The differential amount of work done in moving a charge ( q ) in an electric field ( \\mathbf{E} ) is given by:<\/p>\n\n\n\n
[
dW = q \\mathbf{E} \\cdot d\\mathbf{L}
]<\/p>\n\n\n\n
where:<\/p>\n\n\n\n
- \n
- ( q = 6 ) nC = ( 6 \\times 10^{-9} ) C<\/li>\n\n\n\n
- ( dL = 2 ) \u00b5m = ( 2 \\times 10^{-6} ) m<\/li>\n\n\n\n
- ( \\mathbf{E} = \\frac{1}{z^2} (8xyz \\hat{a}_x + 4x^2z \\hat{a}_y – 4x^2y \\hat{a}_z) ) V\/m<\/li>\n\n\n\n
- ( P(2, -2, 3) )<\/li>\n\n\n\n
- Direction vectors:<\/li>\n<\/ul>\n\n\n\n
- \n
- ( \\hat{a}_L = -\\frac{6}{7} \\hat{a}_x + \\frac{3}{7} \\hat{a}_y + \\frac{2}{7} \\hat{a}_z )<\/li>\n\n\n\n
- ( \\hat{a}_L = \\frac{6}{7} \\hat{a}_x – \\frac{3}{7} \\hat{a}_y – \\frac{2}{7} \\hat{a}_z )<\/li>\n\n\n\n
- ( \\hat{a}_L = \\frac{3}{7} \\hat{a}_x + \\frac{6}{7} \\hat{a}_y )<\/li>\n<\/ol>\n\n\n\n
Let’s calculate the work done for each case.<\/p>\n\n\n\n
Results:<\/h3>\n\n\n\n
The differential work done for each direction is:<\/p>\n\n\n\n
- \n
- Case 1<\/strong> ((-\\frac{6}{7} \\hat{a}_x + \\frac{3}{7} \\hat{a}_y + \\frac{2}{7} \\hat{a}_z)):
[
dW = 1.493 \\times 10^{-13} \\text{ J}
]<\/li>\n\n\n\n- Case 2<\/strong> ((\\frac{6}{7} \\hat{a}_x – \\frac{3}{7} \\hat{a}_y – \\frac{2}{7} \\hat{a}_z)):
[
dW = -1.493 \\times 10^{-13} \\text{ J}
]<\/li>\n\n\n\n- Case 3<\/strong> ((\\frac{3}{7} \\hat{a}_x + \\frac{6}{7} \\hat{a}_y)):
[
dW \\approx 3.73 \\times 10^{-30} \\text{ J} \\approx 0
]<\/li>\n<\/ol>\n\n\n\nExplanation:<\/h3>\n\n\n\n
The work done by an electric field is given by the dot product of (\\mathbf{E}) and the displacement (\\mathbf{dL}). The charge ( q ) scales this value accordingly.<\/p>\n\n\n\n
- \n
- Case 1:<\/strong> The work is positive, meaning the force aligns with the displacement, and energy is transferred to the charge.<\/li>\n\n\n\n
- Case 2:<\/strong> The work is negative, indicating movement against the field, requiring an external force.<\/li>\n\n\n\n
- Case 3:<\/strong> The work is nearly zero, implying movement perpendicular to the force (no energy transfer).<\/li>\n<\/ul>\n\n\n\n
Visualization:<\/h3>\n\n\n\n
Now, let’s generate an image illustrating the movement of the charge in different directions within the electric field.<\/p>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-1696-1010x1024.png\")
<\/figure>\n\n\n\nThe image above illustrates the movement of the charge from the initial point ( P(2, -2, 3) ) in the three different directions given. Each vector represents a displacement of ( 2 \\mu m ) in one of the specified directions:<\/p>\n\n\n\n
- \n
- Blue (Case 1)<\/strong>: Work is positive<\/strong>, indicating motion along the field.<\/li>\n\n\n\n
- Green (Case 2)<\/strong>: Work is negative<\/strong>, indicating motion against the field.<\/li>\n\n\n\n
- Purple (Case 3)<\/strong>: Work is almost zero<\/strong>, indicating perpendicular movement with no significant energy exchange.<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"
Given the electric field E = 1\/z2 (8xyz ax + 4x2z ay – 4x2y az) V\/m, find the differential amount of work done in moving a 6 nC charge a distance of 2 \u00b5m, starting at P(2, -2, 3) and proceeding in the direction: aL = – 6\/7ax+ 3\/7 ay + 2\/7 azaL= + 6\/7ax […]<\/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-195636","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195636","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=195636"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195636\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=195636"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=195636"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=195636"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- Green (Case 2)<\/strong>: Work is negative<\/strong>, indicating motion against the field.<\/li>\n\n\n\n
- Case 2:<\/strong> The work is negative, indicating movement against the field, requiring an external force.<\/li>\n\n\n\n
- Case 2<\/strong> ((\\frac{6}{7} \\hat{a}_x – \\frac{3}{7} \\hat{a}_y – \\frac{2}{7} \\hat{a}_z)):
- Case 1<\/strong> ((-\\frac{6}{7} \\hat{a}_x + \\frac{3}{7} \\hat{a}_y + \\frac{2}{7} \\hat{a}_z)):