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Physics Lecture 3 Electrostatics Electric field (cont.) Conductors in electrostatic equilibrium The oscilloscope Electric flux and Gauss’s law http://www.physics.wayne.edu/~apetrov/PHY2140/ Chapter 15 5/5/2017 1 Lightning Review Last lecture: F ke 1. Coulomb’s law q1 q2 r2 the superposition principle 2. The electric field E F q0 Review Problem: A “free” electron and “free” proton are placed in an identical electric field. Compare the electric force on each particle. Compare their accelerations. 5/5/2017 2 Example: Electric Field Due to Two Point Charges Question: Charges q1=4.9 mC and charge q2=-2.8 mC are placed as shown, 2 cm and 4 cm from the origin. Find the electric field at point P, which has coordinates (0, 0.015) m. q1 (0, 0.02 m) E2 P (0, 0.015 m) 0 E1 5/5/2017 q2 (0, 0.04 m) E1 2 3 Question: Charges q1=4.9 mC and charge q2=-2.8 mC are placed as shown, 2 cm and 4 cm from the origin. Find the electric field at point P, which has coordinates (0, 0.015) m. Observations: First find the field at point P due to charge q1 and q2. Field E1 at P due to q1 is directed away from q1. Field E2 at due to q2 is directed towards q2(to the right). The net field at point P is the vector sum of E1 and E2. The magnitude is obtained with E ke 5/5/2017 q r 2 4 Question: Charges q1=4.9 mC and charge q2=-2.8 mC are placed as shown, 2 cm and 4 cm from the origin. Find the electric field at point P, which has coordinates (0, 0.015) m. Given: q1 = 4.9 mC q2 = -2.8 mC d1 = 0.02 m d2 = 0.04 m d3 = 0.015 m Find: E1 ke d3 4.9.00 10 C 0.7110 N / C 0.02 0.015 m 2.8.00 10 C 0.4 10 N / C 0.025 m d12 d32 8.99 109 8 2 2 8.99 109 2 Nm C2 2 E12, x E1 cos E2 0.83108 N / C E12, y E1 sin 0.57 10 N / C d1 as arctan 53 d 3 5/5/2017 2 6 q1 d 2 d3 Nm2 C2 8 E1+2 = ? d2 6 q1 E2 ke d1 8 2 2 E1 2 1.0 108 N / C 35 5 15.5 Electric Field Lines A convenient way to visualize field patterns is to draw lines in the direction of the electric field. Such lines are called field lines. Remarks: 1. 2. Electric field vector, E, is tangent to the electric field lines at each point in space. The number of lines per unit area through a surface perpendicular to the lines is proportional to the strength of the electric field in a given region. E is large when the field lines are close together and small when far apart. 5/5/2017 6 15.5 Electric Field Lines (2) Electric field lines of single positive (a) and (b) negative charges. a) b) + q 5/5/2017 - q 7 15.5 Electric Field Lines (3) Rules for drawing electric field lines for any charge distribution. 1. 2. 3. 5/5/2017 Lines must begin on positive charges (or at infinity) and must terminate on negative charges or in the case of excess charge at infinity. The number of lines drawn leaving a positive charge or approaching a negative charge is proportional to the magnitude of the charge. No two field lines can cross each other. 8 15.5 Electric Field Lines (4) Electric field lines of a dipole. + 5/5/2017 - 9 Application: Measurement of the atmospheric electric field The electric field near the surface of the Earth is about 100 N/C downward. Under a thundercloud, the electric field can be as large as 20000 N/C. How can such a (large) field be measured? A 5/5/2017 10 A 5/5/2017 11 15.6 Conductors in Electrostatic Equilibrium Good conductors (e.g. copper, gold) contain charges (electron) that are not bound to a particular atom, and are free to move within the material. When no net motion of these electrons occur the conductor is said to be in electro-static equilibrium. 5/5/2017 12 15.6 Conductors in Electrostatic Equilibrium Properties of an isolated conductor (insulated from the ground). 1. 2. 3. 4. 5/5/2017 Electric field is zero everywhere within the conductor. Any excess charge field on an isolated conductor resides entirely on its surface. The electric field just outside a charged conductor is perpendicular to the conductor’s surface. On an irregular shaped conductor, the charge tends to accumulate at locations where the radius of curvature of the surface is smallest – at sharp points. 13 1. Electric field is zero everywhere within the conductor. If this was not true, the field inside would be finite. Free charge there would move under the influence of the field. A current would be induced. The conductor would not be in an electrostatic state. 5/5/2017 14 2. Any excess charge field on an isolated conductor resides entirely on its surface. This property is a direct result of the 1/r2 repulsion between like charges. If an excess of charge is placed within the volume, the repulsive force pushes them as far apart as they can go. They thus migrate to the surface. 5/5/2017 15 3. The electric field just outside a charged conductor is perpendicular to the conductor’s surface. If not true, the field would have components parallel to the surface of the conductor. This field component would cause free charges of the conductor to move. A current would be created. There would no longer be a electro-static equilibrium. 