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Electric Flux •To find the flux you need to see how much is going through perpendicular to the surface E •Multiply by cos En E = EnA= EA cos •For a non-constant electric field, or a curvy surface, you have to integrate over the surface E E dA E cos dA •Usually you can pick your surface so that the integration doesn’t need to be done given a constant field. Quiz: Flux and field A cube with 1.40 m edges is oriented as shown in the figure Suppose the cube sits in a uniform electric field of 10i ? • What is the magnitude of the flux through the whole cube? • What is the magnitude of the flux through the top side? • How many sides have nonzero flux? A) 2q/eoD) 1 B) 4q/4eo C) 0 D) q/6eo Electric Flux •What is electric flux E through surface surrounding a charge q? R charge q ke q 4 R E 4 R 2 4 ke q R E 4 ke q 2 2 E 2 ke q 2 ke q 4 ke q Answer is always 4keq Gauss’s Law •Flux out of an enclosed region depends only on total charge inside E qin e0 charge q A positive charge q is set down outside a sphere. Qualitatively, what is the total electric flux out of the sphere as a consequence? A) Positive B) Negative C) Zero D) It is impossible to tell from the given information Gauss’s Law charge q’’ charge q’ charge q E E dA 4 k q 4 k q ' 4 k q q ' e e 4 ke q '' E E dA 4 ke qin e qin e0 Quiz: Drawing gaussian surfaces charge q charge 2q charge -q charge -2q How do we draw surfaces to contain the +2q charge and have flux?: Zero ? +3q/e0? -2q/e0 ? Quiz: Flux A) q/eo B) -q/eo C) 0 Figure 24-29. •What is the flux through the first surface? •What is the flux through the second surface? •What is the flux through the third surface? •What is the flux through the fourth surface? •What is the flux through the fifth surface? D) 2q/eo Quizzes: Flux II A cube with 1.40 m edges is oriented as shown in the figure Suppose there is a charge situated in the middle of the cube. • What is the magnitude of the flux through the whole cube? • What is the magnitude of the flux through any one side? A) q/eo B) q/4eo C) 0 D) q/6eo Gauss’ Law and Coulumb’s •Suppose we had measured the flux as: E 4 ke q •From Gauss’ law: E 4 R E 4 ke q 2 E R charge q Keq R2 So Gauss’ law implies Coulomb’s law •What if we lived in a Universe with a different number of physical dimensions? Applying Gauss’s Law •Can be used to determine total flux through a surface in simple cases •Must have a great deal of symmetry to use easily Charge in a long triangular channel What is flux out of one side? charge q E q e0 q S 3e 0 S 1 S 2 S 3 E1 E 2 3 S 0 Applying Gauss’s Law L R r •Infinite cylinder radius R charge density •What is the electric field inside and outside the cylinder? •Draw a cylinder with the desired radius inside the cylindrical charge •Electric Field will point directly out from the axis •No flux through end surfaces qin V AE 2 rLE e0 e0 r E 2e 0 r L e0 2 Applying Gauss’s Law L R r •Infinite cylinder radius R charge density •What is the electric field inside and outside the cylinder? •Draw a cylinder with the desired radius outside the cylindrical charge •Electric Field will point directly out from the center •No flux through endcaps qin V AE 2 rLE 2 R e0 e0 E 2e 0 r R L e0 2 Applying Gauss’s Law r E 3e 0 Sphere volume: V = 4a3/3 R r Sphere area: A = 4a2 •Sphere radius R charge density . What is E-field inside? •Draw a Gaussian surface inside the sphere of radius r What is the magnitude of the electric field inside the sphere at radius r? A) R3/3e0r2 B) r2/3e0R C) R/3e0 D) r/3e0 AE 4 r E 2 V e0 4 r 3e 0 3 qin e0 Conductors in Equilbirum •A conductor has charges that can move freely •In equilibrium the charges are not moving •Therefore, there are no electric fields in a conductor in equilibrium F qE ma =0 qin e 0 e 0 E dA =0 •The interior of a conductor never has any charge in it •Charge on a conductor is always on the surface Electric Fields near Conductors •No electric field inside the conductor •Electric field outside cannot be tangential – must be perpendicular •Add a gaussian pillbox that penetrates the surface Area A 0 0 AE Surface charge qin e0 A E nˆ e0 e0 •Electric field points directly out from (or in to) conductor Conductors shield charges No net charge •What is electric field outside the spherical conductor? Charge q •Draw a Gaussian surface •No electric field – no charge •Inner charge is hidden – except Charge -q Charge +q E •Charge +q on outside to compensate •Charge distributed uniformly qrˆ 4e 0 r 2