{"id":1949,"date":"2018-04-19T06:19:06","date_gmt":"2018-04-18T22:19:06","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1949"},"modified":"2023-08-09T05:03:47","modified_gmt":"2023-08-09T05:03:47","slug":"electric-field-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/electric-field-problems-and-solutions.htm","title":{"rendered":"Electric field \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Electric field \u2013 problems and solutions<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><span style=\"color: #000000;\">1. <\/span><span style=\"color: #000000;\">Point A located at the center between two charges. Both charges have the same magnitude but opposite sign and separated by a distance of a. The magnitude of the electric field at point A is 36 N\/C. If point A moved 1\/2a close to one of both <a href=\"https:\/\/gurumuda.net\/physics\/types-of-electric-charges-problems-and-solutions.htm\" target=\"_blank\" rel=\"noopener\">charges<\/a>, what is the magnitude of the <a href=\"https:\/\/gurumuda.net\/physics\/electric-field-problems-and-solutions.htm\" target=\"_blank\" rel=\"noopener\">electric field<\/a> at point A?<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"color: #000000; font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Known :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><span style=\"color: #000000;\">Charge <\/span><span style=\"color: #000000;\">1 (q<\/span><sub><span style=\"color: #000000;\">1<\/span><\/sub><span style=\"color: #000000;\">) = +Q<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><span style=\"color: #000000;\">Charge <\/span><span style=\"color: #000000;\">2 (q<\/span><sub><span style=\"color: #000000;\">2<\/span><\/sub><span style=\"color: #000000;\">) = -Q<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><span style=\"color: #000000;\">The distance between charge 1 and point A <\/span><span style=\"color: #000000;\">(r<\/span><sub><span style=\"color: #000000;\">1A<\/span><\/sub><span style=\"color: #000000;\">) = \u00bd a<\/span><!--more--><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><span style=\"color: #000000;\">The distance between charge 2 and point A <\/span><span style=\"color: #000000;\">(r<\/span><sub><span style=\"color: #000000;\">2A<\/span><\/sub><span style=\"color: #000000;\">) = \u00bd a<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><span style=\"color: #000000;\">The magnitude of the electric field at point A <\/span><span style=\"color: #000000;\">(E<\/span><sub><span style=\"color: #000000;\">A<\/span><\/sub><span style=\"color: #000000;\">) = 36 NC<\/span><span style=\"color: #000000;\"><sup>-1<\/sup><\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><span style=\"color: #000000;\"><u>Wanted:<\/u><\/span><span style=\"color: #000000;\"> The magnitude of the electric field <\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"color: #000000; font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Solution :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><span style=\"color: #000000;\"><b>Step <\/b><\/span><span style=\"color: #000000;\"><b>1<\/b><\/span><span style=\"color: #000000;\">.<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><span style=\"color: #000000;\"><u>The electric field produced by a charge <\/u><\/span><span style=\"color: #000000;\"><u>+Q <\/u><\/span><span style=\"color: #000000;\"><u>at point <\/u><\/span><span style=\"color: #000000;\"><u>A<\/u><\/span><span style=\"color: #000000;\"> :<\/span><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-1975\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/04\/Electric-field-\u2013-problems-and-solutions-1-300x63.png\" alt=\"Electric field \u2013 problems and solutions 1\" width=\"300\" height=\"63\" srcset=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-1-300x63.png 300w, https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-1.png 310w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Test charge is positive and charges 1 is positive so that the direction of the electric field points to charge 2.<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><span style=\"color: #000000;\"><u>The electric charge produced by a charge <\/u><\/span><span style=\"color: #000000;\"><u>-Q <\/u><\/span><span style=\"color: #000000;\"><u>at point <\/u><\/span><span style=\"color: #000000;\"><u>A<\/u><\/span><span style=\"color: #000000;\"> :<\/span><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-1950\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/04\/Electric-field-\u2013-problems-and-solutions-2.png\" alt=\"Electric field \u2013 problems and solutions 2\" width=\"286\" height=\"66\" \/><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Test charge is positive and charges 2 is negative so that the direction of the electric field points to charge 2.<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"color: #000000; font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>The resultant of the electric field at point A :<\/u><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-1951\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/04\/Electric-field-\u2013-problems-and-solutions-3-300x161.png\" alt=\"Electric field \u2013 problems and solutions 3\" width=\"300\" height=\"161\" srcset=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-3-300x161.png 300w, https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-3.png 307w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"color: #000000; font-size: 12pt; font-family: 'times new roman', times, serif;\"><b>Step 2.