{"id":1329,"date":"2018-03-03T17:28:38","date_gmt":"2018-03-03T09:28:38","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1329"},"modified":"2018-03-03T17:28:38","modified_gmt":"2018-03-03T09:28:38","slug":"the-magnitude-and-direction-of-electric-field-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/the-magnitude-and-direction-of-electric-field-problems-and-solutions.htm","title":{"rendered":"The magnitude and direction of electric field – problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

1. Calculate the magnitude and direction of the electric field<\/a> at a point A located at 5 cm from a point charge Q = <\/span><\/span>+10 <\/span><\/span><\/span>\u03bcC. <\/span><\/span><\/span>k = 9 x 10<\/span><\/span>9 <\/span><\/span><\/sup>Nm<\/span><\/span>2<\/span><\/span><\/sup>C<\/span><\/span>\u22122<\/span><\/span><\/sup>, 1 \u03bcC = 10<\/span><\/span>\u22126<\/span><\/span><\/sup> C)<\/span><\/span><\/p>\n

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

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

The distance<\/a> between point A and point charge Q (r<\/span><\/span>A<\/span><\/sub>) = 5 cm = 0.05 m = 5 x 10<\/span><\/span>-2<\/span><\/span><\/sup> m <\/span><\/span><\/p>\n

k = 9 x 10<\/span><\/span>9 <\/span><\/span><\/sup>Nm<\/span><\/span>2<\/span><\/span><\/sup>C<\/span><\/span>\u22122<\/span><\/span><\/sup><\/p>\n

Wanted:<\/u><\/span><\/span> The magnitude and direction of the electric field<\/a> at point <\/span><\/span>A<\/span><\/span><\/p>\n

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

\"The<\/p>\n

The direction of the electric field at point A :<\/u><\/span><\/span><\/p>\n

The electric charge is positive hence the direction of the electric field away from the electrical charge and points A.<\/span><\/span><\/span><\/p>\n\n

2. Calculate the magnitude and direction of the electric field at a point P located at 10 cm from a point charge Q = -2<\/span><\/span>0 <\/span><\/span><\/span>\u03bcC. <\/span><\/span><\/span>k = 9 x 10<\/span><\/span>9 <\/span><\/span><\/sup>Nm<\/span><\/span>2<\/span><\/span><\/sup>C<\/span><\/span>\u22122<\/span><\/span><\/sup>, 1 \u03bcC = 10<\/span><\/span>\u22126<\/span><\/span><\/sup> C.<\/span><\/span><\/p>\n

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

Electric charge (q) = <\/span><\/span>-20<\/span><\/span><\/span> \u03bcC = <\/span><\/span>-20<\/span><\/span><\/span> x 10<\/span><\/span>-6<\/span><\/span><\/sup> C<\/span><\/span><\/p>\n

The distance between point P and electric charge (r<\/span><\/span>P<\/span><\/sub>) = 10 cm = 0.1 m = 1 x 10<\/span><\/span>-1<\/span><\/span><\/sup> m <\/span><\/span><\/p>\n

k = 9 x 10<\/span><\/span>9 <\/span><\/span><\/sup>Nm<\/span><\/span>2<\/span><\/span><\/sup>C<\/span><\/span>\u22122<\/span><\/span><\/sup><\/p>\n

Wanted:<\/u><\/span><\/span> The magnitude and direction of the electric field at point <\/span><\/span>P<\/span><\/span><\/p>\n

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

\"The<\/p>\n

The direction of electric field at point A :<\/span><\/span><\/p>\n

The electric charge is negative hence the direction of the electric field to the electrical charge.<\/span><\/span><\/span><\/p>\n\n

3. Two point charges are separated by a distance of 40 cm. What is the magnitude and direction of the electric field at the point P between the two charges, that is 20 cm from point A?<\/span><\/span><\/p>\n

\"The<\/p>\n

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

Charge A (q<\/span><\/span>A<\/span><\/sub>) = <\/span><\/span>-2 <\/span><\/span><\/span>\u03bcC = <\/span><\/span>-2 <\/span><\/span><\/span>x 10<\/span><\/span>-6<\/span><\/span><\/sup> C<\/span><\/span><\/p>\n

Charge B (q<\/span><\/span>B<\/span><\/sub>) = +<\/span><\/span>4 <\/span><\/span><\/span>\u03bcC = <\/span><\/span>+4 <\/span><\/span><\/span>x 10<\/span><\/span>-6<\/span><\/span><\/sup> C<\/span><\/span><\/p>\n

The distance between charge A and point P (r<\/span><\/span>AP<\/span><\/sub>) = 20 cm = 0.2 m = 2 x 10<\/span><\/span>-1<\/span><\/span><\/sup> m<\/span><\/span><\/p>\n

