{"id":2751,"date":"2018-05-16T13:56:32","date_gmt":"2018-05-16T05:56:32","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=2751"},"modified":"2023-08-06T14:53:10","modified_gmt":"2023-08-06T14:53:10","slug":"electric-potential-at-the-conductor-of-ball-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/electric-potential-at-the-conductor-of-ball-problems-and-solutions.htm","title":{"rendered":"Electric potential at the conductor of ball \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Electric potential at the conductor of ball \u2013 problems and solutions<\/span><\/p>\n

1. A 4-\u03bcC hollow ball conductor has radius of 8-cm. Determine the electric potential<\/a> at the surface of the ball. (k = 9.109<\/sup> N.m2<\/sup>.C-2<\/sup>)<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

The electric charge<\/a> (Q) = 4 \u03bcC = 4 x 10-6<\/sup> C \"Electric<\/span><\/p>\n

The radius of ball (r) = 8 cm = 8 x 10-2<\/sup> m<\/span><\/p>\n

Coulomb’s constant (k) = 9.109<\/sup> N.m2<\/sup>.C-2<\/sup><\/span><\/p>\n

Wanted :<\/u> The electric potential at the surface of the ball (V)<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

V = k Q \/ r<\/span><\/p>\n

V = (9 x 109<\/sup>)(4 x 10-6<\/sup>) \/ (8 x 10-2<\/sup>)<\/span><\/p>\n

V = (36 x 103<\/sup>) \/ (8 x 10-2<\/sup>)<\/span><\/p>\n

V = (36\/8) x 103<\/sup> x 102<\/sup><\/span><\/p>\n

V = 4.5 x 105 <\/sup>Volt<\/span><\/p>\n

2. A spherical conductor <\/span>has radius of <\/span>3-cm <\/span>(1 \u03bcC = 10-6 <\/sup>C and k = 9.109<\/sup> N.m2<\/sup>.C-2<\/sup>). If the conductor is positively charged +1 \u03bcC then the electric potential at point A is …<\/span><\/p>\n

Solution :<\/span><\/p>\n

The electric potential inside the spherical conductor = The electric potential at the surface of the spherical conductor.<\/i><\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

The electric charge (Q) = 1 \u03bcC = 1 x 10-6<\/sup> C \"Electric<\/span><\/p>\n

The radius of the spherical conductor <\/span>(r) = 3 cm = 3 x 10-2<\/sup> m<\/span><\/p>\n

Coulomb’s constant (k) = 9.109<\/sup> N.m2<\/sup>.C-2<\/sup><\/span><\/p>\n

Wanted :<\/u> The electric potential at point A (V)<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

V = k Q \/ r<\/span><\/p>\n

V = (9 x 109<\/sup>)(1 x 10-6<\/sup>) \/ (3 x 10-2<\/sup>)<\/span><\/p>\n

V = (9 x 103<\/sup>) \/ (3 x 10-2<\/sup>)<\/span><\/p>\n

V = (9\/3) x 103<\/sup> x 102<\/sup><\/span><\/p>\n

V = 3 x 105 <\/sup>Volt<\/span><\/p>\n

3. A hollow metal ball <\/span>with <\/span>radius <\/span>of <\/span>9-cm <\/span>has 6.4 x 10<\/span>-9<\/span><\/sup> Coulomb electric charge, as shown in figure below. Distance<\/a> between point O and point <\/span>P = 4 cm; Distance between point P and point Q = 5 cm; Distance between point Q and point R = 18 cm and k = 9.109<\/sup> N.m2<\/sup>.C-2<\/sup>. Determine the electric potential at point P.<\/span><\/p>\n

Solution<\/span><\/p>\n

The electric potential inside the spherical conductor = The electric potential at the surface of the spherical conductor.<\/i><\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

The electric charge (Q) = 6.4 x 10-9<\/sup> C\"Electric<\/span><\/p>\n

The radius of the spherical conductor<\/i><\/span> (r) = OP + PQ = 4 cm + 5 cm = 9 cm = 9 x 10-2 <\/sup>m <\/span><\/p>\n

Coulomb’s constant (k) = 9.109<\/sup> N.m2<\/sup>.C-2<\/sup><\/span><\/p>\n

Wanted :<\/u> The electric potential at point P (V)<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

