Electrostatics PDF

Electrostatics An alpha particle is shot 2.00 x 105 m/s into the middle of parallel plates that have a field intensity 3.30 N/C. If the plates are 10.0 cm long, what is the deflection from the original path? Draw field lines, particle’s path  There were 2 ancient effects that seemed like ‘a...

Electrostatics An alpha particle is shot 2.00 x 105 m/s into the middle of parallel plates that have a field intensity 3.30 N/C. If the plates are 10.0 cm long, what is the deflection from the original path? Draw field lines, particle’s path  There were 2 ancient effects that seemed like ‘action-at-a-distance’ 1. Lodestone effect (magnetism) -Works well on certain metals (Fe, Co, Ni) -Limited effect: these metals only strongly attract certain metals 2. Amber effect (electrostatics)  Electrostatics: History ~600 BCE: Thales noted the ‘Amber Effect’  Amber is semi-transparent, solid, (yellow or brown) petrified/fossilized tree sap  Amber had a property of attraction if rubbed vigorously against a cloth. This is the first recorded evidence of electrostatic attraction. Electro= Greek word for amber Static= stationary Electrostatics: the study of electric charges at rest.  The amber effect works on many substances.  People thought that only certain substances could be electrified by friction (electrics) and others couldn’t (non-electrics)  ‘Non-electric’s were metals.  Later it was found that ‘electrics’ didn’t allow electricity to flow easily (insulators)  ‘Non-electrics’ were conductors  Conductors can be charged if they are insulated from everything else  It seemed that metals couldn’t be charged. Why not? Because they are grounded easily. They have to be insulated from the ground. Extra: 2 types of effects: Resinous and Vitreous Dufay’s 2 Fluid Theory. France (1700 AD)  Dufay proposed that substances were composed of 2 different types of electric fluids (vitreous, resinous).  A positive fluid and a negative fluid.  Neutral objects would contain equal amounts of each fluid and friction would cause an excess of one or the other fluid to accumulate on the object. Father of Electricity  An easier explanation: 1 fluid theory Benjamin Franklin’s 1 Fluid Theory: (USA 1706 -1790)  Charges of vitreous and resinous fluids neutralize each other  Proposed a 1 fluid theory  Neutral objects have a ‘normal’ amount of electric fluid  “Charges” due to transfer of electric fluid when rubbed against each other.  excess of the electric fluid = a positive charge, lack of the fluid = a negative charge.  If an object gains fluid, it becomes positive  As objects lose electric fluid, they become more negative  Fluid composed of tiny (invisible to the eye) particles  If you rub a neutral ebonite rod with neutral fur, the ebonite rod gets a charge  If Ben’s right, the fur should also get charged  Which gained fluid and which lost fluid? Fur Ben’s fluid Modern view: electrons transfer from glass rod to fur. Fur becomes –ve, glass +ve  Ben didn’t know so he arbitrarily decided that:  The charge on glass when rubbed with silk is positive (the glass gained fluid)  The charge on ebonite when rubbed with fur is negative (the ebonite lost fluid)  This convention still holds  (Electron is not discovered until 1897)  Electrons are ‘negative’ because they are repelled by an ebonite rod  Grounding Grounding: providing a path for the charge to flow to/from the ground, leaving the object neutral.  We insulate conductors so they can hold a charge: (pith ball is insulated by floss, charged metal is insulated by insulating stand)  A ground is an object so large that the gain or loss of a few e- has a negligible effect Symbol:  Law of Electrostatic Charges  Like charges repel  Negative opposes Negative; positive opposes positive  Unlike charges attract  Negative attracts positive  Neutral attracted to both types (+ and -)  Law of Conservation of Charge  Discovered by Ben Franklin  The net electric charge of an isolated system is constant.  