A +5.0-\u00b5C point charge is placed at the 0 cm mark of a meter stick and a -4.0-\u00b5C charge is placed at the 50 cm mark. What is the net electric field at the 30 cm mark? (k = 1\/4pe0 = 8.99 \u00d7 10^9 N m\u00b2\/C\u00b2)<\/p>\n\n\n\n
91) __<\/strong><\/p>\n\n\n\n A) 9.0 \u00d7 10^5 N\/C<\/p>\n\n\n\n B) 4.0 \u00d7 10^5 N\/C<\/p>\n\n\n\n C) 1.4 \u00d7 10^6 N\/C<\/p>\n\n\n\n D) 5.0 \u00d7 10^5 N\/C<\/p>\n\n\n\n The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n To find the net electric field at the 30 cm mark due to the two point charges placed at the 0 cm and 50 cm marks, we need to apply Coulomb\u2019s law for electric fields and consider the contributions of both charges.<\/p>\n\n\n\n The electric field (E) at a point due to a point charge is given by:<\/p>\n\n\n\n [ where:<\/p>\n\n\n\n The distance between the 0 cm mark (where the +5.0-\u00b5C charge is located) and the 30 cm mark is 30 cm = 0.30 m. The electric field due to this charge is:<\/p>\n\n\n\n [ The direction of the electric field due to a positive charge is away from the charge. Since the +5.0-\u00b5C charge is to the left of the 30 cm mark, the electric field at the 30 cm mark will point to the right.<\/p>\n\n\n\n The distance between the 50 cm mark (where the -4.0-\u00b5C charge is located) and the 30 cm mark is 20 cm = 0.20 m. The electric field due to this charge is:<\/p>\n\n\n\n [ The direction of the electric field due to a negative charge is toward the charge. Since the -4.0-\u00b5C charge is to the right of the 30 cm mark, the electric field at the 30 cm mark will point to the left.<\/p>\n\n\n\n The net electric field is the vector sum of the fields due to both charges. Since the fields due to the +5.0-\u00b5C charge and the -4.0-\u00b5C charge are in opposite directions, we subtract the magnitudes of the fields:<\/p>\n\n\n\n [ The net electric field at the 30 cm mark is 4.0 \u00d7 10^5 N\/C<\/strong>.<\/p>\n\n\n\n Thus, the correct answer is B) 4.0 \u00d7 10^5 N\/C<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":" A +5.0-\u00b5C point charge is placed at the 0 cm mark of a meter stick and a -4.0-\u00b5C charge is placed at the 50 cm mark. What is the net electric field at the 30 cm mark? (k = 1\/4pe0 = 8.99 \u00d7 10^9 N m\u00b2\/C\u00b2) 91) __ A) 9.0 \u00d7 10^5 N\/C B) 4.0 […]<\/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-207861","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/207861","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=207861"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/207861\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=207861"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=207861"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=207861"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Formula for the Electric Field<\/h3>\n\n\n\n
E = \\frac{k |Q|}{r^2}
]<\/p>\n\n\n\n\n
Step 2: Electric Field Due to the +5.0-\u00b5C Charge at the 0 cm Mark<\/h3>\n\n\n\n
E_1 = \\frac{k \\times |Q_1|}{r_1^2} = \\frac{8.99 \\times 10^9 \\times 5.0 \\times 10^{-6}}{(0.30)^2}
]
[
E_1 = \\frac{8.99 \\times 10^9 \\times 5.0 \\times 10^{-6}}{0.09} = 4.99 \\times 10^5 \\, \\text{N\/C}
]<\/p>\n\n\n\nStep 3: Electric Field Due to the -4.0-\u00b5C Charge at the 50 cm Mark<\/h3>\n\n\n\n
E_2 = \\frac{k \\times |Q_2|}{r_2^2} = \\frac{8.99 \\times 10^9 \\times 4.0 \\times 10^{-6}}{(0.20)^2}
]
[
E_2 = \\frac{8.99 \\times 10^9 \\times 4.0 \\times 10^{-6}}{0.04} = 8.99 \\times 10^5 \\, \\text{N\/C}
]<\/p>\n\n\n\nStep 4: Net Electric Field<\/h3>\n\n\n\n
E_{\\text{net}} = E_2 – E_1 = 8.99 \\times 10^5 – 4.99 \\times 10^5 = 4.0 \\times 10^5 \\, \\text{N\/C}
]<\/p>\n\n\n\nFinal Answer:<\/h3>\n\n\n\n