problemas dielectricos 1, Ejercicios de Física

¡Descarga problemas dielectricos 1 y más Ejercicios en PDF de Física solo en Docsity!

Example 1 A thin parallel-plate capacitor contains two dielectrics of dielectric constant e, and ez as shown in Fig. 1 Neglecting edge electa, find (he capacitance, Let the charge of the capacitor be q. By symmetry, the field in the capacitor is homogeneous (except near the edges). Constructing a Gaussian Frio, 1 Pirst example of a capacitor with two dielectrics. The thickness of the to dl capacitor is exaggerated. surface $ in the shape of a box enclosing the positive plate, and observing that if the edge effects are neglected the only contribution to the integral $D - 48 comes [rom the portion of the Gaussian surface lying directly between the plates, we have from Gauss's law $o.0s=[Dds= na =o where A ¡is Lhe arca of the enclosed plate, Hencc, between the plates, D=2, A The electric field is then, by the displacement law q and in dielectrics 1 and 2, respectively. “he voltage between the plates is r=Je.1- f as E, di Dietectrio 2 > 4 q 4, tod d, sosa or » EA CA ) eANe, * 87) The capacitance is therefore EpA é=Z a VO dale, + bfep) * Example 3 A conducting sphere of radius a carrying a charge 9 is submerged halfivay into a nonconducting liquid of dielectric constant E Tre. 3 — (a) Charged conducting sphere floating in a nonconducting liquid. (b) Field lines of D. (c) Field lines of E. Find the electric field outside the sphere and the charge density on the surlace of the sphere. Constructing a concentric spherical Gaussian surface $ of radius y enclosing the sphere, and applying Gauss's law to this surface, we have Sm .0s — [Dana 48 + | Da, 048 =0, y Ys, ¿sa where $, and $, are the parts of the Caussian surface passing through the liquid and through the air, respectively. The geometry of the problem suggosts that the field is everywhere radial, so that D-4S — Dd8. It also suggests that Diyuja is constant at all points of S, and D,;, is constant at all points of Sa, so that 7) can be factored out from under the integral signs. We can therefore write O 2] ds =4, > . ds. or raid E Da 2ar? = q; where 21? is the area of S, and S¿. Now, by the displacement law Diana = E08Fynaa 20d Day — 89E too Since Ue field is radial, it is tangent to the boundary between the liquid and the air, and hence, by Lhe boundary condition + Exquia = Este The subscripts on £ are then not needed, and we can write Dia — £08F: Dase — Ep. Substituting these expressions into , we obtain syeE — eyE)2nr2 = eyle + 1)E2nr? = o w ol or rn E ate + Da?