<\/b><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"color: #000000; font-size: 12pt; font-family: 'times new roman', times, serif;\">If point A is moved close to charge 1 then :<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><span style=\"color: #000000;\">The distance between charge 1 and point <\/span><span style=\"color: #000000;\">A (r<\/span><span style=\"color: #000000;\"><sub>1A<\/sub><\/span><span style=\"color: #000000;\">) = \u00bc a<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><span style=\"color: #000000;\">The distance between charge 2 and point A<\/span><span style=\"color: #000000;\"> (r<\/span><sub><span style=\"color: #000000;\">2A<\/span><\/sub><span style=\"color: #000000;\">) = \u00be a<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><span style=\"color: #000000;\"><u>The electric field produced by charge <\/u><\/span><span style=\"color: #000000;\"><u>+Q <\/u><\/span><span style=\"color: #000000;\"><u>at point <\/u><\/span><span style=\"color: #000000;\"><u>A<\/u><\/span><span style=\"color: #000000;\"> :<\/span><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-1952\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/04\/Electric-field-\u2013-problems-and-solutions-4-300x59.png\" alt=\"Electric field \u2013a problems and solutions 4\" width=\"300\" height=\"59\" srcset=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-4-300x59.png 300w, https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-4.png 333w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Test charge is positive and charges 1 is positive so that the direction of the electric field points to charge 2.<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><span style=\"color: #000000;\"><u>The electric field produced by charge <\/u><\/span><span style=\"color: #000000;\"><u>-Q <\/u><\/span><span style=\"color: #000000;\"><u>at point A <\/u><\/span><span style=\"color: #000000;\">:<\/span><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-1953\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/04\/Electric-field-\u2013-problems-and-solutions-5-300x61.png\" alt=\"Electric field \u2013 problems and solutions 5\" width=\"300\" height=\"61\" srcset=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-5-300x61.png 300w, https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-5.png 320w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Test charge is positive and charges 2 is negative so that the direction of the electric field points to charge 2.<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"color: #000000; font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>The resultant of the electric field at point A :<\/u><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-1954\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/04\/Electric-field-\u2013-problems-and-solutions-6-300x51.png\" alt=\"Electric field \u2013 problems and solutions 6\" width=\"300\" height=\"51\" srcset=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-6-300x51.png 300w, https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-6.png 489w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">2. Two charges q<sub>A<\/sub> = 1 \u03bcC and q<sub>B<\/sub> = 4 \u03bcC are separated by a distance of 4 cm (k = 9 x 10<sup>9<\/sup> N m<sup>2<\/sup> C<sup>\u22122<\/sup>). What is the magnitude of the electric field at the center between q<sub>A<\/sub> and q<sub>B.<\/sub><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"color: #000000; font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Known :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Charge A (q<sub>A<\/sub>) = 1 \u03bcC = 1 x <span style=\"color: #000000;\">10<\/span><span style=\"color: #000000;\"><sup>\u22126<\/sup><\/span><span style=\"color: #000000;\"> C<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Charge B (q<sub>B<\/sub>) = 4 \u03bcC = 4 x <span style=\"color: #000000;\">10<\/span><span style=\"color: #000000;\"><sup>\u22126<\/sup><\/span><span style=\"color: #000000;\"> C<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">k = 9 x 10<sup>9<\/sup> N m<sup>2<\/sup> C<sup>\u22122<\/sup><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Distance between charge A and B (r<sub>AB<\/sub>) = 4 cm = 0.04 meters<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><span style=\"color: #000000;\">Distance between charge A and the center point <\/span><span style=\"color: #000000;\">(r<\/span><sub><span style=\"color: #000000;\">A<\/span><\/sub><span style=\"color: #000000;\">) = 0.02 meter<\/span><span style=\"color: #000000;\">s<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><span style=\"color: #000000;\">Distance between charge B and the center point <\/span><span style=\"color: #000000;\">(r<\/span><sub><span style=\"color: #000000;\">B<\/span><\/sub><span style=\"color: #000000;\">) = 0.02 meter<\/span><span style=\"color: #000000;\">s<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><span style=\"color: #000000;\"><u>Known:<\/u><\/span> <span style=\"color: #000000;\">The magnitude of the electric field<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"color: #000000; font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Solution :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"color: #000000; font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>The electric field produced by charge A at the center point :<\/u><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-1955\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/04\/Electric-field-\u2013-problems-and-solutions-7-300x30.png\" alt=\"Electric field \u2013 problems and solutions 7\" width=\"300\" height=\"30\" srcset=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-7-300x30.png 300w, https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-7.png 501w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Test charge is positive and charges A is positive so that the direction of the electric field points to charge B.