Physics 3323, Fall 2016 Problem Set 7 due Oct 14, 2016

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1 Physics 333, Fll 16 Problem Set 7 due Oct 14, 16 Reding: Griffiths 4.1 through Electric dipole An electric dipole with p = p ẑ is locted t the origin nd is sitting in n otherwise uniform electric field E = E ẑ. Find the sphericl surfce, centered t the origin, through which no electric field lines pss, tht is for which E ˆn =. Wht is its rdius? The electric field of dipole is given by Griffiths in eqution We will use this form except in our cse, the dipole is p = p ẑ so tht we get E dip = p ) 4π r 3 cos θˆr + sin θˆθ. 1.1) For the uniform field, we just need to replce ẑ with ẑ = cos θˆr sin θˆθ. The totl field becomes E = E p ) 4π r 3 cos θˆr E + p ) 4π r 3 sin θˆθ. 1.) We wnt to find the sphericl surfce where E ˆn =. E ˆn = Ẽ ˆr = E p ) 4π r 3 cos θ = 1.3) r = p π E ) 1/3 1.4) Figure 1: Digrm of dipole nd ion for problem.. Attrctive ions Between ny ion nd neutrl tom, there is force tht rises in the following mnner. The electric field of the ion polrizes the tom, nd the field of tht induced dipole rects on the ion. ) Show tht this force is lwys ttrctive, nd tht it vries with the inverse fifth power of the distnce of seprtion r. b) Derive n expression for the ssocited potentil energy, with zero energy corresponding to infinite seprtion. c) For wht distnce r is this potentil energy of the sme mgnitude s kt t room temperture if the ion is singly chrged q = e) nd the tom is sodium tom? 1

2 ) Let us plce the polrized tom t the origin, nd rrnge our coordinte xes so tht the ion is on the z-xis distnce r wy. We wnt to know the force the polrized tom exerts on the ion, so first we need to know how polrized the tom is. This will depend on the electric field t the origin from the ion, so we strt by clculting the ion s electric field t the origin. E ion ) = 1 q 4π r ẑ.1) Now the induced dipole moment of the tom is given by p = α E. Now we cn clculte the polriztion of the tom. p tom = α E ion = 1 4π αq r ẑ.) We hve the dipole moment of the tom, nd our system is set up in such wy tht it points in the negtive) z direction. We cn use eqution 3.13 from Griffiths with p = αq/4π r nd θ = to write down the electric field from the dipole t the position of the ion. E dip rẑ) = p ) 4π r 3 cos θˆr + sin θˆθ = p 4π r 3 ẑ.3) E dip rẑ) = αq 4π ) r 5 ẑ.4) The force on the ion is then given by F = q E. F = αq 4π ) r 5 ẑ.5) We see tht the force on the ion is directed towrds the tom, i.e. it is ttrctive, nd tht it vries with the inverse fifth power of r. b) We cn get the solution for the potentil energy by integrting the force long pth strting t infinity. U = r F r dr = r αq 4π ) dr.6) r5 αq U = 4π ) r 4.7) 3. Get chrge out of polriztion Griffiths 4.1 ) σ b = P ˆn = P ˆr = krˆr ˆr r=r 3.1) r=r ρ b = P = 1 r r [r P r ] = 1 r r [kr3 ] = k r 3r 3.) σ b ρ b = kr = 3k 3.3) b)we hve sphericl symmetry, so we cn use Guss s Lw in integrl form. Since there is no free chrge, the totl chrge is just the bound chrge. Becuse of the spericl symmetry, E = Eˆr. Inside the Sphere: 4πr E in = 4 ρ 3 πr3 b E in = 4πr3 /3 3k) = kr 3.4) 4π r Outside the Sphere: 4πr E out = 4 ρ 3 πr3 b +4πR σ b = 4πR3 + 4πR3 = 3.5)

3 E in = kr E out = 4. Permnently polrized dielectric shell Griffiths ) For r > b: 4πr E = 4π k/) + 1 b E = 1 [ r k kb ) = k + kb r All together we hve the following solution: kr ) 4πr dr + 4πb k/b) 4.7) ] = 4.8) ) There re two surfces which for which we need to clculte the bound surfce chrge. We lso need to clculte the bound chrge density in between the two. Inner Surfce: σ b, σ b,b = k/ = k/b { k E = rˆr for < r < b otherwise 4.9) Outer Surfce: In between: σ b, = P ˆn = P ˆr = k σ b,b = P ˆr = k b 4.1) 4.) ρ b = P = 1 [ r r k ] = k r r r 4.3) We hve sphericl symmetry so E = Eˆr. We cn use Guss s lw in integrl form to clculte E. For r < : 4πr E = E = 4.4) For < r < b: 4πr E = 4π k/) + 1 E = 1 r [ k r kr ) kr ) 4πr dr 4.5) ] = k r 4.6) b) Since there is no free chrge but we still hve sphericl symmetry, we find from Guss s lw tht D = everywhere. D = E + P = E = 1 P 4.1) { k E = rˆr for < r < b otherwise 5. Hemisphericl cpcitor 4.11) A hemisphericl conducting shell rdius b) is filled with soft plstic, chrcterized by very lrge reltive dielectric constnt ε r nd very smll electricl conductivity σ. A needleshped conductor with hemisphericl tip rdius ) is pressed into the plstic, s shown in the figure, so tht it is concentric with the shell. When the circuit switch is closed, very smll 3

