Physics Lecture Notes - 187-267 Pages PDF

188 Content of the Course ❖ Chapter 1: Physical Quantity, Units and Dimensions (1 week) ❖ Chapter 2: Motion in one Dimension and (2 week) ❖ Chapter 3: Vector Analysis (2 weeks) ❖ Chapter 4: Fluid Mechanics (2 weeks) ❖ Chapter 5: Waves, Oscillations and Sound (2 weeks) ❖ Chapter 6: Heat and Thermodynamic (2 weeks) ❖ Chapter 7: Electricity & Magnetism (2 weeks) Physics Prof. Dr. E. F. Abo Zeid 10/5/2024 1- Electrical charge and coulomb`s law 1-1: PROPERTIES OF ELECTRIC CHARGES A number of simple experiments demonstrate the existence of electric forces and charges. For example, after running a comb through your hair on a dry day, you will ﬁnd that the comb attracts bits of paper. The attractive force is often strong enough to suspend the paper. The same effect occurs when materials such as glass or rubber are rubbed with silk or fur. When materials behave in this way, they are said to be electriﬁed, or to have become electrically charged. 189 Prof. Dr. E. F. Abo Zeid 10/5/2024 In a series of simple experiments, it is found that there are two kinds of electric charges, which were given the names positive and negative by Benjamin Franklin (1706–1790). To verify that this is true, consider a hard rubber rod that has been rubbed with fur and then suspended by a nonmetallic thread. When a glass rod that has been rubbed with silk is brought near the rubber rod, the two attract each other. 190 Prof. Dr. E. F. Abo Zeid 10/5/2024 This observation shows that the rubber and glass are in two different states of electriﬁcation. On the basis of these observations, we conclude that like charges repel one another and unlike charges attract one another. 191 Prof. Dr. E. F. Abo Zeid 10/5/2024 Quick Quiz 1. Object A is attracted to object B. If object B is known to be positively charged, what can we say about object A? (a) It is positively charged. (b) It is negatively charged. (c) It is electrically neutral. (d) Not enough information to answer. 192 Prof. Dr. E. F. Abo Zeid 10/5/2024 COULOMB’S LAW Charles Coulomb (1736–1806) measured the magnitudes of the electric forces between charged objects using the torsion balance. Coulomb’s experiments showed that the electric force between two stationary charged particles r q1 q2 is inversely proportional to the square of the separation r between the particles and directed along the line joining them; is proportional to the product of the charges q1 and q2 on the two particles; is attractive if the charges are of opposite sign and repulsive if the charges have the same sign. 193 Prof. Dr. E. F. Abo Zeid 10/5/2024 The above equation is called Coulomb’s law, which is used to calculate the force between electric charges. In that equation F is measured in Newton (N), q is measured in unit of coulomb (C) and r in meter (m).The constant Ke can be written as where o is known as the Permittivity constant of free space. o = 8.85 x 10-12 C2/N.m2 194 Prof. Dr. E. F. Abo Zeid 10/5/2024 Charge and Matter Particle Symbol Charge Mass Proton P 1.6 x10-19C 1.67 x 10-27Kg Positive Neutron N 0 1.67 x 10-27Kg Neutral Electron E -1.6 x 10-19C 9.11×10-31Kg negative 195 Prof. Dr. E. F. Abo Zeid 10/5/2024 EXAMPLE 2. The Hydrogen Atom The electron and proton of a hydrogen atom are separated (on the average) by a distance of approximately 5.3x10 -11 m. Find the magnitudes of the electric force and the gravitational force between the two particles. Solution: From Coulomb’s law, we ﬁnd that the attractive electric force has the magnitude (G=𝟔. 