{"id":748,"date":"2012-04-28T09:57:13","date_gmt":"2012-04-28T04:27:13","guid":{"rendered":"http:\/\/physicscatalyst.com\/article\/?p=748"},"modified":"2025-07-13T15:45:32","modified_gmt":"2025-07-13T10:15:32","slug":"force-conductor","status":"publish","type":"post","link":"https:\/\/physicscatalyst.com\/article\/force-conductor\/","title":{"rendered":"Force on a conductor in presence of an electric field"},"content":{"rendered":"

We have already learned in our previous discussion that field inside a conductor is zero and the field immediately outside is
\n$E_{n}=\\mathbf{n}\\left (\\frac{\\sigma}{\\epsilon_{0}} \u00a0\\right )$ \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (1)
\nwhere\u00a0n\u00a0<\/strong>is the unit normal vector to the surface of the conductor. We also know that any charge a conductor may carry is distributed on the surface of the conductor.
\nIn presence of an electric field this surface charge will experience a force. If we consider a small area element ?S of the surface of the conductor then force acting on area element is given by
\n$\\Delta\\mathbf{F}=\\left ( \\sigma \\Delta S \\right )\\cdot \\mathbf{E_{0}}$ \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(2)
\nwhere $\\sigma$ is the surface charge density of the conductor , $\\left ( \\sigma \\Delta S \\right )$ is the amount of charge on the area element $\\Delta S$ and\u00a0$\\mathbf{E_{0}}$\u00a0is the field in the region where charge element $\\left ( \\sigma \\Delta S \\right )$ is located.
\nNow there are two fields present $\\mathbf{E_{\\sigma}}$\u00a0and $\\mathbf{E_{0}}$\u00a0and the resultant field both inside and outside the conductor near area element \u00a0$\\Delta S$ would be equal to the superposition of both the fields\u00a0\u00a0$\\mathbf{E_{\\sigma}}$\u00a0and\u00a0$\\mathbf{E_{0}}$\u00a0. Figure below shows the directions of both the fields inside and outside the conductor<\/p>\n

\"\"<\/a><\/div>\n
<\/div>\n
Now field\u00a0$\\mathbf{E_{0}}$\u00a0has same value both inside and outside the conductor and surface element $\\Delta S$ suffers discontinuty because of the charge on the surface and this makes field\u00a0\u00a0$\\mathbf{E_{\\sigma}}$\u00a0<\/sub>on either side pointing away from the surface as shown in the figure given above. Since\u00a0E<\/strong>=0 inside the conductor<\/div>\n

$\\mathbf{E}_{in}=\\mathbf{E}_{0}-\\mathbf{E}_{\\sigma}=0 $
\n<\/strong>$\\mathbf{E}_{0}=\\mathbf{E}_{\\sigma}$
\n<\/strong> Since direction of\u00a0\u00a0E<\/strong>?<\/sub>\u00a0and\u00a0E0<\/sub><\/strong>\u00a0are opposite to each other and outside the conductor near its surface
\n$\\mathbf{E}_{out}=\\mathbf{E}_{0}+\\mathbf{E}_{\\sigma}=2\\mathbf{E}_{0}$<\/p>\n

Thus ,<\/p>\n

$\\mathbf{E}_{0}=\\mathbf{E}\/2$\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (3)
\nEquation (2) thus becomes,regardless of the of<\/p>\n

$\\Delta \\mathbf{F}=\\frac{1}{2}\\left ( \\sigma \\Delta S \\right )\\cdot \\mathbf{E}$ (4)<\/p>\n

From equation 4 , force acting per unit area of the surface of the conductor is<\/p>\n

$\\mathbf{f}=\\frac{1}{2}\\sigma \\mathbf{E}$ \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (5)<\/p>\n

Here\u00a0 is the\u00a0\u00a0$\\textbf{E}_{\\sigma }$\u00a0electric field intensity created by charge on area element $\\Delta S$ at the point very close to this area element. In this region this area element behaves as infinite uniformly charged sheet hence we have,
\n$\\textbf{E}_{\\sigma }=\\frac{\\sigma }{2\\epsilon _{0}}$\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (6)
\nNow,
\n$\\textbf{E}=2\\textbf{E}_{0}=2\\textbf{E}_{\\sigma }=\\frac{\\sigma }{\\epsilon _{0}}\\hat{n}=\\textbf{E}_{n}$
\nwhich is in accordance with equation 1. Hence from equation 5
\n$\\textbf{\\textit{f}}=\\frac{\\sigma ^{2}}{2\\epsilon _{0}} =\\frac{\\epsilon _{0}E^{2}}{2}\\hat{n}$\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (7)
\nThis quantity\u00a0f\u00a0<\/em><\/strong>is known as surface density of force. From equation 7 we can conclude that regardless of the sign of ? and hence direction of\u00a0E\u00a0<\/strong>,\u00a0f\u00a0<\/em><\/strong>is always directed in outward direction of the conductor.<\/p>\n","protected":false},"excerpt":{"rendered":"

Force on a conductor in presence of an electric field
\nIn presence of an electric field this surface charge will experience a force. If we consider a small area element of the surface of the conductor then force acting on area element is given by<\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_uag_custom_page_level_css":"","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":"","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":"set","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":[14],"tags":[],"class_list":["post-748","post","type-post","status-publish","format-standard","hentry","category-physics"],"yoast_head":"\nForce on a conductor in presence of an electric field - physicscatalyst's Blog<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/physicscatalyst.com\/article\/force-conductor\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Force on a conductor in presence of an electric field - physicscatalyst's Blog\" \/>\n<meta property=\"og:description\" content=\"Force on a conductor in presence of an electric field In presence of an electric field this surface charge will experience a force. 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