Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Gauss and All That Jazz ???? Bindell Del Barco The Geometry – Past Chapter?? Define surface charge density s=charge/unit-area (z2+r2)1/2 dq=sdA dA=2prdr dq=s dA = 2psrdr dq cos( ) k 2prsdr z dE z k 2 2 z r z2 r2 z2 r2 R E z 2pksz (z2+r2)1/2 0 z rdr 2 r 2 3/ 2 s z Ez 2 1 z 2 R2 0 When R , s Ez 2 0 1/ 2 s s 2 0 s 2 0 s 2 0 + + + + + + + + + + + + + + Infinite sheet Of charge s 2 0 s 2 0 s 2 0 s 2 0 s 2 0 s E0 s 2 0 s 2 0 s 2 0 s 2 0 s 2 0 s 2 0 + + + + + + + + + + + + + + s s s E 2 0 2 0 0 Infinite sheet Of charge s - Infinite sheet Of charge E0 Field Lines Electric Field know If we where the charges are, we can calculate the Electric Field. QUESTION: If we know the electric field can we determine the charges? Summary Consider a region of space where the ELECTRIC FIELD LINES 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. We won’t use this much Charge Positive or Negative? Charge Positive or Negative? We just showed that So far … The electric field exiting or entering a surface seems to be related to the charge inside. But … what does “exiting a closed surface mean”? How do we really talk about “the electric field exiting” a surface? How do we define such a concept? CAN we define such a concept? Conclusion from Demo An area can be represented by a VECTOR. The Vector is perpendicular to the AREA that it represents. The area itself must be FLAT If not, use a teeny tiny differential area that is small enough that we can consider it flat. The magnitude of this area is dA. The area VECTOR is the unit vector dotted into the area vector: dA ( dA)n where n is a unit vector perpendicular to the surface. Check this out! For this example, since everything is perpendicular to the spherical surface, consider the following calculation: q 1 q 1 q q 2 EdA k dA dA 4 p r 2 2 2 r 4p 0 r 4p 0 r 0 or E n dA qinside 0 But you didn’t derive it! This is E n dA qinside Inside what?? 0 FLUX What the flux? FLUX E ndA qinside 0 That’s MY LAW !! Another QUESTION: Solid Surface Given the electric field at EVERY point on a closed surface, can we determine the charges that caused it?? Types of G-surfaces we can actually use Sphere Charge distribution Must be Centro-symmetric Cylinder Charge must be axially Symmetric. PLANE OF CHARGE Line of Charge Point Charge GENERALLY OF INFINITE EXTENT The “normal component” of the ELECTRIC FIELD E n En E cos( )n En E cos( ) n E n En DEFINITION: FLUX E n En Flux E n A (E n)A or E dA E A cos( ) Types of Surfaces 1. Open Surfaces 2. Closed Surfaces We will be considering CLOSED surfaces En E n cos( ) E n The normal vector to a closed surface is DEFINED as positive if it points OUT of the surface. Remember this definition! “Element” of Flux of a vector E leaving a surface d E dA E NORMAL A also d E dA E ndA For a CLOSED surface: n is a unit OUTWARD pointing vector. This flux is LEAVING the closed surface. Flux is a scalar! How can we ascribe a DIRECTION to a scalar??? Flux leaving is q/0 Flux leaving is zero FLUX is like the “amount” of Electric Field (Light) passing through a surface. This isn’t really a correct statement but it should give you an idea and I will deny I ever said this! “Element” of Flux of a vector E leaving a surface d E dA E NORMAL A also d E dA E ndA For a CLOSED surface: n is a unit OUTWARD pointing vector. Visualizing Flux flux E ndA n is the OUTWARD pointing unit normal. Definition: A Gaussian Surface Any closed surface that is near some distribution of charge and is used for Gauss’s Law Calculations flux E ndA Remember E n E n cos( ) n E A Component of E perpendicular to surface. This is the flux passing through the surface and n is the OUTWARD pointing unit normal vector! Example Cube in a UNIFORM Electric Field Flux is EL2 E Flux is -EL2 L area Note sign E is parallel to four of the surfaces of the cube so the flux is zero across these because E is perpendicular to A and the dot product is zero. Total Flux leaving the cube is zero so there is NO charge in the cube! Simple Example (We already did this!) 1 r q E ndA dA 2 4p0 r Sphere 1 q q 2 4p0 r 1 q dA 4p0 r 2 A q q 2 4pr 2 4p0 r 0 1 Gauss’ Law Flux is total EXITING the Surface. n is the OUTWARD pointing unit normal. qenclosed E ndA 0 q is the total charge ENCLOSED by the Gaussian Surface. qenclosed dA E n 0 Simple Example UNIFORM FIELD LIKE BEFORE E A E E A EA EA q 0 0 Line of Charge Q L charge Q length L Let’s review a little y=L/2 dy Total charge = Q Total Length=L =Q/L dq=dy y O x-axis dq 1 dq 1 r dEx (2) cos( ) (2) 2 2 4p 0 (r y ) 4p 0 (r 2 y 2 ) (r 2 y 2 )1/2 2 r Q dEx dy 2 2 3/2 4p 0 (r y ) L Doing the math (Yuck!) L /2 2 r Q dEx dy 2 2 3/2 4p 0 (r y ) L 2r Q dy dEx 4p 0 L (r 2 y 2 )3/2 2r Q y Ex 4p 0 L r 2 r 2 y 2 2r Q Ex 4p 0 L L/2 r2 L /2 2r Q Ex 4p 0 L dy (r 2 y 2 )3/2 0 From tables of integrals … Ex Q 4p 0 0 2 L r2 4 1 2 L r r2 