che294f04hmwrk11soln

che294f04hmwrk11soln - CHE/MAE 294 F04 Homework 11 Solution...

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CHE/MAE 294 F04 Homework 11 Solution 29.Nov.04 > restart: Problem 1 a) Formulation Polarization (SI Units: C/m2 = C m/m3 = dipoles per unit volume) > eqnPa := proc(kappa,V,dp) (kappa-1)*eo*V/dp end proc: Givens and Solution > readmylib(pchemconstants): eo; .8854 10 -11 > P_1 := eqnPa(1,100,1); P_5 := eqnPa(5,100,1); P_10 := eqnPa(10,100,1); := P_1 0. := P_5 .3541600 10 -8 := P_10 .7968600 10 -8 polarizations are 0 C/m2, 3.54 nC/m2, and 7.97 nC/m2 b) Assume we have a solid with 1 mole of dipoles per m3. The charge on an electron or proton is q. The polarization P is related to the number of dipoles per volume (n/V) and their charge and separation (c and d) by the equation P = (c*d)(n/V). Formulation > eqnPb := proc(c,d,n,V) c*d*n/V end proc: Givens and Solution Assume No dipoles of charge q (proton + electron pair) in 1 m3 of material. > Pb_1 := P_1 = eqnPb(q,d1,No,1); Pb_5 := P_5 = eqnPb(q,d5,No,1); Pb_10 := P_10 = eqnPb(q,d10,No,1); := Pb_1 = 0. 96485.34199 d1 := Pb_5 = .3541600 10 -8 96485.34199 d5 := Pb_10 = .7968600 10 -8 96485.34199 d10

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-- > solve(Pb_1,d1); solve(Pb_5,d5); solve(Pb_10,d10); 0. .3670609366 10 -13 .8258871074 10 -13 separation distances are 0 (zero), 37 fm (femto meters), and 83 fm Problem 2 a) Formulation > eqnvd := proc(mu,V,d) mu*V/d end proc:
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