<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"color: #000000; font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>The electric field produced by charge B at the center point :<\/u><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-1956\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/04\/Electric-field-\u2013-problems-and-solutions-8-300x32.png\" alt=\"Electric field \u2013 problems and solutions 8\" width=\"300\" height=\"32\" srcset=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-8-300x32.png 300w, https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-8.png 450w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Test charge is positive and charge B is positive so that the direction of the electric field points to charge A.<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"color: #000000; font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>The resultant of the electric field at the center point :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E<sub>A<\/sub> and E<sub>B<\/sub> have the opposite direction.<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E = E<sub>B <\/sub>\u2013 E<sub>A<\/sub> = 9 x 10<sup>7 \u2013<\/sup> 2.25 x 10<sup>7 <\/sup>= 6.75 x 10<sup>7 <\/sup><span style=\"color: #000000;\">NC<\/span><span style=\"color: #000000;\"><sup>-1<\/sup><\/span> <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><span style=\"color: #000000;\">3<\/span><span style=\"color: #000000;\">. <\/span><span style=\"color: #000000;\">According to figure below, where the point P is located so that the magnitude of the electric field at point P = 0 ? <\/span>(k = 9 x 10<sup>9 <\/sup>Nm<sup>2<\/sup>C<sup>\u22122<\/sup>, 1 \u03bcC = 10<sup>\u22126<\/sup> C)<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-1957\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/04\/Electric-field-\u2013-problems-and-solutions-9.png\" alt=\"Electric field \u2013 problems and solutions 9\" width=\"198\" height=\"56\" \/><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"color: #000000; font-size: 12pt; font-family: 'times new roman', times, serif;\">Solution<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><span style=\"color: #000000;\"><u>If point P located at the left of <\/u><\/span><span style=\"color: #000000;\"><u>Q<\/u><\/span><sub><span style=\"color: #000000;\"><u>1<\/u><\/span><\/sub><span style=\"color: #000000;\">; <\/span><span style=\"color: #000000;\">the electric field produced by <\/span><span style=\"color: #000000;\">Q<\/span><sub><span style=\"color: #000000;\">1<\/span><\/sub> <span style=\"color: #000000;\">on point P points to leftward <\/span><span style=\"color: #000000;\">(<\/span><span style=\"color: #000000;\">away from <\/span><span style=\"color: #000000;\">Q<\/span><sub><span style=\"color: #000000;\">1<\/span><\/sub><span style=\"color: #000000;\">) <\/span><span style=\"color: #000000;\">and the electric field produced by <\/span><span style=\"color: #000000;\">Q<\/span><sub><span style=\"color: #000000;\">2<\/span><\/sub> <span style=\"color: #000000;\">on point P points to rightward (point to <\/span><span style=\"color: #000000;\">Q<\/span><sub><span style=\"color: #000000;\">1<\/span><\/sub><span style=\"color: #000000;\">). <\/span><span style=\"color: #000000;\">The direction of the electric field is opposite so that the electric field at point P = 0.<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"color: #000000; font-size: 12pt; font-family: 'times new roman', times, serif;\">Known :<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><span style=\"color: #000000;\">Q<\/span><sub><span style=\"color: #000000;\">1<\/span><\/sub><span style=\"color: #000000;\"> = +9<\/span> \u03bcC = +9 x 10<sup>\u22126<\/sup> C <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><span style=\"color: #000000;\">Q<\/span><sub><span style=\"color: #000000;\">2<\/span><\/sub><span style=\"color: #000000;\"> = -4<\/span> \u03bcC = -4 x 10<sup>\u22126<\/sup> C<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">k = 9 x 10<sup>9 <\/sup>Nm<sup>2<\/sup>C<sup>\u22122<\/sup><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Distance between charge 1 and charge 2 = 3 cm<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><span style=\"color: #000000;\">Distance between <\/span><span style=\"color: #000000;\">Q<\/span><sub><span style=\"color: #000000;\">1<\/span><\/sub> <span style=\"color: #000000;\">and point P <\/span><span style=\"color: #000000;\">(r<\/span><sub><span style=\"color: #000000;\">1P<\/span><\/sub><span style=\"color: #000000;\">) = a<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><span style=\"color: #000000;\">Distance between<\/span><span style=\"color: #000000;\"> Q<\/span><sub><span style=\"color: #000000;\">2<\/span><\/sub> <span style=\"color: #000000;\">and point <\/span><span style=\"color: #000000;\">P (r<\/span><sub><span style=\"color: #000000;\">2P<\/span><\/sub><span style=\"color: #000000;\">) = 3 + a<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><span style=\"color: #000000;\"><u>Wanted :<\/u><\/span><span style=\"color: #000000;\"> Position of point P<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"color: #000000; font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Solution :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><span style=\"color: #000000;\">Point P located at leftward of <\/span><span style=\"color: #000000;\">Q<\/span><sub><span style=\"color: #000000;\">1<\/span><\/sub><span style=\"color: #000000;\">.