1 1 W04D2 Exam One Review 2 Exam One Thursday Feb 27 7:30-9:30 pm Conflict Exam Friday Feb 28 8:00-10:00 am in 32-082 10-12 noon in 32-082 3 Exam Rooms Section L01 Walker Memorial Third Floor 50-340 Section L02 26-152 Section L03 32-123 Section L04 26-100 Section L05 34-101 Section L06 26-100 Section L07 32-123 Section L08 34-101 Section L09 Walker Memorial Third Floor 50-3402 4 Exam 1 Topics (1) Fields (visualizations) (2) Electric Field • Discrete Point Charge and Continuous Charge Distributions • Symmetric Distributions – Gaussʼs Law with superposition (3) Electric Potential Difference (4) Electric Dipole: Torque and Force (5) Energy and Force Applications 5 What You Should Study • Review Friday Problem Solving (& Solutions) • Review In Class Problems (& Solutions) • Review Concept Questions (& Solutions) • Review Problem Sets (& Solutions) • Review PowerPoint Presentations • Review Relevant Parts of Course Notes (& Included Examples) • Review Pre-class MITx pages • Do Sample Exams (online under Exam Prep) 6 Exam One: Review Problems Spring 2013 http://web.mit.edu/8.02t/www/materials/ExamPrep/exam1/Exam1_2013Spr.pdf3 7 8 94 10 11 Coulomb's Law Coulombʼs Law: Force on q2 due to interaction between q1 and q2 ke=14!"0= 8.9875 # 109 N m2/C2 !F12= keq1q2r122ˆr12 ˆr12:unit vector from q1 to q2 ˆr12=!r12r12 !!F12= keq1q2r123!r12 !r12:vector from q1 to q2 12 Superposition Principle The electric field due to a collection of N point charges is the vector sum of the individual electric fields due to each charge !E =!E1+!E2+ .. ... =!Eii=1N!5 13 Fields Grass Seeds Know how to read Field Lines Know how to draw • Field line density tells you field strength • Lines have tension (want to be straight) • Lines are repulsive (want to be far from other lines) • Lines begin and end on sources (charges) or infinity 14 Electric Dipole Two equal but opposite charges +q and –q, separated by a distance 2a !p ! q2aˆj = 2qaˆjpoints from negative to positive charge prq -q 2a Dipole moment two charges prDipole moment for neutral charge distribution, N point charges !p ! qii=1i= N"!ri15 Dipole in Uniform Field !E = Eˆi !p = 2qa(cos!ˆi + sin!ˆj) !Fnet=!F++!F!= q!E + (!q)!E = 0Total Net Force: Torque on Dipole: tends to align with the electric field pr !! =!r "!F = (2a)(qE)sin(!) =!p !!E != rF+sin(") = pE sin(!)6 16 Force on a Charge in an electric Field !F = q!EIf you put a charged particle, (charge q), in a field: !F = m!a ! q!E = m!a17 Continuous Sources: Charge Density Length L=LwL Area = A = wLRLdLdQλ=dAdQσ=dVdQρ= != Q / V (uniform) Volume = V =!R2L != Q / A (uniform) != Q / L (uniform)18 ( )?P =ErV Continuous Charge Distributions Q = !qii"Break distribution into parts: !!E = ke!qr2ˆrE field at P due to Superposition: !E = !!E"# d!E$ ! dqV""" ! d!E = kedqr2ˆr !q7 19 Calculating Electric Field for Continuous Distributions 1. Choose integration variables. define vector from origin to 2. Identify 3. Define vector from origin to charge element 4. Choose field point variables. Define vector from origin to field point 5. Calculate source to field point distance 6. Define limits of integral and integrate !!r !r d!q ="d!s#d!a$d!v%&''('' !r -!!r !E(!r) = ked!q (!r -!!r )!r -!!r3sou rce"20 rˆθθ2L−2L+xd′x′xddq′=λs22xsr′+=PjˆiˆLine of Charge ˆr =!rr=!"xˆi + sˆj("x2+ s2)1/2 !E = d!E!= kedqr2ˆrsource! !E(0,s) = ke!d"x (#"xˆi + sˆj)("x2+ s2)("x2+ s2)1/2"x =# L/ 2"x = L/2$Coordinates of point-source dq: Give integration variable, xʻ, a “primed” variable name. Coordinates of field point P: Distance form source to field point: (!x ,0) dq =!d"x (0,s) r = (!x2+ s2)1/221 Case II: E is uniform vector field directed at angle to planar surface S of area A !E= EAco s"Electric Flux !E=!E " d!A## d!A = dAˆnˆnθ8 22 Gaussʼs Law – The Equation Electric flux (the surface integral of E over closed surface S) is proportional to charge enclosed by the volume enclosed by S !E=!E " d!Aclosedsurface S"##=qenclosed$0 ΦE23 Gaussʼs Law: !E ! d!AS"""=qin#0Spherical Symmetry Cylindrical Symmetry Planar Symmetry Gaussian Pillbox 24 Applying Gaussʼs Law 1. Based on the source, identify regions in which to calculate electric field. 2. Choose Gaussian surface S: Symmetry 3. Calculate 4. Calculate qenc, charge enclosed by surface S 5. Apply Gaussʼs Law to calculate electric field: !E ! d!Aclosedsurface S"""=qenclosed#0 !E=!E " d!AS"##9 25 Electrical Work Work done by electrical force moving object 1 from A to B: W1=!F12! d!s12AB"Electrical force on object 1 due to interaction between charged objects 1 and 2: !F12= keq1q2r122ˆr12LINE INTEGRAL 26 Potential Energy Difference !U " UB#UA= #!F12$ d!sAB%= #W = keqsq11rB#1rA&'()*+ Suppose charged object 1 is fixed and located at the origin and charge object 2 moves from an initial position A a distance rA from the origin to a final position B, a distance rB from the origin. The potential energy difference due to the interaction is defined to be the negative of the work done object 2 in moving from A to B: 27 Configuration Energy What is the potential energy stored in a configuration of charged objects? Start with all the charged objects at infinity. Choose (1) Bring in the first charged object. (2) Bring in the second charged object (3) Bring in the third charged object (4) Configuration energy !U2= U12= keq1q2/ r12 U (!) " 0 !U3= U23+ U13= keq2q3/ r23+ keq1q3/ r13 !U1= 0 !U = U12+ U23+ U13= keq1q2/ r12+ keq2q3/ r23+ keq1q3/ r1310 28 Electric Potential Difference !V "!Uqt= #!FqtAB$% d!s = #!EAB$% d!sUnits: Joules/Coulomb = Volts Change in potential energy per test object in moving the test object (charge qt) from A to B: 29 Charged Particle moving Across an Electric Potential Difference !U = q!VTo move a charged particle, (charge q), in a field and the particle does not change its kinetic energy then: !U + !K = 0 " q!V + !K = 030 Potential Created by Pt Charge d!s = drˆr + r d!ˆ! !V = VB" VA= "!E # d!sAB$ !E = kQˆrr2 = ! kQˆrr2" d!sAB#= !kQdrr2AB# = kQ1rB!1rA"#$%&'Take V = 0 at r = ∞: VPoint Charge(r) =kQr11 31 Continuous Charge Distributions Break distribution into infinitesimal charged elements of charge dq. Electric Potential difference
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