V = k Q \/ r<\/span><\/p>\n

V = (9 x 109<\/sup>)(6,4 x 10-9<\/sup>) \/ (9 x 10-2<\/sup>)<\/span><\/p>\n

V = 109 <\/sup>(6,4 x 10-9<\/sup>) \/ 10-2<\/sup><\/span><\/p>\n

V = 6.4 \/ 10-2<\/sup><\/span><\/p>\n

V = 6.4 x 102<\/sup><\/span><\/p>\n

V = 6.4 x 100<\/span><\/p>\n

V = 640 Volt<\/span><\/p>\n

1. Question:<\/strong> What is electric potential?<\/span><\/p>\n

Answer:<\/strong> Electric potential is the electric potential energy per unit charge at a specific point in space.<\/span><\/p>\n

2. Question:<\/strong> How is the electric potential on the surface of a conducting ball defined?<\/span><\/p>\n

Answer:<\/strong> The electric potential on the surface of a conducting ball is uniform and is given by the formula V = kQ\/R, where Q is the charge on the ball and R is its radius.<\/span><\/p>\n

3. Question:<\/strong> Why is the electric potential constant everywhere on the surface of a charged conducting ball?<\/span><\/p>\n

Answer:<\/strong> Conductors in electrostatic equilibrium have an electric field that’s perpendicular to the surface. Hence, no work is done in moving a charge on the surface, keeping the potential constant.<\/span><\/p>\n

4. Question:<\/strong> How does the potential vary inside a conducting ball?<\/span><\/p>\n

Answer:<\/strong> Inside a uniformly charged conducting ball, the electric potential varies, but in the case of a hollow conducting ball, the potential remains constant inside and equals the potential on its surface.<\/span><\/p>\n

5. Question:<\/strong> How does the size of the conducting ball affect its surface potential, given a constant charge?<\/span><\/p>\n

Answer:<\/strong> A larger ball (greater R) would have a smaller surface potential for a given charge Q, based on the formula V = kQ\/R.<\/span><\/p>\n

6. Question:<\/strong> How is electric potential related to electric field?<\/span><\/p>\n

Answer:<\/strong> Electric potential and electric field are related by the equation E = -dV\/dr, where dV is the change in potential and dr is the change in position.<\/span><\/p>\n

7. Question:<\/strong> Why can’t we have an electric field inside a conducting ball in electrostatic equilibrium?<\/span><\/p>\n

Answer:<\/strong> Electrons in a conductor move until they cancel out any external electric fields. Once equilibrium is reached, the electric field inside the conductor becomes zero.<\/span><\/p>\n

8. Question:<\/strong> What would happen to the electric potential of a conducting ball if the amount of charge on it were doubled?<\/span><\/p>\n

Answer:<\/strong> Doubling the charge Q would double the electric potential V, given the relation V = kQ\/R.<\/span><\/p>\n

9. Question:<\/strong> What is the significance of the term ‘k’ in the formula for electric potential?<\/span><\/p>\n

Answer:<\/strong> ‘k’ is Coulomb’s constant, which is approximately 8.99 x 10\u2079 N.m\u00b2\/C\u00b2 in vacuum.<\/span><\/p>\n

10. Question:<\/strong> How is electric potential energy different from electric potential?<\/span><\/p>\n

Answer:<\/strong> Electric potential energy is the energy a charge has due to its position in an electric field, while electric potential is the electric potential energy per unit charge.<\/span><\/p>\n

11. Question:<\/strong> Does a charged conducting ball influence the electric potential of nearby objects?<\/span><\/p>\n

Answer:<\/strong> Yes, any charged object creates an electric field in its surroundings, which affects the electric potential of nearby objects.<\/span><\/p>\n

12. Question:<\/strong> How is the electric potential at a point near a charged conducting ball calculated?<\/span><\/p>\n

Answer:<\/strong> It’s calculated using the formula V = kQ\/r, where r is the distance from the center of the ball to the point.<\/span><\/p>\n

13. Question:<\/strong> If two identical conducting balls, one charged and one uncharged, are brought into contact and then separated, what happens to the potential of each?<\/span><\/p>\n

Answer:<\/strong> The charge will distribute equally between the two balls, making their potentials identical.<\/span><\/p>\n

14. Question:<\/strong> What is the role of the grounding process in affecting the potential of a conducting ball?<\/span><\/p>\n

