{"id":182476,"date":"2025-01-14T06:09:07","date_gmt":"2025-01-14T06:09:07","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=182476"},"modified":"2025-01-14T06:09:10","modified_gmt":"2025-01-14T06:09:10","slug":"coulombs-law-is-f-kq_1q_2-r-2-where-k-9-10-9-in-sl-units","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/14\/coulombs-law-is-f-kq_1q_2-r-2-where-k-9-10-9-in-sl-units\/","title":{"rendered":"Coulomb’s law is F = (kq_{1}q_{2}) \/ (r ^ 2) where k = 9 * 10 ^ 9 in Sl units"},"content":{"rendered":"\n

1) Coulomb’s law is F = (kq_{1}<\/em>q_{2}) \/ (r ^ 2) where k = 9 * 10 ^ 9 in Sl units. Suppose that q * 1 = 6mu*C and q * 2 = 4mu*C and r = 25 cm, calc. the force between these particles. 2) The proton and electron in a H atom both have a charge of 1.6 X 10-19 C They are r = 5 * 10 ^ – 10 meters apart. Calc. the force between them 3) If two charged pith balls have a force of F due to electrostatic attraction, what is the final force on them if the distance between them is doubled? (a) 2F (b) 4F (c) F\/2 (d) F\/4 4) A charge – 2 C is placed at the origin of the xy plane. A charge of 3 C is placed at x = 25 cm. A final charge of -8 C is placed at x = 50 cm. Calc. the force between the – 2 C and 3 C charges (this force will be in the positive x direction or +i) 5) In problem 4, calc. the force between the – 2 Cand-8 C charges. (this force will be in the negative x direction or – i 6) To get the resultant force on the -2 C charge, add the two force vectors you calculated in problems 4 and 5.<\/p>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

Solutions:<\/h3>\n\n\n\n
\n\n\n\n

1. Coulomb\u2019s Law Calculation<\/strong><\/h4>\n\n\n\n

Given:<\/p>\n\n\n\n

    \n
  • ( q_1 = 6 \\, \\mu C = 6 \\times 10^{-6} \\, C )<\/li>\n\n\n\n
  • ( q_2 = 4 \\, \\mu C = 4 \\times 10^{-6} \\, C )<\/li>\n\n\n\n
  • ( r = 25 \\, cm = 0.25 \\, m )<\/li>\n\n\n\n
  • ( k = 9 \\times 10^9 \\, N \\cdot m^2 \/ C^2 )<\/li>\n<\/ul>\n\n\n\n

    [
    F = \\frac{k \\cdot q_1 \\cdot q_2}{r^2}
    ]<\/p>\n\n\n\n

    [
    F = \\frac{(9 \\times 10^9) \\cdot (6 \\times 10^{-6}) \\cdot (4 \\times 10^{-6})}{(0.25)^2}
    ]<\/p>\n\n\n\n

    [
    F = \\frac{(9 \\times 10^9) \\cdot 24 \\times 10^{-12}}{0.0625}
    ]<\/p>\n\n\n\n

    [
    F = \\frac{216 \\times 10^{-3}}{0.0625} = 3.456 \\, N
    ]<\/p>\n\n\n\n

    \n\n\n\n

    2. Force Between Proton and Electron<\/strong><\/h4>\n\n\n\n

    Given:<\/p>\n\n\n\n

      \n
    • ( q_1 = q_2 = 1.6 \\times 10^{-19} \\, C )<\/li>\n\n\n\n
    • ( r = 5 \\times 10^{-10} \\, m )<\/li>\n\n\n\n
    • ( k = 9 \\times 10^9 \\, N \\cdot m^2 \/ C^2 )<\/li>\n<\/ul>\n\n\n\n

      [
      F = \\frac{k \\cdot q_1 \\cdot q_2}{r^2}
      ]<\/p>\n\n\n\n

      [
      F = \\frac{(9 \\times 10^9) \\cdot (1.6 \\times 10^{-19})^2}{(5 \\times 10^{-10})^2}
      ]<\/p>\n\n\n\n

      [
      F = \\frac{(9 \\times 10^9) \\cdot (2.56 \\times 10^{-38})}{25 \\times 10^{-20}}
      ]<\/p>\n\n\n\n

