Gauss Law AP Physics C
lectric Flux A Let's start be defining an area n the surface f an bject. The magnitude is A and the directin is directed perpendicular t the area like a frce nrmal. Flux ( r FLOW) is a general term assciated with a FILD that is bund by a certain ARA. S LCTRIC FLUX is any ARA that has a LCTRIC FILD passing thrugh it. We generally define an ARA vectr as ne that is perpendicular t the surface f the material. Therefre, yu can see in the figure that the ARA vectr and the lectric Field vectr are PARALLL. This then prduces a DOT PRODUCT between the variables that then define flux.
lectric Flux The electric field lines lk like lines f a "fluid". S yu can imagine these lines are flwing (even thugh nthing is really flwing). The wrd FLUX rughly means FLOW. S based n this idea we can define the LCTRIC FLUX as the LCTRIC FILD thrugh a SURFAC ARA. Since the area vectr is defined as perpendicular t the surface and the electric field ges thrugh it, we define this euatin as a dt prduct, similar t the wrk functin. Φ d Φ A da A csθ A differential amunt f flux is the crss prduct between the electric field and a differential amunt f area. Since yu want the ttal flux, yu integrate t sum up all the small areas. Thus the TOTAL FLUX is fund by integrating ver the NTIR SURFAC. The circle n the integratin sign simply means the surface is CLOSD!!.
lectric Flux Visually we can try t understand that the flux is simply the # f electric field lines passing thrugh any given area. In the left figure, the flux is zer. In the right figure, the flux is. When lines pass utward thrugh a clsed surface, the FLUX is psitive When lines g int a clsed surface, the FLUX is negative
lectric Flux What is the electric flux f this cylinder? Φ Φ Φ Φ 0 Φ Φ 1 1 1 Φ A cs0 A A (1) A Φ 3 ( 1) 0, cs180 A cnstant, What des this tell us? 3 cs90 A 1 A This tells us that there are NO surces r sinks INSID the cylindrical bject.
Gauss Law Where des a fluid cme frm? A spring! The spring is the SOURC f the flw. Suppse yu lse the spring with a clsed surface such as a sphere. If yur water accumulates within the sphere, yu can see that the ttal flw ut f the sphere is eual t the rate at which the surce is prducing water. In the case f electric fields the surce f the field is the CHARG! S we can nw say that the SUM OF TH SOURCS WITHIN A CLOSD SURFAC IS UAL TO TH TOTAL FLUX THROUGH TH SURFAC. This has becme knwn as Gauss' Law
Gauss Law The electric flux (flw) is in direct prprtin t the charge that is lsed within sme type f surface, which we call Gaussian. The vacuum permittivity cnstant is the cnstant f prprtinality in this case as the flw can be interrupted shuld sme type f material cme between the flux and the surface area. Gauss Law then is derived mathematically using knwn expressins fr flux. da
Gauss Law Hw des it wrk? Cnsider a POSITIV POINT CHARG,. Step 1 Is there a surce f symmetry? Yes, it is spherical symmetry! Yu then draw a shape in such a way as t bey the symmetry and NCLOS the charge. In this case, we lse the charge within a sphere. This surface is called a GAUSSIAN SURFAC. Step What d yu knw abut the electric field at all pints n this surface? It is cnstant. da The is then brught ut f the integral.
Gauss Law Hw des it wrk? Step 3 Identify the area f the Gaussian surface? In this case, summing each and every da gives us the surface area f a sphere. (4π r ) Step 4 Identify the charge lsed? The charge lsed is! ( 4 r ) 4πr π This is the euatin fr a POINT CHARG!
Gauss & Michael Faraday Faraday was interested in hw charges mve when placed inside f a cnductr. He placed a charge inside, but as a result the charges mved t the utside surface. Then he chse his Gaussian surface t be just inside the bx. da 0 0( A) He verified all f this because he DID NOT get shcked while INSID the bx. This is called Faraday s cage.
Gauss Law and cylindrical symmetry Cnsider a line( r rd) f charge that is very lng (infinite) We can NCLOS it within a CYLINDR. Thus ur Gaussian surface is a cylinder. da λ πr (πrl) λl (πrl) RCALL : λl A cylinder Macr πrl λ L This is the same euatin we gt ding extended charge distributins.
Gauss Law fr insulating sheets and disks A charge is distributed with a unifrm charge density ver an infinite plane INSULATING thin sheet. Determine utside the sheet. Fr an insulating sheet the charge resides INSID the sheet. Thus there is an electric field n BOTH sides f the plane. A da σ A A, σ A A σa This is the same euatin we gt ding extended charge distributins.
Gauss Law fr cnducting sheets and disks A charge is distributed with a unifrm charge density ver an infinite thick cnducting sheet. Determine utside the sheet. 0 Fr a thick cnducting sheet, the charge exists n the surface nly A da σ A σ, A σa
In summary Whether yu use electric charge distributins r Gauss Law yu get the SAM electric field functins fr symmetrical situatins. d d 4π r 4π r Functin uatin da Pint, hp, r Sphere (Vlume) 4π r Disk r Sheet (ARA) insulating and thin σ Line, rd, r cylinder (LINAR) λ π r