{"id":231686,"date":"2025-06-11T10:01:32","date_gmt":"2025-06-11T10:01:32","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=231686"},"modified":"2025-06-11T10:01:34","modified_gmt":"2025-06-11T10:01:34","slug":"consider-the-arrangement-of-three-point-charges-in-a-right-triangle-shown-in-the-figure-which-have-charges-q1-7-5-i%c2%bcc-q2-68-i%c2%bcc-and-q3-35-i%c2%bcc","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/11\/consider-the-arrangement-of-three-point-charges-in-a-right-triangle-shown-in-the-figure-which-have-charges-q1-7-5-i%c2%bcc-q2-68-i%c2%bcc-and-q3-35-i%c2%bcc\/","title":{"rendered":"Consider the arrangement of three point charges in a right triangle shown in the figure, which have charges q1 = 7.5 \u00ce\u00bcC, q2 = -68 \u00ce\u00bcC, and q3 = 35 \u00ce\u00bcC"},"content":{"rendered":"\n

Consider the arrangement of three point charges in a right triangle shown in the figure, which have charges q1 = 7.5 \u00ce\u00bcC, q2 = -68 \u00ce\u00bcC, and q3 = 35 \u00ce\u00bcC. The distance between q1 and q2 is 35 cm and the distance between q2 and q3 is 75 cm. Randomized Variables q1 = 7.5 \u00ce\u00bcC, q2 = -68 \u00ce\u00bcC, q3 = 35 \u00ce\u00bcC, a = 35 cm, b = 75 cm. (a) How much potential energy, in joules, is stored in this configuration of charges? U = -38.782 J (Correct!) (b) Now assume that q1 and q2 are fixed in space at the locations indicated, and q3 is brought into its position from infinity. What is the change in potential energy of the system, in joules, during this process? For b, I got -45.492 J, but it is not correct.<\/p>\n\n\n\n

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

Step-by-step solution:<\/h3>\n\n\n\n

The change in potential energy<\/strong> when a charge q3q_3q3\u200b is brought from infinity to a location in the electric field of other fixed charges is equal to the electrostatic potential energy<\/strong> due to its interaction with the existing charges q1q_1q1\u200b and q2q_2q2\u200b: \u0394U=Ufinal\u2212Uinitial=Uinteraction with q1 and q2\\Delta U = U_{\\text{final}} – U_{\\text{initial}} = U_{\\text{interaction with } q_1 \\text{ and } q_2}\u0394U=Ufinal\u200b\u2212Uinitial\u200b=Uinteraction with q1\u200b and q2\u200b\u200b<\/p>\n\n\n\n

Since the potential energy at infinity is zero, the change in potential energy is simply the sum of the pairwise potential energies between q3q_3q3\u200b and the fixed charges: \u0394U=ke(q1q3r13+q2q3r23)\\Delta U = k_e \\left( \\frac{q_1 q_3}{r_{13}} + \\frac{q_2 q_3}{r_{23}} \\right)\u0394U=ke\u200b(r13\u200bq1\u200bq3\u200b\u200b+r23\u200bq2\u200bq3\u200b\u200b)<\/p>\n\n\n\n

Where:<\/p>\n\n\n\n

    \n
  • ke=8.988\u00d7109\u2009N\\cdotpm2\/C2k_e = 8.988 \\times 10^9 \\, \\text{N\u00b7m}^2\/\\text{C}^2ke\u200b=8.988\u00d7109N\\cdotpm2\/C2<\/li>\n\n\n\n
  • q1=7.5\u00d710\u22126\u2009Cq_1 = 7.5 \\times 10^{-6} \\, \\text{C}q1\u200b=7.5\u00d710\u22126C<\/li>\n\n\n\n
  • q2=\u221268\u00d710\u22126\u2009Cq_2 = -68 \\times 10^{-6} \\, \\text{C}q2\u200b=\u221268\u00d710\u22126C<\/li>\n\n\n\n
  • q3=35\u00d710\u22126\u2009Cq_3 = 35 \\times 10^{-6} \\, \\text{C}q3\u200b=35\u00d710\u22126C<\/li>\n\n\n\n
  • r13r_{13}r13\u200b is the distance between q1q_1q1\u200b and q3q_3q3\u200b \u2014 the hypotenuse of the triangle<\/li>\n\n\n\n
  • r23=0.75\u2009mr_{23} = 0.75 \\, \\text{m}r23\u200b=0.75m<\/li>\n\n\n\n
  • r12=0.35\u2009mr_{12} = 0.35 \\, \\text{m}r12\u200b=0.35m<\/li>\n\n\n\n
  • Use Pythagoras to get r13=0.352+0.752=0.1225+0.5625=0.685\u22480.8277\u2009mr_{13} = \\sqrt{0.35^2 + 0.75^2} = \\sqrt{0.1225 + 0.5625} = \\sqrt{0.685} \\approx 0.8277 \\, \\text{m}r13\u200b=0.352+0.752\u200b=0.1225+0.5625\u200b=0.685\u200b\u22480.8277m<\/li>\n<\/ul>\n\n\n\n

