PHY 184 Exam 2 Study Guide
PHY 184 Exam 2 Study Guide PHY 184
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This 15 page Study Guide was uploaded by Cameron Blochwitz on Sunday October 16, 2016. The Study Guide belongs to PHY 184 at Michigan State University taught by Oscar Naviliat Cuncic in Fall 2016. Since its upload, it has received 5 views. For similar materials see Physics for Scientists and Engineers II in Physics at Michigan State University.
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Date Created: 10/16/16
PHY 184 Week 5 Notes-Capacitors 9/27-9/30 Capacitors o Device that can store energy Many different uses o Camera Flash o Touch Screens Essential in Electronics Capacitor consist of two separated conductors with an insulating layer o Can be rolled up with another insulator to increase the surface and storage A simple geometry of a capacitor each with area A and distance d in a vacuum with charge on one plate of +q and a charge on the other of -q The surface of the conductor is an equipotential surface o Equipotential lines are close to each other near the plate and far away outside of the plates Electric field lines are perpendicular to equipotential lines o Almost uniform between the plates o Far from the plates it acts as a dipole Field outside of the plates is called the fringe field Capacitance The main property of a Capacitor is its Capacitance Capacitance is given by q o C=| | ∆V Capacitance is an intrinsic value of the capacitor o It cannot change Capacitance shows how much charge can be stored at a given Voltage ∆V o q=C ¿) Farad The unit of capacitance is called the Farad 1C o 1F= 1V 1 Farad is a very large capacitance o Most Capacitors range from pF to μF Finite Parallel Plates We already know that σ q o E= ε = Aε o 0 The voltage difference between the plates is ∆ V=−∫Eds o From the negative to positive plate −Eds =¿Ed d o ∆V=− ∫ 0 Capacitance of Parallel Plates is therefore Aε C= q = 0 o ∆V| d Cylindrical Capacitor Consider a capacitor constructed by 2 co-axial conducting cylinders of length L inner radius r a1d outer radius r 2 o Inner cylinder has charge -q and outer cylinder has charge +q From symmetry electric field is radial ∯ E ∙dA=−E ∯ dA=−E 2πrL = −q ε0 q o E= ε 2πrL 0 Voltage difference is f r2 2 q q r o ∆ V=−∫E∙ds=− E∫r= ∫ dr= ln 2 i r1 1ε02πrL 0 2πL r1 Capacitance is then q q ε 2πL C= = = 0 o |∆V q r2 r2 ln ln ε02πL r1 r1 Spherical Capacitor Capacitor made of 2 concentric spheres o Inner sphere charged to -q with radius r 1 o Outer sphere charged to +q with radius r 2 Electric Field is radial and uniform Applying Gauss Law E ∙dA=EA=E(4πr )=2 q o ∯ ε0 q E= 2 4πε 0 Voltage difference is f 2 2 q −q 1 1 o ∆ V=− ∫∙ds=− Ed∫= ∫ 2dr= 4πε r( )− r i 1 14πε 0 0 1 2 Capacitance r r C=4πε 0 1 2 o r2−r1 Capacitance of a Single Sphere The capacitance of a single sphere is obtained from the above result with the outer sphere having an infinite radius C=4πε R o 0 Potential is then q o ∆ V=k r Electric Properties of a Capacitor These configurations use a vacuum to separate the plates o ε d0termines the electrical properties Dielectric Materials 2 layers of a capacitor are separated by and insulating material that is dielectric o 2 kinds of dielectric material, polar and non-polar Polar materials have a permanent dipole moment These dipoles are randomly oriented Electric fields will make them all point the same way Non-polar materials have no permanent dipole moment Can be polarized to align in an electric field o Smaller effect than polar These materials serve several purposes o Provides a way to maintain separation o Provides insulation o Increases the capacitance o Allows capacitor to hold a higher voltage Electric Field in a Dielectric In a dielectric, the Field d resulting from the aligned dipole will reduce E ET=E−E d o Capacitor with Dielectrics Placing a dielectric has the effect of lowering the electric field allowing more charge to be stored It will increase the capacitance of the capacitor by a dielectric constant κ. C=κC o air κ ≡1,κ ≡1 o vaccum air Parallel Plate Capacitor Placing a dielectric between the plates E=E a=r q = q o κ κε0A εA Electric permittivity o ε=κε 0 Making Capacitance εA o C= d Work to Charge a Capacitor When a battery charges a capacitor charge accumulates on the capacitor We have to work to put charges on something already charged Work is therefore dW=∆V ' dq' o If there is a