{"id":1366,"date":"2018-03-05T07:33:48","date_gmt":"2018-03-04T23:33:48","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1366"},"modified":"2018-03-05T07:33:48","modified_gmt":"2018-03-04T23:33:48","slug":"electric-flux-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/electric-flux-problems-and-solutions.htm","title":{"rendered":"Electric flux – problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

1. A uniform electric field<\/a> E = 8000 N\/C passing through a flat square area A = 10 m<\/span><\/span>2<\/span><\/span><\/sup>. <\/span><\/span>Determine the electric flux<\/a>.<\/span><\/span><\/p>\n

\"ElectricKnown :<\/u><\/span><\/span><\/p>\n

The magnitude of the electric field (E) = 8000 N\/C<\/span><\/span><\/p>\n

Area (A) = 10 m<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

\u03b8\u00a0= 0<\/span><\/span>o<\/span><\/span><\/sup> (the angle between the electric field direction and a line drawn a perpendicular to the area)<\/span><\/span><\/p>\n

Wanted:<\/u><\/span><\/span> Electric flux <\/span><\/span>(<\/span><\/span>\u03a6)<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

The formula of electric flux :<\/span><\/span><\/p>\n

\u03a6 <\/span><\/span>= E A cos <\/span><\/span>q<\/span><\/span><\/p>\n

\u03a6 = electric flux (<\/i><\/span><\/span>Nm<\/i><\/span><\/span>2<\/i><\/span><\/span><\/sup>\/C)<\/i><\/span><\/span>, E = electric field (N\/C), A = area (m<\/i><\/span><\/span>2<\/i><\/span><\/span><\/sup>), <\/i><\/span><\/span>q =<\/i><\/span><\/span> angle between electric field line with the normal line.<\/i><\/span><\/span><\/p>\n

Electric flux :<\/span><\/span><\/p>\n

\u03a6 <\/span><\/span>= E A cos <\/span><\/span>q = (8000)(10)(<\/span><\/span>cos 0) = <\/span><\/span>(8000)(10)(1<\/span><\/span>) = <\/span><\/span>80,000 = 8 <\/span><\/span>x 10<\/span><\/span>4<\/span><\/span><\/sup> Nm<\/span><\/span>2<\/span><\/span><\/sup>\/C<\/span><\/span><\/p>\n

<\/p>\n

2. <\/span><\/span>A uniform electric field E = 5000 N\/C passing through a flat square area A = 2 m<\/span><\/span>2<\/span><\/span><\/sup>. <\/span><\/span>Determine the electric flux.<\/span><\/span><\/p>\n

\"ElectricKnown :<\/u><\/span><\/span><\/p>\n

Electric field (E) = 5000 N\/C<\/span><\/span><\/p>\n

Area (A) = 2 m<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

\u03b8 = 60<\/span><\/span>o <\/span><\/span><\/sup>(the angle between the electric field direction and a line drawn perpendicular to the area)<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> Electric flux <\/span><\/span>(<\/span><\/span>\u03a6)<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

Electric flux :<\/span><\/span><\/p>\n

\u03a6 <\/span><\/span>= E A cos <\/span><\/span>q = (5000)(2)(<\/span><\/span>cos 60) = <\/span><\/span>(5000)(2)(0.5<\/span><\/span>) = 5000 <\/span><\/span>= 5 <\/span><\/span>x 10<\/span><\/span>3<\/span><\/span><\/sup> Nm<\/span><\/span>2<\/span><\/span><\/sup>\/C<\/span><\/span><\/p>\n

<\/p>\n

3. <\/span><\/span><\/span>A solid ball with 0.5 meters radius has 10 \u03bcC electric charge in its center. Determine the electrical flux pass through the solid ball. <\/span><\/span><\/span><\/p>\n

\"ElectricKnown :<\/u><\/span><\/span><\/p>\n

Radius of ball (r) = 0.5 m<\/span><\/span><\/p>\n

Electric charge<\/a> (Q) = <\/span><\/span>10 \u03bcC = 10 x 10<\/span><\/span>-6<\/span><\/span><\/sup> C<\/span><\/span><\/p>\n

Wanted : <\/u><\/span><\/span> Electric flux <\/span><\/span>(<\/span><\/span>\u03a6)<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

