Physics 2102Physics 2102Lecture: 03 TUE 26 JANLecture: 03 TUE 26 JANElectric Fields IIElectric Fields IIMichael Faraday (1791-1867)Physics 2102Jonathan DowlingWhat Are We Going to Learn?What Are We Going to Learn?A Road MapA Road Map• Electric charge- Electric force on other electric charges- Electric field, and electric potential• Moving electric charges : current• Electronic circuit components: batteries, resistors,capacitors• Electric currents - Magnetic field- Magnetic force on moving charges• Time-varying magnetic field è Electric Field• More circuit components: inductors.• Electromagnetic waves - light waves• Geometrical Optics (light rays).• Physical optics (light waves)2q!12F1q+21F12rCoulombCoulomb’’s Laws Law2122112||||||rqqkF =2212001085.8with 41mNCk!"==#$#2291099.8CmN!k =For Charges in aVacuumOften, we write k as:E-Field is E-Force Divided by E-ChargeE-Field is E-Force Divided by E-ChargeDefinition ofElectric Field: ! E =! F q |! F 12|=k | q1| | q2|r122+q1–q2 ! F 12P1P2 |! E 12|=k | q2|r122–q2 ! E 12P1P2Units: F!=![N]! =![Newton]; E!=![N/C]! =![Newton/Coulomb]E-ForceonChargeE-FieldatPointForce on a Charge in Electric FieldForce on a Charge in Electric FieldDefinition ofElectric Field: ! E =! F qForce onCharge Due toElectric Field: ! F = q! EForce on a Charge in Electric FieldForce on a Charge in Electric FieldEEPositive ChargeForce in SameDirection as E-Field (Follows)Negative ChargeForce in OppositeDirection as E-Field (Opposes)+ + + + + + + + + + + + + + + + + + – – – – – – – – – – – – – – – – – – – –Electric Dipole in a Uniform FieldElectric Dipole in a Uniform Field• Net force on dipole = 0;center of mass stays whereit is.• Net TORQUE τ : INTO page.Dipole rotates to line up indirection of E.• | τ | = 2(qE)(d/2)(sin θ)= (qd)(E)sinθ= |p| E sinθ= |p x E|• The dipole tends to “align”itself with the field lines.• What happens if the field isNOT UNIFORM??Distance BetweenCharges = dElectric Charges and FieldsElectric Charges and FieldsFirst: Given Electric Charges, WeCalculate the Electric Field UsingE=kqr/r3.Example: the Electric FieldProduced By a Single Charge, or bya Dipole:ChargeProduces E-FieldE-Field ThenProduces Forceon AnotherChargeSecond: Given an Electric Field,We Calculate the Forces onOther Charges Using F=qEExamples: Forces on a SingleCharge When Immersed in theField of a Dipole, Torque on aDipole When Immersed in anUniform Electric Field.Continuous Charge DistributionContinuous Charge Distribution• Thus Far, We Have Only Dealt WithDiscrete, Point Charges.• Imagine Instead That a Charge q IsSmeared Out Over A:– LINE– AREA– VOLUME• How to Compute the Electric Field E?Calculus!!!qqqqCharge DensityCharge Density• Useful idea: charge density• Line of charge:charge per unit length = λ• Sheet of charge:charge per unit area = σ• Volume of charge:charge per unit volume = ρλ = q/Lσ = q/Aρ = q/VComputing Electric FieldComputing Electric Field of Continuous Charge Distribution of Continuous Charge Distribution• Approach: Divide the Continuous ChargeDistribution Into Infinitesimally SmallDifferential Elements• Treat Each Element As a POINT Charge& Compute Its Electric Field• Sum (Integrate) Over All Elements• Always Look for Symmetry to SimplifyCalculation!dqdq = λ dLdq = σ dSdq = ρ dVDifferential Form of CoulombDifferential Form of Coulomb’’s Laws Lawq2 ! E 12P1P2E-FieldatPointdq2 d! E 12P1DifferentialdE-Field atPoint d! E 12=k dq2r122 ! E 12=k q2r122Field on Bisector of Charged RodField on Bisector of Charged Rod• Uniform line of charge+q spread over length L• What is the direction ofthe electric field at apoint P on theperpendicular bisector?(a) Field is 0.(b) Along +y(c) Along +x• Choose symmetricallylocated elements oflength dq!= λdx• x components of E cancelqLaPoyxdx d! E dx d! E ! ELine of Charge: QuantitativeLine of Charge: Quantitative• Uniform line of charge,length L, total charge q• Compute explicitly themagnitude of E at pointP on perpendicularbisector• Showed earlier thatthe net field at P is inthe y direction — let’snow compute this!qLaPoyxLine Of Charge: Field on bisectorLine Of Charge: Field on bisector Distance hypotenuse: r = a2+ x2( )1/ 2dE =k(dq)r2Lq=!Charge per unit length:2/322)()(cosxaadxkdEdEy+==!"qLaPoxdE!dxrcos!=ar=a(a2+ x2)1/2AdjacentOverHypotenuse[C/m]Line Of Charge: Field on bisectorLine Of Charge: Field on bisector!"+=2/2/2/322)(LLyxadxakE#Point Charge Limit: L << a =2k!La 4a2+ L22/2/222LLaxaxak!"#$%&'+=( Ey=2k!La 4a2+ L2"2k!aLine Charge Limit: L >> a Ey=2k!La 4a2+ L2"k!La2=kqa2Integrate: Trig Substitution!Coulomb’sLaw! Nm2C1mCm! " # $ % & =Nm! " # $ % & Units Check!Binomial Approximation from Taylor Series:x<<1 2k!La 4a2+ L2=k!La21+L2a" # $ % & ' 2( ) * + , - .1/ 2/k!La21.12L2a" # $ % & ' 2( ) * + , - /k!La2 ; (L << a)2k!La 4a2+ L2=2k!LaL1+2aL" # $ % & ' 2( ) * + , - .1/ 2/2k!a1.122aL" # $ % & ' 2( ) * + , - /2k!a ; (L >> a) 1± x( )n! 1± nxExample Example —— Arc of Charge: Quantitative Arc of Charge: Quantitative• Figure shows a uniformlycharged rod of charge -Qbent into a circular arc ofradius R, centered at (0,0).• Compute the direction &magnitude of E at the origin.yx450xyθdQ = λRdθdθ!!coscos2RkdQdEdEx==!!==2/02/02coscos)(""##$##$dRkRRdkExRkEx!=RkEy!=RkE!2=λ = 2Q/(πR)E–QE = Ex2+ Ey2Example : Field on Axis of Charged DiskExample : Field on Axis of Charged Disk• A uniformly charged circular disk(with positive charge)• What is the direction of E at pointP on the axis?(a) Field is 0(b) Along +z(c) Somewhere in the x-y planePyxzExample : Arc of ChargeExample : Arc of Charge• Figure shows a uniformly chargedrod of charge –Q bent into acircular arc of radius R, centeredat (0,0).• What is the direction of theelectric field at the origin?(a) Field is 0.(b) Along +y(c) Along -yyx• Choose symmetric elements• x components cancelSummarySummary• The electric field produced by a system ofcharges at any point in space is the force perunit charge they produce at that point.• We can draw field lines to visualize the electricfield produced by electric charges.• Electric field of a point charge: E=kq/r2• Electric field of a dipole: E~kp/r3• An electric dipole in an electric field rotates toalign itself with the field.• Use CALCULUS to find E-field from a continuouscharge
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