capacitance C dW=∆V dq =' q'dq' o C Total work is then qq' 1 q o Wt=∫dW= ∫ dq'= 0C 2 C Energy stored in the Capacitor is then given by 1 q2 1 2 1 o U= = CV = qV 2 C 2 2 Electric Energy Density, u, is defined by potential energy per volume For a Parallel Plate Capacitor U CV 2 1 2 o u= Ad= 2Ad =2ε 0 Circuits with Capacitors Electric Circuits are a set of elements connected with wires These need some form of power source Wires Wires are represented by a line o All points of the wire are at the same potential Equivalent to a point Charging and Discharging a Capacitor When connected to a battery, current will flow until the capacitor is charged Combination of Capacitors Parallel o Potential is the same across all the capacitors V=V =1 =…2V n o Charges are given by q=q +q +…+q =(C +C +…C )V 1 2 n 1 2 n Series o Charge is the same across the series V= q o c eq 1 1 1 1 = + +…+ C eq C 1 C 2 C n PHY 184 Week 6 Notes-Currents and Resistance 10/3-10/6 The battery produces a constant voltage When closed the current is able to flow o Don’t forget only the electrons move In the absence of an electric field electrons move randomly within the conductor o When an electric field is applied electrons get an additional velocity called drift velocity It is 10 times slower than thermal velocity How is it almost instant then? o Closing the current starts an electric field and sets the electrons in motion Charge Carriers Consider a conductor with cross section A and electric field E Suppose there are n positive carriers per unit volume, all evenly distributed in the same dV with drift velocity v d o In time dt the distance traveled is dx=v dt The volume of charges though A is o Adx=Av dt The number of charges is given by nAv dt d If each charge has charge e, the charge dq is o dq=nevdAdt Electric current is therefore dq o i=dt=nev d Charge carriers in real life are electrons o Charge is -e Everything goes to the left Expression remains unchanged due to a double negative By convention current flows from positive to negative The unit of current is called the Ampere or Amp Current Density Consider a current i flowing in a conductor with area A o Current though area is J i=∫J∙dA If the current is uniform and perpendicular to the surface i=JA o J= =nev A d Current Density is parallel to electric field and anti-parallel to the drift velocity Resistivity Some materials conduct electricity better than others If we apply voltage across the conductor we get some current, if the same voltage is applied to an insulator we get less current This is called resistivity o Describes the inability to conduct ρ=E J Units are in Ωm Ohm’s Law There is a ratio between current and voltage ΔV o R= i When a potential difference is applied to a conductor and we ΔV measure the current R= remains constant i Ohm’s Law states o ΔV=iR Units of resistance are in Ohms 1V o 1Ω= 1A Resistance When applying a potential difference across the conductor the resistance will a determine how large the current is o Large R, small i The property of a device that describes an inability to conduct is called the resistance o It is a property of the material Imagine a narrow pipe Depends on material and geometry Consider a homogenous wire of length L and cross section A o Magnitude of electric field is related to the potential difference E= ∆V L o Magnitude of J is J= i A o Resistance ΔV E L ΔV A Ri A A ρ= J i =( i (L)( i (L) ( )L A L R=ρ A Resistivity and Conductivity Resistivity and Conductivity are both inverses of each other 1 o ρ= σ 1 o σ= ρ Temperature dependence of Resistivity The resistivity of a material is determined in part by its temperature In metals, this is a linear dependence over a wide range of temperatures This relation is given by o ρ−ρ 0ρ α0T−T ) o Where ρ is the resistivity at a given temperature ρ 0 is the resitivy at a given reference temperature α is the temperature coefficient of electric resistivity The resistance is then o R−R 0R α0T(T o) Thermistors Thermistors are resistors with a heavy reliance on temperature o They can be used to measure temperature Resistance decreases with increasing temperature Resistors in Series Resistors that are connected such that the same current flows through all the resistors are connected in series Voltage in a series is given by V=V +V +…+V o 1 2 n Current is given by V=i(R 1R +2+R ) n o Equivalent Resistance is given by o ReqR 1R 2…+R n Resistors in