Hung problem # 3 April 10, 2011 () [4 pts.] The electric field points rdilly inwrd [1 pt.]. Since the chrge distribution is cylindriclly symmetric, we pick cylinder of rdius r for our Gussin surfce S. Then S E d = E (2πrL) = Q enc ε 0 E(r) = = Q ε 0 Q ( ˆr). (0.1) 2πε 0 rl (b) [4 pts.] By symmetry, it is esiest to integrte long the line d l = drˆr. The potentil difference V 0 between nd b is V 0 = b V 0 = E d l = + Q 2πε 0 L Q ( b 2πε 0 L ln b dr r ). (0.2) (c) [4 pts.] Now we integrte only to r insted of b. Note tht to void confusion I hve relbeled the integrtion vribles by dummy vribles r. The nswer depends on where you set the zero of potentil, which wsn t specified, so your nswer is correct if it differs from mine by only constnt. Here I set the zero of potentil t the inner rdius : r V (r) = V (r) = E d l = + Q 2πε 0 L Q 2πε 0 L ln ( r r dr r ). (0.3) (d) [4 pts.] (e) [4 pts.] C = Q = 2πε 0L V 0 ln ( ). (0.4) b U = 1 2 QV = ( ) Q2 b 4πε 0 L ln. (0.5) 1

(f) [4 pts.] The energy density is u = 1ε 2 0E 2, which we need to integrte over ll spce where there is nonzero electric field, nmely the spce between the two shells. Tking dv = 2πrLdr we hve E 2 dv = 1 b 2 ε Q 2 0 (2πrLdr) (0.6) 2 U = 1 2 U = Q2 4πε 0 L ln 4π 2 ε 2 0L 2 r ( b This checks with our nswer from (e) (s it should)! ). (0.7) (g) [6 pts.] The upper hlf is hlf cylindricl cpcitor C 1 with dielectric connected in series with n ordinry hlf cylindricl cpcitor C 2. (The cpcitnce of hlf cylinder is hlf the cpcitnce of full cylindricl cpcitor.) The combintion is connected in prllel with hlf-cylindricl cpcitor C 3, which gin hs hlf the cpcitnce of full cylindricl cpcitor. We hve: C 1 = πkε 0L ln ( ), +t (0.8) C 2 = πε 0L ln ( ), b +t (0.9) C 3 = πε 0L ln ( ) b (0.10) The equivlent cpcitnce is ( 1 C eq = + 1 ) 1 + C 3 C 1 C 2 (0.11) C eq = πε 0 L 1 ln ( ) ( +t K + ln b ) + πε 0L ln ( ). (0.12) b +t Common mistkes: (i) I took point off for writing ln r insted of ln ( r ) or ln ( r b), since the quntity inside the logrithm should be dimensionless. (ii) A number of people got prt () wrong, finding the electric field to be constnt or go like 1/r 2, nd hd this mistke propgte through. I gve prtil credit for lter nswers if your pproch got you the correct nswer given your incorrect nswer from (). In the cse tht the student found the electric field to be constnt, I gve lower frctions of the points in lter prts since this mde the clcultions lot esier. In the cse tht there ws just fctor of two or some non-fundmentl mistke tht propgted, I tried to tke off points only once (unless you mde some other mistke). (iii) The most common mistke ws integrting u = 1 2 ε 0E 2 over dr (or some even more cretive things) insted of dv = 2πrLdr for prt (f). u is energy per volume, so be creful bout your units! 2

Problem 4 [15 pts] Rubrics & Common Errors -R/2 dx 0 R/2 x R x r x You got: [6 pts] if you relized tht you should brek the resistors into tiny pieces, nd wrote tht dr = ρ dl ρ l ρ dl A, but not dr = da, dr = da, or ny other wild vrition; [5 pts] if you hve shown cler understnding of the physics nd geometry. Those points were roughly broken s We cnnot use ny formul directly (such s R = ρl/a) to clculte the resistnce of this geometry directly since the cross sectionl re chnges s the current moves inside the conductor. But we cn brek the truncted sphere into tiny discs, ech with resistnce dr, ssume tht the current flows uniformly through these discs, nd clculte the totl resistnce R integrting. So, for ech tiny disc, we hve: dr = ρ dl A where dl is the thickness of the disk, nd A is the cross-sectionl re tht the current sees s it goes from left to right. According to the geometry bove, we hve tht dl = dx, A = π r 2 = π (R 2 x 2 ). The integrtion should go from x = R/2 to x = R/2. ˆ R = dr = ˆ R/2 R/2 ρ dx π (R 2 x 2 ) = 2ρ ˆ R/2 dx π 0 (R 2 x 2 ) Using the given tble of integrls, we get: R = 2ρ π 1 ( ) X=R/2 R + x 2R ln = ρ [ln (3) ln (1)] R x x=0 πr R = ρ πr ln 3 [2 pts] for putting the right limits in the integrl (consistent with your choice of geometry); [3 pts] for correctly setting up the integrl in spce (dx), but not in volume, rdius, or ny other vrible; [2 pts] for giving the right expression for the re; [2 pts] for the integrtion/finl nswer. Some common errors: Hevily flwed nswers got [ 5 pts]. Writing n integrl in volume insted of length showed little understnding of the physics, so [ 6 pts]. Clculting the resistnce directly s R = ρ l A, putting some vlue of A ws lso hevily flwed yielding [4 pts]. Other points: Some students got π r 2 s the expression for the re, but they didn t define r, nd their limits of integrtion ws incomptible with r. I hve seen in some few cses the expression like π (R 2 (l R 2 )2 ) for the re. This expression is perfectly fine if you define l = 0 t the left prt of the resistor. Since dr = ρ dl A, you should integrte in dl, or dx, or whichever vrible tht goes long the slices of your resistor. Some students integrted over dθ. This is wrong: if we hve x = R cos θ, then dx = R sin θ dθ. I didn t took points of for simplifiction. Note tht ln 3 ln ( 1 3) = ln 3 + ln 3 = ln 9 = 2 ln 3. 1