{"id":215583,"date":"2025-05-17T07:43:26","date_gmt":"2025-05-17T07:43:26","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=215583"},"modified":"2025-05-17T07:43:28","modified_gmt":"2025-05-17T07:43:28","slug":"what-is-q-q-if-the-net-electrostatic-force-on-particle-1-is-zero","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/17\/what-is-q-q-if-the-net-electrostatic-force-on-particle-1-is-zero\/","title":{"rendered":"What is q\/Q if the net electrostatic force on particle 1 is zero"},"content":{"rendered":"\n<p>In the figure four particles form a square. The charges are q1 = +Q, q2 = q3 = q, and q4 = -21.0Q. What is q\/Q if the net electrostatic force on particle 1 is zero?<\/p>\n\n\n\n<p>Number<\/p>\n\n\n\n<p>Units<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To find the ratio qQ\\frac{q}{Q} such that the <strong>net electrostatic force on particle 1 is zero<\/strong>, we must analyze the forces acting on it due to the other three charges arranged in a square.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step-by-step Setup<\/strong><\/h3>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Four charges form a square of side length aa.<\/li>\n\n\n\n<li>q1=+Qq_1 = +Q (Top-left corner)<\/li>\n\n\n\n<li>q2=qq_2 = q (Top-right corner)<\/li>\n\n\n\n<li>q3=qq_3 = q (Bottom-left corner)<\/li>\n\n\n\n<li>q4=\u221221Qq_4 = -21Q (Bottom-right corner)<\/li>\n<\/ul>\n\n\n\n<p>We need to find qQ\\frac{q}{Q} such that the net force on charge q1q_1 is zero.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 1: Understand the Geometry<\/strong><\/h3>\n\n\n\n<p>Label the square:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Let\u2019s place the square in the x-y plane with side length aa.<\/li>\n\n\n\n<li>q1q_1 is at (0, a)<\/li>\n\n\n\n<li>q2q_2 is at (a, a)<\/li>\n\n\n\n<li>q3q_3 is at (0, 0)<\/li>\n\n\n\n<li>q4q_4 is at (a, 0)<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 2: Calculate Forces on q1q_1<\/strong><\/h3>\n\n\n\n<p><strong>Force due to q2=qq_2 = q:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Distance: aa<\/li>\n\n\n\n<li>Direction: Left (\u2212x)<\/li>\n\n\n\n<li>Magnitude: F12=kQqa2F_{12} = k \\frac{Qq}{a^2}<\/li>\n<\/ul>\n\n\n\n<p><strong>Force due to q3=qq_3 = q:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Distance: aa<\/li>\n\n\n\n<li>Direction: Down (\u2212y)<\/li>\n\n\n\n<li>Magnitude: F13=kQqa2F_{13} = k \\frac{Qq}{a^2}<\/li>\n<\/ul>\n\n\n\n<p><strong>Force due to q4=\u221221Qq_4 = -21Q:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Distance: 2a\\sqrt{2}a<\/li>\n\n\n\n<li>Direction: Along the diagonal (from (a,0) to (0,a))<\/li>\n\n\n\n<li>The force is attractive, pointing toward q4q_4<\/li>\n\n\n\n<li>Components of the force:\n<ul class=\"wp-block-list\">\n<li>F14=kQ\u22c521Q(2a2)=21kQ22a2F_{14} = k \\frac{Q \\cdot 21Q}{(2a^2)} = \\frac{21kQ^2}{2a^2}<\/li>\n\n\n\n<li>Break into x and y: both components are equal since the diagonal makes a 45\u00b0 angle.<\/li>\n\n\n\n<li>So: F14x=F14y=21kQ22a2\u22c5cos\u2061(45\u2218)=21kQ22a2\u22c512=21kQ222a2F_{14x} = F_{14y} = \\frac{21kQ^2}{2a^2} \\cdot \\cos(45^\\circ) = \\frac{21kQ^2}{2a^2} \\cdot \\frac{1}{\\sqrt{2}} = \\frac{21kQ^2}{2\\sqrt{2}a^2}<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 3: Set Net Force to Zero<\/strong><\/h3>\n\n\n\n<p>We want net force on q1q_1 to be zero in both x and y directions.