Today in Physics 122: Gauss Day Electrostatic ion drive Flux of E Flux of E through closed surfaces, and enclosed charge Gauss s law: an easier way to calculate E for symmetrical charge distributions. The Dawn spacecraft, currently leaving orbit around the asteroid Vesta, on its way to the asteroid Ceres, uses an electrostatic ion drive (Artist s conception: William K. Hartmann and NASA) 13 September 2012 Physics 122, Fall 2012 1 Application: ion drive Even simple, uniform E can be quite useful, as in the ion drive, the latest wrinkle in spacecraft propulsion. Charge up two close parallel plates that have holes in them, much smaller than d the spacing between the plates but large enough for atoms to leak through. As we learned last time, the E 4k zˆ resulting E is uniform and 0 perpendicular to the plates, and zero outside. 13 September 2012 Physics 122, Fall 2012 2 Ion drive (continued) Now let some heavy, easily-ionized gas, like xenon, leak in from the positively-charged side. Ionize the leaked atoms, e.g. with UV light. Then the xenon ions positively charged are accelerated by E, and escape through the holes when they reach the other side. The electrons from ionization move the opposite direction in E. The larger is E, the faster the ions are going when they escape. Xe UV 13 September 2012 Physics 122, Fall 2012 3 (c) University of Rochester 1
Ion drive (continued) The speed of the ions can easily be calculated from what you learned in PHY 121: 1 2 2dmXe F QEma a ee mxe ; d at t 2 ee 2eEd 25 v at m Xe 2.2 10 kg 124 m H mxe If N ions per unit time v escape, they carry away momentum at the rate UV dp 2eEd Nv N zˆ dt mxe Xe 13 September 2012 Physics 122, Fall 2012 4 Ion drive (continued) But momentum is conserved, so the assembly containing the plates, gas supply, UV light, etc. experiences a thrust, dp Fi N dt 2eEd zˆ mxe and thus the assembly (mass M) is propelled in the direction opposite that of the escaping ions, at constant acceleration F i /M which is larger, the larger is E. F i v 13 September 2012 Physics 122, Fall 2012 5 Ion drive (continued) This is the principle behind NASA s, which have been used on the Deep Space 1 and Dawn missions to asteroids and comets 1.3-2.6 times further away from the Sun than Earth (r = 1.3-2.6 AU). The drives are powered by solar-panel generators. Typically NSTAR drives consume 2.3 kw of electrical power and produce 0.02 lb of thrust. Doesn t sound like much? Read on Diagram of Dawn s ion drive (JPL/NASA). 13 September 2012 Physics 122, Fall 2012 6 (c) University of Rochester 2
Ion drive (continued) Ordinary rockets work by heating their fuel to high temperature and pressure, and letting some of this escape to produce thrust. Compared to ion drives, rockets are inefficient, since the ignited fuel atoms move in random directions in the ignition chamber. Thus an ion drive can reach higher speeds than a rocket of the same mass, though it may take longer to get to high speeds. Schematic diagram of the J-2X rocket motor (NASA). 13 September 2012 Physics 122, Fall 2012 7 Ion drive vs. rockets: tale of the tape Let s compare the best rocket with the best ion drive, for taking a 1000 kg satellite out of an orbit around the Sun. Best rocket: the mighty, famous from the Apollo missions. Total mass of satellite plus threestage rocket 2.910 kg. 6 7 Initial thrust 3.910 N 6 8.7 10 lb. 20x Atlas V; 40x Delta II H. Best ion drive: SAFE-400 nuclear reactor (100 kwe), 43 NSTAR drives; total mass 600 kg. Launch of Apollo 11, 1969, on a booster (NASA). 13 September 2012 Physics 122, Fall 2012 8 Ion drive vs. rockets: tale of the tape (continued) Start the two vehicles with the same mass adding Xe to the ion-drive vehicle til its mass is the same as the other and accelerate as long as they are producing thrust. Their mass decreases as they go along, and use up fuel, so the acceleration is a function of time: at Fthrust mt t vt at dt 110 3 0 5 10 15 20 0 Time since start (minutes) t Mass of the rocket-powered vehicle, as x t vtdt the expends and ejects its three stages. 0 13 September 2012 Physics 122, Fall 2012 9 Total vehicle mass s m(t) (kg) 110 7 110 6 110 5 110 4 (c) University of Rochester 3
Ion drive vs. rockets: tale of the tape (continued) Result: the can get anywhere within the Solar system faster than current, but eventually the ion drive goes farther and flies faster. Distance (AU) 110 6 110 5 110 4 110 3 100 Speed/speed of lig ght 110 3 110 5 110 7 110 9 110 11 10 110 13 1 0.1 1 10 100 110 3 110 4 110 15 110 8 110 5 0.01 10 110 4 13 September 2012 Physics 122, Fall 2012 10 Ion drive vs. rockets: tale of the tape (continued) Ion drives have much more room for improvement than rockets. Here are the results for NSTAR drives that can run on 100 W of electrical power: Distance (AU) 110 6 110 5 110 4 110 3 100 10 Speed/speed of lig ght 110 3 110 5 110 7 110 9 110 11 1 0.1 1 10 100 110 3 110 4 110 13 110 8 110 5 0.01 10 110 4 13 September 2012 Physics 122, Fall 2012 11 Ion drive vs. rockets: tale of the tape (continued) If in addition E could be made 10000 times larger, the ion drive s top speed begins to look respectable, at 10% the speed of light, and gets to the nearest stars in about 1500 years. 110 3 0.1 Distance (light yea ars) 10 0.1 110 3 Speed/speed of lig ght 110 3 110 5 110 7 110 9 110 11 110 5 0.1 10 110 3 110 5 110 13 110 8 110 5 0.01 10 110 4 13 September 2012 Physics 122, Fall 2012 12 (c) University of Rochester 4
