Physics 10 Jonthn Dowling Physics 10 Lecture Electric Fields Chrles-Augustin de Coulomb (1736-1806) Jnury 17, 07 Version: 1/17/07
Wht re we going to lern? A rod mp Electric chrge Electric force on other electric chrges Electric field, nd electric potentil Moving electric chrges : current Electronic circuit components: btteries, resistors, cpcitors Electric currents Mgnetic field Mgnetic force on moving chrges Time-vrying mgnetic field Electric Field More circuit components: inductors. Electromgnetic wves light wves Geometricl Optics (light rys). Physicl optics (light wves)
Coulomb s s lw + q 1 F 1 F1 q F = 1 k q 1 1 r q r 1 For chrges in VACUUM k = 8.99 10 9 N m C Often, we write k s: k = 1 with # 4$# 0 0 = 8.85" 10 1 C N m
Electric Fields Electric field E t some point in spce is defined s the force eperienced by n imginry point chrge of +1 C, divided by 1 C. Note tht E is VECTOR. Since E is the force per unit chrge, it is mesured in units of N/C. We mesure the electric field using very smll test chrges, nd dividing the mesured force by the mgnitude of the chrge. Electric field of point chrge +1 C q E R E = k q R
Superposition Question: How do we figure out the field due to severl point chrges? Answer: consider one chrge t time, clculte the field ( vector) produced by ech chrge, nd then dd ll the vectors ( superposition ) Useful to look out for SYMMETRY to simplify clcultions
Emple +q Totl electric field -q 4 chrges re plced t the corners of squre s shown. Wht is the direction of the electric field t the center of the squre? -q y +q () Field is ZERO (b) Along +y (c) Along +
Electric Field Lines Field lines: useful wy to visulize electric field E Field lines strt t positive chrge, end t negtive chrge E t ny point in spce is tngentil to field line Field lines re closer where E is stronger Emple: negtive point chrge note sphericl symmetry
Electric Field of Dipole Electric dipole: two point chrges +q nd q seprted by distnce d Common rrngement in Nture: molecules, ntenne, Note il or cylindricl symmetry Define dipole moment vector p: from q to +q, with mgnitude qd Cncer, Cispltin nd electric dipoles: http://chemcses.com/cisplt/cisplt01.htm
Electric Field ON is of dipole Electric Field ON is of dipole + + = E E E Superposition : " # $ % & ' = + kq E " # $ % & + = ' ' kq E " # $ $ $ $ % & ' ( ) * +, + - ' ( ) * +, - = 1 1 kq E 4 " # $ $ % & ' = kq P -q +q -q +q
Electric Field ON is of dipole E = kq & $ % ' 4 # " = & $ % kp ' 4 # " p = q dipole moment -- VECTOR - + Wht if >>? (i.e. very fr wy) kp E = 4 kp 3 r E E~p/r 3 is ctully true for ANY point fr from dipole (not just on is) r r p 3
Electric Dipole in Uniform Field Net force on dipole = 0; center of mss stys where it is. Net TORQUE τ: INTO pge. Dipole rottes to line up in direction of E. τ = (QE)(d/)(sin θ) = (Qd)(E)sinθ = p E sinθ = p E The dipole tends to lign itself with the field lines. Wht hppens if the field is NOT UNIFORM?? Distnce between chrges = d
Electric chrges nd fields We work with two different kinds of problems, esily confused: Given certin electric chrges, we clculte the electric field produced by those chrges (using E=kqr/r 3 for ech chrge) Emple: the electric field produced by single chrge, or by dipole: Given n electric field, we clculte the forces pplied by this electric field on chrges tht come into the field, using F=qE Emples: forces on single chrge when immersed in the field of dipole, torque on dipole when immersed in n uniform electric field.
Continuous Chrge Distribution Thus fr, we hve only delt with discrete, point chrges. Imgine insted tht chrge Q is smered out over : LINE AREA VOLUME How to compute the electric field E?? Q Q Q Q
Chrge Density Useful ide: chrge density Line of chrge: chrge per unit length = λ Sheet of chrge: chrge per unit re = σ Volume of chrge: chrge per unit volume = ρ λ = Q/L σ = Q/A ρ = Q/V
Computing electric field of continuous chrge distribution Approch: divide the continuous chrge distribution into infinitesimlly smll elements Tret ech element s POINT chrge & compute its electric field Sum (integrte) over ll elements Alwys look for symmetry to simplify life
Emple: Field on Bisector of Chrged Rod Uniform line of chrge +Q spred over length L Wht is the direction of the electric field t point P on the perpendiculr bisector? y P () Field is 0. (b) Along +y (c) Along + Choose symmetriclly locted elements of length d components of E cncel d o L d q
Emple --Line of Chrge: Quntittive Uniform line of chrge, length L, totl chrge Q Compute eplicitly the mgnitude of E t point P on perpendiculr bisector y P Showed erlier tht the net field t P is in the y direction -- let s now compute this o L Q
Line Of Chrge: Field on bisector d d P o L de Distnce de = d = + Chrge per unit length Q k( dq) d = q L de y = de cos" = ( + cos = ( + k( d) ) 1/ ) 3/
Line Of Chrge: Field on bisector Line Of Chrge: Field on bisector " + = / / 3/ ) ( L L y d k E # Wht is E very fr wy from the line (L<<)? Wht is E if the line is infinitely long (L >> )? 4 L L k + = / / L L k " # $ % & ' + = ( k L L k E y = =
Emple -- Arc of Chrge: Quntittive Figure shows uniformly chrged rod of chrge Q bent into circulr rc of rdius R, centered t (0,0). Compute the direction & mgnitude of E t the origin. E kdq de = de cos = cos R = " / " / k( $ Rd# )cos# k$ = cos# d# R R 0 k E = R k E y = R E 0 = y k R y 45 0 dq = λrdθ dθ θ λ = Q/(πR)
Emple : Field on Ais of Chrged Disk A uniformly chrged circulr disk (with positive chrge) Wht is the direction of E t point P on the is? P () Field is 0 (b) Along +z (c) Somewhere in the -y plne z y
Emple : Arc of Chrge Figure shows uniformly chrged rod of chrge -Q bent into circulr rc of rdius R, centered t (0,0). Wht is the direction of the electric field t the origin? () Field is 0. (b) Along +y Choose symmetric elements (c) Along -y components cncel y
Summry The electric field produced by system of chrges t ny point in spce is the force per unit chrge they produce t tht point. We cn drw field lines to visulize the electric field produced by electric chrges. Electric field of point chrge: E=kq/r Electric field of dipole: E~kp/r 3 An electric dipole in n electric field rottes to lign itself with the field. Use CALCULUS to find E-field from continuous chrge distribution.