Phys 4321 Final Exam December 14, 2009
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1 Phys 4321 Finl Exm December 14, 2009 You my NOT use the text book or notes to complete this exm. You nd my not receive ny id from nyone other tht the instructor. You will hve 3 hours to finish. DO YOUR OWN WORK. Express your nswers clerly nd concisely so tht pproprite credit cn be ssigned for ech problem. Full credit for ech problem is equl. Choose nd turn in only 5 of the 6 problems. I will only grde the first 5 problems on your pper. 1) Find the distnce, d, tht chrge, q, nd mss, M, could be suspended t rest below n infinite conducting plte by the grvittionl nd electric fields. Obtin the field by the method of imges. Plce chrge, q, distnce, d, bove the plne. The force on q is due to the chrge, q F = 1 4πǫ 0 q 2 (2d) 2 = 1 16πǫ 0 ( q d )2 The forces blnce when, Mg = 1 16πǫ 0 ( q d )2 This gives; d = q 2 16πǫ 0 Mg 2) A dielectric sphere of rdius,, is polrized with polriztion, P = (κ/r) ˆr. Find the electric field for r > The volume chrge density is given by P ρ = P = (1/r 2 ) r (r2 κ/r) = κ/r 2 The chrge is sphericlly symmetric. Thus clculte the chrge; Q = dτ ( κ/r 2 ) = 4π Q = (4πRκ) R 0 r 2 dr κ/r 2 1
2 The surfce chrge density is σ = κ/r. The totl chrge on the surfce is then; Q s = 4πκR 2 /R = 4πκR The totl chrge on, nd within, the sphere is Q Q s = 0 Then becuse of sphericl symmetry, Guss Lw shows tht there is no field outside the sphere. 3) A non-conducting, sphericl shell of negligible thickness hs rdius, R, nd surfce chrge distribution, σ = σ 0 cos(θ). A grounded, conducting sheet is plced so tht it contins dimeter the shell. Find electric the potentil in the hollow between the sheet nd the shell. z σ = σ o cos( ) θ x y Figure 1: The geometry for problem 3. Solve by seprtion of vribles. The interior solution hs the form; V = A l r l P l (cos(θ)) As we expect only terms involving cos(θ) we hve only terms for l = 1. Therefore; V = A 1 r cos(θ) The potentil vnishes for θ = π/2. The field t the surfce of the sphericl shell is; E = V = A 1 [cos(θ) ˆr sin(θ) ˆθ] The exterior solution is; V = B l r (l+1) P l (cos(θ)) We use only the l = 1 term. Then 2
3 V = (B 1 /r 2 ) cos(θ) E = B 1 /r 3 [2 cos(θ) ˆr + sin(θ) ˆθ] We know the chrge density on the surfce, so tht the difference of the perpendiculr components, E r out E r in, is relted to the surfce chrge density. E r out E r in = (σ/ǫ) cos(θ) 2B 1 /R 3 cos(θ) + A 1 cos(θ) = σ/ǫ 0 cos(θ) Also the tngentil components of E re continuous. B 1 /R 3 sin(θ) = A 1 sin(θ) Solving these we obtin; A 1 = (σ 0 /3)ǫ 0 V = (σ 0 /3ǫ 0 ) r cos(θ) Then E is the grdient of V 4) An infinite conducting chnnel hs width nd sides,, nd is held t zero potentil. A conducting cp covers the chnnel nd is held t potentil, V 0. ) Find stored energy per unit length b) Find the cpcitnce per unit length V = Vo Figure 2: The geometry for problem 4. 3
4 Solve for the potetil by seprtion of vribles in Crtesin coordintes in 2D spce. V = n sin(kx) sinh(ky) k = nπ/ Then evlute the coefficients when y =. V 0 = n 0 sin(nπx/) sinh(nπ/) dxv 0 sin(mπx/) = n A n 0 dx sin(nπx/) sin(mπx/) sinh(nπ) Orthogonlity of the sin function projects out the m th coefficient. Integrtion nd evlution gives; A n = 4V 0 nπ sinh(nπ) Substitution gives V V = ( 4V 0 ) sin(kx) sinh(ky) π n sinh(nπ) The grdient gives E E = V = 4V 0 π The energy density is W = ǫ 0 /2 E 2 W = ǫ 0 /2 dxdy E 2 W = 16V 2 0 ǫ nm 1 [cos(kx)sinh(ky) ˆx + sin(kx)coshky ŷ] sinh(nπ) dxdy ( 1 sinh 2 (nπ) ) [cos(nπx/) cos(mπx/) sinh(nπy/)sinh(mπy/) + sin(nπx/) sin(mπx/) cosh(nπy/)cosh(mπy/)] Integrte over x using orthogonllity W = m W = 4V 2 0 ǫ 0 4V ( 0 2 ǫ 0 sinh 2 (nπ) ) dy [sinh 2 (mπy/) + cosh 2 (mπy/)] C = 8ǫ 0 1 sinh 2 (nπ) 1 sinh 2 (nπ) = (1/2)V 2 C 4
5 5) A circulr loop of wire of rdius,, crries current, I nd hs moment of inerti, I. ) Find the mgnetic moment b) If this loop is plced in uniform mgnetic field, it experiences force tending to lign the moment with the field. In the limit of smll ngulr displcement, show by developing the eqution for the ngle s function of time, tht the dipole oscilltes with simple hrmonic motion bout the field direction if relesed from rest fter displcement. The mgnetic moment is m = Ire. The torque is ; N = m B Newton s eqution of motion is then; I d2 θ dt 2 = mb sin(θ) For smll θ sin(θ) θ then the eqution of motion is I d2 θ dt 2 = mb θ which is the eqution of SHM 6) A long wire of rdius, ρ =, crring current, I, is surrounded by n insultor, rdius, ρ = b, hving mgnetic permebility, µ. Find H, B, M for ρ >. 2 2b Figure 3: The geometry for problem 6. Use Ampere s lw to get H H d l = I 5
6 H = I/2πr Then B = µh M = µ µ 0 µ 0 I 2πr 6
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