2 Calendar for the Weekj Today (Monday) Wednesday, Friday One or two problemsIntroduction to the concept of FLUXWednesday, FridayGauss’s Law & some problemsEXAM DATE WILL BE NEXT WEDNESDAY through Gauss
3 Protons are projected with an initial speed vi = 9 Protons are projected with an initial speed vi = m/s into a region where a uniform electric field E = (-720 j) N/C is present, as shown in Figure P The protons are to hit a target that lies at a horizontal distance of 1.27 mm from the point where the protons cross the plane and enter the electric field in Figure
4 Summary from last week VECTOR VECTOR (Note: I left off the unit vectors in the lastequation set, but be aware that they shouldbe there.)
5 Electric FieldWe will now introduce a convenient way to represent the overall electric field in a region of space.It is kinda sorta a map of the field strength.We will then introduce a new conceptFLUXWe will use this new concept to introduce Gauss’s Law
6 Look at the “Field Lines” of an infinite sheet of charge …
8 Ignore the Dashed Line … Remember last time .. the big plane? s/2e0s/2e0s/2e0s/2e0s/2e0s/2e0E= E=s/e E=0
9 NEW RULES (Bill Maher)Imagine a region of space where the ELECTRIC FIELD LINED HAVE BEEN DRAWN.The electric field at a point in this region is TANGENT to the Electric Field lines that have been drawn.If you construct a small rectangle normal to the field lines, the Electric Field is proportional to the number of field lines that cross the small area.The DENSITY of the lines.
13 Recall… We were given Coulomb’s Law We defined the electric field. Calculated the Electric Field given a distribution of charges using Coulomb’s Law.(Units: N / C)
14 A Question:Given the magnitude and direction of the Electric Field at a point, can we determine the charge distribution that created the field?Is it Unique?Question … given the Electric Field at a number of points, can we determine the charge distribution that caused it?How many points must we know??
15 Another QUESTION: Solid Surface Given the electric field at EVERY pointon a closed surface, can we determinethe charges that caused it??
16 Still another question Given a small area, how can you describe both the area itself and its orientation with a single stroke!
17 The “Area Vector” Consider a small area. It’s orientation can be described by a vector NORMAL to the surface.We usually define the unit normal vector n.If the area is FLAT, the area vector is given by An, where A is the area.A is usually a differential area of a small part of a general surface that is small enough to be considered flat.
18 The normal component of a vector qThe normal vector to a closed surface is DEFINED as positiveif it points OUT of the surface. Remember this definition!
19 ANOTHER DEFINITION: Element of Flux through a surface ENORMALDAEDF=|ENORMAL| | |DA|(a scalar)
20 “Element” of Flux of a vector E leaving a surface n is a unit OUTWARD pointing vector.
23 Visualizing Fluxn is the OUTWARDpointing unit normal.
24 Definition: A Gaussian Surface Any closed surface thatis near some distributionof charge
25 Remember Component of E perpendicular to surface. This is the flux passing throughthe surface andn is the OUTWARDpointing unit normalvector!nEqqA
26 Example Cube in a UNIFORM Electric Field Flux is EL2Flux is -EL2areaLNote signE is parallel to four of the surfaces of the cube so the flux is zero across thesebecause E is perpendicular to A and the dot product is zero.Total Flux leaving the cube is zero
33 Looking at A Cylinder from its END DrunkCircularRectangular
34 Infinite Sheet of Charge cylinderWe got this sameresult from thatugly integration!
35 Materials Conductors THE ELECTRIC FIELD INSIDE A CONDUCTOR IS ZERO! Electrons are free to move.In equilibrium, all charges are a rest.If they are at rest, they aren’t moving!If they aren’t moving, there is no net force on them.If there is no net force on them, the electric field must be zero.THE ELECTRIC FIELD INSIDE A CONDUCTOR IS ZERO!
36 More on ConductorsCharge cannot reside in the volume of a conductor because it would repel other charges in the volume which would move and constitute a current. This is not allowed.Charge can’t “fall out” of a conductor.
37 Isolated Conductor Electric Field is ZERO in the interior of a conductor.Gauss’ law on surface shownAlso says that the enclosedCharge must be ZERO.Again, all charge on aConductor must reside onThe SURFACE.
38 Charged Conductors s E=0 E Charge Must reside on the SURFACE - - - - - Very SMALL Gaussian Surface
39 Charged Isolated Conductor The ELECTRIC FIELD is normal to the surface outside of the conductor.The field is given by:Inside of the isolated conductor, the Electric field is ZERO.If the electric field had a component parallel to the surface, there would be a current flow!
40 Isolated (Charged) Conductor with a HOLE in it. Because E=0 everywhereinside the surface.So Q (total) =0 inside the holeIncluding the surface.
41 A Spherical Conducting Shell with A Charge Inside. A Thinker!
42 Insulators In an insulator all of the charge is bound. None of the charge can move.We can therefore have charge anywhere in the volume and it can’t “flow” anywhere so it stays there.You can therefore have a charge density inside an insulator.You can also have an ELECTRIC FIELD in an insulator as well.
43 Example – A Spatial Distribution of charge. Uniform charge density = r = charge per unit volumerEO(Vectors)A Solid SPHERE
46 E is the same in magnitude EVERYWHERE. The direction is Charged Metal Plates sEAEE is the same in magnitude EVERYWHERE. The direction isdifferent on each side.
47 Apply Gauss’ Law s s + + + + + + + + + + + + + + + + A A E Same result!
48 Negatively Charged ISOLATED Metal Plate -s s-E is in opposite direction butSame absolute value as before
49 Bring the two plates together As the plates come together, all charge on B is attractedTo the inside surface while the negative charge pushes theElectrons in A to the outside surface.This leaves each inner surface charged and the outer surfaceUncharged. The charge density is DOUBLED.
51 VERY POWERFULL IDEA Remember It!! Superposition The field obtained at a point is equal to the superposition of the fields caused by each of the charged objects creating the field INDEPENDENTLY.Remember It!!
52 Problem #1 Trick Question Consider a cube with each edge = 55cm. There is a 1.8 mC chargeIn the center of the cube. Calculate the total flux exiting the cube.NOTE: INDEPENDENT OF THE SHAPE OF THE SURFACE!Easy, yes??
53 Problem #2 (15 from text) +10 mC initial +3 mC added Note: the problem is poorly stated in the text.Consider an isolated conductor with an initial charge of 10 mC on theExterior. A charge of +3mC is then added to the center of a cavity.Inside the conductor.(a) What is the charge on the inside surface of the cavity?(b) What is the final charge on the exterior of the cavity?+10 mC initial+3 mC added
54 Another Problem from the book Charged Sheet+sam,q both given as is aGaussianSurface
55 -2am,q both given as is aaTqE+smgFree body diagram