PHY 184 Exam 1 Study Guide
PHY 184 Exam 1 Study Guide PHY 184
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This 21 page Study Guide was uploaded by Cameron Blochwitz on Saturday September 24, 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 68 views. For similar materials see Physics for Scientists and Engineers II in Physics at Michigan State University.
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Date Created: 09/24/16
PHY 184 Lecture Notes Week 1 8/31-9/2 Fundamental Forces There are four fundamental forces in Physics o Gravitational o Electromagnetic These forces operate at an infinite range o Weak Force o Strong Force These forces only operate within short distances All physical interactions are the cumulative result of the interactions of these forces o Most large scale interactions result from Gravity and Electromagnetism Relative strengths of the forces o If all of the forces are listed by relative strength they would look something like this Strong = 1 -2 Electromagnetism = 10 Weak = 10 -5 Gravity = 10 -38 Clearly Gravity is by far the weakest of all fundamental forces Gravitational Force The force on an object of mass m from a mass M is described by the following equation, Newton’s Law → F =−G Mm u G r2 This is instantaneous action at a distance o If M disappears, F sGops acting on m at once. Electrostatic Force The force on a charge q from a charge Q is described by the following equation, Coulomb’s Law Qq → FE=−K r2 u Acts in a very similar way to gravity o Also instantaneous, and also will disappear if Q disappears Light speed The speed of light, c, is given by: c=299,792,458 m/s o Time for Light to reach Earth from: Sun 8.3 Minutes SN 1987A 167,885 years As we know nothing can go faster than the speed of light o How can Earth suddenly not be bound by gravity but light from the sun still shine for 8.3 if the Sun disappears? Force Mediators(Bosons) Electromagnetic o γ (Photon) e- Weak Force e- o z Boson o w Boson Strong Force o g (Gluon) t - - γ e e n This case is created by the interaction of photons It also tells us that in fact, gravity and electromagnetism don’t act instantly, and that Newton’s/Coulomb’s Laws are incomplete, they do not have any kind of allowance for the finite speeds achievable in our universe All forces in our universe comes from the exchange of particles. PHY 184 Lecture Notes Week 2 9/6-9/8 Electrostatics Quark Structure of matter o Quarks are the smallest unit of matter Atomic Structure Atom Nucleus Nucleons Quark There are 6 different kinds of quarks +2 o Up/Charm/Top all have a charge of 3 e −1 o Down/Strange/Bottom all have a charge of 3 e All matter is made up of these three elementary particles Up/Down/Electron Isolated quarks have never been observed The carry a fractional charge Structure of an Atom Atoms are neutrally charged o Composed of a positive nucleus and negative electrons Nucleus itself is made of positive protons and neutral neutrons Describing an Atom Atomic Number (Z): Number of Protons or Electrons Mass Number (A): Number of Nucleons (Protons and Electrons) Number of Electrons: equal to Z, charge -Ze Number of Protons: equal to Z, Charge Ze Number of Neutrons N: A-Z=N Atomic Mass: Z*M +p*M n Electrons and Protons Electron is an elementary particle o It cannot be further subdivided Proton is not an elementary particle o Composed of 2 Up Quarks and 1 Down Quark o Charge of a Proton 2 2 1 qp=3e+3 e−3e=e Up Up Down Neutron is not an elementary particle o Composed of 2 Down Quarks and 1 Down Quark o Charge of a Proton n = e− e− e=0 3 3 3 UP Dow n Down Quarks are held together by the strong force Electric charge is a fundamental property that a particle or object must have in order to experience any electromagnetic interaction o Without charge the object cannot interact with electromagnetic fields. There are 2 main kinds of charges positive and negative Charged Rods Demonstration o Rubbing the rods will charge them Like charges will repel each other Opposite charges will attract each other Unit of charge is the Coulomb, C o Defined in terms of the Ampere 1 C = 1 A*s (Ampere Second) Unit Charge is defined by the electron o qe=−e −19 o e=1.602×10 C o Proton carries a charge of +e 1 C is a very large amount of charge o Often will be using smaller charges −6 1 μC = 1×10 C 1×10 −9 1 nC = C 1×10−12 1 pC = C Number of Electons to produce 1 C 1C N e −19 o 1.602×10 C The electric charge is quantized o There is a minimum amount of charge a particle can have This charge is e This is too small to be noticed in daily life But the Quarks have fractional charges! o This is possible because quarks are never isolated They are unable to exist