{"id":231690,"date":"2025-06-11T10:07:34","date_gmt":"2025-06-11T10:07:34","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=231690"},"modified":"2025-06-11T10:07:37","modified_gmt":"2025-06-11T10:07:37","slug":"consider-the-arrangement-of-two-point-c-charges-as-shown-in-the-figure","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/11\/consider-the-arrangement-of-two-point-c-charges-as-shown-in-the-figure\/","title":{"rendered":"Consider the arrangement of two point-C charges as shown in the figure"},"content":{"rendered":"\n<p>Consider the arrangement of two point-C charges as shown in the figure. Calculate the total electric field at point a. Calculate the total electric field at point b. Calculate the total electric potential at point a. Calculate the total electric potential at point b. Perpendicular bisector: -12 nC 12 nC 4.0 C 5.0 C 5.0 cm<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/image-367.png\" alt=\"\" class=\"wp-image-231691\"\/><\/figure>\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>Let&#8217;s analyze the problem step by step.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Given:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>q1=+12\u2009nCq_1 = +12\\, \\text{nC}q1\u200b=+12nC located 5.0 cm to the left of point a.<\/li>\n\n\n\n<li>q2=\u221212\u2009nCq_2 = -12\\, \\text{nC}q2\u200b=\u221212nC located 5.0 cm to the right of point a.<\/li>\n\n\n\n<li>Distance from point a to both charges = 5.0 cm = 0.050 m.<\/li>\n\n\n\n<li>Distance from point b to q1q_1q1\u200b: 4.0\u2009cm=0.040\u2009m4.0\\, \\text{cm} = 0.040\\, \\text{m}4.0cm=0.040m; from point b to q2q_2q2\u200b: 9.0\u2009cm=0.090\u2009m9.0\\, \\text{cm} = 0.090\\, \\text{m}9.0cm=0.090m.<\/li>\n\n\n\n<li>k=14\u03c0\u03b50=8.99\u00d7109\u2009Nm2\/C2k = \\frac{1}{4\\pi \\varepsilon_0} = 8.99 \\times 10^9 \\, \\text{Nm}^2\/\\text{C}^2k=4\u03c0\u03b50\u200b1\u200b=8.99\u00d7109Nm2\/C2<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>(a) Total Electric Field at Point a<\/strong><\/h2>\n\n\n\n<p>Electric field due to a point charge:E=k\u2223q\u2223r2E = \\frac{k |q|}{r^2}E=r2k\u2223q\u2223\u200b<\/p>\n\n\n\n<p>At point a, the electric fields due to both charges are equal in magnitude and opposite in direction.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>E1=8.99\u00d7109\u00d712\u00d710\u22129(0.05)2=43176\u2009N\/CE_1 = \\frac{8.99 \\times 10^9 \\times 12 \\times 10^{-9}}{(0.05)^2} = 43176 \\, \\text{N\/C}E1\u200b=(0.05)28.99\u00d7109\u00d712\u00d710\u22129\u200b=43176N\/C (to the right due to +q)<\/li>\n\n\n\n<li>E2=8.99\u00d7109\u00d712\u00d710\u22129(0.05)2=43176\u2009N\/CE_2 = \\frac{8.99 \\times 10^9 \\times 12 \\times 10^{-9}}{(0.05)^2} = 43176 \\, \\text{N\/C}E2\u200b=(0.05)28.99\u00d7109\u00d712\u00d710\u22129\u200b=43176N\/C (to the left due to -q)<\/li>\n<\/ul>\n\n\n\n<p>Net field:Enet&nbsp;at&nbsp;a=E1\u2212E2=0\u2009N\/CE_{\\text{net at a}} = E_1 &#8211; E_2 = 0 \\, \\text{N\/C}Enet&nbsp;at&nbsp;a\u200b=E1\u200b\u2212E2\u200b=0N\/C<\/p>\n\n\n\n<p>\u2705 <strong>Answer (a):<\/strong> 0\u2009N\/C0 \\, \\text{N\/C}0N\/C<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>(b) Total Electric Field at Point b<\/strong><\/h2>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Distance to q1q_1q1\u200b: r=0.04\u2009mr = 0.04\\, \\text{m}r=0.04m<\/li>\n\n\n\n<li>Distance to q2q_2q2\u200b: r=0.09\u2009mr = 0.09\\, \\text{m}r=0.09m<\/li>\n<\/ul>\n\n\n\n<p>E1=k\u22c512\u00d710\u22129(0.04)2=6.74\u00d7104\u2009N\/C&nbsp;(to&nbsp;the&nbsp;left)E_1 = \\frac{k \\cdot 12 \\times 10^{-9}}{(0.04)^2} = 6.74 \\times 10^4 \\, \\text{N\/C} \\text{ (to the left)}E1\u200b=(0.04)2k\u22c512\u00d710\u22129\u200b=6.74\u00d7104N\/C&nbsp;(to&nbsp;the&nbsp;left)E2=k\u22c512\u00d710\u22129(0.09)2=1.33\u00d7104\u2009N\/C&nbsp;(to&nbsp;the&nbsp;left;&nbsp;since&nbsp;q\u2082&nbsp;is&nbsp;negative)E_2 = \\frac{k \\cdot 12 \\times 10^{-9}}{(0.09)^2} = 1.33 \\times 10^4 \\, \\text{N\/C} \\text{ (to the left; since q\u2082 is negative)}E2\u200b=(0.09)2k\u22c512\u00d710\u22129\u200b=1.33\u00d7104N\/C&nbsp;(to&nbsp;the&nbsp;left;&nbsp;since&nbsp;q\u2082&nbsp;is&nbsp;negative)<\/p>\n\n\n\n<p>Both fields point <strong>left<\/strong> at b, so they add:Enet&nbsp;at&nbsp;b=6.74\u00d7104+1.33\u00d7104=8.07\u00d7104\u2009N\/CE_{\\text{net at b}} = 6.74 \\times 10^4 + 1.33 \\times 10^4 = 8.07 \\times 10^4 \\, \\text{N\/C}Enet&nbsp;at&nbsp;b\u200b=6.74\u00d7104+1.33\u00d7104=8.07\u00d7104N\/C<\/p>\n\n\n\n<p>\u2705 <strong>Answer (b):<\/strong> 8.07\u00d7104\u2009N\/C,&nbsp;to&nbsp;the&nbsp;left8.07 \\times 10^4 \\, \\text{N\/C}, \\text{ to the left}8.07\u00d7104N\/C,&nbsp;to&nbsp;the&nbsp;left<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>(c) Total Electric Potential at Point a<\/strong><\/h2>\n\n\n\n<p>Electric potential is scalar:V=kqrV = \\frac{kq}{r}V=rkq\u200bV1=8.99\u00d7109\u22c512\u00d710\u221290.05=2.1576\u00d7104\u2009VV_1 = \\frac{8.99 \\times 10^9 \\cdot 12 \\times 10^{-9}}{0.05} = 2.1576 \\times 10^4 \\, \\text{V}V1\u200b=0.058.99\u00d7109\u22c512\u00d710\u22129\u200b=2.1576\u00d7104VV2=8.99\u00d7109\u22c5(\u221212\u00d710\u22129)0.05=\u22122.1576\u00d7104\u2009VV_2 = \\frac{8.99 \\times 10^9 \\cdot (-12 \\times 10^{-9})}{0.05} = -2.1576 \\times 10^4 \\, \\text{V}V2\u200b=0.058.99\u00d7109\u22c5(\u221212\u00d710\u22129)\u200b=\u22122.1576\u00d7104VVnet&nbsp;at&nbsp;a=V1+V2=0\u2009VV_{\\text{net at a}} = V_1 + V_2 = 0 \\, \\text{V}Vnet&nbsp;at&nbsp;a\u200b=V1\u200b+V2\u200b=0V<\/p>\n\n\n\n<p>\u2705 <strong>Answer (c):<\/strong> 0\u2009V0 \\, \\text{V}0V<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>(d) Total Electric Potential at Point b<\/strong><\/h2>\n\n\n\n<p>V1=8.99\u00d7109\u22c512\u00d710\u221290.04=2.697\u00d7104\u2009VV_1 = \\frac{8.99 \\times 10^9 \\cdot 12 \\times 10^{-9}}{0.04} = 2.697 \\times 10^4 \\, \\text{V}V1\u200b=0.048.99\u00d7109\u22c512\u00d710\u22129\u200b=2.697\u00d7104VV2=8.99\u00d7109\u22c5(\u221212\u00d710\u22129)0.09=\u22121.1987\u00d7104\u2009VV_2 = \\frac{8.99 \\times 10^9 \\cdot (-12 \\times 10^{-9})}{0.09} = -1.1987 \\times 10^4 \\, \\text{V}V2\u200b=0.098.99\u00d7109\u22c5(\u221212\u00d710\u22129)\u200b=\u22121.1987\u00d7104VVnet&nbsp;at&nbsp;b=2.697\u00d7104\u22121.1987\u00d7104=1.4983\u00d7104\u2009VV_{\\text{net