If a positive charge is created, then a negative charge is also created in an isolated system (a free neutron will decay into a proton and an electron after a ½ life of 11 minutes)  Franklin described how excess and deficits of electric fluid account for attraction/repulsion q  q ' Attraction/Repulsion can cause motion Polarization is the separation of charge in an object. Polarization can occur if there is a charged influence.  Modern Explanation: key points ~1800: Dalton’s model of the atom was proposed 1897 electron and 1911 proton discovered Only e- s are transferred, p+ s are trapped in the nucleus  Positive = deficit of electrons  Negative = excess of electrons  Neutral have equal numbers of p+ s and e- s  Electrons are much easier to move than a nucleus Franklin is pretty close. He pictured a positive fluid exchange, not negative electrons Dufay is pretty close. Both +ve and -ve particles exist Materials are categorized according to electrostatic behavior.  Conductors: ‘Free’ electrons In conductors, -ve charges can move easily. Law of electrostatic charges: charges want to be as far apart from like charges and close to unlike  Solids = only e- s move from one atom to the next. Electrons are not bound tightly to an atom  Liquids/solutions/gases = both +ve and -ve parts can drift, causing polarization (eg ions) Conductor: Material in which the e- s of the outermost region of the atom are free to move  Insulators: Localized effects  Movement of charge is difficult.  An excess/deficit of charge remains localized  Electrons bound to atoms/molecules may shift but not leave. Insulators: Material in which the e- s are tightly bound to the nucleus and not free to move within the substance Charge shift: movement of e- s in an atom where one side becomes -ve and the other +ve (polarization).  Extra: Semi conductors: special paths Crystalline structure provide low resistance channels (hallways) for e- flow  Good conductors in some circumstances and good insulators in others Eg: selenium is an insulator in the dark and a conductor in the light. Important component of a photocopier drum. Extra: Superconductors: very chilly Onnes (1911) observed mercury lost resistance when cooled to -269ºC  No measurable resistance at cold temperatures (100% efficient current)  1986 high temperature (-216ºC) superconductors YBa2Cu3O7  Latest: copper alloys (-123ºC)  Imagine if we found room temperature superconductivity!  http://www.youtube.com/watch?v=4VGACLNfZ8s  Ways to charge an object: 1. Friction: energy input causes e- transfer 2. Conduction: transfer of charge by contact  Both objects share the charge of the system  Net charge is conserved  Unequal charge sharing if objects are different shapes/sizes/materials 3. Induction: polarization of charge due to an influence and then grounding  Due to the law of electrostatics, charge (-) moves toward or away from influence charge  Charging by Friction 2 neutral objects: after rubbing which will have an excess of e- s?  When materials are rubbed together, the energy is used to move electrons from one material to the other.  