Physics 102 Lecture 2

Physics 102 Lecture 2 - Electric Fields - why? Problem: -...

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Unformatted text preview: Electric Fields - why? Problem: - If want to assign a unique property in space, force depends on charge: — Thus force is not the right quantity — Show with different forces in same point - Need to define an Electric Field as this unique property. Electric Field — Definition, cont Calculate: . Electric field = force (on qn)l unit charge 1:341qu Electric Field — Definition Required Properties: - Charge -> electric field around it. — charged object is the source particle - Electric field assigns force to charged object in space. What about the test charge? - E is the field produced by the source, not the test charge - E is a propefly of the source, not the test charge - E exists regardless of the test charge - Test charge is only a detector of E Electric Field — Test Particle How to find the Electric Field: - Lets have a test positive particle, qu - The test particle is only a probe to: — (i) find and — (ii) to get the strength of the electric field - The test charge is assumed to be small enough not to disturb the charge distribution responsible for the field Summary, cont. ' Fe=ql§ — This is valid for a point charge only — One of zero size — For larger objects, the field may vary over the size of the object - If q is positive, I?" and T3 are in the same direction - If q is negative,13 and E‘ are in opposite directions Electric Field, Vector Form - Coulomb's Law, point charges: 1-5,:k, ‘13"; r - Then, the electric field will be E 2 a f: E=—“=k if qa r Superposition Example Components: Ex=kelzcose+ke%c059=2k,%c059 r r r _ q~ ‘1- _ Ey—ke7sm9—ke7sm9—0 geometry... a a c059=—= 2 2 r a +y 2a u. EXZke 2 33,2 Note:fory>>a (‘1 +y ) 2aq Ewk, 3 3’ Electric dipole decays faster than a point charge! 10 Superposition with Electric Fields - At any point (x,y,z) the total electric field due to a group of source charges equals the vector sum of electric fields at that point due to all the particles - Method to calculate E for a collection of point charges. Continuous Charge Distribution - If the distances between charges in a group of charges is much smaller than the distance between the group and a point of interest, the system of charges may be modeled as continuous 11 . E=E,+E, Superposition Example Find the electric field Eldue to ql, Find the electric field T32 due to C12: — Reminder: fields add as vectors — The direction of the individual fields is the direction of the force on a positive test charge BTW, this is an electric dipole... n it Continuous Charge Distribution — recipe for field Procedure: — Dlvlde the charge distribution into small elements, each of which contains Aq — calculate the electric field due to m of these elements at point P — calculate the totalfield by / summing the contributions of all [I the charge elements /‘ Continuous Charge Distribution — equations for field - For the individual charge elements _. A ‘ 411494?i rt - Because the charge distribution is continuous E: lim kez A—gikaef d—qf ri 2 Alb—)0 i r 13 Uniform Ring of Charge Continuous charge problem instead of a bunch of discrete charges. Steps: -Discretize the charge and use the eq. for the field -Use any available symmetry -Do integral instead of sum — still superposition principle 16 Charge Densities - Volume charge density distributed evenly throughout a volume - p = Q / V - Surface charge density — when a charge is distributed evenly over a surface area - a = Q IA - Linear charge density — when a charge is distributed along a line —x=Q/| —when a charge is Uniform Ring of Charge 1. Discretize: dE=ke 2. Use available symmetry: (1E, = k and geometry c059=ke c059=£ —3x 3. Integrate d x x E=j ke—gx=ke—3_I dq=ke—3Q r r r More geometry: r=\laz+x2 E:k Uniform Ring of Charge Ring of radius 'a' has total 'Q' spread out uniformly throughout the ring. How much is the electric field at 'P', a distance 'x' from the center? Uniform Ring of Charge x Eike 2 2 3/2Q (0 +x ) Useful limits. How much isEat: N _ nxzo 3)E~ke 3x 2)x>>a a 3) X“ a Harmonic force. e.g. on -q 1)x=o,E=o _ Q Q F——keq—3x a 2) ENke—z 3: Just like a point charge 18 Problem Solving Strategy - Conceptualize — Imagine the type of electric field that would be created by the charges or charge distribution - Categorize — Analyzing a group of individual charges or a continuous charge