Flux of E First, the flux of rain. Suppose rain is falling straight down at a constant rate: call the mass per unit area falling on the ground f f. You have a bucket sitting flat on the ground, with area A at the top. The rate at which the bucket fills up is determined by how big or small the flux fa is. 13 September 2012 Physics 122, Fall 2012 13 f A Flux of E (continued) Now tip the bucket, so that its axis makes an angle with the direction of the raindrops. Clearly it will capture water at a smaller rate than before, because its mouth presents a smaller area Acos to the rain. So the flux, which determines how fast the bucket fills, is more generally facos. Acos f A 13 September 2012 Physics 122, Fall 2012 14 Flux of E (continued) Now, E: Electric field doesn t flow, but the direction and density of lines of E function like the velocity and density of raindrops. For uniform E and planar A, the flux of E is therefore Acos E EAcos A 13 September 2012 Physics 122, Fall 2012 15 (c) University of Rochester 5
Flux of E (continued) Or, inventing a vector A which has magnitude equal to the area A and direction perpendicular to the area, we can write the flux in a vector shorthand that will turn out to be convenient: EA Flux is the scalar ( dot ) product of the vectors E and A. A E 13 September 2012 Physics 122, Fall 2012 16 1 2 1) 2) 13 September 2012 Physics 122, Fall 2012 17 1 2 1) 2) 13 September 2012 Physics 122, Fall 2012 18 (c) University of Rochester 6
1 2 1) 2) 13 September 2012 Physics 122, Fall 2012 19 1 2 1) 2) 13 September 2012 Physics 122, Fall 2012 20 1 2 1) 2) 13 September 2012 Physics 122, Fall 2012 21 (c) University of Rochester 7
1 2 1) 2) 13 September 2012 Physics 122, Fall 2012 22 Nonuniform field, irregular surface Even if the field varies in strength with position, and the surface is irregular, one can always go to the location of each infinitesimal area element in the surface and find the local value of E dfi define an area vector da for the area element. Then the total flux through that surface is the sum of the fluxes through all the infinitesimal elements: EdA surface Don t worry, we ll only be using simple surfaces that lead to do-able integrals 13 September 2012 Physics 122, Fall 2012 23 Closed surfaces and fields from charges Use your intuition. Electric field lines from charges originate at + charges and terminate at charges. So which of these spheres has a non-zero flux through it? 1) Left 2) Right 3) Both 13 September 2012 Physics 122, Fall 2012 24 (c) University of Rochester 8
Closed surfaces and fields from charges (continued) This sphere contains an electric dipole, but the + (red) charge is closest to its inner surface. The flux through it is 1) positive 2) negative 3) zero 13 September 2012 Physics 122, Fall 2012 25 Closed surfaces and fields from charges (continued) Evidently, the flux of E through a closed surface depends upon how much charge it contains, since lines of E can only start and finish on charges. For such a flux to be positive (negative), it needs to contain an net positive (negative) charge. If it contains no net charge, every flux line that enters the surface has to leave it too. + - 13 September 2012 Physics 122, Fall 2012 26 Closed surfaces and fields from charges (continued) We can write this in the form of our integral for flux (page 23) as EdA some Qenclosed constant where the symbol means an integral over the closed surface for which da is an infinitesimal area element. da points outward from convex surfaces. 13 September 2012 Physics 122, Fall 2012 27 (c) University of Rochester 9
Closed surfaces and fields from charges (continued) To find the constant, consider a point charge at the center of an imaginary sphere. Since for spheres the surface is perpendicular to the radius, cos =1 for all elements. Since all the points on the sphere lie a distance r away from the charge, E is the same for all elements. Thus Q Q 2 EdA k da k 2 2 4 r r r r 4kQ Q 0 13 September 2012 Physics 122, Fall 2012 28 Gauss s Law So now we know the constant, and can write EdA 4 kqenclosed Qenclosed 0 This result, called Gauss s Law, is the first of the four fundamental equations, called the Maxwell equations, that form the most compact expression of the theory of electromagnetism. It also represents a very powerful technique for the calculation of electric fields, which can be used whenever there are identifiable symmetries in the charge and electric field distribution. 13 September 2012 Physics 122, Fall 2012 29 Example of a Gauss s Law field calculation Infinite line charge, density : what s the field a distance r away from the wire? We know that the field must point perpendicularly away from the wire and be cylindrically symmetrical. (How could it be otherwise?) So if we construct an imaginary cylinder of radius r and height h coaxial with the line, the flux through the circular ends is zero and E is uniform on the cylinder walls. r h E 13 September 2012 Physics 122, Fall 2012 30 (c) University of Rochester 10
Example of a Gauss s Law field calculation (continued) Furthermore, as with the sphere, cos = 1 everywhere on the cylinder walls, so EdA E da 2 rhe cyl. walls The cylinder encloses a charge Qenclosed h So the hs cancel it doesn t matter how long the cylinder is and Same 2k result as 2rhE 4kh E last time, r but easier. 13 September 2012 Physics 122, Fall 2012 31 r h E Why Gauss Day? Gauss s law, one of the major scientific results of the 1800s, was one of many huge achievements by Carl Friedrich Gauss who, when he was what we would now call a graduate student, discovered and named the asteroid Vesta (the fourth asteroid to be found) which is now being explored by the ion-drive-powered Dawn satellite. Gauss (Wikimedia Commons) 13 September 2012 Physics 122, Fall 2012 32 (c) University of Rochester 11