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MIT 8 02T - Exam Two Review

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1 W07D2 Exam Two ReviewAnnouncements Exam 2 Thursday March 20 7:30 - 9:30 pm: L01 50-340 Walker Memorial L02 26-152 L03 32-123 L04 26-100 L05 34-101 L06 26-100 L07 32-123 L08 34-101 L09 50-340 Walker Memorial Conflict Friday March 21 8-10 am and 10-12 noon in 32-082 2Exam Two Topics • Determining Potential Difference from Electric Fields • Conductors • Capacitance • Stored Energy in Electric Fields • Dielectrics • Current, Resistance, and Ohm’s Law • Magnetic Field • Magnetic Force • Conceptual problems involving the Biot-Savart lawExam Two Practice Problems See W07D2 Daily Webpage for six practice problems http://web.mit.edu/8.02t/www/coursedocs/coursematerial/day.htm?SEC=L07;DAY=W07D25 W04D1 Electric Potential and Gauss’ Law Equipotential Lines Today’s Reading Assignment Course Notes: Sections 3.3-3.4, 4.4-4.6. 4.10.56 Using Gauss’s Law to find Electric Potential from Electric Field If the charge distribution has a lot of symmetry, we use Gauss’s Law to calculate the electric field and then calculate the electric potential V using BBAAVV d−=−⋅∫Es7 Group Problem: Coaxial Cylinders A very long thin uniformly charged cylindrical shell (length h and radius a) carrying a positive charge +Q is surrounded by a thin uniformly charged cylindrical shell (length h and radius a ) with negative charge -Q , as shown in the figure. You may ignore edge effects. Find V(b) – V(a).8 Worked Example: Spherical Shells These two spherical shells have equal but opposite charge. Find for the regions (i) b < r (ii) a < r < b (iii) 0 < r < a Choose V (∞) = 0 V (r) − V (∞)9 Electric Potential for Nested Shells E =Q4πε0r2ˆr, a < r < b0, elsewhere⎧⎨⎪⎩⎪⎪From Gauss’s Law V (r) − V (∞)= 0= − 0∞r∫dr = 0 VB− VA= −EAB∫⋅ dsUse Region 1: r > b r No field à No change in V!10 Electric Potential for Nested Shells Region 2: a < r < b V (r) − V (r = b)= 0  = − drQ4πε0r2br∫ =Q4πε0rr = br =14πε0Q1r−1b⎛⎝⎜⎞⎠⎟r Electric field is just a point charge. Electric potential is DIFFERENT – surroundings matter11 Electric Potential for Nested Shells Region 3: r < a V (r) − V (a)= kQ1a−1b⎛⎝⎜⎞⎠⎟= − dr 0ar∫= 0 V (r) = V (a) =Q4πε01a−1b⎛⎝⎜⎞⎠⎟r Again, potential is CONSTANT since E = 0, but the potential is NOT ZERO for r < a.12 Gradient (del) operator: ∇ ≡∂∂xˆi +∂∂yˆj +∂∂zˆkIf we do all coordinates: E = −∇VDeriving E from V E = −∂V∂xˆi +∂V∂yˆj +∂V∂zˆk⎛⎝⎜⎞⎠⎟ = −∂∂xˆi +∂∂yˆj +∂∂zˆk⎛⎝⎜⎞⎠⎟V13 Equipotentials14 Equipotential Curves: Two Dimensions All points on equipotential curve are at same potential. Each curve represented by V(x,y) = constant15 Properties of Equipotentials E field lines point from high to low potential E field lines perpendicular to equipotentials E field has no component along equipotential The electrostatic force does zero work to move a charged particle along equipotentialConcept Questions with Answers 8.02 W04D117 W04D1 Electric Potential and Gauss’ Law Equipotential Lines18 Concept Question: E from V The graph above shows a potential V as a function of x. The magnitude of the electric field for x > 0 is 1. larger than that for x < 0 2. smaller than that for x < 0 3. equal to that for x < 0 4. I don’t know19 Concept Question: E from V The above shows potential V(x). Which is true? 1. Ex > 0 is positive and Ex < 0 is positive 2. Ex > 0 is positive and Ex < 0 is negative 3. Ex > 0 is negative and Ex < 0 is negative 4. Ex > 0 is negative and Ex < 0 is positive20 Concept Question: E from V Consider the point-like charged objects arranged in the figure below. The electric potential difference between the point P and infinity and is V (P) = − kQ aFrom that can you derive E(P)? 1. Yes, its kQ/a2 (up) 2. Yes, its kQ/a2 (down) 3. Yes in theory, but I don’t know how to take a gradient 4. No, you can’t get E(P) from V(P)21 Conductors and Insulators Conductor: Charges are free to move Electrons weakly bound to atoms Example: metals Insulator: Charges are NOT free to move Electrons strongly bound to atoms Examples: plastic, paper, wood22 W05D1 Conductors and Insulators Capacitance & Capacitors Energy Stored in Capacitors W05D1 Reading Assignment Course Notes: Sections 3.3, 4.5, 5.1-5.423 Conductors in Equilibrium Conductor Placed in External Electric Field 1) E = 0 inside 2) E perpendicular to surface 3) Induced surface charge distribution24 Electric Field on Surface of Conductor 1) E perpendicular to surface 2) Excess charge on surface Apply Gauss’s Law EsurfaceA =σAε0⇒ Esurface=σε025 Conductors are Equipotential Surfaces 1) Conductors are equipotential objects 2) E perpendicular to surface26 Capacitors and Capacitance Our first of 3 standard electronics devices (Capacitors, Resistors & Inductors)27 Capacitors: Store Electric Charge Capacitor: Two isolated conductors Equal and opposite charges ±Q Potential difference between them. Units: Coulombs/Volt or Farads C =QΔVC is Always Positive ΔV28 Calculating E (Gauss’s Law) E ⋅ dAS∫∫=qinε000εεσAQE == E AGauss( )=σAGaussε0Note: We only “consider” a single sheet! Doesn’t the other sheet matter?29 Parallel Plate Capacitor C depends only on geometric factors A and d ΔV = −E ⋅ dSbottomtop∫= Ed =QAε0d C =QΔV=ε0Ad30 Energy To Charge Capacitor 1. Capacitor starts uncharged. 2. Carry +dq from bottom to top. Now top has charge q = +dq, bottom -dq 3. Repeat 4. Finish when top has charge q = +Q, bottom -Q31 Change in stored energy to move dq is: Total energy to charge to Q Stored Energy in Charging Capacitor dU = dqV = dqqC=1Cq dq U = dU∫=1Cq dq0Q∫=1CQ2232 Energy Stored in Capacitor C =QΔVSince 2221212VCVQCQU Δ=Δ==Where is the energy stored???33 Energy Stored in Capacitor uE=εoE22Parallel-plate capacitor: C =εoAdand V = EdEnergy stored in the E field! U =12CV2=12εoAdEd( )2=εoE22× ( Ad ) = uE× (volume)Energy density [J/m3]34 W05D1 Concept Questions Conductors and Insulators Capacitance & Capacitors Energy Stored in Capacitors35 Concept Question: Point Charge in Conductor A point charge +q is placed inside a hollow cavity of a


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MIT 8 02T - Exam Two Review

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