The material that holds electrons loosely becomes +ve, the material that holds e- tightly becomes –ve. Electrostatic Series  Charging by Conduction Transfer from area of high concentration to lower Law of electrostatic charges helps us predict the distribution of charge: the goal of the electrons is maximum personal space  Charging by Conduction There is perfect charge sharing if the conductors have the same size/shape/material  Studying Electrostatics: electroscope metallic knob & stem (CONDUCTOR) Insulating rubber stopper glass box gold foil leaves Neutral Charged electroscope electroscope All conducting parts share the charge thus the 2 foils repel  Charge Migration: Polarization Conductor! Electrons shift away Insulator Everything stuck in place Law of Electrostatic Charges and category of material helps us explain the electron story  Movement of charge Charge migration: movement of e- s in a neutral object where one side of the object becomes positive and the other negative The object is in a POLARIZED state Grounding: the process of transferring of charge to and from the Earth  Use ‘electron stories’ to explain phenomena  Where/why are the electrons moving from/to?  Use a known charge on a pith ball or ebonite rod to determine if another object is +ve or –ve. Bring the object close to the pith ball, attract or repel?  Remember that polarization is very common.  Why does a balloon stick to a wall?  Sometimes a neutral pith ball attracts to a charged object and sticks to it for a while before repelling. Why?  Charging by Induction 17-01 Charging an object by polarizing it by induction, then retain the charge by grounding  Extra: Electrophoresis 17-03 How does the metal plate attain a charge?  Rub the resin block with fur, resin becomes –ve  Put metal disk on resin and ground the metal  Remove the metal disk and test its charge with a pith ball (metal has +ve charge)  Repeat steps and keep getting more charge without losing original –ve charge on resin! Whilmhurst 17-04  Process of induction is driven by the law of electrostatic charges. Your tools: the electroscope neutral electroscope charged electroscope  Lab: Develop a procedure to charge an electroscope: Negative  Process #1: Conduction  Process #2: Induction  You must start at the beginning. Choose the correct equipment (begins neutral)  Show the charge movement in the diagrams  The electron story: Use proper terminology to explain what happens and why  Formation of Lightning 17-11 Electrostatic 1.Frictional discharge can effects be FAST 3.Electrons Eg lightning flow down Electrostatic 2.Polarization discharge can Induced be SLOW Eg lightning rods  Coulomb: France, 1785  Joseph Priestly had proposed that the formula for electric force between 2 charges would be similar to Newton’s Universal Law of Gravitation: Gm1m2 Fg  2 r  Priestly proposed that the size of Fe would depend on 3 things: F e  q1q2  Size of charges  Distance between charges 1 Fe  2  Medium r  Coulomb  Coulomb used a torsion balance to find the electric force constant k. kq q F  e 1 2 2 r  Coulomb did this 13 years before Cavendish used a torsion balance to find G  G is much harder to measure than k, because Fg is about 1000 times weaker than Fe. http://ffden-2.phys.uaf.edu/104_2012_web_projects/cicely_shankle/Images/coulomb%20torsion%20balance.gif retrieved Feb 28, 2014  Torsion Balance  Use small charges (“point charges”) so you use the distance between their centers. Remember that charge accumulates on the outside of objects. Fe (N) Fe (N) Fe (N) r (m) 1/r (m-1) 1/r2 (m-2)  Torsion Balance  When q is manipulated, it’s hard to keep r constant. Fe (N) Fe (N) q1 q2  Electrostatic Force: Charles de Coulomb Attractive or Repulsive Coulomb’s Law: the strength of electrostatic interaction depends on: F e  q1q2 1 Fe  2 kq1q2 r Fe  2 r  k = 8.99 x 109 Nm2/C2  k is a proportionality constant. It makes the units work properly, so you must use the correct units.  