* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Chapter 16: Electric Forces and Fields
History of quantum field theory wikipedia , lookup
Electron mobility wikipedia , lookup
Casimir effect wikipedia , lookup
Magnetic monopole wikipedia , lookup
History of electromagnetic theory wikipedia , lookup
Anti-gravity wikipedia , lookup
Elementary particle wikipedia , lookup
Electrical resistivity and conductivity wikipedia , lookup
Weightlessness wikipedia , lookup
Introduction to gauge theory wikipedia , lookup
Speed of gravity wikipedia , lookup
Maxwell's equations wikipedia , lookup
Aharonov–Bohm effect wikipedia , lookup
Fundamental interaction wikipedia , lookup
Electromagnetism wikipedia , lookup
Field (physics) wikipedia , lookup
Lorentz force wikipedia , lookup
Chapter 16: Electric Forces and Fields •Electric Charge •Conductors & Insulators •Coulomb’s Law •Electric Field •Motion of a Point Charge in a Uniform E-field •Conductors in Electrostatic Equilibrium •Gauss’s Law 1 As reported by the Greek philosopher Thales of Miletus around 600 BC Greeks found that if they rub an amber (or “elektron”) rod with animal fur the rod attracts small light weight objects. They “charged” the amber rod. Conclusion: There is a force between amber rod and particles! It is a fundamental force like gravity and called the electric force. Other combinations of materials like glass and silk are showing the same effect. 2 3 Electric Charge Amber-Amber Repulsion Amber – Glas Attraction “Opposites attracts” Like charges repel each other (++ or --) Unlike charges repel each other (+ -) 1747 Benjamin Franklin (Besides he helped to draw up the Declaration of Independence) There are two kinds of electric charge: positive and negative. A body is electrically neutral if the sum of all the charges in a body is zero. 4 Hollywood’s Viewpoint … 5 The elementary unit of charge is e = 1.602×10-19 C. The charge on the electron is -1e. The charge on the proton is +1e. The charge on the neutron is 0e. Particle Representati on Relati ve charg e Relati ve mass electron -1 1 1800 proton +1 1 neutron o 1 The electron is part of a family of fundamental particles known as leptons The proton and the neutron are not fundamental particles. Each consists of three quarks. The value of a charge has to be an integer number (…is quantized) of e (+/-e, +/-2e,+/-3e………. ) 6 It can never be a decimal like 1.5e So what is charge actually? Charge is a property of fundamental particles that causes them to exert forces on each other. The unit of electrical charge is the Coulomb [C] = [1A 1s] (Electric charge is a conserved property of some subatomic particles, which determines their electromagnetic interactions. Electrically charged matter is influenced by, and produces, electromagnetic fields. The interaction between charge and field is the source of one of the four fundamental forces, the electromagnetic force.) 7 Law of conservation of net charge The amount of charge produced in any process is zero Or in other words: Charge is just there. You cannot create nor destroy it but you can separate or “transfer” it. Mechanism: Caused by rubbing the amber rod with animal fur it transfers electrons from the fur to the rod. Electrons are at the shell of an atom. They can transferred in most processes rather than protons An object is…… Positive charged: Lack of electrons Negative charged: Surplus of electrons If the object possesses no net charge it is said to be neutral. An atom is normally neutral, because it possesses an equal number of electrons and protons. However, if one or more electrons are removed or added, an ion is formed, which is charged. 8 Conductors and Insulators A conductor is made of material that allows electric charge to move through it easily. An insulator is made of material that does not allow electric charge to move through it easily. This is measured by the conductivity, a quantity how freely charge can move in a material. Unit: [Ω-1 – Ohm-1] Conductivity high: Conductor (i.e. metals) Conductivity very low: Insulator (i.e. ceramics, plastics) Semiconductor with an intermediate conductivity 9 Example: Amber rod, Insulator Metal sphere, Conductor -- 10 This body is electrically neutral. - - + + + + + - An object can become polarized if the charges within it can be separated. + + Why can we pickup/attract + + + + + + light weight with a zero net + charge? + That is what we call “Polarization” of an object charged If bring the positive charged rod close to the surface of our particles nuclei/protons were repelled, electrons attracted. neutral & polarized 11 How to tell if you are kissing correctly … 12 Coulomb’s Law The magnitude of the force between two point charges is: F= k q1 q2 r2 where q1 and q2 are the charges and r is the separation between the two charges. k = 8.99 ×10 Nm /C 9 where k = 1 4πε 0 2 2 and ε 0 = 8.85 ×10 −12 C 2 /Nm 2 and ε0 is called the permittivity of free space. 