<\/p>\n\n\n\n<p><strong>X-direction:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Leftward force from q2q_2: F12x=\u2212kQqa2F_{12x} = -k\\frac{Qq}{a^2}<\/li>\n\n\n\n<li>Rightward component from q4q_4: F14x=21kQ222a2F_{14x} = \\frac{21kQ^2}{2\\sqrt{2}a^2}<\/li>\n<\/ul>\n\n\n\n<p>\u2212kQqa2+21kQ222a2=0\u21d2\u2212Qq+21Q222=0\u21d2q=21Q22-k\\frac{Qq}{a^2} + \\frac{21kQ^2}{2\\sqrt{2}a^2} = 0 \\Rightarrow -Qq + \\frac{21Q^2}{2\\sqrt{2}} = 0 \\Rightarrow q = \\frac{21Q}{2\\sqrt{2}}<\/p>\n\n\n\n<p><strong>Y-direction:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Downward force from q3q_3: F13y=\u2212kQqa2F_{13y} = -k\\frac{Qq}{a^2}<\/li>\n\n\n\n<li>Upward component from q4q_4: F14y=21kQ222a2F_{14y} = \\frac{21kQ^2}{2\\sqrt{2}a^2}<br>Same equation as x-direction, confirming consistency.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Final Answer:<\/strong><\/h3>\n\n\n\n<p>qQ=2122\u22487.42(unitless)\\frac{q}{Q} = \\frac{21}{2\\sqrt{2}} \\approx \\boxed{7.42} \\quad \\text{(unitless)}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation (300 words):<\/strong><\/h3>\n\n\n\n<p>This problem involves balancing vector forces from Coulomb\u2019s Law to determine when the net force on a charge is zero. Each particle interacts with every other, and because the configuration is a square, symmetry helps simplify the math. Charge q1=+Qq_1 = +Q experiences repulsive forces from other positive charges and attractive forces from the negative one. The aim is to set the <strong>vector sum<\/strong> of the electrostatic forces acting on q1q_1 to zero, considering both x and y components.<\/p>\n\n\n\n<p>The force from each adjacent charge q2q_2 and q3q_3 is directly horizontal or vertical due to the square arrangement. Their forces are straightforward: equal in magnitude (since they are same charge qq and at equal distance aa), directed along the axes. However, the force from q4=\u221221Qq_4 = -21Q is along the diagonal and must be broken into x and y components using trigonometry (cos(45\u00b0) or sin(45\u00b0) = 1\/\u221a2).<\/p>\n\n\n\n<p>By equating the sum of x-components and y-components of the forces to zero, we derive two identical equations involving qq and QQ. Solving for q\/Qq\/Q gives the ratio that ensures balance. This means that the magnitude of force from the very strong negative charge q4q_4 balances the combined repulsive forces of the smaller positive charges q2q_2 and q3q_3.<\/p>\n\n\n\n<p>Hence, the ratio qQ=2122\u22487.42\\frac{q}{Q} = \\frac{21}{2\\sqrt{2}} \\approx 7.42 ensures that the vector sum of all electrostatic forces on charge 1 is zero, meaning it is in equilibrium.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>In the figure four particles form a square. The charges are q1 = +Q, q2 = q3 = q, and q4 = -21.0Q. What is q\/Q if the net electrostatic force on particle 1 is zero? Number Units The Correct Answer and Explanation is: To find the ratio qQ\\frac{q}{Q} such that the net electrostatic force [&hellip;]<\/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-215583","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/215583","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=215583"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/215583\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=215583"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=215583"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=215583"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}