Electric field :<\/span><\/span><\/p>\n

E = k q\/r<\/span><\/span>2 <\/span><\/span><\/sup><\/p>\n

E <\/span><\/span>= (9 x 10<\/span><\/span>9<\/span><\/span><\/sup> Nm<\/span><\/span>2<\/span><\/span><\/sup>\/C<\/span><\/span>2<\/span><\/span><\/sup>)(<\/span><\/span>10 x 10<\/span><\/span>-6<\/span><\/span><\/sup> C) \/ <\/span><\/span>0.5<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

E = (90 x 10<\/span><\/span>3<\/span><\/span><\/sup>) <\/span><\/span>\/ <\/span><\/span>0,25<\/span><\/span><\/p>\n

E = 360 x 10<\/span><\/span>3<\/span><\/span><\/sup><\/p>\n

E = 3.60 x 10<\/span><\/span>5<\/span><\/span><\/sup> N\/C<\/span><\/span><\/p>\n

Surface area :<\/span><\/span><\/p>\n

A = 4 <\/span><\/span>\u03c0 r<\/span><\/span>2 <\/span><\/span><\/sup>= 4 (3.14)(0.5)<\/span><\/span>2 <\/span><\/span><\/sup>= (12.56)(0.25) = 3.14 m<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

Electric flux :<\/span><\/span><\/p>\n

The electric field lines perpendicular to area, so that <\/span><\/span>the angle between the electric field direction and a line drawn perpendicular to the area, is <\/span><\/span>0<\/span><\/span>o<\/span><\/span><\/sup>.<\/span><\/span><\/p>\n

\u03a6 <\/span><\/span>= E A cos <\/span><\/span>q<\/span><\/span><\/p>\n

\u03a6 = (<\/span><\/span>3.60 x 10<\/span><\/span>5<\/span><\/span><\/sup>)(<\/span><\/span>3.14<\/span><\/span>)(<\/span><\/span>cos 0)<\/span><\/span><\/p>\n

\u03a6 <\/span><\/span>= (11.304 x <\/span><\/span>10<\/span><\/span>5<\/span><\/span><\/sup>)(1)<\/span><\/span><\/p>\n

\u03a6 <\/span><\/span>= 11.304 x <\/span><\/span>10<\/span><\/span>5<\/span><\/span><\/sup><\/p>\n

\u03a6 <\/span><\/span>= 1.13 x <\/span><\/span>10<\/span><\/span>6<\/span><\/span><\/sup> Nm<\/span><\/span>2<\/span><\/span><\/sup>\/C<\/span><\/span><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

1. A uniform electric field E = 8000 N\/C passing through a flat square area A = 10 m2. Determine the electric flux. Known : The magnitude of the electric field (E) = 8000 N\/C Area (A) = 10 m2 \u03b8\u00a0= 0o (the angle between the electric field direction and a line drawn a perpendicular … Read more<\/a><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_seopress_titles_title":"","_seopress_titles_desc":"","_seopress_robots_index":"","_seopress_robots_follow":"","_seopress_robots_imageindex":"","_seopress_robots_snippet":"","_seopress_robots_primary_cat":"","_seopress_robots_breadcrumbs":"","_seopress_robots_freeze_modified_date":"","_seopress_robots_custom_modified_date":"","_seopress_robots_canonical":"","_seopress_social_fb_title":"","_seopress_social_fb_desc":"","_seopress_social_fb_img":"","_seopress_social_fb_img_attachment_id":0,"_seopress_social_fb_img_width":0,"_seopress_social_fb_img_height":0,"_seopress_social_twitter_title":"","_seopress_social_twitter_desc":"","_seopress_social_twitter_img":"","_seopress_social_twitter_img_attachment_id":0,"_seopress_social_twitter_img_width":0,"_seopress_social_twitter_img_height":0,"_seopress_redirections_value":"","_seopress_redirections_enabled":"","_seopress_redirections_enabled_regex":"","_seopress_redirections_logged_status":"","_seopress_redirections_param":"","_seopress_redirections_type":0,"_seopress_analysis_target_kw":"Electric flux - problems and solutions","_seopress_news_disabled":"","_seopress_video_disabled":"","_seopress_video":[],"_seopress_pro_schemas_manual":[],"_seopress_pro_rich_snippets_disable_all":"","_seopress_pro_rich_snippets_disable":[],"_seopress_pro_schemas":[],"footnotes":""},"categories":[3],"tags":[],"class_list":["post-1366","post","type-post","status-publish","format-standard","hentry","category-solved-problems-in-basic-physics"],"gt_translate_keys":[{"key":"link","format":"url"}],"_links":{"self":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1366","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/comments?post=1366"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1366\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=1366"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=1366"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=1366"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}