4 b cpcitnce of two concentric spheres filled with dielectric. V = b D = 4πr ˆr E = E r dr = 4πε r For the hemisphere: b dr r = 4πε r rˆr 5.1) 4πε r C = V = 4πε b r b 1 1 ) b 5.) 5.3) Figure : Cpcitor for problem 5. current I flows nd chrges + nd pper on the conductors. ) Wht is the cpcitnce of this rrngement? b) Find the mgnitude E of the electric field in the volume between the two conductors s function of the distnce r from the center of the conductors. c) FInd the surfce density of bound chrge on the inner r = ) nd the outer r = b surfces of the dielectric. d) Wht is the surfce density of bound chrge on the flt surfce of the dielectric? Explin. ) Since full sphericl cpcitor is just two hemisphericl cpcitors in prllel, hemisphericl cpcitor hs hlf the cpcitnce of sphericl cpcitor. We cn clculte the b C = πε r b 5.4) Becuse the dielectric hs lrge dielectric constnt, we cn ignore the edge effects due to the hemisphere ending nd breking sphericl symmetry. b) The electric field ws found in prt ) in Eq.??). c) E = 4πε r r 5.5) σ b = P ˆn = D Ẽ) ˆn 5.6) σ b, = P ˆr = ) 1 4π 1εr 5.7) σ b,b = P ˆr = ) 1 4πb 1εr 5.8) d) There will be no surfce bound chrge on the flt prt of the dielectric since E, D, nd P ll point rdilly. 4

5 6. Cylindricl cpcitor in dielectric oil Griffiths 4.8 Griffiths 4.64 on pge 195 gives generl expression for the force exerted on moveble dielectric. F = 1 V dc dx 6.1) We cn clculte the force per unit re of the oil in the cpcitor, nd equte it to the pressure in fluid t depth x = h to find h. P = ρgh 6.) First though, we need to find the cpcitnce of the cpcitor so we cn tke its derivtive. Suppose tht before the oil rises there is long) length L oil under the oil nd similrly long) length L ir bove the oil. Wht is the cpcitnce of the cylindricl cpcitor? We cn find the electric field inside the cpcitor s function of, the chrge on the inner cylinder, using Gus s lw nd then integrte it to get the potentil difference between the two cylinders. πrle = E ε = /L ŝ 6.3) πs V = b /L ŝ dsŝ = /L ln πs π b ) 6.4) Thus we find tht the cpcitnce of the prt of the cpcitor tht is in ir. C ir = ir V = π ln b/ L ir x) 6.5) Here ir = L ir x) /L. The prt of the cpcitor in the oil with suceptibility χ e hs different cpcitnce. C L = 1 + χ e) π ln b/ L oil + x) 6.6) Now we cn clculte dc dx. dc dx = π ln b/ χ e) = πχ e ln b/ F = 1 V dc dx = πχ e V ln b/ 6.7) 6.8) Now we equte the pressure to the force per unit re. A = πb ). 7. Current Loop P = ρgh = F A = π χ e V πb ) ln b/ h = χ e V ρgb ) ln b/ 6.9) 6.1) A circulr current loop with rdius lies in the xy plne, crries current I, nd is centered t the origin. Appliction of the Biot-Svrt lw in Griffiths Exmple 5.6 gives the mgnetic field of the current loop long the z-xis in the form B z z)ẑ. ) Evlute + B z dz. b) Now use Ampère s lw in the form B l for the closed pth tht extends long the z xis from z = to z = +, nd then returns long semicirculr rc t r = +. Wht re the ingredients tht go into the integrl rc B l in the limit tht r. Find the limiting vlue of the integrl over the rc. 5

6 ) The mgnetic field long the xis is given in Griffiths Exmple 5.6 Eqution 5.38). B z z) = µ I + z ) 3/ 7.1) B z dz = µ I + z dz 7.) ) 3/ Eq.??) cn be evluted with the chnge of vribles z = tn θ, dz = sec θdθ. B z dz = µ I π/ = µ I π/ π/ 3 sec θ tn dθ 7.3) θ) 3/ π/ cos θ dθ 7.4) B z dz = µ I 7.5) b) The current enclosed by this pth is I. By Ampère s lw, this integrl should be µ I. We cn show this by breking the integrl into two prts; one long the xis nd nother round the semicirculr rc t r =. B d l = lim r [ r B z dz + r rc ] B d l 7.6) B z dz + lim B d r rc l 7.7) Now note tht B goes s 1/r 3 t r since the current source hs dipole term. The integrl pth hs length proportionl to r. The integrl rc B l hs leding term O1/r ). In the limit r, this portion of the integrl vnishes. B d l = µ I 7.8) 8. Current sheet Griffiths 5.7 previously 5.6) Find the vector potentil bove nd below the x-y plne with uniform surfce current density K = kˆx. Exmple 5.8 gives the mgnetic field for the plne surfce current. B = { +µ /)Kŷ for z < µ /)Kŷ for z > 8.1) Becuse the current source is infinite, we cn t just integrte it to get the vector potentil s would be suggested by Griffiths Eqution However, we cn use the properties of the vector potentil to get t it. First, the vector potentil generlly points in the direction of the current so we would expect A = A xˆx. Also, A = B, so we cn use Eq.??) to derive differentil eqution for A x. Below the plne we get: A = z ŷ y ẑ = B 8.) z = µ K A x = µ Kz Above the plne we hve something similr: z = µ K y = 8.3) 8.4) y = 8.5) 6

7 A x = µ Kz The vector potentil is 8.6) A = µ K z ˆx 8.7) We cn check this solution by tking its curl nd seeing tht it gives the expression for the mgnetic field. This expression for A is not unique, since we cn dd to it the grdient of ny function nd its curl will not chnge. 7

Problems for HW X. C. Gwinn. November 30, 2009

Problems for HW X. C. Gwinn. November 30, 2009 Problems for HW X C. Gwinn November 30, 2009 These problems will not be grded. 1 HWX Problem 1 Suppose thn n object is composed of liner dielectric mteril, with constnt reltive permittivity ɛ r. The object