Transcript Slide 1
Today’s agenda: Electric potential energy (continued). You must be able to use electric potential energy in work-energy calculations. Electric potential. You must be able to calculate the electric potential for a point charge, and use the electric potential in work-energy calculations. Electric potential and electric potential energy of a system of charges. You must be able to calculate both electric potential and electric potential energy for a system of charged particles (point charges today, charge distributions next lecture). The electron volt. You must be able to use the electron volt as an alternative unit of energy. Remember conservation of energy from Physics 23? An object of mass m in a gravitational field has potential energy U(y) = mgy and “feels” a gravitational force FG = GmM/r2, attractive. y If released, it gains kinetic energy and loses potential energy, but mechanical energy is conserved: Ef=Ei. The change in potential energy is Uf - Ui = -(Wc)if. The gravitational force does + work. Ui = mgyi yi x Uf = 0 What force does Wc? Force due to gravity. graphic “borrowed” from http://csep10.phys.utk.edu/astr161/lect/history/newtongrav.html A charged particle in an electric field has electric potential energy. ++++++++++++++ It “feels” a force (as given by Coulomb’s law). It gains kinetic energy and loses potential energy if released. The Coulomb force does positive work, and mechanical energy is conserved. + E F ------------------- Now your deep philosophical question for the day… If you have a great big nail to drive, are you going to pound it with a dinky little screwdriver? Or a hammer? Ef Ei Wother if “The hammer equation.”—©Prof. R. J. Bieniek Here is another important Physics 23 Starting Equation, which you may need for tomorrow’s homework… The Work-Energy Theorem: Wnet if K Wnet is the total work, and includes work done by the conservative force (if any) and all other forces (if any). Notation: Wab = Wa-Wb = [W]ba Example: two isolated protons are constrained to be a distance D = 2x10-10 meters apart (a typical atom-atom distance in a solid). If the protons are released from rest, what maximum speed do they achieve, and how far apart are they when they reach this maximum speed? 2.63x10 m/s 4 To be worked at the blackboard in lecture… 2.63x104 m/s Example: two isolated protons are constrained to be a distance D = 2x10-10 meters apart (a typical atom-atom distance in a solid). If the protons are released from rest, what maximum speed do they achieve, and how far apart are they when they reach this maximum speed? 2.63x10 m/s 4 We need to do some thinking first. What is the proton’s potential energy when they reach their maximum speed? How far apart are the protons when they reach their maximum speed? Example: two isolated protons are constrained to be a distance D = 2x10-10 meters apart (a typical atom-atom distance in a solid). If the protons are released from rest, what maximum speed do they achieve, and how far apart are they when they reach this maximum speed? 2.63x10 m/s 4 v=0 +e Initial v +e +e v=0 ri=2x10-10 m Final rf= +e v There is an unasked conservation of momentum problem buried in here, isn’t there! v=0 +e Initial v +e v=0 ri=2x10-10 m Final +e +e rf= Ef Ei Wother if 0 0 0 Kf Uf Ki Ui Wother if K f Ui v v=0 +e Initial v +e +e v=0 ri=2x10-10 m Final rf= +e v K f Ui How many objects are moving in the final state? Two. How many Kf terms are there? Two. How many pairs of charged particles in the initial state? One. How many Ui terms are there? One. v=0 +e Initial v +e v=0 ri=2x10-10 m Final +e +e rf= K f Ui e e 1 2 2 m p v k ri 2 v 2 ke m p ri 9 10 1.6 10 1.6710 2 10 -19 2 9 - 27 -10 m 2.6310 s 4 Is that fast, or slow? v Another way to calculate electrical potential energy. UE UEf UEi WE if WE if FE d rf rf ri ri k q1q 2 dr 2 r12 The subscript “E” is to remind you I am talking about electric potential energy. After this slide, I will drop the subscript “E.” Move one of charges from ri to rf, in the presence the other charge. The minus sign in this equation comes from the definition of change in potential energy. The sign from the dot product is “automatically” correct if you include the signs of q and q0. U E q1 rf ri rf f kq 2 dr q1 E 2dr q1 E 2 d 2 ri i r12 Move q1 from ri to rf, in the presence of q2. A justification, but not a mathematically “legal” derivation. Generalizing: f Uf Ui q E d i When a charge q is moved from one position to another in the presence of an electric field due to one or more other charged particles, its change in potential energy is given by the above equation. I’ve done something important here. I’ve generalized from the specific case of one charged particle moving in the presence of another, to a charged particle moving in the electric field due to all the other charged particles in its “universe.” “i” and “f” refer to the two points for which we are calculating the potential energy difference. You could also use “a” and “b” like your text does, or “0” and “1” or anything else convenient. I use “i” and “f” because I always remember that (anything) = (anything)f – (anything)i. So far in today’s lecture… I reminded you of some energy concepts from Physics 23: U Uf Ui Wconservative if definition of potential energy Wexternal if Wconservative if true if kinetic energy is constant Ef Ei Wother if Wnet if K everybody’s favorite Phys. 23 equation work-energy theorem You mastered all of the above equations in Physics 23. So far in today’s lecture… Then I “derived” an equation for the electrical potential energy of two point charges q1q 2 1 q1q 2 U r12 k . r12 40 r12 I also derived an equation (which we haven’t used yet) for the change in electrical potential energy of a point charge that moves in the presence of an electric field f Uf Ui q E d i Above is today’s stuff. So far. Lots of lecturing for only two equations.