BMU#2  Unit of Charge: coulomb  A coulomb is the laboratory developed standard unit of charge  Symbol for coulomb: C  Proton p+ = +1.60 x 10-19 C  Electron e- = -1.60 x 10-19 C  Alpha particle α2+ = +3.20 x 10-19 C How many electrons are contained in 1.0 C? Elementary charge = 1.60 x 10-19 C/e- , so: 6.25 x 1018 e/C How much charge is contained in an 1. SO42- 2. Fe3+ 3. Fe atom 4. Fe2(SO4)3  2 spheres each have a 1.0 C charge and are 1.00 m apart. Find the force between them. Fe = kq1q2/r2 Fe   8.99 x 10 9 Nm 2 /C 2  1.0 C 1.0 C  = 9.0 x 109 N repel (1 m) 2  This is a massive force! These would be killer spheres, they’d repel through walls, people, etc  1.0 C is a ridiculously large static charge!! In a hydrogen atom, the e- and the p+ are separated by 5.29 x 10-11 m. What is the electric force on the electron? Coulomb’s law gives the strength of the interaction. Since the charges are opposite in sign, they will attract. q e = -1.60 x 10-19 C, q p = +1.60 x 10-19 C, r = 5.29 x 10-11 m q1q2 Fe  k 2 r Fe   8.99 x 10 9 Nm 2 /C 2 1.60 x 10 -19 C 1.60 x 10 -19 C (5.29 x 10-11m) 2 Fe = 8.22 x 10-8 N attraction How does this compare to gravitational force?  The Fe between 2 charges is 3.00 x 10-6 N repulsion, what will be the force if:  The charge on each is doubled?  1.20 x 10-5 N repulsion  The distance is halved?  1.20 x 10-5 N repulsion  The charge on one is doubled and the distance is tripled?  6.67 x 10-7 N repulsion  Could these particles be (p+, e), (p+, alpha), or (p+, neutron)?  Solving Ratio Problems  Find the equation relating the variables  Set up the new (prime ‘ ) as the numerator  Set up the old/original as the denominator  Cancel the variables that don’t change q1q2 Fe  k 2 r q1 ' q2 ' 2q1 2q2 Fe ' k ' 2 k' 2  r'  r' 4 Fe q1q2 q1q2 k 2 k 2 r r The Fe between 2 charged objects is 400 N at distance r apart. If the objects have their charges reduced to ½ their original charge, and the distance between them is reduced to 1/3 r, find the new force between them q1q2 Fe  k 2 r 1 1 q1 q2 k' 2 2 2 q1 ' q2 ' 1  Fe ' k '  r  r '2   3   9 400 N q1q2 q1q2 4 k 2 k 2 r r 9 F e '  (400 N )  900 N 4  Vector nature of Fe (net force must be determined) The charged spheres below are touched together and then returned to their position. Find the electrostatic force on A. A B 0.500m -3.00μC +4.00 μC  Solution: according to the Conservation of charge, once contact made between the spheres, the net charge will be shared evenly. After contact they will have the same charge and a repulsive force will result (Law of Electrostatics). Coulomb’s law will tell us the magnitude Net charge = -3.00 μC + +4.00 μC = +1.00 μC A B 0.500m +0.50 μC +0.50μC kq1q2 Fe  2 r Fe   9 8.99 x 10 Nm /C2 2   0.500 μC  2 [repulsion]  8.99 x 10-3 N left   0.500 m  2 Where should a charge be placed so it would remain stationary? 4.00 m A B C X=?? A=+3.00C B=+9.00C  Solution: in order to remain stationary Fnet on c = 0, Principle: Balanced Forces. Fe  Fe ac bc kq a q c kq b q c =  4.00 m - x  2 2 x Choose the mathematical path to make things easier. Fe  Fe ac bc kq a q c kq b q c =  4.00 m - x  2 2 x qa qb = cross multiply and  4.00 m - x  2 2 x qa x2 =  4.00 m - x  2 qb 3.00C x  9.00C  4.00 m - x  x = 1.46 m from A or 2.54 m from B  Find Fnet on B -5.00 C -2.00 C +3.00 C A 2.00 m 3.00 m C B  Use Coulomb’s law to find the magnitude of each force  Use the Law of Electrostatics to find the directions.  