Parallel When 2+ resistors are connected such that they share the same potential difference they are connected in parallel From Ohm’s Law i = V o a R a Since potential difference each resistor is equal to the potential difference provided by the battery the currents are given by V V o i1= and 2 = R1 R2 Current is from the source must equal the sum i=i +i +…+i o 1 2 n Therefore the equivalent resistance is 1 1 1 1 o R eq R1+ R2+…+ R n Electric Energy and Power When introducing the fundamental concepts, we have seen that the work of external forces required to move a charge across a potential is related to dU o dU=dqΔV i= dq dt we can rewrite the differenctial as o dU=idtΔV Power is defined as P= dU =iΔV o dt Unit of Power is the Watt Electric bills are based on energy used o Energy is Power X Time Energy in a Resistor Using Ohm’s Law 2 V 2 o P=i ΔV=i R= r PHY 184 Week 7 Notes-DC Circuits 10/10-10/13 Most practical circuits cannon be resolved into series or parallel systems of capacitors and resistors o To handle this we use Kirchhoff’s rules A Junction is a place where 3+ wires connect o Each connection between 2 junctions is called a branch Each branch has the same current within the branch A Loop in a current is any set of connected wires forming a closed path o You can get back to where you started Kirchhoff’s Junction Rule The sum of currents entering a junction equals the sum of currents leaving the junction Positive terms enter, negative terms exit n i =0 o k=1k Consider a Junction with current i e1tering and i an2 i lea3ing i =i +i o 1 2 3 This is a direct result of conservation of electric charge o Junctions cannot store charge Kirchhoff’s Loop Rule The potential difference of a complete loop is equal to 0 o Direction is irrelevant All that matter is that you follow the same conventions around the loop Resistors o Along Current -iR o Opposite Current +iR Battery o Same as emf +V o Opposite emf -V Emf goes from negative to positive The direction you travel around the loop and the direction of the current are totally arbitrary o Will not change final result outside of a minus sign General Observations Voltage drop is a linear relation Kirchhoff’s Junction Rule established linear relations between currents in the junctions o Coefficients are +1 or -1 If a circuit contains only batteries and resistors Kirchhoff’s Loop Rule established other linear relations For Linear circuits we can determine the voltage by knowing the current in each branch A circuit with n junction has n-1 independent equations A circuit with 4 junctions and three loops have 3 equations from junctions and 3 from loops o Choose the smallest loops to work on Analysis of Linear Circuits Choose a direction for all currents Choose loops and a way to travel them Write the n-1 relations for currents Look at ΔV along the selected loops and write the loop and write the loop rule Finally, you have N equations and N unknowns o You can solve the equation Devices Ammeter measures current o Is in series and has low resistance A Voltmeter Measures Voltage o Is in parallel and has high resistance RC Circuits An RC circuit has both a resistor and a capacitor o They will vary over time Charging a Capacitor We start with the capacitor uncharged When the switch is closed current begins to flow building the charge on the capacitor o This creates a potential difference in the capacitor When the capacitor is fully charged no current will flow o Potential from the capacitor and the battery are equal When charged the charge on the plates is q=CV To Calculate the current in the circuit we use the KLR for a single loop Start at a given point with arbitrary current and loop directions Applying KLR V−i t)R− q(t=0 o C By definition dq(t) o (t= dt So we can rewrite KLR as V− dq(t− q(t=0 o dt C Which can be rearranged to dq(t) 1 q o ( ) = − ( ) () dt R c Integrating this and solving for q gives −t o qt)=qmax(1−eRC)¿ o τ=RC qmaxCV o Current flowing though the circuit is the time derivative −t (t= V eRC o r i0= V R Potential difference across a capacitor can be written RC o V t)=V (1−e ) Discharging a Capacitor Consider the circuit containing one resistor and one capacitor o When disconnected from a battery the charged capacitor will begin to discharge until depleted Vr+V i0 q(t) (t)R+ C =0 Solution is therefo−t RC o q(t=q 0 Current is q o it)= RC
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