𝟔𝟕 × 𝟏𝟎−𝟏𝟏 N.m2/Kg2) Fe= 8.2 x 10-8 N Fg= 3.6 x 10-47 N 196 Prof. Dr. E. F. Abo Zeid 10/5/2024 EXAMPLE 3. Object A has a charge of 2 μC, and object B has a charge of 6 μ C. Which statement is true? (a) FAB = -3FBA (b) F AB = - FBA (c) 3FAB = FBA 197 Prof. Dr. E. F. Abo Zeid 10/5/2024 Electric force between two electric charges The same value The opposite sign 198 Prof. Dr. E. F. Abo Zeid 10/5/2024 EXAMPLE 4. Calculate the value of two equal charges if they repel one another with a force of 0.1N when situated 50 cm apart in a vacuum. Solution Since q1=q2 q = 1.7x10-6C = 1.7μC 199 Prof. Dr. E. F. Abo Zeid 10/5/2024 Electric Fields 206 Review of gravitational fields Electric field vector Electric fields for various charge configurations Field strengths for point charges and uniform fields Work done by fields & change in potential energy Potential & equipotential surfaces Capacitors, capacitance, & voltage drops across capacitors Millikan oil drop experiment Excess Charge Distribution on a Conductor Prof. Dr. E. F. Abo Zeid 10/5/2024 Gravitational Fields: Review Recall that surrounding any object with mass, or collection of objects with mass, is a gravitational field. Any mass placed in a gravitational field will experience a 207 gravitational force. We defined the field strength as the gravitational force per unit mass on any “test mass” placed in the field: g = F / m. g is a vector that points in the direction of the net gravitational force; its units are N / kg. F is the vector force on the test mass, and m is the test mass, a scalar. g and F are always parallel. The strength of the field is independent of the test mass. For example, near Earth’s surface mg / m = g = 9.8 N / kg for any mass. Some fields are uniform (parallel, equally spaced fields lines). Nonuniform fields are stronger where the field lines are closer together. uniform field 10 kg nonuniform field Earth 9.8 N F Earth’s surface m Prof. Dr. E. F. Abo Zeid 10/5/2024 Electric Fields: Intro Surrounding any object with charge, or collection of objects with charge, is a electric field. Any charge placed in an electric field will experience a electrical force. We 208 defined the field strength as the electric force per unit charge on any “test charge” placed in the field: E = F / q. E is a vector that points, by definition, in the direction of the net electric force on a positive charge; its units are N / C. F is the vector force on the test charge, and q is the test charge, a scalar. E and F are only parallel if the test charge is positive. Some fields are uniform (parallel, equally spaced fields lines) such as the field on the left formed by a sheet of negative charge. Nonuniform fields are stronger where the field lines are closer together, such as the field on the right produced by a sphere of negative charge. uniform field +q nonuniform field - F F + -------------- q Prof. Dr. E. F. Abo Zeid 10/5/2024 Overview of Fields Charge, like mass, is an intrinsic property of an object. Charges produce electric 209 fields that affect other charges; masses produce gravitational fields that affect other masses. Gravitational fields lines always point toward an isolated mass. Unlike mass, though, charges can be positive or negative. Electric field lines emanate from positive charges and penetrate into negative charge. We refer to the charge producing a field as a field charge. A group of field charges can produce very nonuniform fields. To determine the strength of the field at a particular point, we place a small, positive test charge in the field. We then measure the electric force on it and divide by the test charge: E = F / q. Prof. Dr. E. F. Abo Zeid 10/5/2024 Electric & Gravitational Fields Compared 210 Field Intrinsic Gravity: Force SI units strength Property Electric g = W / m N / kg Force: E = FE / q N/C Field strength is given by force per unit mass or force per unit charge, depending on the type of field. Field strength means the magnitude of a field vector. Ex #1: If a +10 C charge is placed in an electric field and experiences a 50 N force, the field strength at the location of the charge is 5 N/C. The electric field vector is given by: E = 5 N/C, where the direction of this vector is parallel to the force vector (and the field lines). Ex #2: If a -10 C charge experiences a 50 N force, E = 5 N/C in a direction opposite the force vector (opposite the direction of the field lines). Prof. Dr. E. F. Abo Zeid 10/5/2024 Coulomb’s Law Review The force that two point charges, Q and q, separated by a 215 distance r, exert on one another is given by: K Qq where K = 9  109 Nm2/C2 (constant). F= 2 r This formula only applies to point charges or spherically symmetric charges. Suppose that the force two point charges are exerting on one another is F. What is the force when one charge is tripled, the other is doubled, and the distance is cut in half ? Answer: 24 F Prof. Dr. E. F. Abo Zeid 10/5/2024 Field Strengths: Point Charge; Point Mass Suppose a test charge q is placed in the electric field produced by a point-like field charge Q. From the definition of electric field and Coulomb’s law Note that the field strength is independent of the charge placed in it. Suppose a test mass m is placed in the gravitational field produced by a point-like field mass M. From the definition of gravitational field and Newton’s law of universal gravitation Again, the field strength is independent of the mass place in it. Prof. Dr. E. F. Abo Zeid 10/5/2024 Electric Field of a Continuous Charge Distribution 1 Δ𝑞 ❑ Find an expression for dq: Δ𝐸 = 𝑟Ƹ ◼ dq = λdl for a line distribution 4𝜋𝜀0 𝑟 2 ◼ dq = σdA for a surface distribution ◼ dq = ρdV for a volume distribution 𝐸≈ 1 Δ𝑞𝑖 ෍ 2 𝑟𝑖Ƹ ❑ Represent field contributions at P due 4𝜋𝜀0 𝑖 𝑟𝑖 to point charges dq located in the distribution. Use symmetry, 𝑑𝑞 𝑑𝐸 = 𝑟Ƹ 1 Δ𝑞𝑖 1 𝑑𝑞 4𝜋𝜀0 𝑟 2 𝐸= lim ෍ 2 𝑟𝑖Ƹ = න 2 𝑟Ƹ 4𝜋𝜀0 Δ𝑞→0 𝑖 𝑟𝑖 4𝜋𝜀0 𝑟 ❑ Add up (integrate the contributions) over the whole distribution, varying the displacement as needed, 𝐸 = න𝑑𝐸 10/5/2024 Example: Electric Field Due to a Charged Rod ❑ A rod of length L has a uniform positive charge per unit length λ and a total charge Q. Calculate the electric field at a point P that is located along the long axis of the rod and a distance a from one end. ❑ Start with 𝑑𝑞 = 𝜆𝑑𝑥 1 𝑑𝑞 1 𝜆𝑑𝑥 𝑑𝐸 = 2 = 4𝜋𝜀0 𝑥 4𝜋𝜀0 𝑥 2 ❑ then, 𝑙+𝑎 𝑙+𝑎 𝑙+𝑎 𝜆 𝑑𝑥 𝜆 𝑑𝑥 𝜆 1 𝐸=න = න 2= − 4𝜋𝜀0 𝑥 2 4𝜋𝜀0 𝑥 4𝜋𝜀0 𝑥 𝑎 𝑎 𝑎 ❑ So ❑ Finalize 1 𝑄 1 1 𝑄 ◼ l => 0 ? 𝐸= − = 4𝜋𝜀0 𝑙 𝑎 𝑙 + 𝑎 4𝜋𝜀0 𝑎(𝑙 + 𝑎) ◼ a >> l ? 10/5/2024 Motion of a Charged Particle in a Uniform Electric Field 𝐹Ԧ = 𝑞𝐸 ❑ If the electric field E is uniform (magnitude and direction), the electric force F on the particle is constant. 𝐹Ԧ = 𝑞𝐸 = 𝑚𝑎Ԧ 𝑞𝐸 ❑ If the particle has a positive charge, its 𝑎Ԧ = 𝑚 acceleration a and electric force F are in the direction of the electric field E. ❑ If the particle has a negative charge, its acceleration a and electric force F are in the direction opposite the electric field E. 10/5/2024 Prof. Dr. E. F. Abo Zeid 219 Quiz1. If the electric field E is uniform (magnitude and direction), the electric force F on the particle is( constant, decreases, increases) Quiz2. If the particle has a positive charge, its acceleration a and electric force F are in the (same direction- opposite direction- perpendicular direction) of the electric field E. Quiz3. Calculate the acceleration a of a particle moving in an electric field E? 220 Prof. Dr. E. F. Abo Zeid 10/5/2024 Work-Energy Example Here the E field is to the right and approximately uniform. The applied force is FA to the left, as is the displacement. 