<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u><span style=\"color: #000000;\">The electric field produced by <\/span><span style=\"color: #000000;\">Q<\/span><sub><span style=\"color: #000000;\">1<\/span><\/sub> <span style=\"color: #000000;\">at point <\/span><span style=\"color: #000000;\">P :<\/span><\/u><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-1958\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/04\/Electric-field-\u2013-problems-and-solutions-10-300x48.png\" alt=\"Electric field \u2013 problems and solutions 10\" width=\"300\" height=\"48\" \/><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Test charge is positive and Q<sub>1<\/sub> is positive so that the direction of the electric field to leftward.<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><span style=\"color: #000000;\"><u>The electric field produced by <\/u><\/span><span style=\"color: #000000;\"><u>Q<\/u><\/span><sub><span style=\"color: #000000;\"><u>2<\/u><\/span><u> <\/u><\/sub><span style=\"color: #000000;\"><u>at point <\/u><\/span><span style=\"color: #000000;\"><u>P<\/u><\/span><span style=\"color: #000000;\"> :<\/span><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-1959\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/04\/Electric-field-\u2013-problems-and-solutions-11.png\" alt=\"Electric field \u2013 problems and solutions 11\" width=\"279\" height=\"46\" \/><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Test charge is positive and Q<sub>2<\/sub> is negative so that the direction of the electric field to rightward.<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"color: #000000; font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Resultant of the electric field at point A :<\/u><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-1960\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/04\/Electric-field-\u2013-problems-and-solutions-12.png\" alt=\"Electric field \u2013 problems and solutions 12\" width=\"230\" height=\"213\" \/><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"color: #000000; font-size: 12pt; font-family: 'times new roman', times, serif;\">Use quadratic formula to find a :<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-1961\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/04\/Electric-field-\u2013-problems-and-solutions-13-300x243.png\" alt=\"Electric field \u2013 problems and solutions 12\" width=\"300\" height=\"243\" srcset=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-13-300x243.png 300w, https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-13.png 324w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><span style=\"color: #000000;\">Distance between <\/span><span style=\"color: #000000;\">Q<\/span><sub><span style=\"color: #000000;\">2<\/span><\/sub> <span style=\"color: #000000;\">and point <\/span><span style=\"color: #000000;\">P (r<\/span><span style=\"color: #000000;\"><sub>2P<\/sub><\/span><span style=\"color: #000000;\">) = 3 + a = 3 \u2013 1.8 = 1.2 cm <\/span><span style=\"color: #000000;\">or 3 + a = 3 \u2013 9 = -6 cm.<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><span style=\"color: #000000;\">Distance between <\/span><span style=\"color: #000000;\">Q<\/span><sub><span style=\"color: #000000;\">1<\/span> <\/sub><span style=\"color: #000000;\">and point <\/span><span style=\"color: #000000;\">P (r<\/span><span style=\"color: #000000;\"><sub>1P<\/sub><\/span><span style=\"color: #000000;\">) = a = -9 cm or -1.8 cm.<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><span style=\"color: #000000;\">Point P located at <\/span><span style=\"color: #000000;\">1.2 cm <\/span><span style=\"color: #000000;\">rightward of <\/span><span style=\"color: #000000;\">Q<\/span><span style=\"color: #000000;\"><sub>2<\/sub><\/span><span style=\"color: #000000;\">.<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">4. Charge q<sub>3<\/sub> located at 5 cm rightward of q<sub>2<\/sub>, as shown in the figure below. What is the magnitude of the electric field at charge q<sub>3<\/sub> (1 \u00b5C = 10<sup>-6<\/sup> C).<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-1962\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/04\/Electric-field-\u2013-problems-and-solutions-14.png\" alt=\"Electric field \u2013 problems and solutions 14\" width=\"257\" height=\"62\" \/><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Solution :<\/u><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-1963\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/04\/Electric-field-\u2013-problems-and-solutions-15-300x54.png\" alt=\"Electric field \u2013 problems and solutions 15\" width=\"300\" height=\"54\" srcset=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-15-300x54.png 300w, https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-15.png 304w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Charge q<sub>3<\/sub> is positive so that the direction of the electric field at charge q<sub>3<\/sub> points to the minus charge q<sub>2<\/sub> (E<sub>2<\/sub>) and away from the plus charge q<sub>1 <\/sub>(E<sub>1<\/sub>). The resultant of the electric field is the sum of the electric field E<sub>1<\/sub> and E<sub>2<\/sub>. <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Known :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Charge q<sub>1<\/sub> = 5 \u00b5C = 5 x 10<sup>-6<\/sup> Coulomb<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Charge q<sub>2<\/sub> = 5 \u00b5C = -5 x 10<sup>-6<\/sup> Coulomb<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Distance between charge q<sub>1<\/sub> and charge q<sub>3<\/sub> (r<sub>1<\/sub>) = 15 cm = 0.15 m = 15 x 10<sup>-2<\/sup> meters<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Distance between charge q<sub>2<\/sub> and