at b}} = 2.697 \\times 10^4 &#8211; 1.1987 \\times 10^4 = 1.4983 \\times 10^4 \\, \\text{V}Vnet&nbsp;at&nbsp;b\u200b=2.697\u00d7104\u22121.1987\u00d7104=1.4983\u00d7104V<\/p>\n\n\n\n<p>\u2705 <strong>Answer (d):<\/strong> 1.50\u00d7104\u2009V1.50 \\times 10^4 \\, \\text{V}1.50\u00d7104V<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\ud83d\udcda <strong>Explanation <\/strong><\/h3>\n\n\n\n<p>In electrostatics, the electric field and potential are fundamental concepts describing the influence of charges on their surroundings. The electric field is a vector quantity, meaning direction matters, while electric potential is scalar, meaning only magnitude is considered.<\/p>\n\n\n\n<p>At point <strong>a<\/strong>, which lies equidistant between two equal and opposite charges, the electric fields due to each charge are equal in magnitude but opposite in direction. The positive charge pushes the field away, and the negative pulls toward it \u2014 both pointing in opposite directions. Therefore, the net electric field cancels out to <strong>zero<\/strong> at that midpoint. Likewise, since electric potential is a scalar, and both charges have the same magnitude but opposite sign at equal distances, their potentials cancel out, yielding <strong>zero net potential<\/strong>.<\/p>\n\n\n\n<p>At point <strong>b<\/strong>, closer to the positive charge and farther from the negative charge, both electric fields from the charges point in the <strong>same direction<\/strong>\u2014toward the negative and away from the positive\u2014resulting in a <strong>strong net field to the left<\/strong>. The potentials from each charge do not cancel here because the distances are unequal. The closer positive charge contributes more, but the negative charge still subtracts some potential. This leads to a <strong>positive net potential<\/strong> at point <strong>b<\/strong>.<\/p>\n\n\n\n<p>Understanding these principles is key in fields like electronics, physics, and engineering, where managing electric forces and potentials governs device behavior.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner4-948.jpeg\" alt=\"\" class=\"wp-image-231692\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Consider the arrangement of two point-C charges as shown in the figure. Calculate the total electric field at point a. Calculate the total electric field at point b. Calculate the total electric potential at point a. Calculate the total electric potential at point b. Perpendicular bisector: -12 nC 12 nC 4.0 C 5.0 C 5.0 [&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-231690","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231690","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=231690"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231690\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=231690"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=231690"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=231690"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}