Physics 1120 ReviewWhat follows is a rough outline of physics 1120 material, it is by no means complete but is meant tohighlight basic ideas and topics. It is certainly not meant to be used as a “crib shee t” for your exam. As usualthe equations by themselves a re meaningless so please read the text of this document to aid in understandingtheir meaning. Where vector arrows are omitted for vector quantities it denotes the amplitude only. Thereader must determine the direction.1. ElectrostaticsA. Electric Fields and Forc esThe electric field is a vector field which obeys the laws of superposition, i.e. each contributing charge orcharge distribution can be considered separately and then the vector s um of all the contributions can betaken to give the total field. Electric field lines are a too l for visualizing a field. The strength of the fieldis proportional to the density of field lines. Lines originate a t positive charge s and terminate on negativecharges.Electric Field of a point charge:!E =kqr2ˆrForce on a charged particle due to an electric field:!Fe= q!EForce between two charged particles (combine last two eqns.):!Fe= ±k|q1||q2|r2ˆrIn the last equation the plus or minus sign is chosen so that like charges repel and opposite charges attract.A couple of examples of electric fields for specific charge distributions follow:Electric field for a single infinite sheet of charge: E =σ2#oElectric field of two infinite planes of opposite charge (b e tween the plates): E =σ#oThis last result can be seen from the superpo sition of two infinite plate fields, where each plate has anopposite charge (but equal magnitudes). Of cours e the field o utside of two oppositely charged parallel platesis z e ro (why?). All of the expressions for electric fields that you will see can be derived from Gauss’s Law(see below). Electric flux is defined to beΦe=!!E · d!Awhere since we are integrating over an area the double integral is implied. Flux is interpreted as the amount offield lines penetrating a surface. If the surface is clos ed (ie a sphere) then the outward normal is traditionallytaken as the positive. For a non-closed surface the choice for the direction of the area vector lies with you.For a uniform electric field and a flat surfa ce the above equation reduces toΦe=!E ·!A = EA cos (θ) (uniform field and flat area only)Here θ is the angle between the area vector and the electric field.To find the e le c tric field for a charge distribution with high symmetry a useful tool is Gauss’s law:Φe="!E · d!A =qenc#oHere qencis of course the amount of charge enclosed by our (closed) imaginary Gaussian surface. Themost common Gaussian surfa c e s are spheres, cylinders, and b oxes. Without high symmetry Gauss’s law isessentially useless. It is only useful in cases where we can choose a surface that is perp e ndicula r to the E1field everywhere, and everywhere along that surface the field has the same magnitude. Then we can back theelectric field out of the integral, leaving a simple integral over dA which just gives the area of our gaussiansurface (know your area formulas). It is then a simple matter to solve for the electric field.Conductors in electrostatic equilibrium have a few properties:• The e le c tric field inside a conductor is zero. (make s ure to know why)• On the surface of a conductor the electric field points perpendicular to the surface (or parallel to thenormal). This is due to the fact that the entire conductor is an equipotential (see below).B. Electric potential and Electric potential energySince the electric field is a vector quantity, it is mathematically a pain in the ass to deal with at times. Avery useful concept which is both convenient and offers a bridge into the world of electronics is electricpotential field. As with gravitational potential (with which you are hopefully more familiar) the zero po intof the potential can be chosen anywhere we want as long as we are consistent after that choice. Since thismeans that the absolute value of the potential has no meaning, the only thing that is meaningful are changesin potential. In terms of the electric field, change in electric potential is defined to be∆V = −!!E · d!lThis is a path integral, the dot product serves to pick out c omponents of our path that are along theelectric field (through the cosine dependence). Thus if we move perpendicularly to the field we move alongan equipotential line. Instead of drawing field lines, we can draw equip otential lines (or surfaces if youare a talented artist) to represe nt a field. By the preceding argument, field lines are always perpendicularto equipotential surfaces. In the gravity analogy, the gravitational force acts downward, and eq uipotentiallines/surfaces are then planes of constant height. If we are in a uniform field, like in a parallel plate capacitor,and choose a path along the direction of the field, the a bove integral reduces to∆V = −E∆lFor non-infinite charge distributions we normally take the zero point for the potential to be at infinity.With this convention we can writeElectric potential due to a point charge: V =kqrNotice that potential can be either positive or neg ative depending on the sig n of the charge. Since weintegrate the field, the electric potential for a point charges falls off like 1/r in contrast to the electric fieldthat falls off like 1/r2.Just as an object with mass in a gravitational field can have gravitational p otential energy, a object withcharge can have electric potential energy (EPE).EP E = qV for a point chargeThis is the potential energy associated with a charged particle (q) being placed in a electric potential fieldcreated by other charge(s). Moving a charged particle in an electric field can change its potential. The workrequired and the corresponding change in potential are related by the following:∆V = −Workelecq=Workextq=∆EP EqThe subscripts “elec” and “e xt” represent the works done by the field itself and an ex ternal agent respectively.The minus sign is important, convince yourself why. Notice that if we move along an equipo tential, ∆EP Eis zero and so no work is done by either the field or the external agent. Recall that Work =!F ·!d, the dotproduct of the force and the displacement. Ano ther useful result is the work-kinetic energy theorem:Worknet= ∆KE2where net means the total work done on an object, by both cons ervative and non- conservative forces. Theenergy density
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