charge q<sub>3<\/sub> (r<sub>2<\/sub>) = 5 cm = 0.05 m = 5 x 10<sup>-2<\/sup> meters<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">k = 9 x 10<sup>9<\/sup> N m<sup>2 <\/sup>C<sup>-2<\/sup><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Wanted :<\/u> The electric field at charge q<sub>3<\/sub><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Solution :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>The electric field 1 :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E<sub>1<\/sub> = k q<sub>1<\/sub> \/ r<sub>1<\/sub><sup>2<\/sup> <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E<sub>1<\/sub> = (9 x 10<sup>9<\/sup>)(5 x 10<sup>-6<\/sup>) \/ (15 x 10<sup>-2<\/sup>)<sup>2<\/sup> <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E<sub>1<\/sub> = (45 x 10<sup>3<\/sup>) \/ (225 x 10<sup>-4<\/sup>) <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E<sub>1<\/sub> = 0.2 x 10<sup>7 <\/sup>N\/C <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>The electric field 2 :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E<sub>2<\/sub> = k q<sub>2<\/sub> \/ r<sub>2<\/sub><sup>2<\/sup> <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E<sub>2<\/sub> = (9 x 10<sup>9<\/sup>)(5 x 10<sup>-6<\/sup>) \/ (5 x 10<sup>-2<\/sup>)<sup>2<\/sup> <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E<sub>2 <\/sub>= (45 x 10<sup>3<\/sup>) \/ (25 x 10<sup>-4<\/sup>) <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E<sub>2<\/sub> = 1.8 x 10<sup>7 <\/sup>N\/C <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Resultant of the electric field :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">The resultant of the electric field at charge q<sub>3<\/sub> :<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E = E<sub>2<\/sub> \u2013 E<sub>1<\/sub> = (1.8 x 10<sup>7<\/sup>) \u2013 (0.2 x 10<sup>7<\/sup>) = 1.6 x 10<sup>7 <\/sup>N\/C <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">The direction of the electric field points to leftward (same direction as E<sub>2<\/sub>). <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">5. Two charges are separated as shown in figure below. What is the electric field at point P (k = 9 x 10<sup>9<\/sup> N m<sup>2 <\/sup>C<sup>-2<\/sup>)<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Solution<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-1964\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/04\/Electric-field-\u2013-problems-and-solutions-16-300x89.png\" alt=\"Electric field \u2013 problems and solutions 16\" width=\"300\" height=\"89\" srcset=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-16-300x89.png 300w, https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-16.png 366w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Known :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Charge q<sub>A<\/sub> = +2.5 C <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Charge q<sub>B<\/sub> = -2 C <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Distance between charge q<sub>A<\/sub> and point P (r<sub>A<\/sub>) = 5 m <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Distance between charge q<sub>B<\/sub> and point P (r<sub>B<\/sub>) = 2 m<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">k = 9 x 10<sup>9<\/sup> N m<sup>2 <\/sup>C<sup>-2<\/sup><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Wanted :<\/u> the magnitude of the electric field at point P.<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Solution :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>The electric field A :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E<sub>A<\/sub> = k q<sub>A<\/sub> \/ r<sub>A<\/sub><sup>2<\/sup> <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E<sub>A<\/sub> = (9 x 10<sup>9<\/sup>)(2.5) \/ (5)<sup>2<\/sup> <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E<sub>A<\/sub> = (22.5 x 10<sup>9<\/sup>) \/ 25 <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E<sub>A<\/sub> = 0.9 x 10<sup>9 <\/sup>N\/C <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>The electric field B :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E<sub>B<\/sub> = k q<sub>B<\/sub> \/ r<sub>B<\/sub><sup>2<\/sup> <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E<sub>B<\/sub> = (9 x 10<sup>9<\/sup>)(2) \/ (2)<sup>2<\/sup> <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E<sub>B<\/sub> = (18 x 10<sup>9<\/sup>) \/ 4 <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E<sub>B <\/sub>= 4.5 x 10<sup>9 <\/sup>N\/C <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Resultant of the electric field :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Resultant of the electric field at point P :<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E = E<sub>B<\/sub> &#8211; EA = (4.5 \u2013 0.9) x 10<sup>9 <\/sup>= 3.6 x 10<sup>9<\/sup> N\/C <\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">The direction to leftward (same direction as E<sub>B<\/sub>).<\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">6. Two charges Q<sub>1<\/sub> = -40 \u00b5C and Q<sub>2<\/sub> = +5 \u00b5C as shown in figure below (k = 9 x 10<sup>9<\/sup> N.m<sup>2<\/sup>.C<sup>-2<\/sup> and 1 \u00b5C = 10<sup>-6<\/sup> C),. What is the magnitude of the electric field at point P.