Homework Assignment 6 Solution Set


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1 Homework Assignment 6 Solution Set PHYCS 440 Mrch, 004 Prolem (Griffiths 4.6 One wy to find the energy is to find the E nd D fields everywhere nd then integrte the energy density for those fields. We know tht D depends only on the free chrge nd is therefore continuous cross the dielectric oundry with vcuum. Thus, the energy is W = D E = llspce Q = Q 8πɛ 0 ɛ(4πr dτ + ( ( + χ e ( + χ e = Q 8πɛ 0 + χ e +. Q ɛ 0 (4πr dτ Prolem (Griffiths 4. There is only one free chrge: q. So inside nd outside the dielectric we hve Inside the dielectric, then, we know nd So the ound chrges re E = D = P = ɛ 0 χ e E = σ = P ˆr r=r = q 4πr ˆr. q 4πɛ 0 ( + χ e r ˆr χ e q 4π( + χ e r ˆr. χ e q 4π( + χ e R ρ = P = 0 (except t r = 0. The totl ound chrge on the surfce is just q χe +χ e, nd it is compensted y the volume chrge tht is loclized immeditely round the implnted chrge.
2 Note lso tht the field outside the dielectric is the sme s if the dielectric were not present. The field is screened inside the dielectric ut is otherwise unffected once you re outside the dielectric gin. Prolem (Griffiths 4.  optionl The ehvior of E is known t the interfce from the oundry conditions tht rise due to the reltionships etween E, D nd the free nd ound chrges. We know tht the prllel component of E should e continous cross the oundry, ut the perpendiculr component hs discontinuity due to the surfce chrges there. However, we know tht D only depends on free chrges, so the perpendiculr component of D is continuous cross the interfce. Thus, tn θ tn θ = = E, E, E, E, D, ɛ D, ɛ = ɛ ɛ. In the region 0 < θ < π, tn θ is everywhere positive nd grows with incresing θ. Thus, electric field lines ehve opposite to the nlogous light wves in medi  they end towrd the norml when going from regions of lrge ɛ to smll ɛ. Therefore, convex lens would e diverging lens for electric field lines (i.e., they would defocus the field. Prolem 4 (Griffiths optionl This is oundry vlue prolem tht is very similr to severl others tht we hve done. This time, however, we dd nother oundry nd few more conditions sed on the properties of dielectrics. The generl solution is V i (r, θ = ( A li r li + B l i r li+ P li (cos θ l i where i denotes the region of interest (i =,, for inside the conductor, inside the dielectric, nd outside respectively with the following oundry conditions: V (r E 0 r cos θ ( V = 0 (for simplicity ( V (r = = V (r = = 0 ( V (r = = V (r = (4 ɛ V r r= = ɛ 0 V r r= (5
3 From ( we see tht ll A l vnish except for l =, for which A = E 0. Condition (4 then tells us tht nd Thus, B l l+ B l l+ A l l = 0 (for l 0 l = l. (B l B l l+ A l l = 0. (6 Now, condition (5 gives ( ɛ r (l A l l (l + B l l+ P l (cos θ = E 0 cos θ + (l + B l l+ P l (cos θ l l nd so, l + l+ (ɛ rb l B l ɛ r l l A l = 0 (for l 0. (7 Now, tke condition ( which gives nd comine it with results (6 & (7 ove to get A l = B l l+ (8 ( B l = B l + l+ l+ B l = B l ɛ r ( l + l+ + ll l+ (9 (0 B l = B l = A l = 0 (for l 0 ( So, ll tht s left is to find A nd B (we ll get B in the process. Go ck to conditions (4 nd (5 nd write them gin, this time with l = (sustituting for A vi (8. We hve E 0 + B = B E 0 + B = B ɛ r B = ( ( + E 0 ɛ r ( + + A = B = E 0 ɛ r ( + + ( ( (4 (5
4 THEREFORE... nd, so, E( r = V ( r V (r, θ = E 0 = ɛ r ( + + E 0 ɛ r ( + + (r r (( + r cos θˆr cos θ ( r sin θˆθ. Apprently the screening y the dielectric mkes it so tht the electric field in the dielectric hs some θ dependence s you move wy from r =. Thus, the ending of E 0 is weker thn if we hd just plced the conducting sphere there without ny dielectric. Prolem 5 (Griffiths 5. We cn just strt from the result of Exmple 5. y(t = C cos ωt + C sin ωt + E B t + C z(t = C cos ωt C sin ωt + C 4 nd pply the given initil conditions to solve for the unknown constnts. In ech cse we hve x(0 = y(0 = z(0 = 0 which tells us C = C C = C 4. dy dt t=0 = E B dz dt t=0 = 0 C = 0 C = 0 y(t = E B t z(t = 0. In this cse the mgnetic nd electric forces exctly cncel nd the trjectory is stright line in the ŷ direction. 4
5 dy dt t=0 = E B dz dt t=0 = 0 C = E ωb C = 0 y(t = E ωb sin ωt + E B t z(t = E cos ωt + E ωb ωb. Now the velocity in the ŷ direction is not enough to cuse the mgnetic force to cncel with the electric force, so there is some rolling like in Exmple 5.. The trjectory is sketched elow, with z s the verticle xis nd y s the horizontl xis, oth in units of E B with Q m such tht ω = πsec c dy dt t=0 = E B dz dt t=0 = E B C = 0 C = E ωb y(t = E ωb cos ωt + E B t + E ωb z(t = E sin ωt. ωb 5
6 This is nother cycloid, ut centered long the y xis. The sketch is elow in the sme units s the previous plot Prolem 6 (Griffiths 5.6 K(r = σ(r v(r = σωrˆθ J(r, θ = ρ v(r, θ = ρωr sin θ ˆφ Prolem 7 (Griffiths 5.8 The field due to finite stright segment of current is given in eq. 5.5 s B(s = µ 0I πs (sin θ sin θ where s is the perpendiculr distnce from the current to the point nd θ ndθ re the ngles to the ends of the current segment reltive to the perpendiculr s. The direction is determined y the right hnd rule. At the center of squre loop the contriution to the mgnetic field from ech of the four sides points in the sme direction (norml to the plne of the loop, so the totl field t the center of squre of side R is just B = µ ( ( 0I π ( sin sin π πr 4 4 µ0 I = πr. Generlizing to more sides just gives since sin θ sin( θ = sin θ. B = Nµ 0I 4πR sin π N 6
7 c As N we get µ 0 I N µ 0 I sin x lim sin θ = lim N πr π x 0 R x = µ 0I R. This grees with eq. 5.8, the field nywhere on the xis of circulr loop, when z = 0. Prolem 8 (Griffiths 5. From Ampere s lw we get quite esily B inside = 0 B outside = µ 0I πs ˆθ B inside = µ 0 πs I encl = µ 0 πs = Cµ 0s ˆθ, s 0 Cs πs ds ut how do we find C? We know tht the totl current is I, so This gives nd, of course, 0 Cs πs ds = I B inside = µ 0s C = π I. π I = µ 0Is π ˆθ B outside = µ 0I πs ˆθ. 7
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