226 The work done by FA is + FA d. The work done by the field is WF = - q E d. The change in electric potential energy is U = - WF = + q E d. Since FA > q E, the applied force does more positive work than the field does negative work. The difference goes into kinetic energy and heat. The work done by friction is Wfric < 0. So, Wnet = FA d - q E d - |Wfric| = K by the work-energy theorem. d + - + - + FA qE - + + - + q - 10/5/2024 Prof. Dr. E. F. Abo Zeid Capacitors - Overview A capacitor is a device that stores electrical charge. 227 A charged capacitor is actually neutral overall, but it maintains a charge separation. The charge storing capacity of a capacitor is called its capacitance. An electric field exists inside a charged capacitor, between the positive and negative charge separation. A charged capacitor store electrical potential energy. Capacitors are ubiquitous in electrical devices. They’re used in power transmission, computer memory, photoflash units in cameras, tuners for radios and TV’s, defibrillators, etc. Prof. Dr. E. F. Abo Zeid 10/5/2024 capacitor Parallel Plate Capacitor -Q +Q 228 Area, A The simplest type of capacitor is a parallel plate capacitor, which consists d of two parallel metal plates, each of area A, separated by a distance d. As a V whole wire battery As a whole, the capacitor remains neutral, but we say it now has a charge Q, the amount of charge moved from C one plate to the other. As soon as the voltage drop across the capacitor (the potential difference between its plates) is the same as +Q -Q that of the battery, V, the charging ceases. The capacitor can remain charged even when i disconnected from the battery. Note the symbols used in V the circuit diagram to the right. Prof. Dr. E. F. Abo Zeid 10/5/2024 Parallel Plate Capacitor: E & U 229 Outside the plates the field is very weak. The strength of the E field inside is proportional to how much charge is on the capacitor and inversely proportional to how the area of the capacitor. Touching a charged capacitor will allow it to discharge quickly and will result in a shock. Once discharged, the electric field vanishes, and the potential energy is converted to some other form. - - - - - - - - + + + Prof. Dr. E.+F. Abo Zeid+ 10/5/2024 + Capacitance Capacitance, C, is the capacity to store charge. The amount of charge, Q, stored on given capacitor depends on the 230 potential difference between its plates, V, and its capacitance C. In other words, Q is directly proportional to V, and the constant of proportionality is C: Q = CV Ex: A 12 V battery will cause a capacitor to store twice as much charge as a 6 V battery. Also, if C capacitor #2 has twice the capacitance of capacitor #1, then #2 will store twice as much charge as #1, provided they are charged by the same battery. C Q depends on the type of capacitor. For a parallel plate capacitor, C is proportional to the area, A and inversely proportional to the plate separation, d. V Prof. Dr. E. F. Abo Zeid 10/5/2024 Capacitance: SI Units The SI unit for capacitance is the farad, named for the famous 231 19th century scientist Michael Faraday. Its symbol is F. From the defining equation for capacitance, define a farad: Q = CV implies 1 F = (1 C)/(1 V) So, a farad is a coulomb per volt. This means a capacitor with a capacitance of 3 F could store 30 C of charge if connected to a 10 V battery. Thus a farad is a large amount of capacitance. Many capacitors have capacitances measured in pF or fF (pico or femto farads). m: milli = 10 , μ: micro = 10 , n: nano = 10 , -3 -6 -9 p: pico = 10-12, f: femto = 10-15 Prof. Dr. E. F. Abo Zeid 10/5/2024 Capacitance Problem: A parallel plate capacitor is fully charged by a 20 V battery, acquiring a charge of 1.62 nC. The area of each plate is 3.5 cm2 and the gap 232 between them is 1.3 mm. What is the capacitance of the capacitor? From Q = C V, C = Q / V = (1.62  10-9 C) / (20 V) = 8.1  10-11 F = 81  10-12 F = 81 pF. The gap and area are extraneous. - 1.62 μC C = ε A/ d + 1.62 μC 3.5 cm2 Where ε the permittivity of air, 1.3 mm A the area, d the distance 20 V between the plates Prof. Dr. E. F. Abo Zeid 10/5/2024 V = Ed As argued on the slide entitled “Potential,” in a uniform field, V = E d. This argument was based on an analogy with gravity and applies only to uniform fields: gravitational: U = m g h electric: U = q E d  U / q = E d  V=Ed Since E is uniform inside a parallel plate capacitor, the voltage drop across it is equal to the magnitude of the electric field times the distance between the plates. d Prof. Dr. E. F. Abo Zeid 10/5/2024 Electric Current and Direct-Current Circuits Current and Resistance Whenever there is a net movement of charge, there exists an electrical current. If a charge Q moves perpendicularly through a “surface” of area A in a time t, then there is a current I: Q I = t The unit of current is the Ampere (A): 1 A = 1 C/s. By convention, the direction of the current is the direction of the flow of positive charges. The actual charge carriers are electrons; hence they move in the opposite direction to I. 235 Prof. Dr. E. F. Abo Zeid 10/5/2024 Resistance and Ohm’s Law In order for a current I to flow there must be a potential difference, or voltage V, across the conducting material. We define the resistance, R, of a material to be: V R= I The unit of resistance is Ohms (W): 1 W = 1 V/A For many materials, R is constant (independent of V). Such a material is said to be ohmic, and we write Ohm’s Law: V = IR 237 Prof. Dr. E. F. Abo Zeid 10/5/2024 Resistivity An object which provides resistance to current flow is called a resistor. The actual resistance depends on: properties of the material the geometry (size and shape) The symbol for a resistor is For a conductor of length L and area A, the resistance is L R =  of the material. where  is called the resistivityA 238 Prof. Dr. E. F. Abo Zeid 10/5/2024 Temperature Dependence and Superconductivity In general, the resistivity of most materials will depend on the temperature. For most metals, resistivity increases linearly with temperature:  = 0 [1 +  (T - T0 )] Some materials, when very cold, have a resistivity which abruptly drops to zero. Such materials are called superconductors. 239 Prof. Dr. E. F. Abo Zeid 10/5/2024 Ex. A bird lands on a copper wire carrying a current of 32 A. The wire is 8 gauge, which means that its cross-sectional area is 0.13 cm2. (a) Find the difference in potential between the bird’s feet, assuming they are separated by a distance of 6.0 cm. (b) Will your answer to part (a) increase or decrease if the separation between the bird’s feet increases? Solution: Where I=32 A, L= 6 cm = 6x10-2 m, A= 0.13 cm2 = 0.13x10-4 m2 ρ for copper = 1.7x10-8 Ω m. Hence, and V = IR L R= A -2 L 6x10 m R =  = 1.7x10 W m -8 -4 2 = 78.5 x10-6 W A 0.13x10 m V = IR = 32 Ax 78.5 x10 -6 W = 2510.7 x10 -6 V = 2.510 x10 -3V = 2.510 mV Prof. Dr. E. F. Abo Zeid 10/5/2024 Energy and Power in Electric Circuits Resistance is like an internal friction; energy is dissipated. The energy dissipated per unit time is the power P: P =U/ t =(Q/t)V = IV SI unit: watt, W Using Ohm’s Law, V=IR, power can be rewritten as: P = I2R = V2/R Energy Usage: 1 kilowatt-hour = (1000 W)(3600 s) = (1000 J/s)(3600 s) = 3.6106 J 243 