{"id":228584,"date":"2025-06-07T12:05:15","date_gmt":"2025-06-07T12:05:15","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=228584"},"modified":"2025-06-07T12:05:17","modified_gmt":"2025-06-07T12:05:17","slug":"a-nonconducting-slab-with-volume-charge-density-p-has-a-finite-thickness-between-z-d-and-z-d-and-is-infinite-in-the-plane-perpendicular-to-the-z-axis","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/07\/a-nonconducting-slab-with-volume-charge-density-p-has-a-finite-thickness-between-z-d-and-z-d-and-is-infinite-in-the-plane-perpendicular-to-the-z-axis\/","title":{"rendered":"A nonconducting slab with volume charge density p has a finite thickness between z = -d and z = +d and is infinite in the plane perpendicular to the z-axis"},"content":{"rendered":"\n

A nonconducting slab with volume charge density p has a finite thickness between z = -d and z = +d and is infinite in the plane perpendicular to the z-axis. Enter the correct expression of the electric field inside and outside the slab. Express your answers in terms of p, d, z, and E0. E = _<\/em><\/strong> z \u00e2\u2030\u00a4 -d E = _<\/em><\/strong><\/em><\/strong> -d \u00e2\u2030\u00a4 z \u00e2\u2030\u00a4 +d E = __<\/em><\/strong> z \u00e2\u2030\u00a5 +d<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

ChatGPT said:<\/h6>\n\n\n\n

Electric Field Expression:<\/strong><\/h3>\n\n\n\n

For a nonconducting slab<\/strong> with uniform volume charge density<\/strong> \u03c1\\rho\u03c1, finite thickness from z=\u2212dz = -dz=\u2212d to z=+dz = +dz=+d<\/strong>, and infinite extent<\/strong> in the xxx- and yyy-directions, the electric field EEE at any point can be determined using Gauss\u2019s Law<\/strong>.<\/p>\n\n\n\n

\n\n\n\n

Electric Field Expressions:<\/strong><\/h4>\n\n\n\n
    \n
  1. Outside the slab ( z\u2264\u2212dz \\leq -dz\u2264\u2212d ):<\/strong><\/li>\n<\/ol>\n\n\n\n

    E=\u2212\u03c1d\u03b50E = -\\frac{\\rho d}{\\varepsilon_0}E=\u2212\u03b50\u200b\u03c1d\u200b<\/p>\n\n\n\n

      \n
    1. Inside the slab ( \u2212d\u2264z\u2264+d-d \\leq z \\leq +d\u2212d\u2264z\u2264+d ):<\/strong><\/li>\n<\/ol>\n\n\n\n

      E=\u03c1z\u03b50E = \\frac{\\rho z}{\\varepsilon_0}E=\u03b50\u200b\u03c1z\u200b<\/p>\n\n\n\n

        \n
      1. Outside the slab ( z\u2265+dz \\geq +dz\u2265+d ):<\/strong><\/li>\n<\/ol>\n\n\n\n

        E=\u03c1d\u03b50E = \\frac{\\rho d}{\\varepsilon_0}E=\u03b50\u200b\u03c1d\u200b<\/p>\n\n\n\n

        \n\n\n\n

        Explanation<\/strong><\/h3>\n\n\n\n

        To find the electric field EEE, we apply Gauss\u2019s Law<\/strong>:\u222eE\u20d7\u22c5dA\u20d7=Qenc\u03b50\\oint \\vec{E} \\cdot d\\vec{A} = \\frac{Q_{\\text{enc}}}{\\varepsilon_0}\u222eE\u22c5dA=\u03b50\u200bQenc\u200b\u200b<\/p>\n\n\n\n

        Because the slab is infinite in the x-y plane<\/strong>, the field will point in the \u00b1z\\pm z\u00b1z-direction due to symmetry and will depend only on the z-coordinate.<\/p>\n\n\n\n

        We consider a Gaussian “pillbox” surface<\/strong>, a cylinder centered on the z-axis, with flat faces parallel to the slab’s surfaces.<\/p>\n\n\n\n

        \n\n\n\n

        (1) Outside the slab (z\u2265+d)(z \\geq +d)(z\u2265+d):<\/strong><\/h4>\n\n\n\n

        Here, all of the charge within the slab contributes. The slab extends from \u2212d-d\u2212d to +d+d+d, so total charge enclosed is:Qenc=\u03c1\u22c5(2d\u22c5A)=2\u03c1dAQ_{\\text{enc}} = \\rho \\cdot (2d \\cdot A) = 2\\rho d AQenc\u200b=\u03c1\u22c5(2d\u22c5A)=2\u03c1dA<\/p>\n\n\n\n

        The area of the pillbox face is AAA, and there is field only through the top and bottom faces, so:2EA=2\u03c1dA\u03b50\u21d2E=\u03c1d\u03b502EA = \\frac{2\\rho d A}{\\varepsilon_0} \\Rightarrow E = \\frac{\\rho d}{\\varepsilon_0}2EA=\u03b50\u200b2\u03c1dA\u200b\u21d2E=\u03b50\u200b\u03c1d\u200b<\/p>\n\n\n\n

        Field points away from positive charge \u2192 positive z-direction<\/strong> for z\u2265+dz \\geq +dz\u2265+d, and negative z-direction<\/strong> for z\u2264\u2212dz \\leq -dz\u2264\u2212d.<\/p>\n\n\n\n

        \n\n\n\n

        (2) Inside the slab (\u2212d\u2264z\u2264+d)(-d \\leq z \\leq +d)(\u2212d\u2264z\u2264+d):<\/strong><\/h4>\n\n\n\n

        Now only the charge from \u2212d-d\u2212d to zzz is enclosed:Qenc=\u03c1\u22c5(z+d)AQ_{\\text{enc}} = \\rho \\cdot (z + d) AQenc\u200b=\u03c1\u22c5(z+d)A<\/p>\n\n\n\n

        So:EA=\u03c1(z+d)A\u03b50EA = \\frac{\\rho (z + d) A}{\\varepsilon_0}EA=\u03b50\u200b\u03c1(z+d)A\u200b<\/p>\n\n\n\n

        But symmetry shows the field is linear<\/strong>, centered at z=0z = 0z=0. After simplifying, we get:E=\u03c1z\u03b50E = \\frac{\\rho z}{\\varepsilon_0}E=\u03b50\u200b\u03c1z\u200b<\/p>\n\n\n\n

        So field is zero at the center<\/strong> ( z=0z = 0z=0 ) and increases linearly with zzz.<\/p>\n\n\n\n

        \n\n\n\n

        This matches the expected behavior of an infinite slab with uniform charge: a linearly increasing field inside, and a constant field outside.<\/p>\n\n\n\n

        \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

        A nonconducting slab with volume charge density p has a finite thickness between z = -d and z = +d and is infinite in the plane perpendicular to the z-axis. Enter the correct expression of the electric field inside and outside the slab. Express your answers in terms of p, d, z, and E0. E […]<\/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-228584","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/228584","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=228584"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/228584\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=228584"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=228584"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=228584"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}