    Now calculate the energy: \u0394U=(8.988\u00d7109)((7.5\u00d710\u22126)(35\u00d710\u22126)0.8277+(\u221268\u00d710\u22126)(35\u00d710\u22126)0.75)\\Delta U = (8.988 \\times 10^9) \\left( \\frac{(7.5 \\times 10^{-6})(35 \\times 10^{-6})}{0.8277} + \\frac{(-68 \\times 10^{-6})(35 \\times 10^{-6})}{0.75} \\right)\u0394U=(8.988\u00d7109)(0.8277(7.5\u00d710\u22126)(35\u00d710\u22126)\u200b+0.75(\u221268\u00d710\u22126)(35\u00d710\u22126)\u200b)<\/p>\n\n\n\n

    Break it into two terms:<\/p>\n\n\n\n

      \n
    1. (7.5)(35)0.8277=262.50.8277\u2248317.15\\frac{(7.5)(35)}{0.8277} = \\frac{262.5}{0.8277} \\approx 317.150.8277(7.5)(35)\u200b=0.8277262.5\u200b\u2248317.15<\/li>\n\n\n\n
    2. (\u221268)(35)0.75=\u221223800.75=\u22123173.33\\frac{(-68)(35)}{0.75} = \\frac{-2380}{0.75} = -3173.330.75(\u221268)(35)\u200b=0.75\u22122380\u200b=\u22123173.33<\/li>\n<\/ol>\n\n\n\n

      Now multiply by 8.988\u00d710\u221238.988 \\times 10^{-3}8.988\u00d710\u22123 (since \u03bcC\u00d7\u03bcC=10\u221212\\mu C \\times \\mu C = 10^{-12}\u03bcC\u00d7\u03bcC=10\u221212, and we factor 10\u2212310^{-3}10\u22123 into the result): \u0394U=8.988\u00d710\u22123\u00d7(317.15\u22123173.33)=8.988\u00d710\u22123\u00d7(\u22122856.18)\u2248\u221225.66\u2009J\\Delta U = 8.988 \\times 10^{-3} \\times (317.15 – 3173.33) = 8.988 \\times 10^{-3} \\times (-2856.18) \\approx -25.66 \\, \\text{J}\u0394U=8.988\u00d710\u22123\u00d7(317.15\u22123173.33)=8.988\u00d710\u22123\u00d7(\u22122856.18)\u2248\u221225.66J<\/p>\n\n\n\n

      \n\n\n\n

      Final Answer:<\/h3>\n\n\n\n

      (b) The change in potential energy is:<\/strong> \u221225.66\u2009J\\boxed{-25.66 \\, \\text{J}}\u221225.66J\u200b<\/p>\n\n\n\n

      \n\n\n\n

      Book-style explanation<\/h3>\n\n\n\n

      When dealing with the potential energy of a system of point charges, it’s essential to understand that energy is stored in the configuration due to electrostatic interactions. For a system of three point charges, the total potential energy is the sum of the potential energy of each unique pair of charges. In this case, charges q1q_1q1\u200b and q2q_2q2\u200b are fixed, and the third charge q3q_3q3\u200b is brought in from infinity. Since electric potential energy is defined as the work required to assemble the configuration from infinitely separated charges, we only consider the interactions involving the incoming charge, q3q_3q3\u200b, because the potential energy at infinity is taken as zero.<\/p>\n\n\n\n

      The energy associated with bringing q3q_3q3\u200b into position is given by the sum of its interactions with both q1q_1q1\u200b and q2q_2q2\u200b, using Coulomb’s law for potential energy: U=ke(q1q3r13+q2q3r23)U = k_e \\left( \\frac{q_1 q_3}{r_{13}} + \\frac{q_2 q_3}{r_{23}} \\right)U=ke\u200b(r13\u200bq1\u200bq3\u200b\u200b+r23\u200bq2\u200bq3\u200b\u200b)<\/p>\n\n\n\n

      Here, r13r_{13}r13\u200b is calculated using the Pythagorean theorem since q1,q2,q_1, q_2,q1\u200b,q2\u200b, and q3q_3q3\u200b form a right triangle. The sign of each charge plays a vital role: if charges have opposite signs, the potential energy is negative (indicating an attractive interaction), while like charges contribute positive potential energy (repulsion).<\/p>\n\n\n\n

      After plugging in the known values and evaluating, the total change in potential energy for this process is found to be \u201325.66 J<\/strong>. This negative value indicates that energy is released when q3q_3q3\u200b is brought into position, meaning the system becomes more stable due to attractive forces dominating over repulsive ones.<\/p>\n\n\n\n

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

      Consider the arrangement of three point charges in a right triangle shown in the figure, which have charges q1 = 7.5 \u00ce\u00bcC, q2 = -68 \u00ce\u00bcC, and q3 = 35 \u00ce\u00bcC. The distance between q1 and q2 is 35 cm and the distance between q2 and q3 is 75 cm. Randomized Variables q1 = 7.5 […]<\/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-231686","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231686","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=231686"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231686\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=231686"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=231686"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=231686"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}