Prof. Dr. E. F. Abo Zeid 10/5/2024 Example It costs 2.6 cents to charge a car battery at a voltage of 12 V and a current of 15 A for 120 minutes. What is the cost of electrical energy per kilowatt-hour at this location? Prof. Dr. E. F. Abo Zeid 10/5/2024 Example A 75-W light bulb operates on a potential difference of 95 V. Find the current in the bulb and its resistance. Prof. Dr. E. F. Abo Zeid 10/5/2024 Equivalent Resistance R1 R1 R2 a b b a R2 Series Combination Parallel Combination The current I is the same in both The current may be different in resistors, so the voltage Vba must each resistor, but the voltage satisfy: Vba is the same across each Vba= IR1 + IR2 = I(R1 + R2) resistor and the total current is conserved: I = I1 + I2 Req = R1+ R2 1 1 1 = + Req R1 R2 247 Prof. Dr. E. F. Abo Zeid 10/5/2024 Problem What is the equivalent resistance? 252 Prof. Dr. E. F. Abo Zeid 10/5/2024 Problem The current in the 13.8 W resistor is 0.750 A. Find the current in the other resistors in the circuit and the total resistance and the potential difference of the battery. Hent: Rq=32.25Ω, I=2.17A and ε=70.05V 253 Prof. Dr. E. F. Abo Zeid 10/5/2024 Problem How much current flows through each battery when the switch is (a) closed and (b) open? (c) With the switch open, suppose that point A is grounded. What is the potential at point B? B A 259 Prof. Dr. E. F. Abo Zeid 10/5/2024 Circuits containing Capacitors Capacitors are used in electronic circuits. The symbol for a capacitor is + - We can also combine separate capacitors into one effective or equivalent capacitor. For example, two capacitors can be combined either in parallel or in series. Series Parallel Combination Combination C1 C1 C2 C2 260 Prof. Dr. E. F. Abo Zeid 10/5/2024 Capacitors in series (a) Three capacitors, C1, C2, and C3, connected in series. Note that each capacitor has the same magnitude of charge on its plates. (b) The equivalent capacitance, 1 1 1 1 = + + Ceq C1 C2 C3 has the same charge as the original capacitors. 262 Prof. Dr. E. F. Abo Zeid 10/5/2024 Parallel vs. Series Combination Parallel Series charge Q1 , Q2 charge on each is Q total Q = Q1 + Q2 total charge is Q voltage on each is V voltage V1 , V2 Q1= C1V Q = C1V1 Q2= C2V Q = C2V2 Q = CeffV Q = Ceff(V1+V2) Ceff = C1+C2 1/Ceff = 1/C1+1/C2 263 Prof. Dr. E. F. Abo Zeid 10/5/2024 Problem A 15 V battery is connected to three capacitors in series. The capacitors have the following capacitance: 4.5 F, 12 F, and 32 F. Find the voltage across the 32 F capacitor. 264 Prof. Dr. E. F. Abo Zeid 10/5/2024 RC Circuits We can construct circuits with more than just a resistor. For example, we can have a resistor, a capacitor, and a switch: R  C S When the switch is closed the current will change. The capacitor acts like an open circuit: no charge flows across the gap. However, when the switch is closed, current can flow from the negative plate of the capacitor to the positive plate. 265 Prof. Dr. E. F. Abo Zeid 10/5/2024 Capacitor Charging Assume that at time t = 0, the capacitor is uncharged, and we close the switch. It can be shown that the charge on the capacitor at some later Charge versus time for an RC circuit time t is: q = qmax(1 – e-t/t) The time constant t = RC, and qmax is the maximum amount of charge Current versus time for an RC that the capacitor will acquire: circuit qmax=C The current is given by 267 I = (/R)e-t/t What happens after the switch is closed? The capacitor is initially uncharged. 