Vector add the forces to find Fnet This is like the planetary questions from gravitation in P20:  Steps: find the interaction between each charge  Use law of electrostatics to find direction and draw free body diagrams  Vector add the forces to find Fnet  Find Fnet on B -5.00 C -2.00 C +3.00 C A B C kqa qb kqc qb Fe  2 F e  ab r bc r2  8.99 x 10 Nm /C  (3.00 C)(5.00 C) F e  9 2 2  8.99 x 10 9 Nm 2 /C 2  (3.00 C)(2.00 C) Fe  bc   2  2.00 m  ab 2 3.00 m Fe  5.99333 x 109 N  attraction  F e  3.37125 x 1010 N [attraction] bc ab Fnet on B = 2.77 x 1010 N [left]  Find the Fnet on B Solution: Find the size of N each force using qA = -2.10 μC Coulomb’s Law. Draw qB = +1.50 μC a vector diagram of the forces using the law of qC = +1.80 μC rab = 2.00 cm electrostatics. Find the A B resultant FNet rbc = 3.00 cm C  Evaluate Fnet on B Fe   8.99 x 10 9 Nm 2 /C 2  2.10 x 10 -6 C 1.50 x 10 -6 C  2.00 x 10 m AB -2 2 B Fe  70.79... N attraction west AB Fe   8.99 x 10 9 Nm 2 /C 2 1.80 x 10-6 C 1.50 x 10-6 C   3.00x10 m cb -2 2 Fe  26.97 N repulsion north A cb F e netB  26.97 N 2  70.80 N 2  75.8 N  26.97 N    tan 1    20.9° nt of wt  70.80 N  The net Electric force experience on B is 75.8N 20.9º nt of wt  Analyzing amount of Charge  Stationary Fnet = 0 balanced forces  Thus Fg + Fe + FT = 0 We know the Fg = mg We can see the angle, separation and length of string. Equal masses of 2.00 g are attached to the ends of a 1.00 m long thread. When charged equally, they diverge 30.0º. Find the charge on each.  Use the triangle to calculate r  Calculate Fg and then use triangle to calculate Fe  Apply law to find q FT Fe Fg = mg  Use ½ of the triangle for a right triangle  Balanced Forces!! You absolutely need this principle and the vector diagram!! FT ɵ Fg Fe 1 Employ ½ of the triangle for a right triangle r = (sin15.0)(0.50 m) = 0.129...m, then r = 0.25888 m 2 m = 2.0 g = 2.0 x 10-3 kg F g  m g   2.0 x 10-3 kg  (9.81 m/s 2 ) = 0.01962 N F e  tan15.0 F g  tan15.0(0.01962 N) =0.0052573 N since q1 =q 2 kq1q2 Fe  r2 kq 2 Fe  2 r  0.0052573 N  0.25888 m  2 2 Fe r q  = 1.9792 x 10-7 C k  8.99 x 10 9 Nm 2 /C 2  q  0.20  C How is Fe transmitted between charged objects? Theories:  Action at a distance: Fe appears at 2nd object without travelling through intervening space: magic?  Electrical Ether: there are no vacuums and Fe is transmitted by stresses and strains in the electrical ether. Problem: ether has never been detected.  Modern version: particle exchange  Electrical Fields: what we’ll use. You don’t have to believe in electric fields, but the equations give the right answers.  Field theory: Faraday (1791-1867)  Electric field were proposed by Faraday in 1840ish  Faraday invented the idea of fields to explain how Fe is transmitted  Field: a region of influence surrounding an object  Einstein used field concept for gravity: mass warps the space around it, if a 2nd body enters the field, the field puts a force on the 2nd object. BMU-03  Review: Gravitational Field  Gravitational field surrounding a large mystery mass = Fg on a small test mass 𝐹Ԧ𝑔  𝑔Ԧ = 𝑚𝑡𝑒𝑠𝑡  Value of g represents how warped the gravitational space is. The more warped, the higher the acceleration  Units for g: m/s2 or N/kg  Electric Field Electrical field surrounding a large mystery charge = Fe on a small positive test charge Unit for Field intensity = N/C kq1q2 Fe  2 r Fe kqsource  2  E qtest r  Quantifying Electric field strength Fe E  This equation can be used in all cases qtest kqsource E  2 can only be used for r electric field surrounding a point charge Field Direction: same direction as 𝐹Ԧ𝑒 would be on an imaginary positive ‘test’ charge. Field direction is the same as the direction of 𝐹Ԧ𝑒 on a positive test charge. Unit for Field intensity = N/C or V/m Field lines DON’T cross each other!!  Electric Field Lines 1. Field is stronger where field lines are closer together. 2. Direction of electric field lines is the same direction as Fe would be on a positive test charge. 