{"id":220705,"date":"2025-05-28T16:10:28","date_gmt":"2025-05-28T16:10:28","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=220705"},"modified":"2025-05-28T16:10:30","modified_gmt":"2025-05-28T16:10:30","slug":"figure-shows-a-gaussian-surface-in-the-shape-of-a-cube-with-edge-length-1-40-m","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/28\/figure-shows-a-gaussian-surface-in-the-shape-of-a-cube-with-edge-length-1-40-m\/","title":{"rendered":"Figure shows a Gaussian surface in the shape of a cube with edge length 1.40 m."},"content":{"rendered":"\n<p>Figure shows a Gaussian surface in the shape of a cube with edge length 1.40 m. What are:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>The net flux F through the surface<\/li>\n\n\n\n<li>The net charge qenc enclosed by the surface if E = (3.00yj) N\/C, with y in meters<\/li>\n\n\n\n<li>F<\/li>\n\n\n\n<li>qenc if E = [\u2013 4.00i + (6.00 + 3.00y)j] N\/C<\/li>\n<\/ol>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-2-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To solve this problem, we need to use <strong>Gauss\u2019s Law<\/strong>, which states: \u03a6E=qenc\u03b50\\Phi_E = \\frac{q_{\\text{enc}}}{\\varepsilon_0}<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u03a6E\\Phi_E is the net electric flux through a closed surface,<\/li>\n\n\n\n<li>qencq_{\\text{enc}} is the net charge enclosed by the surface,<\/li>\n\n\n\n<li>\u03b50=8.854\u00d710\u221212\u00a0C2\/N\\cdotpm2\\varepsilon_0 = 8.854 \\times 10^{-12}\\ \\text{C}^2\/\\text{N\u00b7m}^2 is the permittivity of free space.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Part 1: E = (3.00y) j N\/C<\/strong><\/h3>\n\n\n\n<p>This is a <strong>non-uniform electric field<\/strong> in the <strong>y-direction<\/strong> that increases linearly with yy. The cube has edge length L=1.40&nbsp;mL = 1.40\\ \\text{m}, and we assume its sides are aligned with the axes and it&#8217;s centered in space.<\/p>\n\n\n\n<p>Only the <strong>top<\/strong> and <strong>bottom faces<\/strong> contribute to the net flux since the electric field is in the <strong>y-direction<\/strong>. The field through side faces in xx and zz directions contributes zero net flux.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Bottom face (y = 0):<\/strong> Ebottom=0E_{\\text{bottom}} = 0<\/li>\n\n\n\n<li><strong>Top face (y = 1.40 m):<\/strong> Etop=3.00\u00d71.40=4.20\u00a0N\/CE_{\\text{top}} = 3.00 \\times 1.40 = 4.20\\ \\text{N\/C}<\/li>\n<\/ul>\n\n\n\n<p>Each face has area: A=(1.40)2=1.96&nbsp;m2A = (1.40)^2 = 1.96\\ \\text{m}^2<\/p>\n\n\n\n<p>So, net flux: \u03a6=Etop\u22c5A\u2212Ebottom\u22c5A=4.20\u22c51.96\u22120=8.232&nbsp;N\\cdotpm2\/C\\Phi = E_{\\text{top}} \\cdot A &#8211; E_{\\text{bottom}} \\cdot A = 4.20 \\cdot 1.96 &#8211; 0 = 8.232\\ \\text{N\u00b7m}^2\/\\text{C}<\/p>\n\n\n\n<p>Net enclosed charge: qenc=\u03a6\u22c5\u03b50=8.232\u22c58.854\u00d710\u221212\u22487.29\u00d710\u221211&nbsp;Cq_{\\text{enc}} = \\Phi \\cdot \\varepsilon_0 = 8.232 \\cdot 8.854 \\times 10^{-12} \\approx \\boxed{7.29 \\times 10^{-11}\\ \\text{C}}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Part 2: E = [\u20134.00 i + (6.00 + 3.00y) j] N\/C<\/strong><\/h3>\n\n\n\n<p>This field has <strong>x- and y-components<\/strong>. However:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The <strong>x-component<\/strong> is constant (\u22124.00\u00a0N\/C)(-4.00\\ \\text{N\/C}), so the flux entering and leaving the sides cancel.<\/li>\n\n\n\n<li>The <strong>y-component<\/strong> is variable: Ey=6.00+3.00yE_y = 6.00 + 3.00y<\/li>\n<\/ul>\n\n\n\n<p>As before, compute flux only through <strong>top and bottom<\/strong> faces:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>At bottom (y = 0): Ebottom=6.00E_{\\text{bottom}} = 6.00<\/li>\n\n\n\n<li>At top (y = 1.40): Etop=6.00+3.00\u22c51.40=10.20E_{\\text{top}} = 6.00 + 3.00 \\cdot 1.40 = 10.20<\/li>\n<\/ul>\n\n\n\n<p>So: \u03a6=Etop\u22c5A\u2212Ebottom\u22c5A=(10.20\u22126.00)\u22c51.96=4.20\u22c51.96=8.232&nbsp;N\\cdotpm2\/C\\Phi = E_{\\text{top}} \\cdot A &#8211; E_{\\text{bottom}} \\cdot A = (10.20 &#8211; 6.00) \\cdot 1.96 = 4.20 \\cdot 1.96 = 8.232\\ \\text{N\u00b7m}^2\/\\text{C}<\/p>\n\n\n\n<p>Same as before. qenc=8.232\u22c58.854\u00d710\u221212=7.29\u00d710\u221211&nbsp;Cq_{\\text{enc}} = 8.232 \\cdot 8.854 \\times 10^{-12} = \\boxed{7.29 \\times 10^{-11}\\ \\text{C}}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Summary Answers<\/strong><\/h3>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th>Case<\/th><th>Net Flux \u03a6E\\Phi_E (N\u00b7m\u00b2\/C)<\/th><th>Enclosed Charge qencq_{\\text{enc}} (C)<\/th><\/tr><\/thead><tbody><tr><td>1: E=3.00yj^E = 3.00y \\hat{j}<\/td><td>8.23\\boxed{8.23}<\/td><td>7.29\u00d710\u221211\\boxed{7.29 \\times 10^{-11}}<\/td><\/tr><tr><td>2: E=\u22124.00i^+(6.00+3.00y)j^E = -4.00\\hat{i} + (6.00 + 3.00y)\\hat{j}<\/td><td>8.23\\boxed{8.23}<\/td><td>7.29\u00d710\u221211\\boxed{7.29 \\times 10^{-11}}<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>In both cases, the net flux and charge enclosed are the same, because only the y-component of the field varies within the cube and contributes to net flux.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/learnexams-banner4-72.jpeg\" alt=\"\" class=\"wp-image-220706\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Figure shows a Gaussian surface in the shape of a cube with edge length 1.40 m. What are: The Correct Answer and Explanation is: To solve this problem, we need to use Gauss\u2019s Law, which states: \u03a6E=qenc\u03b50\\Phi_E = \\frac{q_{\\text{enc}}}{\\varepsilon_0} Where: Part 1: E = (3.00y) j N\/C This is a non-uniform electric field in the [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-220705","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220705","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=220705"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220705\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=220705"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=220705"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=220705"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}