5/5/2017 16 4. On an irregular shaped conductor, the charge tends to accumulate at locations where the radius of curvature of the surface is smallest – at sharp points. Consider, for instance, a conductor fairly flat at one end and relatively pointed at the other. Excess of charge move to the surface. Forces between charges on the flat surface, tend to be parallel to the surface. Those charges move apart until repulsion from other charges creates an equilibrium. At the sharp ends, the forces are predominantly directed away from the surface. There is less of tendency for charges located at sharp edges to move away from one another. Produces large fields (and force) near sharp edges. - 5/5/2017 - 17 Remarks Property 4 is the basis for the use of lightning rods near houses and buildings. (Very important application) 5/5/2017 Most of any charge on the house will pass through the sharp point of the lightning rod. First developed by B. Franklin. 18 Faraday’s ice-pail experiment +++++ + + + + + - - - + +++++ - + + + + + + + + - + + + + + + + + + + + + + + + + + + + + Demonstrates that the charge resides on the surface of a conductor. 5/5/2017 19 Mini-quiz Question: Suppose a point charge +Q is in empty space. Wearing rubber gloves, we sneak up and surround the charge with a spherical conducting shell. What effect does this have on the field lines of the charge? ? + q 5/5/2017 + 20 Question: Suppose a point charge +Q is in empty space. Wearing rubber gloves, we sneak up and surround the charge with a spherical conducting shell. What effect does this have on the field lines of the charge? Answer: Negative charge will build up on the inside of the shell. Positive charge will build up on the outside of the shell. There will be no field lines inside the conductor but the field lines will remain outside the shell. + + + + + - - + q - + - - - + - - + + 5/5/2017 + - - + + 21 Mini-Quiz Question: Is it safe to stay inside an automobile during a lightning storm? Why? 5/5/2017 22 Question: Is it safe to stay inside an automobile during a lightning storm? Why? Answer: Yes. It is. The metal body of the car carries the excess charges on its external surface. Occupants touching the inner surface are in no danger. SAFE 5/5/2017 23 15.8 The Van De Graaff Generator Read Textbook + Discuss in Lab. 5/5/2017 24 15.9 The oscilloscope Changing E field applied on the deflection plate (electrodes) moves the electron beam. V2 d L 5/5/2017 V1 25 Oscilloscope: deflection angle (additional) V2 d L V1 vx electron gun between plates vy ayt vy ay eE eV2 me me d L vx t eV2 L me d vx tan 5/5/2017 2eV1 me vy vx eV2 L eV2 Lme V2 L me d vx 2 me d 2eV1 d 2eV 26 15.10 Electric Flux and Gauss’s Law Discuss a technique introduced by Karl F. Gauss (17771855) to calculate electric fields. Requires symmetric charge distributions. Technique based on the notion of electrical flux. 5/5/2017 27 15.10 Electric Flux To introduce the notion of flux, consider a situation where the electric field is uniform in magnitude and direction. Consider also that the field lines cross a surface of area A which is perpendicular to the field. The number of field lines per unit of area is constant. The flux, F, is defined as the product of the field magnitude by the area crossed by the field lines. F EA 5/5/2017 Area=A E 28 15.10 Electric Flux Units: Nm2/C in SI units. Find the electric flux through the area A = 2 m2, which is perpendicular to an electric field E=22 N/C F EA Answer: F = 44 Nm2/C. 5/5/2017 29 15.10 Electric Flux If the surface is not perpendicular to the field, the expression of the field becomes: F EA cos Where is the angle between the field and a normal to the surface. N 5/5/2017 30 15.10 Electric Flux Remark: When an area is constructed such that a closed surface is formed, we shall adopt the convention that the flux lines passing into the interior of the volume are negative and those passing out of the interior of the volume are positive. 5/5/2017 31 Example: Question: Calculate the flux of a constant E field (along x) through a cube of side “L”. y 1 2 E x z 5/5/2017 32 Question: Calculate the flux of a constant E field (along x) through a cube of side “L”. Reasoning: Dealing with a composite, closed surface. Sum of the fluxes through all surfaces. Flux of field going in is negative Flux of field going out is positive. E is parallel to all surfaces except surfaces labeled 1 and 2. So only those surface contribute to the flux. y 1 2 E x z 5/5/2017 33 Question: Calculate the flux of a constant E field (along x) through a cube of side “L”. Reasoning: Dealing with a composite, closed surface. Sum of the fluxes through all surfaces. Flux of field going in is negative Flux of field going out is positive. E is parallel to all surfaces except surfaces labeled 1 and 2. So only those surface contribute to the flux. Solution: F1 EA1 cos 1 EL2 F 2 EA2 cos 2 EL2 F net EL2 EL2 0 y 1 2 E x z 5/5/2017 34 15.10 Gauss’s Law The net flux passing through a closed surface surrounding a charge Q is proportional to the magnitude of Q: Fnet EA cos Q In free space, the constant of proportionality is 1/eo where eo is called the permittivity of of free space. 1 1 eo 4 ke 4 8.99 109 Nm2 / C 2 5/5/2017 35 15.10 Gauss’s Law The net flux passing through any closed surface is equal to the net charge inside the surface divided by eo. F net EA cos 5/5/2017 Q eo 36