The distance between charge B and point P (r<\/span><\/span>BP<\/span><\/sub>) = 20 cm = 0.2 m = 2 x 10<\/span><\/span>-1<\/span><\/span><\/sup> m<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> The magnitude and direction of electric field at point P.<\/span><\/span><\/p>\n

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

\"The<\/p>\n

Charge A is negative so that the direction of the electric field points toward Q<\/span><\/span><\/span>A<\/span><\/span><\/span><\/sub> (to the left).<\/span><\/span><\/span><\/p>\n

\"The<\/p>\n

Charge B is positive so that the direction of the electric field points away from Q<\/span><\/span><\/span>B<\/span><\/span><\/span><\/sub> (to the left).<\/span><\/span><\/span><\/p>\n

The total electric field at point A :<\/u><\/span><\/span><\/span><\/p>\n

E = E<\/span><\/span>A<\/span><\/sub> + E<\/span><\/span>B<\/span><\/sub><\/p>\n

E = (4.5 x 10<\/span><\/span>5<\/span><\/span><\/sup>) + (9 x 10<\/span><\/span>5<\/span><\/span><\/sup>)<\/span><\/span><\/p>\n

E = 13.5 x 10<\/span><\/span>5 <\/span><\/span><\/sup>N\/C<\/span><\/span><\/p>\n

The direction of the electric field points toward Q<\/span><\/span><\/span>A<\/span><\/span><\/sub> (to the left).<\/span><\/span><\/span><\/p>\n

<\/p>\n

4. The magnitude of the electric field is zero at…<\/span><\/span><\/p>\n

\"The<\/p>\n

The charge A is positive and the charge B is positive so that the magnitude of the electric field is zero located <\/span><\/span><\/span>at point P, <\/span><\/span><\/span>between both charges. <\/span><\/span><\/span><\/p>\n

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

Charge <\/span><\/span><\/span>A (q<\/span><\/span><\/span>A<\/span><\/span><\/sub>) = +<\/span><\/span><\/span>20 \u03bcC = +20 x 10<\/span><\/span><\/span>\u22126<\/span><\/span><\/sup><\/span> C<\/span><\/span><\/span><\/p>\n

Charge <\/span><\/span><\/span>B (q<\/span><\/span><\/span>B<\/span><\/span><\/sub>) = +40 <\/span><\/span><\/span>\u03bcC = +40 x 10<\/span><\/span><\/span>\u22126<\/span><\/span><\/sup><\/span> C<\/span><\/span><\/span><\/p>\n

k = 9 x 10<\/span><\/span><\/span>9<\/span><\/span><\/sup><\/span> Nm<\/span><\/span><\/span>2<\/span><\/span><\/sup><\/span>C<\/span><\/span><\/span>\u22122<\/span><\/span><\/sup><\/span><\/p>\n

The distance between charge <\/span><\/span><\/span>A <\/span><\/span><\/span>and the charge <\/span><\/span><\/span>B = 20 cm <\/span><\/span><\/span><\/p>\n

The charge between charge A and point P <\/span><\/span><\/span>(r<\/span><\/span><\/span>AP<\/span><\/span><\/sub>) = a<\/span><\/span><\/span><\/p>\n

The distance between charge B and point P <\/span><\/span><\/span>(r<\/span><\/span><\/span>BP<\/span><\/span><\/sub>) = 20 – a<\/span><\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span><\/span> The magnitude of the electric field is zero located at….<\/span><\/span><\/span><\/p>\n

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

The magnitude of the electric field produced by charge A at point P<\/u><\/span><\/span><\/span><\/p>\n

\"The<\/p>\n

Charge A is positive so that the direction of the electric field points away from charge A (to the right). <\/span><\/span><\/p>\n

The magnitude of the electric field produced by charge B at point P :<\/u><\/span><\/span><\/span><\/p>\n

\"The<\/p>\n

Charge B is positive so that the direction of the electric field points away from charge B (to the left). <\/span><\/span><\/p>\n

The total electric field at point P = 0 :<\/u><\/span><\/span><\/span><\/p>\n

\"The<\/p>\n

We use the quadratic formula to determine a. <\/span><\/span><\/span><\/p>\n

\"The<\/p>\n

The magnitude of the electric field is zero located at 8 cm from charge A or 12 cm from charge B. <\/span><\/span><\/p>\n\n

5. Based on the figure below, w<\/span><\/span><\/span>here is the point P so that the electric field at point P is zero?<\/span><\/span><\/span> (k = 9 x 10<\/span><\/span>9 <\/span><\/span><\/sup>Nm<\/span><\/span>2<\/span><\/span><\/sup>C<\/span><\/span>\u22122<\/span><\/span><\/sup>, 1 \u03bcC = 10<\/span><\/span>\u22126<\/span><\/span><\/sup> C)<\/span><\/span><\/p>\n

