{"id":11680,"date":"2025-08-26T13:52:30","date_gmt":"2025-08-26T18:52:30","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/analisisnumerico\/?p=11680"},"modified":"2025-12-13T07:36:16","modified_gmt":"2025-12-13T12:36:16","slug":"2eva2025paoi_t3-edp-eliptica-distribucion-de-potencial","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-2eva30\/2eva2025paoi_t3-edp-eliptica-distribucion-de-potencial\/","title":{"rendered":"2Eva2025PAOI_T3 EDP El\u00edptica, distribuci\u00f3n de potencial"},"content":{"rendered":"\n<h2 class=\"wp-block-heading\">2da Evaluaci\u00f3n 2025-2026 PAO I. 26\/Agosto\/2025<\/h2>\n\n\n\n<p><strong>Tema 3<\/strong> (30 puntos)&nbsp;Un cuadrado diel\u00e9ctrico de 2 cm de lado donde los bordes est\u00e1n a tierra, 0 Voltios, y el&nbsp;v\u00e9rtice opuesto est\u00e1 a 80V. Calcular la distribuci\u00f3n de potencial, suponiendo que la densidad de carga f(x,y) es nula.<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> \\frac{\\partial^2 \\phi}{\\partial x^2} + \\frac{\\partial^2 \\phi}{\\partial y^2} =0<\/span>\n\n\n\n<p>Condiciones de contorno se muestran junto con la ecuaci\u00f3n diferencial parcial<\/p>\n\n\n\n<div class=\"wp-block-columns is-layout-flex wp-container-core-columns-is-layout-28f84493 wp-block-columns-is-layout-flex\">\n<div class=\"wp-block-column is-layout-flow wp-block-column-is-layout-flow\"><span class=\"wp-katex-eq katex-display\" data-display=\"true\"> \\phi (x,0) = \\phi(0,y)=0 <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> \\phi (x,2) = 40x <\/span>\n<\/div>\n\n\n\n<div class=\"wp-block-column is-layout-flow wp-block-column-is-layout-flow\"><span class=\"wp-katex-eq katex-display\" data-display=\"true\"> \\phi (2,y) = 40y <\/span>\n<\/div>\n<\/div>\n\n\n\n<p>Suponiendo que se satisface la ley de Ohm, considere \u0394x=\u0394y=1\/4<br>Utilice diferencias finitas para las variables independientes x,y<\/p>\n\n\n\n<p>a. Plantee las ecuaciones discretas a usar un m\u00e9todo num\u00e9rico en un nodo i,j<\/p>\n\n\n\n<p>b. Realice la gr\u00e1fica de malla, detalle los valores de i, j, x<sub>i<\/sub>, y<sub>j<\/sub><\/p>\n\n\n\n<p>c. Desarrolle y obtenga el modelo discreto para \u03a6(xi,yj)<\/p>\n\n\n\n<p>d. Determine el valor de Lambda \u03bb<\/p>\n\n\n\n<p>e. Adjunte los archivos del algoritmo.py, resultados.txt, gr\u00e1ficas.png<\/p>\n\n\n\n<p><strong>R\u00fabrica<\/strong>: Selecci\u00f3n de diferencias finitas divididas (5 puntos), literal b (5 puntos), literal c (10 puntos), literal d (5 puntos), literal e (5 puntos)<\/p>\n\n\n\n<p><strong>Referencia<\/strong>: Chapter 13: Partial Differential Equations (Part 2 - Elliptic PDEs). Lindsey Westover. 18 Marzo 2021.<\/p>\n\n\n\n<figure class=\"wp-block-embed is-type-video is-provider-youtube wp-block-embed-youtube wp-embed-aspect-4-3 wp-has-aspect-ratio\"><div class=\"wp-block-embed__wrapper\">\n<iframe loading=\"lazy\" title=\"Chapter 13: Partial Differential Equations (Part 2 - Elliptic PDEs)\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/0eI5zrhtEjE?start=633&feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe>\n<\/div><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>2da Evaluaci\u00f3n 2025-2026 PAO I. 26\/Agosto\/2025 Tema 3 (30 puntos)&nbsp;Un cuadrado diel\u00e9ctrico de 2 cm de lado donde los bordes est\u00e1n a tierra, 0 Voltios, y el&nbsp;v\u00e9rtice opuesto est\u00e1 a 80V. Calcular la distribuci\u00f3n de potencial, suponiendo que la densidad de carga f(x,y) es nula. Condiciones de contorno se muestran junto con la ecuaci\u00f3n diferencial [&hellip;]<\/p>\n","protected":false},"author":8043,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"wp-custom-template-entrada-mn","format":"standard","meta":{"footnotes":""},"categories":[22],"tags":[57],"class_list":["post-11680","post","type-post","status-publish","format-standard","hentry","category-mn-2eva30","tag-edp"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/11680","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/users\/8043"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/comments?post=11680"}],"version-history":[{"count":6,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/11680\/revisions"}],"predecessor-version":[{"id":17574,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/11680\/revisions\/17574"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=11680"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=11680"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=11680"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}