{"id":259747,"date":"2025-07-19T05:20:25","date_gmt":"2025-07-19T05:20:25","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=259747"},"modified":"2025-07-19T07:09:33","modified_gmt":"2025-07-19T07:09:33","slug":"state-whether-the-triangles-could-be-proven-congruent-if-possible-by-sss-or-sas-include-a-congruency-statement-for-all-congruent-triangles-2-3-5","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/19\/state-whether-the-triangles-could-be-proven-congruent-if-possible-by-sss-or-sas-include-a-congruency-statement-for-all-congruent-triangles-2-3-5\/","title":{"rendered":"A conducting sphere of 3.0 cm radius is charged with – 40 nC."},"content":{"rendered":"\n

A conducting sphere of 3.0 cm radius is charged with – 40 nC. In what follows, the electric potential is taken to be zero at infinity. a) Calculate the electric potential of the sphere. (2 points) b) Calculate the electric potential at 6.0 cm from the center of the sphere. (2 points)<\/a><\/p>\n\n\n\n

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

Solution<\/strong><\/h3>\n\n\n\n

Given:
Radius r1=3.0\u2009cm=0.03\u2009mr_1 = 3.0 \\, \\text{cm} = 0.03 \\, \\text{m}r1\u200b=3.0cm=0.03m
Charge Q=\u221240\u2009nC=\u221240\u00d710\u22129\u2009CQ = -40 \\, \\text{nC} = -40 \\times 10^{-9} \\, \\text{C}Q=\u221240nC=\u221240\u00d710\u22129C
k=14\u03c0\u03b50=9.0\u00d7109\u2009N\\cdotpm2\/C2k = \\dfrac{1}{4 \\pi \\varepsilon_0} = 9.0 \\times 10^9 \\, \\text{N\u00b7m}^2\/\\text{C}^2k=4\u03c0\u03b50\u200b1\u200b=9.0\u00d7109N\\cdotpm2\/C2<\/p>\n\n\n\n

\n\n\n\n

(a) Electric potential on the sphere surface<\/strong><\/h4>\n\n\n\n

For a conducting sphere, the potential at its surface is: V=kQrV = \\dfrac{k Q}{r}V=rkQ\u200b<\/p>\n\n\n\n

Substitute: V=(9.0\u00d7109)(\u221240\u00d710\u22129)0.03V = \\dfrac{(9.0 \\times 10^9)(-40 \\times 10^{-9})}{0.03}V=0.03(9.0\u00d7109)(\u221240\u00d710\u22129)\u200b V=\u22123600.03=\u221212,000\u2009VV = \\dfrac{-360}{0.03} = -12{,}000 \\, \\text{V}V=0.03\u2212360\u200b=\u221212,000V V=\u22121.2\u00d7104\u2009V\\boxed{ V = -1.2 \\times 10^4 \\, \\text{V} }V=\u22121.2\u00d7104V\u200b<\/p>\n\n\n\n

\n\n\n\n

(b) Electric potential at 6.0 cm from center<\/strong><\/h4>\n\n\n\n

At distance r2=6.0\u2009cm=0.06\u2009mr_2 = 6.0 \\, \\text{cm} = 0.06 \\, \\text{m}r2\u200b=6.0cm=0.06m, the sphere behaves like a point charge: V=kQr2V = \\dfrac{k Q}{r_2}V=r2\u200bkQ\u200b V=(9.0\u00d7109)(\u221240\u00d710\u22129)0.06V = \\dfrac{(9.0 \\times 10^9)(-40 \\times 10^{-9})}{0.06}V=0.06(9.0\u00d7109)(\u221240\u00d710\u22129)\u200b V=\u22123600.06=\u22126,000\u2009VV = \\dfrac{-360}{0.06} = -6{,}000 \\, \\text{V}V=0.06\u2212360\u200b=\u22126,000V V=\u22126.0\u00d7103\u2009V\\boxed{ V = -6.0 \\times 10^3 \\, \\text{V} }V=\u22126.0\u00d7103V\u200b<\/p>\n\n\n\n

\n\n\n\n

Explanation (\u2248300 words)<\/strong><\/h3>\n\n\n\n

The electric potential due to a charged conducting sphere depends on the distribution of charge and the distance from its center. For a conducting sphere, all the charge resides on its outer surface because like charges repel and move as far apart as possible. This makes the sphere behave as if all its charge were concentrated at the center when calculating potential outside the sphere.<\/p>\n\n\n\n

The formula for electric potential at a distance rrr from a point charge is: V=kQrV = \\dfrac{k Q}{r}V=rkQ\u200b<\/p>\n\n\n\n

where k=9.0\u00d7109\u2009N\\cdotpm2\/C2k = 9.0 \\times 10^9 \\, \\text{N\u00b7m}^2\/\\text{C}^2k=9.0\u00d7109N\\cdotpm2\/C2, QQQ is the charge, and rrr is the distance from the center.<\/p>\n\n\n\n

Inside the conducting sphere, the potential is constant and equal to the potential at the surface. Therefore, to calculate the potential at the surface, we substitute the sphere\u2019s radius into the formula. With a radius of 0.03 m and a charge of -40 nC, the potential at the surface is: V=\u22121.2\u00d7104\u2009VV = -1.2 \\times 10^4 \\, \\text{V}V=\u22121.2\u00d7104V<\/p>\n\n\n\n

The negative sign indicates that the potential is lower than zero since the charge is negative.<\/p>\n\n\n\n

For the point outside the sphere at 6.0 cm, the same formula applies because the sphere acts like a point charge beyond its radius. Using r=0.06\u2009mr = 0.06 \\, \\text{m}r=0.06m, we find the potential to be: V=\u22126.0\u00d7103\u2009VV = -6.0 \\times 10^3 \\, \\text{V}V=\u22126.0\u00d7103V<\/p>\n\n\n\n

This shows that the potential decreases as the distance increases, following an inverse relationship with distance. The concept is crucial in electrostatics because it explains how charged objects influence their surroundings. These calculations illustrate how potential varies with distance and why conductors maintain uniform potential on their surface.<\/p>\n","protected":false},"excerpt":{"rendered":"

A conducting sphere of 3.0 cm radius is charged with – 40 nC. In what follows, the electric potential is taken to be zero at infinity. a) Calculate the electric potential of the sphere. (2 points) b) Calculate the electric potential at 6.0 cm from the center of the sphere. (2 points) The Correct Answer […]<\/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-259747","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/259747","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=259747"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/259747\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=259747"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=259747"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=259747"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}