13 The electric force is directed between the centers of the two point charges. q1 F12 F21 q2 Attractive force between q1 and q2. r Repulsive force between q1 and q2. F12 q1 q2 F21 r If charges at rest we call the situation “electrostatic” Newton’s 3rd Law F12 = -F21 The electric force is an example of a long-range or field force, just like the force of gravity. 14 There may be many charges around ÖSuperposition to find the resulting force acting on a point charge (Principle of Superposition) 15 Example: What is the net force on the charge q1 due to the other two charges? q1 = +1.2 μC, q2 = -0.60 μC, and q3 = +0.20 μC. F12 θ F13 The net force on q1 is Fnet = F12 + F13 16 Example continued: The magnitudes of the forces are: F12 = k q1 q2 F13 = k q1 q3 ( 9 ×10 = Nm 2 /C 2 (1.2 ×10 −6 C)(0.60 ×10 −6 C) (1.2 m) 2 + (0.5 m) 2 9 Nm 2 /C 2 (1.2 ×10 −6 C)(0.20 ×10 −6 C) (1.2 m) 2 r212 = 3.8 ×10 −3 N (9 ×10 = r312 = 1.5 ×10 −3 N ) 9 ) 17 Example continued: The components of the net force are: Fnet , x = F13, x + F12 , x = − F13 + F12 cosθ = 2.1 × 10 N −3 Fnet , y = F13, y + F12, y = 0 + F12 sin θ = 1.4 ×10 −3 N 1 .2 m = 0.92 1.3 m 0 .5 m sin θ = = 0.38 1.3 m cos θ = Where from the figure 18 Example continued: The magnitude of the net force is: Fnet = Fnet2 , x + Fnet2 , y = 2.5 × 10−3 N The direction of the net force is: Fnet , y tan φ = = 0.66 Fnet , x φ = 33° 19 q1 q 2 Fe = k 2 r12 Electric Force m1m2 Fg = G 2 r12 Gravitational force The electric force… …varies directly with the magnitude of each charge …varies inversely with the square of the distance between charges …is directed along the line joining the charges …can be either attractive or repulsive depending on the signs of the charges Gravitational field constant G ≈ 7 10-11 N m2 kg-2 Electric field constant k ≈ 9 109 N m2 C-2 Fe is much stronger than Fg Fe can be attractive or repulsive; Fg is always attractive q must be an integer number of e 20 Example: What is the ratio of the electric force and gravitational force between a proton and an electron separated by 5.3×10-11 m (the radius of a Hydrogen atom)? Fe = k q1 q2 r2 Gm1m2 Fg = r2 q1 = q2 = e m1 = m p = 1.67 ×10 − 27 kg m2 = me = 9.11× 10 −31 kg Fe k q1 q2 ke 2 The ratio is: = = = 2.3 × 1039 Fg Gm1m2 Gme m p 21 The Electric Field Michael Faraday conceived the concept of electric field Recall : Fg = mg Fe = qE Where g is the strength of the gravitational field. Similarly for electric forces we can define the strength of the electric field E. 22 E-field applet In other words: The force experienced by any charge is then viewed as the interaction between the charge and the electric field. Electric field vector is equal to the quotient of force vector a positive point charge q0 (so called test charge) would feel at a certain point and the test charge itself. r r FA E= q0 23 For a point charge of charge Q, the magnitude of the force per unit charge at a distance r (the electric field) is: Fe k Q E= = 2 q r The electric field at a point in space is found by adding all of the electric fields present. E net = ∑ Ei i Be careful! The electric field is a vector! 24 E points outward for a positive point charge E points inward for a negative point charge 25 Example: Find the electric field at the point P. P x q1 = +e q2 = -2e x = 0m x = 1m x = 2m E is a vector. What is its direction? Place a positive test charge at the point of interest. The direction of the electric field at the location of the test charge is the same as the direction of the force on the test charge. 26 Example continued: q1 = +e P x q2 = -2e Locate the positive test charge here. P x q1 = +e q2 = -2e Direction of E due to charge 2 Direction of E due to charge 1 27 Example continued: The net electric field at point P is: E net = E1 + E 2 The magnitude of the electric field is: Enet = E1 − E2 28 Example continued: E1 = k q1 E2 = k q2 r2 r2 ( 9 ×10 = ( 9 ×10 = ) 9 Nm 2 / C 2 (1.6 ×10 −19 C) −10 = 3 . 6 × 10 N/C 2 (2 m) 9 Nm 2 / C 2 (2 *1.6 ×10 −19 C) −9 = 2 . 9 × 10 N/C 2 (1 m) ) −9 Enet = E1 − E2 = −2.5 ×10 N/C The net E-field is directed to the left> 29 Electric field lines Electric field lines are a useful way to indicate what the magnitude and direction of an electric field is in space. Rules: Field lines… - point always in the direction of E (E is always tangential) - start at positive charges or at infinity - end at negative charges or at infinity - are more dense when E has a greater magnitude - the number of field lines entering or leaving a charge is proportional to the magnitude of the charge - field lines always end/start perpendicular to the surface - never cross or touch each other 30 Pictorial representation of the rules on the previous slide: 31 32 Charged Plate Parallel plate capacitor Field lines parallel to each other