Answer:<\/strong> Grounding a conductor allows it to exchange charge with the Earth until its potential becomes zero.<\/span><\/p>\n

15. Question:<\/strong> Why is the electric potential zero inside a grounded conducting ball?<\/span><\/p>\n

Answer:<\/strong> When grounded, a conductor exchanges charges with the Earth until the electric field inside it is nullified, making the potential zero.<\/span><\/p>\n

16. Question:<\/strong> Does the shape of a conductor affect its surface potential?<\/span><\/p>\n

Answer:<\/strong> For a given charge, the distribution may vary with shape, but in electrostatic equilibrium, the potential remains constant across the surface of any conductor.<\/span><\/p>\n

17. Question:<\/strong> How does the distribution of charge vary on a non-spherical conductor?<\/span><\/p>\n

Answer:<\/strong> On a non-spherical conductor, charge tends to accumulate more at sharp points or edges due to the concentration of electric field lines.<\/span><\/p>\n

18. Question:<\/strong> Why can birds sit on high voltage power lines without getting electrocuted?<\/span><\/p>\n

Answer:<\/strong> Birds don’t get electrocuted because they are not completing a circuit. The electric potential is constant all over their body as they touch only one wire, so there’s no potential difference.<\/span><\/p>\n

19. Question:<\/strong> How does the presence of a dielectric material near a charged conducting ball affect its electric potential?<\/span><\/p>\n

Answer:<\/strong> A dielectric material affects the electric field near the conductor, which can influence the potential. Typically, dielectrics reduce the effective electric field.<\/span><\/p>\n

20. Question:<\/strong> What is the electric potential at a point infinitely far from a charged conducting ball?<\/span><\/p>\n

Answer:<\/strong> The electric potential approaches zero as one moves infinitely far from the charged source.<\/span><\/p>\n

Understanding the nuances of electric potential, especially with conductors like balls, provides insights into many phenomena in electrostatics and real-world applications.<\/span><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

Electric potential at the conductor of ball \u2013 problems and solutions 1. A 4-\u03bcC hollow ball conductor has radius of 8-cm. Determine the electric potential at the surface of the ball. (k = 9.109 N.m2.C-2) Known : The electric charge (Q) = 4 \u03bcC = 4 x 10-6 C The radius of ball (r) = … Read more<\/a><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_seopress_titles_title":"","_seopress_titles_desc":"","_seopress_robots_index":"","_seopress_robots_follow":"","_seopress_robots_imageindex":"","_seopress_robots_snippet":"","_seopress_robots_primary_cat":"","_seopress_robots_breadcrumbs":"","_seopress_robots_freeze_modified_date":"","_seopress_robots_custom_modified_date":"","_seopress_robots_canonical":"","_seopress_social_fb_title":"","_seopress_social_fb_desc":"","_seopress_social_fb_img":"","_seopress_social_fb_img_attachment_id":0,"_seopress_social_fb_img_width":0,"_seopress_social_fb_img_height":0,"_seopress_social_twitter_title":"","_seopress_social_twitter_desc":"","_seopress_social_twitter_img":"","_seopress_social_twitter_img_attachment_id":0,"_seopress_social_twitter_img_width":0,"_seopress_social_twitter_img_height":0,"_seopress_redirections_value":"","_seopress_redirections_enabled":"","_seopress_redirections_enabled_regex":"","_seopress_redirections_logged_status":"","_seopress_redirections_param":"","_seopress_redirections_type":0,"_seopress_analysis_target_kw":"Electric potential at the conductor of ball \u2013 problems and solutions","_seopress_news_disabled":"","_seopress_video_disabled":"","_seopress_video":[],"_seopress_pro_schemas_manual":[],"_seopress_pro_rich_snippets_disable_all":"","_seopress_pro_rich_snippets_disable":[],"_seopress_pro_schemas":[],"footnotes":""},"categories":[3],"tags":[],"class_list":["post-2751","post","type-post","status-publish","format-standard","hentry","category-solved-problems-in-basic-physics"],"gt_translate_keys":[{"key":"link","format":"url"}],"_links":{"self":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/2751","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/comments?post=2751"}],"version-history":[{"count":2,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/2751\/revisions"}],"predecessor-version":[{"id":8550,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/2751\/revisions\/8550"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=2751"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=2751"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=2751"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}