<\/span><\/p>\n<p lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-1965\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/04\/Electric-field-\u2013-problems-and-solutions-17-300x87.png\" alt=\"Electric field \u2013 problems and solutions 17\" width=\"300\" height=\"87\" srcset=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-17-300x87.png 300w, https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-17.png 348w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Known :<\/u><\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Charge q<sub>1<\/sub> = -40 \u00b5C = -40 x 10<sup>-6<\/sup> C <\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Charge q<sub>2<\/sub> = +5 \u00b5C = +5 x 10<sup>-6<\/sup> C <\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Distance between q<sub>1<\/sub> and point P (r<sub>1<\/sub>) = 40 cm = 0.4 m = 4 x 10<sup>-1<\/sup> m<\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Distance between q<sub>2<\/sub> and point P (r<sub>2<\/sub>) = 10 cm = 0.1 = 1 x 10<sup>-1<\/sup> m<\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">k = 9 x 10<sup>9<\/sup> N m<sup>2 <\/sup>C<sup>-2<\/sup><\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Wanted :<\/u> the magnitude of the electric field at point P.<\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Solution :<\/u><\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>The electric field 1 :<\/u><\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E<sub>1 <\/sub>= k q<sub>1<\/sub> \/ r<sub>1<\/sub><sup>2<\/sup> <\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E<sub>1<\/sub> = (9 x 10<sup>9<\/sup>)(40 x 10<sup>-6<\/sup>) \/ (4 x 10<sup>-1<\/sup>)<sup>2<\/sup> <\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E<sub>1<\/sub> = (360 x 10<sup>3<\/sup>) \/ (16 x 10<sup>-2<\/sup>) <\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E<sub>1<\/sub> = 22.5 x 10<sup>5 <\/sup>N\/C <\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>The electric field 2 :<\/u><\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E<sub>2<\/sub> = k q<sub>2<\/sub> \/ r<sub>2<\/sub><sup>2<\/sup> <\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E<sub>2<\/sub> = (9 x 10<sup>9<\/sup>)(5 x 10<sup>-6<\/sup>) \/ (1 x 10<sup>-1<\/sup>)<sup>2<\/sup> <\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E<sub>2<\/sub> = (45 x 10<sup>3<\/sup>) \/ 1 x 10<sup>-2<\/sup> <\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E<sub>2<\/sub> = 45 x 10<sup>5 <\/sup>N\/C <\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Resultant of the electric field :<\/u><\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">The resultant of the electric field at point P :<\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E = E<sub>2<\/sub> \u2013 E1 = (45 \u2013 22.5) x 10<sup>5 <\/sup>= 22.5 x 10<sup>5<\/sup> N\/C <\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E = 2.25 x 10<sup>6<\/sup> N\/C <\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">The direction of the electric field points to rightward (same direction as E<sub>2<\/sub>).<\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">7. Two point charges as shown in figure below.<\/span><\/p>\n<p lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-1966\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/04\/Electric-field-\u2013-problems-and-solutions-18-300x66.png\" alt=\"Electric field \u2013 problems and solutions 18\" width=\"300\" height=\"66\" srcset=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-18-300x66.png 300w, https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-18.png 336w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Where is point P located so that the magnitude of the electric field at point P = 0. k = 9.10<sup>9<\/sup> Nm<sup>2<\/sup>.C<sup>-2<\/sup>, 1 \u00b5C = 10<sup>-6<\/sup> C.<\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Known :<\/u><\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Charge 1 (q<sub>1<\/sub>) = -9 \u00b5C = -9.10<sup>-6<\/sup> Coulomb <\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Charge 2 (q<sub>2<\/sub>) = 1 \u00b5C = 1.10<sup>-6<\/sup> Coulomb <\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Distance between q<sub>1<\/sub> and q<sub>2 <\/sub>(r<sub>12<\/sub>) = 1 cm<\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">k = 9.10<sup>9<\/sup> Nm<sup>2<\/sup>.C<sup>-2<\/sup> <\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Wanted :<\/u> Position of point P<\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Solution :<\/u><\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E<sub>1<\/sub> = the magnitude of the electric field produced by q<sub>1 <\/sub>at point P<\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><i>The direction of <\/i><i>E<\/i><sub><i>1<\/i><i> <\/i><\/sub><i>to <\/i><i>q<\/i><sub><i>1<\/i><i> <\/i><\/sub><i>because <\/i><i>q<\/i><sub><i>1<\/i><\/sub><i> <\/i><i>is negative.<\/i><\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E<sub>2<\/sub> = the magnitude of the electric field produced by q<sub>2<\/sub> at point P<\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><i>The direction of <\/i><i>E<\/i><sub><i>2<\/i><\/sub><i> <\/i><i>away from <\/i><i>q<\/i><sub><i>2<\/i><\/sub><i> <\/i><i>because <\/i><i>q<\/i><sub><i>2<\/i><\/sub><i> <\/i><i>is positive.<\/i><\/span><\/p>\n<p lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-1967\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/04\/Electric-field-\u2013-problems-and-solutions-19-300x96.png\" alt=\"Electric field \u2013 problems and solutions 19\" width=\"300\" height=\"96\" srcset=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-19-300x96.png 300w, https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-19.png 402w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-1968\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/04\/Electric-field-\u2013-problems-and-solutions-20.png\" alt=\"Electric field \u2013 problems and solutions 20\" width=\"277\" height=\"145\" \/><\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">The electric field at point = 0.