere made of the same material but double the radius? b) Total flux (outwards) through Gaussian surface 𝐴2 = _______________ Total charge enclosed by Gaussian surface 𝐴2 = _______________ Applying Gauss’s Law: c.f Notes on gravitation – what is the g-field inside Earth. 38 Electric Field inside conductor at Electrostatic Equilibrium I [Giancoli pg 577, 598-600] Electrostatic equilibrium – The electric field inside a conductor is zero when the conductor is in electrostatic equilibrium - charges are at rest. If there is a non-zero electric field within the conductor, there would be a force on the free electrons. The electrons would move until they reached positions where the electric field, hence electric force on them, become zero. A few consequences at electrostatic equilibrium: 1. The electric field is zero everywhere inside the conductor (hollow or solid). 2. If conductor is isolated, any net charge will distribute itself on the surface. − − Recall − − − − − − − 𝐸=0 − − − conductor Charges deposited Charges redistributed on on conductor conductor such that E- field in conductor is zero Google homework: Faraday cage. 3. If a charge (say +𝑄) is surrounded by an isolated uncharged conducting spherical shell, applying Gauss’s law with the Gaussian surface inside the conductor, We can infer that there is Since the shell is neutral, Though no field exists inside conductor, an E-field exists outside of it, as if the conducting spherical shell were not there. 39 Electric Field inside conductor at Electrostatic Equilibrium II [Giancoli pg 577, 598-600] A few consequences at electrostatic equilibrium: 4. The electric field near the surface of the conductor points perpendicular to the surface. If the electric field 𝐸 at the surface of a conductor had a component parallel to the surface 𝐸|| it would exert a force on the electrons at the surface causing electrons to move along surface until the net force on the system of free charges become zero. Only the component of the electric field perpendicular to the surface 𝐸Ԧ⊥ will not vanish. 𝐸⊥ ≠ 0 𝐸|| = 0 No in electrostatic equilibrium In electrostatic equilibrium 5. For irregularly shaped conductor, the surface density is greatest at locations where the radius of curvature of the surface is smallest. http://www.kshitij-iitjee.com/electric-potential-due-to- charged-conductor 40 Giancoli pg 577 Conceptual Example 21-14: Shielding, and safety in a storm. A neutral hollow metal box is placed between two parallel charged plates as shown. What is the field like inside the box? Three possible cases whereby 𝚽𝐄 = 𝟎 For 𝐴2 Here, Φ𝐸 = ර 𝐸 ∙ d𝐴റ = 0 Φ𝐸 = ර 𝐸 ∙ d𝐴റ = Φ𝐸 = ර 𝐸 ∙ d𝐴റ = 0 and here 𝐸 and here 𝐸 and here 𝐸 Net charge enclosed = Net charge enclosed = Net charge enclosed = 41 Electric Field of Parallel plates 42