distribution? — Think about symmetry 19 Electric Field Lines - Field lines good to visualize the electric field - The electric field vector T3 is tangent to these lines - The number of lines per unit area through a surface perpendicular to the lines is proportional to the magnitude of the electric field in that region 22 Problem Solving Hints, cont - Analyze — Group of Individual charges: use the superposition principle - The resultant field is the vector sum of the individual fields — Continuous charge distributions: the vector sums for evaluating the total electric field at some point must be replaced with vector integrals - Divide the charge distribution into infinitesimal pieces, calculate the vector sum by integrating over the entire charge distribution 20 Electric Field Lines, General - density of lines through surface A > through surface B ' IEA|>|EB| \\ v/ - In this example the lines at ‘1'\<. different locations point in i _ I: r ‘I differentdirections 7 — Al“ \x 7 ’ _ — This indicates the field is non- fl - ' _ " uniform _ 1 ‘ j , \,\ , \\ \\ A\\ B 23 Problem Solving Hints, final - Analyze, cont — Symmetry: take advantage of any symmetry in the system - Finalize — Check to see if your field is consistent with the mental representation and reflects any symmetry — Imagine varying parameters to see if the result changes in a reasonable way 21 Electric Field Lines, Positive Point Charge - The field lines radiate outward in all directions — In three dimensions, the distribution is spherical — A positive test charge would be repelled away from the positive source charge (21) Electric Field Lines, _ _ _ _ Electric Field Lines — Negative Point Charge Elecmc He'd L'neS — DIPOIG Like Charges - The field lines radiate - Equal but opposite inward in all directions V charges — A positive test charge would y _ be attracted toward the r " ' The number 0f flaid negative source charge lines leaving the positive charge equals the number of lines terminating on the “v “ negative charge - equal and positive charges - The same number of lines leave each charge since they are equal in magnitude - At a great distance, the field is approximately equal to that of a single charge of 2q - Limits of small and big __ distance it.) M ‘ r N (b) V ’ 25 26 27 Electric Field Lines, Electric Field Lines — Unequal Charges Rules for Drawing Motion of Charged Particles - The positive charge is tyvice the a - Lines start on positive and end on negative - Charged particle in an electric field —> electrical magnitude of the negative ' y/ charge force charge ,/ g . Two lines leave the positive - in the case Of an EXCEISSIOF one type 0' CharQE. some - Force will cause the particle to accelerate charge for each line that x.,\\ lines wrll begin or end Infinitely far away according to Newton's Second Law tehrminate on the negative - The number of lines is proportional to the , 1?" =qE=ma c arge . . magnitude of the charge _ e _ _ - At a great distance, the field - If E Is uniform, then a Is constant would be approximatelythe - Field lines cannot intersect same as that due to asingle charge of +q 28 29 3|) Uniform field example Electron in a Motion of Particles. cont Uniform Field, Example Initially at rest particle: , , - Posltlve char e —> acceleration is along the - The electron Is prolected direction of thg field 1'C°"Sta"t f°'°e . horizontally into a I h I _ _ I h 2.Accelerates to the right uniform electric field ( ' 349% gtfirgf‘? I: acce eratlon '5 mt e 3.Can use usual equations of - The electron undergoes a "wee '0'" of e_ 'e t t th k_ f motion under constant force to —d°"f’r:war:d away?!“ , ’ ' ’ ’ ’ ’ ' ’ ’ ' “ - acce era Ion Is cons an , e Inema Ic solveforfinalveloci - ec argels neg Ive,5_o ‘.—7,’m' - . l L equations can be used I _ ty figfif'era‘m" '5 °PP°5"e ‘ .\ ' - Describe its motion... ‘ 7 i++r++++n+h\ i y Fy=may=qu=—efi ay=—eEy/me Fx=0 EX. char E between arallel IBIES . g p p 32 ax =0 :constant speed nwtwn 33 31 Cathode Ray Tube (CRT) - ex.: oscilloscopes, radar systems, televisions anode - The electrons are m “:33” ' ' ' ' “ ml / (lento. Hm rlrllet Hm gttaitziatizal°.2ie:y“"° 375 i p C I rr/ // A ‘\ why!) 4; [I / b Elixir" ~ Flumed mi st‘l‘i‘t‘n 34 ...
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This note was uploaded on 11/11/2011 for the course ENGR 231 taught by Professor Olehtretiak during the Fall '10 term at Drexel.

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