EECE 380 Problems, Spring 2008

ele | For the E-fields shown at the interface of two dielectrics in the figure below, relate [PROBLEM 1 | the angles 6; and 6). [SOLUTION _| S OEUTION The boundary conditions for the tangential component of E: Ey, = Ey, (1) The boundary conditions for the normal component of E: (with n from medium / to 2) Dy, — Dy =P; (2) 1) Since we have two dielectrics with no free charges at the interface ps=0 Equation (1) gives E, cos, = E, cos, Equation (2) with 2, = 0 gives &,(-E, sin@, )= ¢,(—E, sind) By division, we get & tan, = &, tan ate TS | Write a set of boundary conditions for the normal and tangential components of the current density vector at the interface of two conducting media having conductivities o; and o>. SOLUTION wo, . From the continuity equation v-y-— at But p, = 0, and V-J=0 The integral form gives fJ-ds=0 s which in turn implies Jn = Jn Since J = of, and the tangential component E 1s continuous, Jy Jin Oo, oO Express the surface charge density p, at the mterface of Problem 2 mn terms of Ey ar Ey and the media constants. From Problem 1, we get Ds, -D,, = Ps 6B) ~ En =P, From Problem 2, we get Jn = Jin OnE, = OE, Hence, 0) O; P, -{« ~& = Ey) =| 8) —— 8 fe oO; 2 Determine the capacitance of the configuration shown in the figure below. +f 1 II Il J- The capacitance of a parallel-plate capacitor is + & € ao | given by + ~ A + x 7 C=eE aq 0 1 2 3 Capacitances in series obey Hence, filled with a dielectric, ¢= 10g. A voltage is applied between the conductors to establish an electric field E=(10°/r)a, (V/m) between the cylinders. Calculate the A pair of 200-mm-long concentric cylindrical conductors of radii 50 and 100mm is energy stored. SOLUTION | SOLUTION ta The stored energy is given by w,-4 [ae'av -£/ i i lo") rdvdddz 2 2 ny jun rabos r 10x10" W, =———x 0.221 x10" n2=38.5J 361x2 Determine the capacitance of a parallel-plate capacitor filled with two dielectrics, as shown below. The capacitance of a parallel-plate capacitor is C=et d >C, =6, “ and b- C, = 6, | cla Because the capacitances are in parallel, C=C,+C, = Flee, -¢,)+ 56] 1 ima | For the spherical conducting shells filled with two dielectrics shown in the figure below, determine the potential distribution in the region c <r < a. SOLUTION The spherically symmetric Laplace’s equation is = R = =-0 dR dR Integrating with respect to R once, we have WG dR R A second integration yields V= Gy, Cy R Thus, in regions 1 and 2, respectively, we have C. ha +0, forb<R<a Cc Vy=-3 Cy forb<R<a where the constants C), C2, C3 and Cz are to be evaluated from the boundary conditions. AtR=c, V2 =0 gives Cc. C,=4 (1) c AtR =a, V; = Voyields C V,=-—++0¢, (2) a At R = b, we have V; = V>, so that Cc, Cc. -1+0C,=-+¢, (3) b “ b Finally, at R = d, (at the interface of two dielectrics), Dy, =D, ln bE = &E p> dV, dV, &—} =6,— AR | »-5 AR | x £,C, = €,C, (4) Solving (1)-(2)-(3)-(4) for the constants and substituting in the equations of /; and V7 yields i= Ya-VYR +1¥. (b< R<a (e, / 2, (1/c-1/b)+(1/b-1/a) y, = ic-U/R y, <R<b) (1/c-1/b)+(e, /e, (1+ b-1/a)