268 Prof. Dr. E. F. Abo Zeid 10/5/2024 Walker Problem The capacitor in an RC circuit (R = 120 W, C = 45 F) is initially uncharged. Find (a) the charge on the capacitor and (b) the current in the circuit one time constant (t = RC) after the circuit is connected to a 9.0 V battery. Prof. Dr. E. F. Abo Zeid 10/5/2024 Capacitor Discharging Consider this circuit with a charged capacitor at time t = 0: Current versus time in an RC circuit R +Q C -Q S It can be shown that the charge on the capacitor is given by: q(t) = Q e-t/t The time constant t = RC. 271 Prof. Dr. E. F. Abo Zeid 10/5/2024 Measuring the current in a circuit An ammeter is device for measuring currents in electrical circuits. To measure the current flowing between points A and B in (a) an ammeter is inserted into the circuit, as shown in (b). An ideal ammeter would have zero resistance. 272 Prof. Dr. E. F. Abo Zeid 10/5/2024 Measuring the voltage in a circuit A voltmeter measures voltage differences in electrical circuits. The voltage difference between points C and D can be measured by connecting a voltmeter in parallel to the original circuit. An ideal voltmeter would have infinite resistance. 273 Prof. Dr. E. F. Abo Zeid 10/5/2024 ELECTRICITY & MAGNETISM 10/5/2024 Lecture 6: Gauss’s Law 275 Summary:  The Electric Field is related to E= F Coulomb’s Force by Q0  Thus knowing the field we can calculate the force on a charge F = QE  The Electric Field is a vector field  Using superposition we thus find  Field lines illustrate the strength & 1 Qi direction of the Electric field E=  2 rˆi 4 0 i | ri | Prof. Dr. E. F. Abo Zeid 10/5/2024 Today 276  Electric Flux  Gauss’s Law  Examples of using Gauss’s Law  Properties of Conductors Prof. Dr. E. F. Abo Zeid 10/5/2024 ELECTRIC FLUX 10/5/2024 Electric Flux: 278 Field Perpendicular For a constant field perpendicular to a surface A Electric Flux is defined as E A  =| E | A Prof. Dr. E. F. Abo Zeid 10/5/2024 Electric Flux: For a constant field Non perpendicular NOT perpendicular 279 to a surface A Electric Flux is E defined as A  =| E | A cos  Prof. Dr. E. F. Abo Zeid 10/5/2024 Electric Flux: 280 Relation to field lines  =| E | A E A Field line  | E | density Field line density A | E | A × Area FLUX Number of flux lines N  Prof. Dr. E. F. Abo Zeid 10/5/2024 Quiz 281  What is the electric flux through a cylindrical surface? The electric field, E, is uniform and perpendicular to the surface. The cylinder has radius r and length L  A) E 4/3  r3 L  B) E r L  C) E  r2 L  D) E 2  r L  E) 0 Prof. Dr. E. F. Abo Zeid 10/5/2024 GAUSS’S LAW Relates flux through a closed surface to 10/5/2024 charge within that surface Flux through a sphere from a point 283 charge The electric field 1 Q | E |= around a point charge 4 0 | r1 |2 E Area Thus the 1 Q r1 flux on a =  4 | r | 2  2 1 sphere is E 4 0 | r1 | × Area Cancelling Q = we get 0 Prof. Dr. E. F. Abo Zeid 10/5/2024 Flux through a sphere from a point charge Now we change the radius The electric field around a point charge | E |= E 1 Q 4 0 | r1 |2 Area of sphere Thus the 284 1 Q r1 flux on a =  4 | r1 |2 sphere is E 4 0 | r1 | 2 × Area Cancelling Q = we get 0 1 Q | E |= r2 4 0 | r2 | 2 1 Q 2 =  4 | r | 2 4 0 | r2 | 2 2 Q The flux is Q 2 = the same  2 = 1 = 0 0 as before Prof. Dr. E. F. Abo Zeid 10/5/2024 Flux lines & Flux 285 Just what we would expect because the number of field lines passing through each N  N sphere is the same and number of lines passing Q  S =  2 = 1 = through each sphere is the same 0 In fact the number of flux 1 lines passing through any surface surrounding this 2 charge is the same even when a line out passes in and out s in out of the surface it crosses out once 10/5/2024 more than in Prof. Dr. E. F. Abo Zeid Principle of superposition: What is the flux from two charges? 