<\/span><\/p>\n<p class=\"western\" lang=\"id-ID\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><img width=\"53\" height=\"20\" align=\"absmiddle\" \/><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Use quadratic formula :<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-1970\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/04\/Electric-field-\u2013-problems-and-solutions-22-220x300.png\" alt=\"Electric field \u2013 problems and solutions 22\" width=\"220\" height=\"300\" srcset=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-22-220x300.png 220w, https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-22.png 237w\" sizes=\"auto, (max-width: 220px) 100vw, 220px\" \/><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Distance between P and q<sub>2 <\/sub>= x = 0.5 cm.<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Point P located at 0.5 cm rightward q<sub>2<\/sub> or 0.25 cm leftward q<sub>1<\/sub>. <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">8. According to the figure below, if the magnitude of the electric field at point P = k Q\/x<sup>2<\/sup>, then x = \u2026. <\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-1971\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/04\/Electric-field-\u2013-problems-and-solutions-23.png\" alt=\"Electric field \u2013 problems and solutions 23\" width=\"238\" height=\"67\" \/><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Known :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E<sub>P<\/sub> = k Q \/ x<sup>2<\/sup><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Wanted :<\/u> x<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Solution :<\/u><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-1972\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/04\/Electric-field-\u2013-problems-and-solutions-24-300x116.png\" alt=\"Electric field \u2013 problems and solutions 24\" width=\"300\" height=\"116\" srcset=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-24-300x116.png 300w, https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-24.png 410w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">E<sub>2<\/sub> = The magnitude of the electric field at point P by charge +32Q <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">r<sub>2 <\/sub>=Distance between charge +32Q and point P = a + x <\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-1973\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/04\/Electric-field-\u2013-problems-and-solutions-25-131x300.png\" alt=\"Electric field \u2013 problems and solutions 25\" width=\"131\" height=\"300\" srcset=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-25-131x300.png 131w, https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-25.png 179w\" sizes=\"auto, (max-width: 131px) 100vw, 131px\" \/><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Use quadratic formula :<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-1974\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/04\/Electric-field-\u2013-problems-and-solutions-26-260x300.png\" alt=\"Electric field \u2013 problems and solutions 26\" width=\"260\" height=\"300\" srcset=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-26-260x300.png 260w, https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/04\/Electric-field-\u2013-problems-and-solutions-26.png 282w\" sizes=\"auto, (max-width: 260px) 100vw, 260px\" \/><\/span><\/p>\n<ol>\n<li style=\"text-align: justify;\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>What is an electric field?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer<\/strong>: An electric field is a region around a charged object where electric forces can be experienced by other charged objects. It is a vector field, meaning it has both magnitude and direction at every point.<\/span><\/li>\n<\/ul>\n<\/li>\n<li style=\"text-align: justify;\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>How is the strength of an electric field determined?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer<\/strong>: The strength or magnitude of an electric field <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">E<\/span><\/span><\/span><\/span><\/span> at a point is defined as the force <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">F<\/span><\/span><\/span><\/span><\/span> experienced by a positive test charge <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mord mathnormal\">q<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><sub><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">0<\/span><\/span><\/span><\/sub><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> placed at that point, divided by the magnitude of the test charge itself: <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">E<\/span><span class=\"mrel\">=<\/span><\/span><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"msupsub\"><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathnormal mtight\">F\/q<sub><span class=\"msupsub\"><span class=\"sizing reset-size3 size1 mtight\">0<\/span><\/span><\/sub><\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>.<\/span><\/li>\n<\/ul>\n<\/li>\n<li style=\"text-align: justify;\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>How does the electric field due to a point charge vary with distance?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer<\/strong>: The electric field <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">E<\/span><\/span><\/span><\/span><\/span> due to a point charge <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">Q<\/span><\/span><\/span><\/span><\/span> is inversely proportional to the square of the distance <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">r<\/span><\/span><\/span><\/span><\/span> from the charge. The relationship is given by <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">E<\/span><span class=\"mrel\">=<\/span><\/span><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathnormal mtight\">k<\/span><span class=\"mord mathnormal mtight\">Q<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b\/r<sup>2<\/sup><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>, where <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">k<\/span><\/span><\/span><\/span><\/span> is Coulomb&#8217;s constant.