* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Chapter 2 Motion Along a Straight Line Position, Displacement
Survey
Document related concepts
Gibbs free energy wikipedia , lookup
Renormalization wikipedia , lookup
Negative mass wikipedia , lookup
Electric charge wikipedia , lookup
Internal energy wikipedia , lookup
Density of states wikipedia , lookup
Anti-gravity wikipedia , lookup
Casimir effect wikipedia , lookup
Introduction to gauge theory wikipedia , lookup
Aharonov–Bohm effect wikipedia , lookup
Conservation of energy wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
Potential energy wikipedia , lookup
Transcript
The contour lines show points on the mountain that are atElectric the same elevation. Potential, Energy, Capacitance The yellow contour line connects points that are 16-2 Equipotential Surfaces 14,200 feet above sea level. Since myus weight 180 when I am map: standing on that Most of have is seen a lbs, topographical contour line I have a potential energy of UG = mgh = (180)(14200) = 2,556,000 ft•lbs FYI: If the altitude of Milwaukee is 600 ft, my potential energy in Milwaukee is U = (180)(600) = 108,000 ft•lbs, making U = 2,448,000 ft•lbs. Where did this energy come from? T or F: The gravitational field vector g is perpendicular to every point on the contour line. Topic 6.2 Extended E – Equipotential surfaces The 2D contour lines define a 3D surface: Here's the 3D view from the top of Mount Elbert: Electric Potential, Energy, Capacitance 16-2 Equipotential Surfaces It is easier to visualize how these equipotential surfaces fit together to form a 3D image of the actual surface if we concentrate on the vicinity of the crest of Mount Elbert: Let's look at the equipotential surfaces from 14,200 feet up to the top of Mount Elbert: 14400 14360 14320 14280 14240 14200 Rotation, tilting, and stacking of these equipotential surfaces will produce what appears to be a 3D image of Mount Elbert: Rotation, tilting, and stacking of these equipotential surfaces will produce what appears to be a 3D image of Mount Elbert: Topic 6.2 Extended E – Equipotential surfaces Of course, on the planetary scale the equipotential surfaces will be spherical, not flat. And the contour lines will look like this: g g g T or F: The gravitational field vector g is perpendicular to every point on the contour line. Question: Why are the surfaces farther from the center farther apart? Topic 6.2 Extended E – Equipotential surfaces The negative point charge acts as a planet, setting up equipotential surfaces in the same way: And the contour lines about a negative point charge will look like this: E E E T or F: The electric field vector E is perpendicular to every point on the equipotential surface. Question: Why are the surfaces farther from the center farther apart? Topic 6.2 Extended E – Equipotential surfaces Suppose we have two parallel plates separated by 15 mm and charged by a 9 V battery. (a) How far apart would the equipotential surfaces be between the plates, if their potential difference was to be 0.10 V? Since the electric field is constant between the plates the equipotential surfaces will be evenly spaced (unlike + those around a point charge). + V = Ed first we find the + 9 = E(0.015) value of the electric + field... + E = 600 V/m + V = Ex now we find the + distance x 0.10 = 600(x) + between the x = 1.6710-4 m + surfaces... + 0.015 m Topic 6.2 Extended E – Equipotential surfaces Suppose we have two parallel plates separated by 15 mm and charged by a 9 V battery. (b) If we assign the value of 0 V to the negative plate, where is the equipotential surface with a potential of +3.75 V located? We can use the formula V = Ex: 0.00625 m 3.75 = 600x: x = 0.00625 m - + + + + + + + + + + 0.015 m Topic 6.2 Extended E – Equipotential surfaces If we know the potentials and the geometry of the equipotential surfaces surrounding a charged object we can find the value of the E-field using V Electric Field E = x From Potential For example, suppose we are given that the voltage difference between the negative plate and the red equipotential surface is 3.75 V (it is) then the above formula gives us V = - V - V0 E = x - x0 x 3.75 - 0 = .00625 - 0 = -600 V/m 0.00625 m - + + + + + + + + + + 0.015 m this is the expected value FYI: The negative in the above formula gives the correct direction of E. FYI: If you are given Topic the equipotential you can construct the E6.2 surfaces, Extended field lines. E – Equipotential surfaces Sketch in the equipotential surfaces between the two charges. + - FYI: The definition ofTopic the eV uses an electron. But any particle that has a 6.2 Extended charge which is an integral number can use electron volts as an energy E – Equipotential surfaces quantity. THE VOLT eVNOT an SI unit of energy, but it is used often for FYI:ELECTRON The electron volt is describing Recall the relationship between voltage (electric the energy of atomic-sized particles. potential) and potential energy: U = qV. If we place an charge in an electric field, it will accelerate, changing its kinetic energy. From energy considerations we have K + U = 0 K = -U K = -qV The Kinetic Energy Change of a Charge We define the electron volt (eV) as the kinetic energy gained by an electron accelerated through a potential difference of exactly 1 volt: K = -qV 1 eV = -(-1.610-19 C)(1.000 V) 1 eV = 1.610-19 J Electron Volt