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• Prices starting at $9 per page • On-time delivery • Writers are native English speakers • Free title and reference pages • Offers and discounts! • Free revisions • 100% plagiarism-free papers ## Our services Have you found yourself running up against deadlines on a repeated basis? Are 3 A.M. mornings becoming a normal part of your life? It’s time to take action! We’re prepared to help you create and refine your essay # Coaxial Cylindrical Capacitor A cylindrical capacitor is made up from two concentric cylinders ($R$and$2R$). Inside the 2R cylinder there is a dielectric from 2R to 1.5R of constant$k=3$. The charges are Q on the R cylinder and -Q on the 2R cylinder. a) Please find the electric field between r=0 and r=3R. b) Please find the capacitance. The condition for the electric field at interface between two mediums 1 and 2 is$E1n*\epsilon1 =E2n*\epsilon2$, for the normal component (the normal polarization vector is constant)$E1t = E2t =0$, for the tangential component (the tangential component of field is zero, otherwise the field will move existent virtual electric charges on the interface surface) Consider the entire capacitor empty. Consider a cylindrical gaussian surface of radius r and length (cylinder height) L. The Gauss law is$E*S =Q(inside)/\epsilon0$(1) In all cases$S = 2*\pi*r*L$a) For$0<r<RQ(inside) =0$From (1) we get$E(r) =0$b) For$R <r<2RQ(inside) = +Q$From (1) we get$E(r) = Q/(2\pi*r*L\epsilon0)$(2) The relation (2) in real case is true for$R<r<1.5R$Outside the dielectric when$r=1.5 RE(1.5*R) =Q/(3\pi*R*L*\epsilon0)$For$1.5R<r<2R$we have$E(r) = 3*E'(r)Q/(2*\pi*r*L*\epsilon0) = 3*E'(r)E'(r) = Q/(6\pi*r L\epsilon0)$value of the field inside the dielectric for$1.5R<r<2R$Inside the dielectric when$r=1.5*RE'(1.5*R) =Q/(9*\pi*R*L*\epsilon0)$c) For$r> 2RQ(inside) =0$From (1) we get$E(r) =0$Now for the capacitance. For a cylindrical capacitance of inner radius$a$and outer radius$b$the capacitance is$C =(2\pi\epsilon0*L)/ln(b/a)$Because of the dielectric there are two cylindrical capacitor in series. One with radius R and 1.5R (and pemittivity$\epsilon0$) and the second with radius$1.5R$and$2R$(and permittivity$3*\epsilon0$)$C1/L = (2\pi\epsilon0)/ln(1.5) =4.932*\pi*\epsilon0C2/L =(2\pi*3\epsilon0) / ln(2/1.5) =20.856*\pi*\epsilon0Ctot =C1*C2/(C1+C2)Ctot/L=3.989*\pi\epsilon0 = 111 pF/m\$

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