\"The<\/p>\n

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

To calculate the electric field strength at point P, assumed at point P there is a positive test charge. Q<\/span><\/span><\/span>1<\/span><\/span><\/sub> is positive and Q<\/span><\/span><\/span>2<\/span><\/span><\/sub> is negative, therefore point P must be on the right of Q<\/span><\/span><\/span>2 <\/span><\/span><\/sub>or left of Q<\/span><\/span><\/span>1<\/span><\/span><\/sub>. If point P is to the left of Q<\/span><\/span><\/span>1<\/span><\/span><\/sub>; the electric field generated by Q<\/span><\/span><\/span>1 <\/span><\/span><\/sub>at the point P is to the left (away from Q<\/span><\/span><\/span>1<\/span><\/span><\/sub>) and the electric field generated by Q<\/span><\/span><\/span>2 <\/span><\/span><\/sub>at the point P to the right (towards Q<\/span><\/span><\/span>1<\/span><\/span><\/sub>). The direction of the electric field is opposite so that the two electric fields eliminate each other so that the electric field strength at point P is zero.<\/span><\/span><\/span><\/p>\n

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

Q<\/span><\/span><\/span>1<\/span><\/span><\/sub> = +9<\/span><\/span><\/span> \u03bcC = +9 x 10<\/span><\/span>\u22126<\/span><\/span><\/sup> C <\/span><\/span><\/p>\n

Q<\/span><\/span><\/span>2<\/span><\/span><\/sub> = -4<\/span><\/span><\/span> \u03bcC = -4 x 10<\/span><\/span>\u22126<\/span><\/span><\/sup> C<\/span><\/span><\/p>\n

k = 9 x 10<\/span><\/span>9 <\/span><\/span><\/sup>Nm<\/span><\/span>2<\/span><\/span><\/sup>C<\/span><\/span>\u22122<\/span><\/span><\/sup><\/p>\n

Distance between charge 1 and charge 2 = 3 cm<\/span><\/span><\/p>\n

Distance between Q<\/span><\/span><\/span>1<\/span><\/span><\/sub> and point P (r<\/span><\/span><\/span>1P<\/span><\/span><\/sub>) = a<\/span><\/span><\/span><\/p>\n

Distance between Q<\/span><\/span><\/span>2<\/span><\/span><\/sub> and point P (r<\/span><\/span><\/span>2P<\/span><\/span><\/sub>) = 3 + a<\/span><\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span><\/span> location of point P so that <\/span><\/span><\/span>the electric field at point P is zero<\/span><\/span><\/span><\/span><\/p>\n

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

Point P is on the left of Q<\/span><\/span><\/span>1<\/span><\/span><\/sub>.<\/span><\/span><\/span><\/p>\n

The electric field produced by Q<\/span><\/span><\/span>1<\/span><\/span><\/sub> at point P:<\/span><\/span><\/span><\/p>\n

\"The<\/p>\n

The test charge is positive and Q<\/span><\/span>1<\/span><\/sub> is positive so that the direction of an electric field to leftward.<\/span><\/span><\/p>\n

The electric field produced by Q<\/span><\/span><\/span><\/span>2<\/span><\/span><\/span><\/sub> at point P:<\/span><\/span><\/span><\/span><\/p>\n

\"The<\/p>\n

The test charge is positive and Q<\/span><\/span>2<\/span><\/span><\/sub> is negative so that the direction of an electric field to rightward.<\/span><\/span><\/p>\n

Net electric field at point A :<\/u><\/span><\/span><\/span><\/p>\n

\"The<\/p>\n

Use the quadratic formula to determine a :<\/span><\/span><\/span><\/p>\n

a = -1.25, b = -13.5, c = -20.25<\/span><\/span><\/span><\/p>\n

\"The<\/p>\n

Distance between Q<\/span><\/span><\/span>2<\/span><\/span><\/sub> and point P (r<\/span><\/span><\/span>2P<\/span><\/span><\/sub>) = 3 + a = 3 \u2013 1.8 = 1.2 cm.<\/span><\/span><\/span><\/p>\n

Point P is on the 1.2 cm right of Q<\/span><\/span><\/span>1<\/span><\/span><\/sub>.<\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

1. Calculate the magnitude and direction of the electric field at a point A located at 5 cm from a point charge Q = +10 \u03bcC. k = 9 x 109 Nm2C\u22122, 1 \u03bcC = 10\u22126 C) Known : Electric charge (Q) = +10 \u03bcC = +10 x 10-6 C The distance between point A … Read more<\/a><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"open","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":"The magnitude and direction of electric field - 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-1329","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\/1329","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=1329"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1329\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=1329"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=1329"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=1329"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}