3. Field lines don’t cross 4. Field direction at a point is tangent to the curved line Direction of Electric Field around a Point Charge 17-07 17-08 Convention: think “how would a positive test charge move?”  Drawing Electric Field Lines 17-10 Convention: field direction is the same as the direction of Fe on a positive test charge. Field lines don’t cross.  Electric Fields point away from positives and toward negatives The number of field lines indicates the strength of the field at that point. Notice in the diagram above how the field lines are more dense close to each charge and become spread further apart away from the charges. Draw the electric field lines for 2 identical positive point charges which are a short distance apart. Direction of electric field is by definition the same direction as the Fe would be on an imaginary positive test charge A -2.0 μC charge experiences Fe of 6.0 x 10-2 N left. Find the electric field at that point  Direction: if a –ve charge experiences Fe left, a +ve charge would experience Fe to the right E = Fe ==  6.0 x 10 N  2 = 3.0 x 10 4 N/C q  2.0 x 10 C 6  What is the electric field 300 cm to the right of a -3.0 μC point charge? Field strength intensity is equal at a radius of 300 cm. This is a non-uniform field because the field gets weaker as r increases. kq emitter E = r2 E=  8.99 x 10 9 Nm 2 /C 2  3.0 x 10 6 C (3.00 m)2 3 E = 3.0 x 10 N/C left A 60 μC charge is placed into an electric field. If the mass of the charge is 0.25 mg and it experiences an acceleration of 1.25 x 104 m/s2, what is the electric field strength at that point?  Remember Newton’s 2nd law tells us Fnet = ma Assuming Fnet = Fe  The 6.0 x 10-5 C is the charge trapped in the field Calculate the force on the particle at that instant F e  F net  m a   2.5 x 10 kg  (1.25 x 10 m/s ) -7 4 2 F e = 3.125 x 10-3 N calculate the electric field strength at that point Fe3.125 x 10-3 N E   -6 = 52 N/C q 60 x 10 C  2.4 x 1020 excess electrons are loaded on to the surface of a sphere with a diameter of 4.0 cm. What is the electric field intensity at a distance of 16 cm from the surface of the sphere? (Hint: What assumption is made for Coulomb's law?) Charge is concentrated In a point charge R measured center to center +  (1.1 x 1013 N/C) What is the net field strength at the center of a 16.0 cm2 square? Red + 2.00 C Blue -3.00 C Key: Distance between the center and a corner = 2√2 cm The 2 opposing positive corners cancel therefore this is a linear problem Electric field is a vector! Vector addition applies. 2 2cm = 0.02828427 m r 2 =  0.00080 m 2  E red kq red = 2 =  8.99 x 10 Nm /C   2.00 C  9 2 2 13 = 2.2475 x 10 N/C toward blue r  0.00080 m  2 E blue kq blue = 2 =  8.99 x 10 Nm /C   3.00 C  9 2 2 13 = 3.37125 x 10 N/C toward blue r  0.00080 m  2 Since field from each are in the same direction, E net is the sum E = (2.2475 + 3.37125) x 1013 N/C = 5.6 x 1013 N/C towards the negative (blue  Special Rules for Charged Conductors 1. Field lines meet the surface of a charged conductor at 90° to the surface 2. Charge concentrates at sharp points on a conductor!! 3. There is no charge and no electric field inside a charged solid conductor. All the charge resides on the outside of the conductor.  Charged Irregular Shaped Conducting Object and Electric Field Lines 17-09 Charge accumulates at points on objects. The non-pointy side has a larger SA. A larger total charge on that side (when evenly spread out) causes charges to migrate toward the smaller SA Result: the strong 𝐸 near the point causes neutral molecules to polarize, attract, touch, transfer charge and repel. Charge ‘leaks off’ of points. If excess electrons evenly distributed, huge charge on the side with the larger SA. Charges on the large SA side repel and move closer together on the