THE ELECTRIC FIELDQ1Q2Q3pEELECTRIC FIELD at point p is the forcethat would be exerted on a unit chargeplaced at point p. The Electric fieldexists at every point in space.rrFQE== rrEFQ== rE is a vector. It has bothMAGNITUDE and DIRECTIONThe convention for the direction of rEat a point is the direction of the force ona + charge placed at the point.Units of E:newtons/coulombN/CQprWhat is rE at point p due to thecharge Q?Put a test charge qt at point p. Theforce on qt due to Q is FkQqrt==2and EFqkQrt====2Thus the magnitude of the electric fieldat point p due to the charge Q is EkQr==2NOTE:Q produces E. The effect of E onanother charge ′′Q is a force rrFQE==′′NOTE: k ====××1491009ππεεExample:Q=+2 C r=3 m+2 C3 mEpWe find: ENC==××××==××910 2921099/Electric field at a point p due to severalpoint charges? Use thePrinciple of Superposition:Determine the electric field at the point pdue to each of the charges. Then, the netelectric field at p is the vector sum ofthese fields.Example:+2 C+2 C3 m3 m 3 mxy30°Each charge produces a field ENC==××××==××910 2921099/Add these two fieldsvectoriallyxy30°30°yEEComponents of E :Horizontal: E sin 30°= 2××××109×××× 12 = 1××××109 N/CVertical: E cos 30°= 2××××109×××× 32= 1.73××××109 N/CThe components of the total field are:Ex = 1××××109-1××××109 = 0Ey = 2××××1.73××××109 = 3.46××××109 N/CELECTRIC FIELDSproduced byCONTINUOUS CHARGE DISTRIBUTIONS dEdQ== kr2 rrEdE==∫∫Example:A charge Q is uniformly distributed on a thinring of radius a. What is the field on the axis?The charge per unit length on the ring is λλππ==Qa2dlrzθadE dE⊥⊥⊥⊥ dEx====dE cosθθθθ EEkrxa======∫∫∫∫dEldlcos cosθθθθππ202E ======++(())∫∫kQarkQrkQzzaa220222232ππθθθθππcos cosdlFor z>>a EkQz==2 (Field of point charge)Example: A charge Q is uniformlydistributed on a thin disk of radius R. What isthe field on the axis?daRzadE σσσσ====QππππR2For the ring of charge,dQ=2ππππ a σσσσ daand dE==++(())==++(())kzzazazadQ da2202232322σσεε EzazazzRR==++(())==−−++∫∫σσεεσσεε221022002232daIn the limit z<<R (points very close to thedisk), the field becomes E ==σσεε20and is independent of z!Example:What is the electric field at the origin due to acharge Q uniformly distributed on a quartercircle?xyθθθθRRdθdE dEx====dEcosθθθθ dEy====dEsinθθθθ λλππππ====QRQR122 dQ d d====λλππRQθθθθ2dEdQd====kRkQR222ππθθ EkQRkQRx==−−==−−∫∫222022ππθθππππsin/dθθ EkQRkQRy==−−==−−∫∫222022ππθθππππcos/dθθ EEEkQRxy==++==22222ππ tanφφ====EEyx1 φφ==°°°°45 225 ,ELECTRIC FIELD LINES(Lines of force)a) Start on + charges or at ∞.b) End on - charges or at ∞.c) rE at a point is a vector tangent to theelectric field line at the point. It'sdirection is the same as that of thefield line.d) At every point the density of fieldlines crossing a surface at right anglesto the direction of the field isproportional to the magnitude of thefield.e) The number of field lines that start or endon a point charge is proportional to thecharge.Example:Electric Field Lines for a Point Charge+++++++++−−−−++++ELECTRIC DIPOLE+−−−−dpEEF+F−−−−××××ττττφ+q−q rrpqd== ττφφφφφφφφ==++====qEdqEdqdE pE22sin sin sin sin rrrττ==××pE ττφφ== pE sinEnergy of a dipole in an electric fieldWork done by an external agent to rotate adipole from φφφφ1 to φφφφ2 is WpE pE======−−−−[[]]∫∫ ∫∫ττφφφφφφφφφφφφφφddφφφφ121221sin cos cosThe potential energy of the dipole isU=WWe choose the zero of the potential energy tobe when rrpE⊥⊥ (φφφφ1 =90˚). Then the potentialenergy at any other angle φφφφ2≡≡≡≡ φφφφ is UpE