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Physics 2102 Lecture 2 Physics 2102 Jonathan Dowling PPhhyyssicicss 22110022 LLeeccttuurree 22 Charles-Augustin de Coulomb EElleeccttrriicc FFiieellddss (1736-1806) January 17, 07 Version: 1/17/07 WWhhaatt aarree wwee ggooiinngg ttoo lleeaarrnn?? AA rrooaadd mmaapp • 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 Electric Field • More circuit components: inductors. • Electromagnetic waves light waves • Geometrical Optics (light rays). • Physical optics (light waves) CoulombCoulomb’’ss lawlaw +q1 F12 F21 !q2 r12 For charges in a k | q || q | VACUUM | F | 1 2 12 = 2 2 N m r k = 8.99 !109 12 C 2 Often, we write k as: 2 1 !12 C k = with #0 = 8.85"10 2 4$#0 N m EEleleccttrricic FFieieldldss • Electric field E at some point in space is defined as the force experienced by an imaginary point charge of +1 C, divided by Electric field of a point charge 1 C. • Note that E is a VECTOR. +1 C • Since E is the force per unit q charge, it is measured in units of E N/C. • We measure the electric field R using very small “test charges”, and dividing the measured force k | q | by the magnitude of the charge. | E |= R2 SSuuppeerrppoossititioionn • Question: How do we figure out the field due to several point charges? • Answer: consider one charge at a time, calculate the field (a vector!) produced by each charge, and then add all the vectors! (“superposition”) • Useful to look out for SYMMETRY to simplify calculations! Example Total electric field +q -2q • 4 charges are placed at the corners of a square as shown. • What is the direction of the electric field at the center of the square? -q +2q y (a) Field is ZERO! (b) Along +y (c) Along +x x Electric Field Lines • Field lines: useful way to Example: a negative point visualize electric field E charge — note spherical • Field lines start at a symmetry positive charge, end at negative charge • E at any point in space is tangential to field line • Field lines are closer where E is stronger EEleleccttrricic FFieieldld ooff aa DDipipoolele • Electric dipole: two point charges +q and –q separated by a distance d • Common arrangement in Nature: molecules, antennae, … • Note axial or cylindrical symmetry • Define “dipole moment” vector p: from –q to +q, with magnitude qd Cancer, Cisplatin and electric dipoles: http://chemcases.com/cisplat/cisplat01.htm EEleleccttrricic FFieieldld OONN aaxxisis ooff ddipipoolele -q +q P a x Superposition : E = E+ + E! kq kq E E + = 2 ' = ' 2 & a # & a # $ x ' ! $ x + ! % 2 " % 2 " & # $ 1 1 ! 2xa $ ! = kq E = kq 2 - 2 2 2 $ a a ! & 2 a # , ) , ) $ x ! $* x - ' * x + ' ! $ ' ! %+ 2 ( + 2 ( " % 4 " EEleleccttrricic FFieieldld OONN aaxxisis ooff ddipipoolele 2xa 2kpx p = qa “dipole moment” E = kq 2 = 2 & a2 # & a2 # -- VECTOR $ x2 ! $ x2 ! $ ' ! $ ' ! % 4 " % 4 " - + What if x>> a? (i.e. very far away) 2kpx 2kp r pr E ! 4 = 3 E ! x x r3 E~p/r3 is actually true for ANY point far from a dipole (not just on axis) EEleleccttrricic DDipipoolele inin aa UUnnififoorrmm FFieieldld • Net force on dipole = 0; Distance between charges = d center of mass stays where it is. • Net TORQUE τ: INTO page. Dipole rotates to line up in direction 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 is NOT UNIFORM?? EEleleccttrricic cchhaarrggeess aanndd ffieieldldss We work with two different kinds of problems, easily confused: • Given certain electric charges, we calculate the electric field produced by those charges (using E=kqr/r3 for each charge) Example: the electric field produced by a single charge, or by a dipole: • Given an electric