smaller area  Charged Solid Conducting Sphere and Electric Field Lines Net Field inside any closed conductor is ZERO  Charged Hollow Conducting Object and Electric Field Lines  Minds on p. 559 Pearson Famous demonstration: Faraday’s ice pail  Complete charge transfer http://dev.physicslab.org/img/5c2413e2-bc46-4ee8-9f6a-8296a8b3c796.gif Retrieved March 3, 2014 http://upload.wikimedia.org/wikipedia/commons/thumb/0/09/Faradays_ice_pail_experiment_-_electric_fields.png/330px-Faradays_ice_pail_experiment_- _electric_fields.png Retrieved March 2, 2014 Faraday’s Ice Pail Faraday Cage Charged Solid Flat Conductor Plate and Electric Field Lines  Homework:  Check out Howstuffworks.com Research the effects of electricity on the human body Compare and contrast (application, effects, magnitude of the power involved) 1. TASER, 2. Defibrillator, 3. TENS, 4. Pacemakers Charged Parallel Plates, Electric Field Lines The behavior of a charge in the uniform field provides: Fe E q Even distribution of Charge results in a UNIFORM FIELD, Uniform Force Uniform Acceleration (kinematics is useful) What would an acceration-time graph look like (diploma exam)  Applications for parallel plates Parallel plates give charges Ek to do work or other energy conversions Particle accelerators are used for:  CRTs (TV’s)  Electron microscope  Electrostatic filters  Diodes (LED etc)  Research  X-ray tubes  Fluorescent bulbs  Smoke precipitator A 5.25 mg positive dust particle with a +13.0 μC charge is projected into uniform field strength of 2.50 kN/C right. The particle has an initial velocity of 50.0 m/s at the positive plate (ignore gravitational effects )  Find the Fe  Find the acceleration  Find the final lateral speed after 2.00 ms  How long does it take to travel 1.00 m?  A positive charge will accelerate in the same direction as the field. The parallel plates have a uniform field intensity Unbalanced forces: 𝐹റ𝑒 = 𝐸 q = 3.25 x 10-2 N 𝐹റ𝑛𝑒𝑡 = 𝐹റ𝑔 +𝐹റ𝑒 But 𝐹റ𝑔 =mg ≈ 5 x 10-5 N 𝐹റ𝑛𝑒𝑡 𝐹റ𝑒 𝑎റ = = = 6.19 x 103 m/s2 along the field 𝑚 𝑚 𝑣𝑓 −𝑣𝑖 𝑎റ = 𝑣റ𝑓 = 𝑣റ𝑖 + 𝑎𝑡 റ = 62.4 m/s 𝑡 𝑣𝑓2 = 𝑣𝑖2 + 2𝑎𝑑 , vf = 121.98 m/s 𝑣𝑓 −𝑣𝑖 𝑡= = 1.16 x 10-2 s 𝑎  F e  E q =  2.50 x 103 N/C  13.0 μC  = 3.25 x 10-2 N A positive charge will accelerate in the same direction as the field. The 2nd Newton's parallel law: Fplates have net =Fe +F g a uniform field intensity but Fg  5 x 10-5 N and perpendicular then Fnet horizontal = Fe F net 3.25 x 10-2 N a   -6 = 6.19 x 10 3 m/s 2 [along the field] m 5.25 x 10 kg v f  vi kinematics : a  t v f  vi  at  50.0 m/s + (6.19 x 103 m/s 2 )  2.00 x 10-3 s  = 62.4 m/s use v 2f  vi2  2ad to find v f at 1.00 m v f = vi 2 +2ad  (50 m/s) 2 +2(6.19 x 103 m/s 2 ) 1.00 m  v f =121.98 m/s v f  vi 121.98 m/s - 50 m/s t  3 2 = 1.16 x 10 -2 s a (6.19 x 10 m/s )  An alpha particle is shot 2.00 x 105 m/s into the middle of parallel plates that have a field intensity 3.30 N/C. If the plates are 10.0 cm long, what is the deflection from the original path? Draw field lines, particle’s path  Is gravity significant? Draw the forces acting on the particle.  Projectile motion can be analyzed using kinematics and dynamics once Fnet and acceleration are calculated Negative vi positive Solution: vix is perpendicular to the electric field. The Fe acting on the particle will be vertical and therefore the viy = 0. This is a horizontal projectile question. Road map  Draw electric field lines  Draw free body diagram for the particle  Draw the path of the particle  Calculate Fe, Fg, Fnet, a  Sort info into vertical and horizontal components  Solve the projectile problem with the acceleration  Calculate dy  𝐸 3.30 N/C Given: m = 6.65 x 10-27 kg Fg = mg ≈ 10-26 N dn q = 3.20 x 10-19 C Fe ≈ 10-18 N up Fg