on their own Charges of macroscopic objects are neutral o If it carries a negative it has extra electrons o If it carries a positive charge it is missing electrons ONLY ELECTRONS MOVE!!! Charges are able to be moved around o They can never be created o In any process the net charge is always a constant The electron structure of materials determines their ability to conduct electricity o Conductors conduct very well o Insulators conduct very poorly Electric Forces by Polarization Demonstration (Rotating Rod/Hanging Insulator) o There is an electric force between a charged rod and an uncharged one Why does this occur o The charged rod induces polarization it attracts the electrons and produces a net force + + + + + + + + + + + + - + + + + + + + + + + - 2 Point Charges Consider a single point charge q 1 Add a second point charge q 2 r q2 We can use Coulomb’s Law to find the force q1q2 F 1→2k 2 ^ r k is known as Coulomb’s constant 9N m2 8.99×10 2 C r^ is a unit vector pointing from q to 1 2 o This direction is fixed; the direction of the force will be determined by the signs on the charges o k and r are always positive Signs of charge are opposite the attract Signs of charge are the same they repel Coulomb’s Law contains the Law of Charges Comparing Electromagnetism and Gravity qe m e F ek 2 F gG 2 r r r will cancel and so 2 2 F ekq e F gG m e F e 42 =4.20×10 F g The electric interaction is vastly stronger than gravity o So much stronger that we are able to ignore gravity System of Charges When working with a system of charges you will need to consider each interaction of all the charges The force from system of charges (q , q , … , q 1 on2Q is n q1 q2 qn r r^1+r r^2+…+ r ^n o 1 2 n F=kQ¿ o This is known as the superposition principle It shows a vector sum of forces Equilibrium Position Two positive charges q and q are1placed a2 shown, where will a third charge q keep the system in equilibrium 3 o We know that the F 1 on 3must be equal to F 2 on 3 o When added the forces net to zero. o We therefore have the equation q q 1 = 2 (x −x ) 2 (x −x ) 2 3 1 2 3 Solve for x 3 Note the charge of q is irrele3ant! This method is the basis for all equilibrium problems PHY 184 Week 3 Notes 9/12-9/15 Electric Fields and Gauss’s Law We have seen that a point charge in space seems to affect the space We know this because of the force experienced by a second charge, a force given by Coulomb’s Law A field is used to designate a quantity that varies in space from one given point to another o For example, temperature field in France is a scalar field o And a wind field is France is a vector field Definition of the Electric Field The electric field produced by a charge q at point P,1located at a distance r is1given by E =k q1r^ o 1 r1 The field depends only on the charge and the distance from the charge If the field is positive the field lines moves along the unit vector o It the field is negative it is opposite the unit vector If there are several different charges we just calculate the field generated from all the charges q1 q2 q3 qn o E=k( r1 + ^2+ ^3+…+ ^n) r1 r2 r3 rn The force on a charge therefore is F=q E o 2 o This is simply a different way of writing Coulomb’s Law We are able to see then that a charge “creates” a field and that a second charge placed in that field is “sensitive” to that field The definition of the force relative to the field is determined by the sign of the charge that is placed at a point P in that field Field Lines Used to represent the electric fields produced by a charge Tangent of the field line gives the direction of the force on a charge placed there o If the charge is negative the force points opposite the field lines o If the charge is positive the force points along the field line Field lines from a positive point charge radiate outwards from the charge Field lines from a negative point charge radiate inwards toward the charge Electric Dipole An electric diploe is a system of two charges, one positive and one negative The vector sum of the electric field at any given point gives the electric field of the dipole at that point Electric Field of an Electric Dipole We know that the vector sum of the electric field of the two point charges gives the electric field at any given point o If we then preform these calculations when each charge is equal magnitude, we get the following result qd E=2k x3 Where q is charge, d distance between charges, x distance to point we are calculating One charge “screens” the other, but only along the x- axis Electric Dipole Moment The electric diploe moment of a pair of charges having the same magnitude but having opposite signs is defined to be → → o p=qd E=2k