and perpendicular to the surface ⇒E= Q 2ε 0 A Two parallel conducting plates with opposite charge, separated by a distance d. The electric field is uniform between the plates (except at the edges). Q ⇒E= ε0 A 33 Motion of a Point Charge in a Uniform E-Field A region of space with a uniform electric field containing a particle of charge q (q>0) and mass m. 34 FBD for the charge q y Fe x Apply Newton’s 2nd Law and solve for the acceleration. ∑F x = Fe = ma Fe = qE = ma q a= E m One could now use the kinematic equations to solve for distance traveled in a time interval, the velocity at the end of a time interval, etc. 35 Example: What electric field strength is needed to keep an electron suspended in the air? y FBD for the electron: Fe x w To get an upward force on the electron, the electric field must be directed toward the Earth. 36 Example continued: Apply Newton’s 2nd Law: ∑F y = Fe − w = 0 Fe = w qE = eE = mg mg E= = 5.6 ×10 −11 N/C e 37 Example: A horizontal beam of electrons moving 4.0×107 m/s is deflected vertically by the vertical electric field between two oppositely charged parallel plates. The magnitude of the field is 2.00×104 N/C. (a) What is the direction of the field between the plates? From the top plate to the bottom plate 38 Example continued: (b) What is the charge per unit area on the plates? Q σ E= = ε0 A ε0 This is the electric field between two charged plates. Note that E here is independent of the distance from the plates! σ = Eε 0 = (2.00 ×10 4 N/C )(8.85 ×10 −12 C 2 /Nm 2 ) = 1.77 ×10 −7 C/m 2 39 Example continued: (c) What is the vertical deflection d of the electrons as they leave the plates? y FBD for an electron in the beam: Fe x w Apply Newton’s 2nd Law and solve for the acceleration: ∑F y = Fe − w = ma y Fe − w Fe qE ay = = −g = − g = 3.52 × 1015 − 9.8 m/s m m m ( ) 40 Example continued: What is the vertical position of the electron after it travels a horizontal distance of 2.0 cm? 1 02 x = x0 + vox t + a x t 2 t= x − x0 0.02 m −10 = = 5 . 0 × 10 sec 7 4.0 × 10 m/s v0 x Time interval to travel 2.00 cm horizontally 0 1 2 y = y0 + voy t + a y t 2 y − y0 = d = 1 2 a y t = 4.4 × 10 − 4 m 2 Deflection of an electron in the beam 41 Conductors in Electrostatic Equilibrium Conductors are easily polarized. These materials have free electrons that are free to move around inside the material. Any charges that are placed on a conductor will arrange themselves in a stable distribution. This stable situation is called electrostatic equilibrium. When a conductor is in electrostatic equilibrium, the E-field inside it is zero. => Electrostatic Shielding Any net charge must reside on the surface of a conductor in electrostatic equilibrium. 42 Just outside the surface of a conductor in electrostatic equilibrium the electric field must be perpendicular to the surface. If this were not true, then any surface charge would have a net force acting on it, and the conductor would not be in electrostatic equilibrium. Fe 43 Any excess charge on the surface of a conductor will accumulate where the surface is highly curved (i.e. a sharp point). 44 Gauss’s Law Enclose a point charge +Q with an imaginary sphere. +Q Here, E-Field lines exit the sphere. 45 Look at a small patch of the surface of the imaginary sphere. With a positive charge inside the sphere you would see electric field lines leaving the surface. 46 number of field lines Recall that E ∝ A so that the number of field lines ∝ EA It is only the component of the electric field that is perpendicular to the surface that exits the surface. E θ Surface 47 Define a quantity called flux, which is related to the number of field lines that cross a surface: flux = Φ e = E⊥ A = (E cos θ )A E This picture defines the value of θ. θ Flux > 0 when field lines exit the surface and flux < 0 when field lines enter the surface. 48 Φ=EA Φ=0 Φ = E A cosθ 49 Example: Find the electric flux through each side of a cube of edge length a in a uniform electric field of magnitude E. A cube has six sides: The field lines enter one face and exit through another. What is the flux through each of the other four faces? 50 Example continued: There is zero electric flux though the other four faces. The electric field lines never enter/exit any of them. The flux through the left face is –EA. The flux through the right face is +EA. The net flux through the cube is zero. 51 Gauss’s Law Consider an arbitrary surface (Gaussian surface) enclosing a total charge q. The electric flux through the surface is Flux is positive if field lines leave an enclosed volume. Flux is negative if field lines enter an enclosed volume. Φ= q ε0 Example 1: ε 0 = 4πk = 8.85×10 C /N ⋅ m −12 1 2 2 ÖCharge must be enclosed by the surface, else Φ=0! q Φ = EA = k 2 ⋅ 4πr 2 = 4π k q r 2 1 −12 C ε0 = = 8.85 ⋅ 10 Nm 2 4πk Permitivity of free space q ⇒ Φ = [Nm 2 / C ] Gauss Law ε0 52