Recall: The Electric Field § A charge creates an electric field around itself and the other charge feels that field. + + Test charge q 3 Review - Point Charge The electric field created by a point charge, as a function of position r, The force exerted by an electric field on a point charge q located at position r … direction: tangent to the field line 42 Electric field points away from the positive charge E +q 3 Electric field points toward the negative charge E -q 4 Continuous Distribution of charges § Two methods: • Method #1: Divide the distribution into infinitesimal elements dE and integrate to get the full electric field • Method #2: If there is some special symmetry of the distribution, use Gauss’ Law to derive the field from 46 Calculating E-field from Coulomb’s law § From a point charge kdq § From continuos distribution of charges dE = rˆ r2 dq Integrate over volume E = k rˆ r2 Z 6 E-field from a uniformly charged rod 7 E-field from a uniformly charged rod 8 E-field from a uniformly charged rod § Simmetry: 9 Electric Field from a Ring of Charge Total charge Q is distributed homogeneously over a ring of radius R. Find E-field on the axis at distance z. 79 Take section dφ dφ dφ Its charge is dq = Q 2⇡ It produces E-field dq dφ Q E = k = k dq R2 + z2 2⇡ R2 + z2 From symmetry: only Ez is important dφ Q z Ez,dq = Edq cos ✓ = k 2⇡ R2 + z2 pR2 + z2 11 dφ Q z Ez,dq = Edq cos ✓ = k 2⇡ R2 + z2 pR2 + z2 Qz Integrate over dφ E = k 2 2 3/2 E (R + z ) 0.3 2 0.2 1/z / 0.1 z/R 0.5 1.0 1.5 2.0 2.5 3.0 12 Field on the axis from a charged disk Qx from previous: E = k x (a2 + x2)3/2 dQx E = k x (a2 + x2)3/2 dQ =2⇥ada a=R z E = k 2⇥a da = x (a2 + z2)3/2 Z0 1 2πk⇥ 1 − 1+(R/x)2 ! p 13 Field on the axis from a charged plate 1 Ex =2πk⇥ 1 − 1+(R/x)2 ! R p !1 Ex =2πk⇥ 14 Gauss’ Law § Two methods fro calculating E-field from continuous distribution of charges: • Method #1: Divide the distribution into infinitesimal elements dE and integrate to get the full electric field • Method #2: If there is some special symmetry of the distribution, use Gauss’ Law to derive the field from Gauss’ Law The flux of electric field through a closed surface is proportional to the charge First we need to enclosed by the surface. define the concept of flux 46 Water Flux § Let’s imagine that we put a ring with area A perpendicular to a stream of water flowing with velocity v § The product of area times velocity, Av, gives the volume of water passing through the ring per unit time • The units are m3/s § If we tilt the ring at an angle θ, then the projected area is Acosθ, and the volume of water per unit time flowing through the ring is Av cosθ. 47 Electric Flux (1) § We call the amount of water flowing through the ring the “flux of water” ! § We can make an analogy with electric field lines from a constant electric field and flowing water ! ! ! § We call the density of electric field lines through an area A the electric flux 48 Electric Flux (2) § Consider a constant electric field E passing through a given area A § The angle θ is the angle between the electric field vector and the area vector § The density of electric field vectors passing through a given area A is called the electric flux N.B.: Nothing is actually flowing! 49 Surfaces and Normal Vectors 50 Electric Flux for Closed Surface § Assume an electric field and a closed surface, rather than the open surface associated with our ring analogy ! § In this closed-surface case, the total electric flux through the surface is given by an integral over the closed surface ! ! § The differential area vectors always point out of the closed surface 51 Gauss’s Law § We can now formulate Gauss’s Law as § Flux through a closed area equals the enclosed charge § q is the total charge inside a closed surface • We call this surface a Gaussian surface • This surface could be our imaginary box • This surface could be any closed surface § Usually we chose this closed surface to have symmetries related to the problem we are trying to study 55 Gauss’s Law (5) § Gauss’s Law (named for German mathematician and scientist Johann Carl Friedrich Gauss, 1777-1855) states ! • (q = net charge enclosed by Gaussian surface S). § If we add the definition of the electric flux we get another expression for Gauss’ Law q ! E dA = ! · ✏ Z 0 § Gauss’s Law: the electric field flux through S is proportional to the net charge enclosed by S 56 Gauss’s Law and Coulomb’s Law Are Equivalent § Let’s derive Coulomb’s Law from Gauss’s Law § We start with a point charge q § We construct a spherical surface with radius r surrounding this charge • This is our “Gaussian surface” 57 Gauss’ Law and Coulomb’s Law (2) § The electric field from a point charge is radial, and thus is perpendicular to the Gaussian surface everywhere ! q ! E dA = ! · ✏0 ! Z § The electric field has the same magnitude anywhere on the surface, directed radially, along the local normal 58 Gauss’ Law and Coulomb’s Law (3) § Now we are left with a simple integral over a spherical surface § So for Gauss’ law related to a point charge we get ! ! § Which gives ! 59 End of lecture 26.