q3 E=2k p3 Therefore, x can be written x Dipole in Electric Fields If there is a diploe placed in an electric field where each charge has the same magnitude but the opposite charge o There is no net force but there is a Torque τ=qEdsinθ=ptsinθ=P×E ´ Charge Distribution We have seen that the electric field of several point charges is simply the vector sun of all the electric fields from all the point charges. o How do we then calculate the field when the charge is continuous? o We need to integrate over the charge distribution E=k ∫ d2^ r o This can be done for a line, surface, or volume dq=λdx dq=σdA dq=ρdV Linear Charge Distribution Before doing anything try to anticipate the direction of the field For a straight horizontal line, all non-vertical component will cancel during vector addition o Preforming the calculations gives you E= 2kλ y However, this calculation is long and difficult, it is much easier using Gauss’s Law The magnitude of an electric field at a point is E=k q2 o r Now over a line it is the derivative λ o E=2k y And for a surface o E=Constant! Flux Consider a disk A perpendicular to “something” flowing with velocity v The product of Av gives the volume of that “something” that flows through the disk per unit time Multiply Av by the density we get the “mass” passing though the disk through the disk per unit time If the disk is tilted by θ the projected area of the disk perpendicular to the velocity is Acosθ The “something” though the disk per unit time is given as Avcosθ The volumetric flux of something though a surface A is defined to be Flu(φ=Avcosθ o o φ=V ∙A Note that flat surfaces are ambiguous By analogy, Electric flux is o φ=EAcosθ φ=E ∙A o Closed Surfaces If we have a closed surface, we can define a perpendicular vector always! It will always point OUT of the surface The total electric flux though a closed surface is defined as φ=∯ E∙d A o The scalar product is to be taken locally at any point on the surface. Electric Permittivity of Space We know the value of Coulomb’s constant k o k can also be written as 1 k= 4πε0 ε -12 2 2 0 is equal to 8.85 x 10 C /Nm The flux and electric field though a closed surface is φ= q o ε 0 The surface that we use to perform this integral is known as a Gaussian surface to make work nice and easy If there is no charge, there is no flux Gauss’s Law makes no assumptions about the charge distribution within the volume o This makes it far more general than Coulomb’s Law It can be used to get Coulomb’s Law Planar Charge Distribution Apply Gauss’s Law to calculate the electric field with a charge density σ We need to calculate the flux q σA o φ= ∯ E∙d A=EA+EA= = ε0 ε0 E= σ o 2ε0 It is a constant! Two Metal Plates Calculate the electric field between two metal plates Q σ= 2 o L o Solving for E gives σ E= ε 0 Apply Gauss’s Law to calculate the electric field produced by a spherical shell of radius r carrying a charge q We know from symmetry that the electric field must the same at all points of a concentric sphere Pick 2 Gaussian surfaces with one radius smaller than r and one greater than r Using Gauss’s Law for the largest sphere is 1 q E= 2 o 4πε 0r2 The same as a point charge! Using Gauss’s Law for the smallest sphere is o E=0 Therefore, a shell acts a point charge from outside and has no charge from the inside Shielding In a conductor, electrons are free to move and when influenced by an electric field, according to the force The Electrons move upstream and leave behind positive charges generating an electric field o It compensates for the external electric field Net electric field inside is 0 Uniformly Charged Sphere Calculate the electric field for a total charge q distrituted with positive density on a radius r Choose 2 surfaces, one smaller and one bigger that the radius of the charges sphere E = ρr1 o insid3ε0 k q Eoutside2t o r2 Polarization of an Uncharged Sphere Suppose we have an uncharged conducting sphere with q 1 inside and q on2the surface o q w1ll add to q dur2ng calculations PHY 184 Week 4 Notes-Electric Potential 9/19-9/22 We have seen that gravity and electromagnetism are very similar in their behavior even going so far as to have almost identical equations 2 o Both have forces that are determined by 1/r These forces are all conservative forces Doesn’t depend on path taken o Work is independent Consider an electrostatic force acting in charge o The work is produced between i and f The workfintegral is ´ i F∙ds The variation of potential energy, ΔU, is the negative of work done by the electrostatic force o It is also the work done against the electrostatic force Like gravity we need to select a reference point is defined Potential