286 Since the flux is related to the Q1 Q2 number of field lines passing S = + through a surface the total flux is 0 0 the total from each charge In general Qi Q1 S =  0 For any Q2 s surface Gauss’s Law Prof. Dr. E. F. Abo Zeid 10/5/2024 287 Quiz What flux is passing through each of these surfaces? 1 -Q/0 0 +Q/0 +2Q/0 1 2 2 Q1 3 3 Prof. Dr. E. F. Abo Zeid 10/5/2024 What is Gauss’s Law? 288 Gauss’s Law does not tell us anything new, it is NOT a new law of physics, but another way of expressing Coulomb’s Law Gauss’s Law is sometimes easier to use than Coulomb’s Law, especially if there is lots of symmetry in the problem Prof. Dr. E. F. Abo Zeid 10/5/2024 Prof. Dr. E. F. Abo Zeid 289 EXAMPLES OF USING GAUSS’S LAW 10/5/2024 Using the Symmetry Example of using Gauss’s Law 1 290 oh no! I’ve just forgotten Coulomb’s Law! Not to worry I remember Gauss’s Law q consider spherical surface Q = centred on charge 0 r2 By symmetry E is ⊥ to surface Q Q Q  =| E | A = =| E | 4r 2 = 0 0 1 Q 1 Q 1 qQ | E |= = F= 4r 2  0 4 0 r 2 F=qE 4r 2  0 Prof. Dr. E. F. Abo Zeid 10/5/2024 Phew! Example of using Gauss’s Law 2 291 What’s the field around a charged spherical shell? Again consider spherical Q surface centred on charged shell Q  out = Outside 0  in  out So as e.g. 1 1 Q | E |= 4 0 r 2 Inside charge within surface = 0  in = 0 E=0 Prof. Dr. E. F. Abo Zeid 10/5/2024 Examples 292 Gauss’s Law Gauss’s Law and a uniform and a line of sphere charge Gauss’s Law around a point charge Prof. Dr. E. F. Abo Zeid 10/5/2024 Quiz 293  In a model of the atom the nucleus is a uniform ball of +ve charge of radius R. At what distance is the E field strongest?  A) r=0  B) r = R/2  C) r = R  D) r = 2 R  E) r = 1.5 R Prof. Dr. E. F. Abo Zeid 10/5/2024 PROPERTIES OF CONDUCTORS Using Gauss’s Law 10/5/2024 Properties of Conductors 295 For a conductor in electrostatic equilibrium 1. E is zero within the conductor 2. Any net charge, Q, is distributed on surface (surface charge density =Q/A) 3. E immediately outside is ⊥ to surface 4.  is greatest where the radius of curvature is smaller 2 1  1   21 Prof. Dr. E. F. Abo Zeid 10/5/2024 1. E is zero within conductor 296 If there is a field in the conductor, then the free electrons would feel a force and be accelerated. They would then move and since there are charges moving the conductor would not be in electrostatic equilibrium Thus E=0 Prof. Dr. E. F. Abo Zeid 10/5/2024 2. Any net charge, Q, is distributed on 297 surface Consider surface S below surface of conductor Since we are in a conductor in equilibrium, rule 1 says E=0, thus =0 Gauss’s Law  = EA =  q /  0 So, net charge within qi thus  qi /  0 = 0 the surface is zero As surface can be drawn arbitrarily close to surface of conductor, all net charge must be distributed on surface Prof. Dr. E. F. Abo Zeid 10/5/2024 3. E immediately outside is ⊥ to surface 298 E⊥ Consider a small cylindrical surface at the surface of the conductor If E|| >0 it would cause surface charge q to move thus E|| it would not be in electrostatic equilibrium, thus E|| =0 cylinder is small enough that E is constant Gauss’s Law  = EA = q /  thus E = q / A E⊥ =  /  Prof. Dr. E. F. Abo Zeid 10/5/2024 Summary: Lecture 6  Electric Flux  Properties of Conductors  E is zero within the conductor  Gauss’s Law =| E | A cos  Any net charge, Q, is distributed on surface (surface Qi charge density =Q/A) S =   Examples of using Gauss’s  0  E immediately outside is ⊥ to Law surface  Isolated charge   is greatest where the radius of curvature is smaller  Charged shell  Line of charge  Uniform sphere 10/5/2024 Prof. Dr. E. F. Abo Zeid 299 300 Thanks for your attention Physics I Prof. Dr. E. F. Abo Zeid 10/5/2024

Physics Lecture Notes - 187-267 Pages PDF

Document Details

Tags

Related

Summary

Full Transcript