      [
      F = \\frac{2.304 \\times 10^{-28}}{25 \\times 10^{-20}} = 9.216 \\times 10^{-9} \\, N
      ]<\/p>\n\n\n\n

      \n\n\n\n

      3. Final Force When Distance Doubles<\/strong><\/h4>\n\n\n\n

      Coulomb\u2019s law states force is inversely proportional to ( r^2 ). If ( r ) doubles, the force becomes:<\/p>\n\n\n\n

      [
      F’ = \\frac{F}{2^2} = \\frac{F}{4}
      ]<\/p>\n\n\n\n

      Answer: ( (d) \\, F\/4 )<\/p>\n\n\n\n

      \n\n\n\n

      4. Force Between (-2 \\, C) and (3 \\, C)<\/strong><\/h4>\n\n\n\n

      Given:<\/p>\n\n\n\n

        \n
      • ( q_1 = -2 \\, C ), ( q_2 = 3 \\, C )<\/li>\n\n\n\n
      • ( r = 25 \\, cm = 0.25 \\, m )<\/li>\n\n\n\n
      • ( k = 9 \\times 10^9 \\, N \\cdot m^2 \/ C^2 )<\/li>\n<\/ul>\n\n\n\n

        [
        F = \\frac{k \\cdot q_1 \\cdot q_2}{r^2}
        ]<\/p>\n\n\n\n

        [
        F = \\frac{(9 \\times 10^9) \\cdot (-2) \\cdot (3)}{(0.25)^2}
        ]<\/p>\n\n\n\n

        [
        F = \\frac{-54 \\times 10^9}{0.0625} = -864 \\, N \\, (\\text{positive x-direction, } +\\hat{i})
        ]<\/p>\n\n\n\n

        \n\n\n\n

        5. Force Between (-2 \\, C) and (-8 \\, C)<\/strong><\/h4>\n\n\n\n

        Given:<\/p>\n\n\n\n

          \n
        • ( q_1 = -2 \\, C ), ( q_3 = -8 \\, C )<\/li>\n\n\n\n
        • ( r = 50 \\, cm = 0.5 \\, m )<\/li>\n<\/ul>\n\n\n\n

          [
          F = \\frac{k \\cdot q_1 \\cdot q_3}{r^2}
          ]<\/p>\n\n\n\n

          [
          F = \\frac{(9 \\times 10^9) \\cdot (-2) \\cdot (-8)}{(0.5)^2}
          ]<\/p>\n\n\n\n

          [
          F = \\frac{144 \\times 10^9}{0.25} = 576 \\, N \\, (\\text{negative x-direction, } -\\hat{i})
          ]<\/p>\n\n\n\n

          \n\n\n\n

          6. Resultant Force<\/strong><\/h4>\n\n\n\n

          Net force on (-2 \\, C) charge:<\/p>\n\n\n\n

          [
          F_{\\text{net}} = F_{14} + F_{15}
          ]<\/p>\n\n\n\n

          [
          F_{\\text{net}} = 864 \\, N \\, (+\\hat{i}) – 576 \\, N \\, (-\\hat{i}) = 288 \\, N \\, (+\\hat{i})
          ]<\/p>\n\n\n\n

          \n\n\n\n

          Explanation<\/strong><\/h3>\n\n\n\n

          Coulomb’s law calculates the electrostatic force between charged particles based on their charges, separation distance, and a constant ( k ). Problems 1 and 2 demonstrate simple applications, while problem 3 reinforces the inverse square law of distance on force. Problems 4\u20136 combine vector addition for charges in 1D. Understanding this principle is foundational for electromagnetism.<\/p>\n\n\n\n

          <\/p>\n","protected":false},"excerpt":{"rendered":"

          1) Coulomb’s law is F = (kq_{1}q_{2}) \/ (r ^ 2) where k = 9 * 10 ^ 9 in Sl units. Suppose that q * 1 = 6mu*C and q * 2 = 4mu*C and r = 25 cm, calc. the force between these particles. 2) The proton and electron in a H atom […]<\/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-182476","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182476","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=182476"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182476\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=182476"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=182476"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=182476"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}