is 0 when the charges are infinitely far apart The variation of the Potential Energy reads like ∆ U=U −f=U−W e,inf o o The negative shows Positive work for U<0 Negative work for U>0 Consider a charge q located in an electric field moving on d where d is at an angle θ to the electric field Force on the charge ´ ´ o F=q∙E W=F ∙d´ o Work done is ´ o W=qE∙d=qEd cosθ If displacement is along the direction of the field o Positive charge loses potential energy If displacement is opposite the field o Positive charge gains potential energy Electric Potential Electric potential is defined to be as the “electric potential energy per unit charge” V= U o q It describes a property of space and therefore is a scalar The unit of electric potential is the Volt, V 1 J o 1V= 1C Therefore, the units of an electric field are able to be given in V/m Variation of Electric Potential V f V i ∆U −W e ∆ V=V fV =i q − q= q = q ∆U f ∆ V= =− ∫∙d ´ s q i Energy of a Proton A Proton is between 2 plates o Potential difference is 150V o Proton is released from positive plate What is the K pt the negative plate? o We are able to use energy conservation to solve this problem Solving for gives Kf=−q∆V Acceleration though a system is able to be calculated again from energy conservation o The particle starts at rest so we can calculate acceleration in two stages ∆ K=|∆U = q∆V |+| 2=K K=e∆V +6 e∆V =7e∆V We are then able to insert numerical values o We are able to use the electron-Volt, eV, instead of a joule -12 1 eV is 1.602 x 10 Joules When an electron “falls” across a 1.5 Volt difference the energy gained by that electron is 1.5 eV Equipotential Lines-Gravity Consider a landscape with mountains o On a topographical map they appear as a series of rings set at a certain height All these lines have the same gravitational potential They are equipotential lines! Equipotential Surfaces An electric charge generates a field in space o Electric potential has a value everywhere in space o Those points that have the same electric potential form a surface o Charges that move along these surfaces have no change in potential energy When moving along Equipotential Surfaces there is no work done by the electric force Conversely, if the line integral of the field is 0 the electric potential must be constant o Therefore, moving a charge perpendicular to the force there is no work The charge moves along equipotential lines Any conductor forms an equipotential surface because the electric force is always perpendicular to the surface Finding Potential Difference in a Uniform Field f ∆ V=− E∙d´ s ∫i Preforming this integral gives the solution ∆ V=−Ed o If at position z 00 and the potential is V we hove o ∆ V=V (z−V =0E(z−z ) 0 o V z)=V 0Ez If potential at infinity is 0 f V r)−V ∞=V (r=− E∙d´ s o ∫i Projection away from charge along field V r)= kq o r Distance of Closest Approach A charge is fixed at x=0. A particle of mass m and charge q is 2 fired with an initial velocity v tow0rd the charge at from a distance d . 1 kq1 o V ( ) x kq q o U ( )q 2 x( ) 1 2 x Using energy conservation, we are able to find that mv2 1 = 1+ o o d f di 2 k 1 2 System of Charges The potential of a system of charges is given by superposition Minimum Potential Given 2 positive charges q and q a1 what p2sition is electric potential minimized Find V, set derivative to 0 and solve o Solving the equation gives x= r2 q2+1 √ q1 Potential Energy of a System In a system of charges each charge has equal weight If two charges have the same sign potential is positive, if opposite negative For a system greater o Consider every pair in the system and sum Electric Field from Electric Potential We have seen so far how to calculate the electric potential given the electric field It is important to know that we can go the other way too, using potential to get the electric field This is done using partial derivatives −∂V ∂V ∂V o E= ( ,− ,− )−∇V ∂x ∂y ∂z o If we have the potential, we are able to derive the electric field at any point Continuous Charge Distribution For a system of charges, the potential is just the sum of the potentials of the charges in the system To determine the electric potential due to a continuous distribution we divide the charge into charge elements and treat them as point charges V =k dq o p ∫ r To find the potential at a point in a semicircle o We know that there is a symmetry giving us the variable θ o The length can be written as dL=Rdθ o Electric charge can be written dq=λdL Solving the integral gives V=πλk If we do the same for a Linear horizontal charge with length 2a and distance d we find √a +d +a o V=kλln a +d −a √
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