Eperiment #4 Gauss s Law Prela Hints This la an prela will make etensive use f Ptentials an Gauss s Law, an using calculus t recast the electric fiel in terms f ptential The intent f this is t prvie sme insight an assistance t help yu slve the prlems fr this prela. Prela Questin : In Prela questins an 3, the electric fiels preicte y Gauss Law fr the infinite planar an infinite cylinrical change symmetries will e given t yu. Using these results, yu will recast the electric fiels an charge ensities as ptentials. Eplain why this recasting is necessary. Frm La 3, we pltte ut the Electric Fiel using the carn paper; it is very easy t measure the Electric Ptential using a DMM ut ifficult t irectly measure the Electric Fiel. We in turn use the plt f the Electric Ptential lines t effectively raw in the Electric Fiel. imilarly, if we can recast everything in terms f the Electric Ptential, it will make the eperiment easier t perfrm an cmpare t a theretical value. Prela Questin : Cnsier an infinite plane with a unifrm surface charge ensity f. Frm Gauss Law, the electric fiel f an infinite plane is: E nˆ First, I ha t re-raw the picture, ecause the ne ave was a little cnfusing t me in the sense f hw is that an infinite plane?. Bair
Ntice that the electric fiel cnveniently pints in the ŷ an ŷ irectin. Als, ntice that we cnveniently chse the Gaussian urface t have fur f its surfaces with nrmals perpenicular t the electric fiel an tw f its surfaces with nrmals in all the same irectin as the electric fiel (i.e., parallel). If we raw in the Gaussian urface (in lue), the charge plane (in lack) an a crinate ais frame, we can start t escrie the electric fiel eing generate. (LMAO an I was just cmplaining aut a pictures ) Using Gauss Law: Q E A IN First, let s cnsier the ight Han ie (H) f Gauss Law: (Ntice that since is a charge ensity it has units f [C/m ], s we nee t multiply it times the area we are cnsiering t get simply the charge enclse in that area.) Q IN A Net, let s cnsier the Left Han ie (LH) f Gauss Law: E A First, if we lk simply at the Gaussian urface, we can see that there are 6 effective areas that we nee t cnsier fr the LH f Gauss Law. Bair
This means that the LH is asically: E A E A E A 3 E A Let s start with Areas -5, since thse are easy : 3 4 E A 4 5 E A 5 6 E A 6 Fr Areas -5, ntice that the area segment nrmal fr each f these areas is pinting perpenicular t the irectin f the electric fiel. This means that the angle etween theses area segment nrmals an the electric fiel will always e 9 egrees. [PAY ATTENTION TO DIECTION: Base upn ur crinate frame ave, n ˆ Area is pinting in the ˆ irectin, n ˆ Area 3 is pinting in the ẑ irectin, n ˆ Area 4 is pinting in the ˆ irectin, an n ˆ Area 5 is pinting in the ẑ irectin.] This is cnvenient, ecause then the t pruct will always e zer fr all these surfaces. If we cnsier an area segment f surface, we see that the infinitesimal area is given as: A z* y ; y : y h; z : z w Likewise, fllwing in suit, fr surfaces 3-5, we see that their infinitesimal areas are given as: A * y ; : l; y : y h 3 A4 A 5 z* y ; y : y h; z : z w * y ; : l; y : y h This means we can nw evaluate the integrals fr surface () thrugh surface 5 (5): E 3 4 h w A ˆ ˆ E y z y cs E cs9 y z E A E h w l w l w 3 E y h w yˆ zˆ cs E cs9 s z A ˆ ˆ 4 E y z y cs E cs9 y z h w Bair 3