field, we calculate the forces applied by this electric field on charges that come into the field, using F=qE Examples: forces on a single charge when immersed in the field of a dipole, torque on a dipole when immersed in an uniform electric field. CCoonnttiinnuuoouuss CChhaarrggee DDiissttrriibbuuttiioonn • Thus far, we have only dealt Q with discrete, point charges. • Imagine instead that a charge Q Q is smeared out over a: – LINE Q – AREA – VOLUME • How to compute the electric field E?? Q CChhaarrggee DDeennssiittyy λ = Q/L • Useful idea: charge density • Line of charge: charge per unit length = λ σ = Q/A • Sheet of charge: charge per unit area = σ • Volume of charge: charge per unit volume = ρ ρ = Q/V ComputingComputing electricelectric fieldfield ofof continuouscontinuous chargecharge distributiondistribution • Approach: divide the continuous charge distribution into infinitesimally small elements • Treat each element as a POINT charge & compute its electric field • Sum (integrate) over all elements • Always look for symmetry to simplify life! Example:Example: FieldField onon BisectorBisector ofof ChargedCharged RodRod • Uniform line of charge +Q spread over length L • What is the direction of the P electric field at a point P on the perpendicular bisector? y a (a) Field is 0. x (b) Along +y (c) Along +x q • Choose symmetrically dx o dx located elements of length dx L • x components of E cancel EExxaammppllee ----LLiinnee ooff CChhaarrggee:: QQuuaannttiittaattiivvee P • Uniform line of charge, length L, total charge Q y • Compute explicitly the a magnitude of E at point P x on perpendicular bisector • Showed earlier that the net Q field at P is in the y o direction -- let’s now L compute this! LineLine OfOf Charge:Charge: FieldField onon bisectorbisector 2 2 Distance d = a + x dE q Charge per unit length ! = P L k(dq) a dE = 2 d! d dx k(! dx)a dEy = dE cos" = 2 2 3/ 2 Q (a + x ) o x a L cos! = (a2 + x2 )1/ 2 LineLine OfOf Charge:Charge: FieldField onon bisectorbisector L / 2 L / 2 dx ' x $ Ey = k# a = k( a ! (a2 x2 )3/ 2 % 2 2 2 " "L / 2 + &a x + a #!L / 2 2k!L = a 4a2 + L2 What is E very far away from the line (L<<a)? What is E if the line is infinitely long (L >> a)? 2k!L 2k! Ey = = a L2 a EExxaammppllee ---- AArrcc ooff CChhaarrggee:: QQuuaannttiittaattiivvee y • Figure shows a uniformly charged rod of charge −Q bent into a circular arc of radius R, centered at (0,0). 450 • Compute the direction & x magnitude of E at the origin. kdQ y dQ = λRdθ dEx = dE cos! = cos! R2 dθ / 2 / 2 " k($Rd# )cos# k$ " Ex = = cos#d# θ ! R2 R ! 0 0 x k! k! k! E = E = E = 2 λ = 2Q/(πR) x R y R R EExxaammppllee :: FFiieelldd oonn AAxxiiss ooff CChhaarrggeedd DDiisskk • A uniformly charged circular disk (with positive charge) P • What is the direction of E at point P on the axis? (a) Field is 0 z (b) Along +z (c) Somewhere in the x-y plane y x EExxaammpplele :: AArrcc ooff CChhaarrggee y • Figure shows a uniformly charged rod of charge -Q bent into a circular arc of radius R, centered at (0,0). • What is the direction of the electric field at the origin? x (a) Field is 0. (b) Along +y • Choose symmetric elements (c) Along -y • x components cancel SSuummmmaarryy • The electric field produced by a system of charges at any point in space is the force per unit charge they produce at that point. • We can draw field lines to visualize the electric field 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 to align itself with the field. • Use CALCULUS to find E-field from a continuous charge distribution. .
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