4 What happens when L gets really big compared to r? (Infinite line of charge) Ex Q 1 4p 0 L2 r r 4 Q L 1 k 4p 0 2 L r Ex Q 1 4p 0 L2 r 4 2Q 1 2 Q1 4p 0 rL 4p 0 L r 2 k E r Line of Charge – The Easy Way! GAUSS q E dA n 0 h E 2prh 0 2 2k E 2p0 r 4p0 r r From SYMMETRY E is Radial and Outward By the way .. What is a Cylindrical Surface?? PONDER Looking at A Cylinder from its END Circular Rectangular Drunk Infinite Sheet of Charge s h E cylinder sA EA EA 0 sA 2 EA 0 We got this same s E result from that 2 0 ugly integration! Gauss – Part 2 WELCOME BACK TO THE TWILIGHT ZONE CORRECTION Gauss was NOT wealthy, as I said last week. I lied. He was born into a poor family but was very bright and went into politics where he passed Gauss’s Law. It only works in Germany where Gauss was born. Gauss is now dead. LAST TIME: We defined FLUX Flux E dA EdA cos We “derived” Gauss’s Law Flux E dA q 0 Gauss’s LAW Flux E dA q 0 Only certain problems can be solved in THIS class and only for certain shapes. The electric field E must be either perpendicular to the surface or parallel to it. E needs to be uniform over the surface that you are using. SHAPES Small area NOPE! Only for small areas W Flux=-EA E A’ L’ L Flux=0 Flux=EA’cos Flux EA' cos EL'W cos E A' WL' L L' cos L L' cos LW cos EA cos PROBLEM dA 2prdr r R sin dr Rd d E 2pR 2 sin d Q 4p 0 R FROM THE TOP r 2 2 p R 2 sin d 0 Q Q Q sin d cos | (1 cos ) 2 0 0 2 0 2 0 0 Materials Conductors ◦ ◦ ◦ ◦ ◦ 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! More on Conductors Charge 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. Isolated Conductor Electric Field is ZERO in the interior of a conductor. Gauss’ law on surface shown Also says that the enclosed Charge must be ZERO. Again, all charge on a Conductor must reside on The SURFACE. Charge Must reside on the SURFACE of a conductor in Equilibrium Charged Conductors sA EA 0 - E=0 - - or E s Very SMALL Gaussian Surface (cylinder) s E 0 Charged Isolated Conductor The ELECTRIC FIELD is normal to the surface outside of the conductor. The field is given by: s E 0 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! Isolated (Charged) Conductor with a HOLE in it. E n dA 0 Q 0 Because E=0 everywhere inside the conductor. So Q (total) =0 inside the hole Including the surface. A Spherical Conducting Shell with A Charge Inside. An interesting example +s Infinite sheet of positive charge s E 2 0 s - E + 2 0 + + + + - Metal + s + E 2 0 + - s + E - + 2 0 + E s 2 0 Hey- Even an idiot knows the electric field in the metal is zero! 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. Example – A Spatial Distribution of charge. Uniform charge density = r charge per unit volume q E dA n 0 rV 4 3 1 E 4pr r pr 0 3 0 E rr (Vectors) E 3 0 2 O r A Solid INSULATING SPHERE Outside The Charge En dA R q 0 r 4 3 Q E 4pr pR 0 3 0 2 r E O or 1 Q E 4p0 r 2 Old Coulomb Law! Graph E R r Charged Metal Plate s E + + + + + + + + s + + + + + + + + A A E E is the same in magnitude EVERYWHERE. The direction is different on each side. Apply Gauss’ Law s s + + + + + + + + + + + + + + + + Top A E A s EA 0 A EA A 0 s E 0 Bottom s EA EA 2 EA 2 A 0 Same result! Bring the two plates together s1 s1 A e s1 s1 B e As the plates come together, all charge on B is attracted To the inside surface while the negative charge pushes the Electrons in A to the outside surface. This leaves each inner surface charged and the outer surface Uncharged. The charge density is DOUBLED. Result is ….. E=0 2s1 2s1 s s E A E=0 B e e sA 0 s 2s E 1 0 0 EA VERY POWERFULL IDEA 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. Problem #1 Trick Question Consider a cube with each edge = 55cm. There is a 1.8 C charge In the center of the cube. Calculate the total flux exiting the cube. 6 1.8 10 5 2 2 . 03 10 Nm /C 12 0 8.85 10 q NOTE: INDEPENDENT OF THE SHAPE OF THE SURFACE! Easy, yes?? Note: the problem is poorly stated in the text it was stolen from. Consider an isolated conductor with an initial charge of 10 C on the Exterior. 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 C initial +3 C added Another Problem Charged Sheet a m,q both given as is a s Gaussian Surface Gauss sA 2 EA 0 s E 2 0 -2 a m,q both given as is a a T qE s T cos(a ) mg mg qs T sin( a ) qE 2 0 Free body diagram -3 Divide qs Tan(a ) 2 0 mg and 2 0 mg tan(a ) s q The charge density r for this solid sphere is non uniform and is given by rAr2 . What is the Electric field as a function of distance from the center for r>R? R The TOTAL charge on the sphere is given by: rR R R R 5 R 2 2 4 Q rdV ( Ar )( 4pr dr ) ( 4 A)(pr dr ) 4 Ap 5 0 0 0 1 5 5 5 5 1 R 1 1 R 1 1 R AR E 4pAR AR AR 2 2 2 4p 0 r 5 0 r 5 0 r 5 5 0 r 2 Inside the slab of stuff : q rV r (2 x) A 2AE d r 0 0 r ( x) E 0 E(x) E(x) A x Outside the slab of stuff : q rAd 2 AE 0 E Uniformly charged 0 rd 2 0 0 STOLEN SLIDE! Oh, Oh! ENOUGH!