5 E A l w l w 5 E y Net, let s cntinue with Areas an 6: yˆ zˆ cs E cs9 s z Fr Areas an 6, ntice that the area segment nrmal fr each f these areas is pinting parallel (an in the same irectin) t the irectin f the electric fiel. This means that the angle etween theses area segment nrmals an the electric fiel will always e egrees. [PAY ATTENTION TO DIECTION: Base upn ur crinate frame ave, n ˆ Area is pinting in the ŷ irectin, an n ˆ Area 6 is pinting in the ŷ irectin.] This is cnvenient, ecause then the t pruct will always e ne fr all these surfaces. If we cnsier an area segment f surface, we see that the infinitesimal area is given as: A z* ; : l; z : z w Likewise, fllwing in suit, fr surface 6, we see that its infinitesimal area is als given as: A6 z* ; : l; y : z w This means the integrals are practically ientical, an we can nw evaluate the integrals fr surface () an surface 6 (6): 6 E E l w A E z yˆ yˆ cs E cs y z E l w l w l w A E z yˆ yˆ 6 cs E cs y z E l w This means that the LH f Gauss Law is simply: E A E A E A Nw, setting the LH equal t the H: 3 E A 3 4 E A E A E A E A6 4 l w 5 E A EA 5 6 E A 6 Bair 4
A EA Thus putting everything tgether an slving fr the Electric Fiel: E nˆ OK, this lks g let s cntinue with the rest f the prlem. The unifrmly charge infinite plane has equiptential surfaces that are parallel t it. ince we can pick anywhere we want t e zer ptential, we will call the equiptential at = the = equiptential, as shwn in the figure elw: a.) Using the relatinship etween ptential an electric fiel: E s fin the ptential ifference etween the pints = an =. First, let s plug in the electric fiel we fun ave an the infinitesimal isplacement (pinting in the same irectin as the electric fiel) [IMPOTANT: Nte that if we went with the uns f fr the integral, instea f (as we have) then the isplacement vectr wuln t have the same sign an wul e pinting in the ppsite irectin as the electric fiel! This wul result in a - fr the t pruct; ut still gives us the same answer. ]: Bair 5
Bair 6 n n n n n ˆ ˆ Net, the trick is t use the negative sign t flip the rati : n ( This means we can etermine the ptential at any aritrary istance frm the surface f the charge plane..) The ptential yu fun in part (a) has the frm: m where m an epen n. Use the fact that at =, = t fin m an in terms f instea f. In this la, yu will cmpare the theretical values f m an t the eperimentally measure values f m an. It s easy t see that the frm mentine here an the frm we fun match, where: m an m If we apply the unary cnitin at = : Nw, slve fr an plug this ack int ur generic () equatin:
This means, nw: m an We have nw successfully recast the electric fiel f an infinite charge plane int ptential (smething we can measure an cmpare in the la). Prela Questin 3: Cnsier an infinite cyliner f raius a with a unifrm charge per length L f. Frm Gauss s Law, the electric fiel f an infinite cyliner at a raial istance r frm its center is: k E rˆ r First, I ha t re-raw the picture, again, ecause the ne ave was a little cnfusing t me in the sense f hw is that an infinite cyliner?. WHO DAW THEE PICTUE?!? Bair 7
Ntice that the electric fiel irectin is given as raiating ff f the infinite cyliner. Als, ntice that we cnveniently chse the Gaussian urface t have tw f its surfaces with nrmals perpenicular t the raiating electric fiel an ne f its surfaces with nrmals in all the same irectin as the electric fiel (i.e., parallel). If we raw in the Gaussian urface (in lue), the charge cyliner (in lack) an a crinate ais frame, we can start t escrie the electric fiel eing generate. If we take an ege n view f this, it is very easy t see that the electric fiel sits in the Y-Z plane an raiates away frm the cyliner (this is given as the irectin rˆ ). Bair 8
