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MIT 8 02T - Conductors and Insulators

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1 1 W05D1 Conductors and Insulators Capacitance & Capacitors Energy Stored in Capacitors W05D1 Reading Assignment Course Notes: Sections 3.3, 4.5, 5.1-5.4 2 Outline Conductors and Insulators Conductors as Shields Capacitance & Capacitors Energy Stored in Capacitors 3 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, wood2 4 Charge Distribution and Conductors The Charged Metal Slab Applet: Non-zero charge placed in metal slab http://web.mit.edu/viz/EM/visualizations/electrostatics/CapacitorsAndCondcutors/chargedmetalslab/chargedmetalslab.htm Charges move to surface (move as far apart as possible) Electric field perpendicular to surface, zero inside slab 5 Induced Charge Distribution in External Electric Field Charging by Induction, Exterior of a Neutral Metallic Box http://web.mit.edu/viz/EM/visualizations/electrostatics/ChargingByInduction/chargebyinductionBox/chargebyinductionBox.htm Induced charges move to surface Electric field perpendicular to surface, zero inside slab 6 Conductors in Equilibrium Conductor Placed in External Electric Field 1) E = 0 inside 2) E perpendicular to surface 3) Induced surface charge distribution3 7 Hollow Conductors: Applet Charge placed OUTSIDE induces charge separation ON OUTSIDE. Electric field is zero inside. http://web.mit.edu/viz/EM/visualizations/electrostatics/ChargingByInduction/shielding/shielding.htm 8 Electric Field on Surface of Conductor 1) E perpendicular to surface 2) Excess charge on surface Apply Gaussʼs Law EsurfaceA =!A"0# Esurface=!"09 Conductors are Equipotential Surfaces 1) Conductors are equipotential objects 2) E perpendicular to surface4 10 Group Problem: Metal Spheres Connected by a Wire Two conducting spheres 1 and 2 with radii r1 and r2 are connected by a thin wire. What is the ratio of the charges q1/q2 on the surfaces of the spheres? You may assume that the spheres are very far apart so that the charge distributions on the spheres are uniform. 11 Concept Question: Point Charge in Conductor A point charge +q is placed inside a hollow cavity of a conductor that carries a net charge +Q. What is the total charge on the outer surface of the conductor? 1. Q. 2. Q + q. 3. q. 4. Q - q. 5. Zero. 12 Hollow Conductors: Applet Charge placed INSIDE induces balancing charge ON INSIDE. Electric field outside is field of point charge. http://web.mit.edu/viz/EM/visualizations/electrostatics/ChargingByInduction/shielding/shielding.htm5 13 Capacitors and Capacitance Our first of 3 standard electronics devices (Capacitors, Resistors & Inductors) 14 Capacitors: Store Electric Charge Capacitor: Two isolated conductors Equal and opposite charges ±Q Potential difference between them. Units: Coulombs/Volt or Farads C =Q!VC is Always Positive !V15 Parallel Plate Capacitor; Applet Oppositely charged plates: Charges move to inner surfaces http://web.mit.edu/viz/EM/visualizations/electrostatics/CapacitorsAndCondcutors/capacitor/capacitor.htm Electric field perpendicular to surface, zero inside plates6 16 Calculating E (Gauss’s Law) !E ! d!AS"""=qin#000εεσAQE == E AGauss( )=!AGauss"0Note: We only “consider” a single sheet! Doesnʼt the other sheet matter? 17 Superposition Principle !E =!E++!E!= !"2#0ˆj !"2#0ˆj = !"#0ˆjBetween the plates: Above the plates: Below plates: !E =!E++!E!= +"2#0ˆj !"2#0ˆj =!0 !E =!E++!E!= !"2#0ˆj +"2#0ˆj =!018 Parallel Plate Capacitor C depends only on geometric factors A and d !V = "!E # d!Sbottomtop$= Ed =QA%0d C =Q!V="0Ad7 19 Group Problem: Spherical Shells A spherical conductor of radius a carries a charge +Q. A second thin conducting spherical shell of radius b carries a charge –Q. Calculate the capacitance. 20 Concept Question: Isolated Spherical Conductor What is the capacitance of an isolated spherical conductor of radius a? 1. Capacitance is not well defined. 2. Capacitance is . 3. Capacitance is infinite. 4. Capacitance is zero. 4!"0a21 Demonstration: Capacitance of Van der Graaf Generator C = 4!"0a !V =keQa E =keQa2! V = Eahttp://tsgphysics.mit.edu/front/?page=demo.php&letnum=D%207&show=08 22 Capacitance of Earth For an isolated spherical conductor of radius a: aC04πε=mF1085.8120−×=εm104.66×=amF7.0F1074=×=−CA Farad is REALLY BIG! We usually use pF (10-12) or nF (10-9) 23 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 -Q 24 Stored Energy in Charging Capacitor At some point top plate has +q, bottom has –q Potential difference is V = q / C Change in stored energy done lifting another dq is dU = dq V9 25 So 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!=1CQ2226 Energy Stored in Capacitor C =Q!VSince 2221212VCVQCQU Δ=Δ==Where is the energy stored??? 27 Energy Stored in Capacitor uE=!oE22Parallel-plate capacitor: C =!oAdand V = EdEnergy stored in the E field! U =12CV2=12!oAdEd( )2=!oE22" ( Ad ) = uE" (vo lume)Energy density [J/m3]10 28 Demonstration: Changing Distance Between Circular Capacitor Plates E4 http://tsgphysics.mit.edu/front/?page=demo.php&letnum=E%204&show=0 29 Concept Question: Changing Dimensions A parallel-plate capacitor is charged until the plates have equal and opposite charges ±Q, separated by a distance d, and then disconnected from the charging source (battery). The plates are pulled apart to a distance D > d. What happens to the magnitude of the potential difference V and charge Q? 1. V, Q increases. 2. V increases, Q is the same. 3. V increases, Q decreases. 4. V is the same, Q increases. 5. V is the same, Q is the same. 6. V is the same, Q decreases. 7. V decreases, Q increases. 8. V decreases, Q is the same. 9. V decreases, Q decreases. 30 Concept Question: Changing Dimensions A parallel-plate capacitor is charged until the plates have equal and opposite charges ±Q, separated by a distance d. While still connected to the charging source, the plates are pulled apart to a distance D > d. What happens to the magnitude of the potential difference V and charge Q? 1. V, Q increases.


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