<\/span><\/li>\n<\/ul>\n<\/li>\n<li style=\"text-align: justify;\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>What is the direction of the electric field due to a positive charge?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer<\/strong>: The electric field due to a positive charge points radially outward from the charge. For a negative charge, the field points radially inward, towards the charge.<\/span><\/li>\n<\/ul>\n<\/li>\n<li style=\"text-align: justify;\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>How can you represent electric fields graphically?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer<\/strong>: Electric fields can be represented graphically using field lines (or lines of force). The direction of the field at any point is tangent to the field line at that point, and the density of the lines indicates the magnitude of the field.<\/span><\/li>\n<\/ul>\n<\/li>\n<li style=\"text-align: justify;\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>What happens to the electric field inside a conductor in electrostatic equilibrium?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer<\/strong>: Inside a conductor in electrostatic equilibrium, the electric field is zero. This is because any external field causes free electrons in the conductor to redistribute, cancelling the external field inside.<\/span><\/li>\n<\/ul>\n<\/li>\n<li style=\"text-align: justify;\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>How do electric field lines behave near a sharp edge of a conductor?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer<\/strong>: Near a sharp edge or pointed tip of a conductor, the electric field lines are more concentrated, leading to a stronger electric field in that region. This is the basis for the operation of devices like the lightning rod.<\/span><\/li>\n<\/ul>\n<\/li>\n<li style=\"text-align: justify;\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>How do superposition principles apply to electric fields?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer<\/strong>: The electric field due to multiple charges at a point is simply the vector sum of the electric fields due to each individual charge. This is known as the superposition principle.<\/span><\/li>\n<\/ul>\n<\/li>\n<li style=\"text-align: justify;\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>How is the work done by an external agent related to the electric field when moving a charge within the field?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer<\/strong>: The work <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">W<\/span><\/span><\/span><\/span><\/span> done by an external agent in moving a charge <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">q<\/span><\/span><\/span><\/span><\/span> from one point to another in an electric field is equal to the negative of the change in electric potential energy, which is <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">W<\/span><span class=\"mrel\">=<\/span><\/span><span class=\"base\"><span class=\"mord mathnormal\">q<\/span><span class=\"mbin\">\u00d7<\/span><\/span><span class=\"base\"><span class=\"mord\">\u0394<\/span><span class=\"mord mathnormal\">V<\/span><\/span><\/span><\/span><\/span>, where <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\">\u0394<\/span><span class=\"mord mathnormal\">V<\/span><\/span><\/span><\/span><\/span> is the change in electric potential.<\/span><\/li>\n<\/ul>\n<\/li>\n<li>\n<p style=\"text-align: justify;\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Can electric field lines ever cross each other?<\/strong><\/span><\/p>\n<ul>\n<li style=\"text-align: justify;\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer<\/strong>: No, electric field lines cannot cross each other. If they did, it would imply that at the point of intersection, there are two different directions of the electric field, which is not possible.<\/span><\/li>\n<\/ul>\n<\/li>\n<\/ol>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"<p>Electric field \u2013 problems and solutions 1. Point A located at the center between two charges. Both charges have the same magnitude but opposite sign and separated by a distance of a. The magnitude of the electric field at point A is 36 N\/C. If point A moved 1\/2a close to one of both charges, &#8230; <a title=\"Electric field \u2013 problems and solutions\" class=\"read-more\" href=\"https:\/\/gurumuda.net\/physics\/electric-field-problems-and-solutions.htm\" aria-label=\"Read more about Electric field \u2013 problems and solutions\">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 field \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-1949","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\/1949","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=1949"}],"version-history":[{"count":2,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1949\/revisions"}],"predecessor-version":[{"id":8679,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1949\/revisions\/8679"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=1949"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=1949"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=1949"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}