{"id":8157,"date":"2023-05-01T02:25:21","date_gmt":"2023-05-01T02:25:21","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=8157"},"modified":"2024-05-25T06:49:51","modified_gmt":"2024-05-25T06:49:51","slug":"potential-difference-equation","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/potential-difference-equation.htm","title":{"rendered":"Potential difference equation","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"<p align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">3 questions about Potential difference equation<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">1. An electric charge is moved in a homogeneous electric field with a force of 2\u221a3 N a distance of 20 cm. If the direction of the force is at an angle of 30<sup>o<\/sup> to the displacement of the electric charge, what is the difference in the electric potential energy at the initial and final positions of the electric charge.<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Known:<\/u><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"color: #000000; font-size: 12pt; font-family: 'times new roman', times, serif;\">Force (F) = 2\u221a3 N<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"color: #000000; font-size: 12pt; font-family: 'times new roman', times, serif;\">Distance (s) = 20 cm = 0.2 m<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><span style=\"color: #000000;\">Angle (\u03b8) = 30<\/span><span style=\"color: #000000;\"><sup>o<\/sup><\/span><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Wanted:<\/u> Electric potential difference<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Solution:<\/u><!--more--><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">The work of transferring charge +q from a to b is equal to the electric potential energy difference at points a and b.<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">\u0394EP = \u0394W<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">If the direction of the force F with respect to the direction of charge transfer +q is angled \u03b8, then the work of charge transfer +q from a to b is:<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">\u0394W = F \u0394s cos <span style=\"color: #000000;\">\u03b8 <\/span><span style=\"color: #000000;\">= (<\/span><span style=\"color: #000000;\">2<\/span><span style=\"color: #000000;\">\u221a<\/span><span style=\"color: #000000;\">3<\/span><span style=\"color: #000000;\">)(0,2)(cos 30) = (0,4<\/span><span style=\"color: #000000;\">\u221a<\/span><span style=\"color: #000000;\">3<\/span><span style=\"color: #000000;\">)(0,5<\/span><span style=\"color: #000000;\">\u221a<\/span><span style=\"color: #000000;\">3<\/span><span style=\"color: #000000;\">) = (0,2)(3) = 0,6 Joule<\/span><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">2. If the charge and capacity of the capacitor are known to be 5 \u00b5C and 20 \u00b5F, respectively, determine the potential difference of the capacitors.<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Known:<\/u><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Electric charge (q) = 5 \u00b5C = 5 x 10<sup>-6<\/sup> C<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Capacitor capacity (C) = 20 \u00b5F = 20 x 10<sup>-6<\/sup> F <\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Wanted:<\/u> Capacitor potential difference (V)<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Solution:<\/u><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">The formula for capacitor potential difference, capacitor capacity and charge:<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">V = q \/ C = (5 x 10<sup>-6<\/sup>) \/ 20 x 10<sup>-6<\/sup> = 5\/20 = 0,25 Volt <\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">3. Two point charges QA = -4 \u00b5C and QB = 8 \u00b5C are 16 cm apart. Determine the electric potential at a point halfway between the two charges.<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Known:<\/u><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Electrical charge 1 (Q<sub>A<\/sub>) = 4 x 10<sup>-6<\/sup> C<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Electrical charge 2 (Q<sub>B<\/sub>) = 8 x 10<sup>-6<\/sup> C<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Distance (r) = 8 cm = 0,08 m<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Coulomb constant (k) = 9 \u00d7 10<sup>9<\/sup> Nm<sup>2<\/sup>\/C<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Wanted:<\/u> The electric potential at C (V<sub>C<\/sub>)<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Solution:<\/u><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">V<sub>A<\/sub> = k Q \/ r = (9 \u00d7 10<sup>9<\/sup>)(4 x 10<sup>-6<\/sup>) \/ 0,08 = (36 \u00d7 10<sup>3<\/sup>) \/ 0,08 = -450 Volt<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">V<sub>B<\/sub> = k Q \/ r = (9 \u00d7 10<sup>9<\/sup>)(8 x 10<sup>-6<\/sup>) \/ 0,08 = (72 \u00d7 10<sup>3<\/sup>) \/ 0,08 = 900 Volt<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">The electric potential at C (V<sub>C<\/sub>) = 900 \u2013 450 = 450 Volt<\/span><\/p>\n<p align=\"justify\">\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"<p>3 questions about Potential difference equation 1. An electric charge is moved in a homogeneous electric field with a force of 2\u221a3 N a distance of 20 cm. If the direction of the force is at an angle of 30o to the displacement of the electric charge, what is the difference in the electric potential &#8230; <a title=\"Potential difference equation\" class=\"read-more\" href=\"https:\/\/gurumuda.net\/physics\/potential-difference-equation.htm\" aria-label=\"Read more about Potential difference equation\">Read more<\/a><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"closed","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":"3","_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":"Potential difference equation","_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-8157","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\/8157","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=8157"}],"version-history":[{"count":7,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/8157\/revisions"}],"predecessor-version":[{"id":9844,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/8157\/revisions\/9844"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=8157"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=8157"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=8157"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}