Using Gauss Law: Q E A IN First, let s cnsier the ight Han ie (H) f Gauss Law: Q IN L Net, let s cnsier the Left Han ie (LH) f Gauss Law: E A First, if we lk simply at the Gaussian urface, we can see that there are 3 effective areas that we nee t cnsier fr the LH f Gauss Law. This means that the LH is asically: E A E A E A 3 E A 3 Let s start with Areas an 3, since thse are easy : Fr Areas an 3, ntice that the area segment nrmal fr each f these areas is pinting perpenicular t the irectin f the electric fiel. This means that the angle etween theses area segment nrmals an the electric fiel will always e 9 egrees (fr area 3 r 7 egrees fr area ). [PAY ATTENTION TO DIECTION: Base upn ur crinate frame ave, n ˆ Area is pinting in the ˆ irectin an n ˆ Area 3 is pinting in Bair 9
the ˆ irectin.] This is cnvenient, ecause then the t pruct will always e zer fr th surfaces. If we cnsier an area segment f surface (which is ientical t surface 3 just pinting in the ppsite irectin), we see that the infinitesimal area is given as: A,3 r* ; r : r ; : ; This means we can nw evaluate the integrals fr surface () an surface 3 (3): E 3 A ˆ ˆ E r r cs E cs7 r E A ˆ ˆ 3 E r r cs E cs9 r Net, let s cntinue with Area : Bair
Fr Area, ntice that fr the Gaussian urface we selecte, the area segment nrmal (n the surface arun the cyliner) is always pinting in the same irectin as the electric fiel. [AGAIN, PAY ATTENTION TO DIECTION: Base upn ur crinate frame ave, n ˆ Area is pinting in the rˆ irectin.] This means that the angle etween the area segment nrmal an the electric fiel will always e. This is cnvenient, ecause then the t pruct will always e ne fr the surface. If we cnsier an area segment f surface, we see that the infinitesimal area is given as: A l* r ; l : l L; : ; This means we can nw evaluate the integrals fr surface (): L E A E l r cs rˆ rˆ E r cs l L E A E r l E r L L r r E L Phew, nw we case see that the LH is simply: E A E A E A E A r E L r E L 3 3 Nw, setting the LH equal t the H: L r E L Bair
Thus putting everything tgether an slving fr the Electric Fiel: E r rˆ Using the relatinship that: 4k E 4 k k rˆ rˆ r r OK, this lks g let s cntinue with the rest f the prlem. The unifrmly charge infinite cyliner has equiptential surfaces that are als cyliners an cncentric with the charge cyliner. We will call the equiptential at r = the () = an at r = a the (a) = : a.) Using the relatinship etween ptential an electric fiel: E s fin the ptential ifference etween the pints r = an r =. Bair
First, let s plug in the electric fiel we fun ave an the infinitesimal isplacement (pinting in the same irectin as the electric fiel) [IMPOTANT: Nte that if we went with the uns f fr the integral, instea f (as we have) then the isplacement vectr wuln t have the same sign an wul e pinting in the ppsite irectin as the electric fiel! This wul result in a - fr the t pruct; ut still gives us the same answer. ]: k r r rˆ rˆ k r k r r r Net, the trick is t use the negative sign t flip the lg : r k ln ln k ln ln k ln k (ln This means we can etermine the ptential at any aritrary istance frm the surface f the charge r. k r k ln r.) The ptential yu fun in part a has the frm: = A ln (Br) where A epens n lama an B is a gemetric parameter. Use the fact that at r = a, (a) = t fin A an B in terms f, a an instea f lama. In this la, yu will cmpare the theretical values f A an B t the eperimentally measure values f A an B. It s easy t see that the frm mentine here an the frm we fun match, where: Aln Br k ln Bair 3
A k an B If we apply the unary cnitin at = a: a a k ln Nw, slve fr k an plug this ack int ur generic () equatin: k a ln a ln Aln Br ln This means, nw: A a ln an B We have nw successfully recast the electric fiel f an infinite charge cyliner int ptential (smething we can measure an cmpare in the la). ~~ TOLD YOU THI WOULD NOT BE EAY!! ~~ Bair 4