{"id":217473,"date":"2022-07-12T16:17:03","date_gmt":"2022-07-12T10:47:03","guid":{"rendered":"https:\/\/www.aakash.ac.in\/blog\/?p=217473"},"modified":"2023-05-03T11:58:34","modified_gmt":"2023-05-03T06:28:34","slug":"gausss-law-and-electrostatic-potential-neet-physics-chapter-11-notes","status":"publish","type":"post","link":"https:\/\/www.aakash.ac.in\/blog\/gausss-law-and-electrostatic-potential-neet-physics-chapter-11-notes\/","title":{"rendered":"Gauss&#8217;s law and Electrostatic Potential: NEET Physics Chapter 11 Notes"},"content":{"rendered":"<p><span style=\"font-weight: 400;\">Electrostatics is the analysis of static or slow-moving electrically charged particles. Electrostatics can be seen in the following phenomena: laser machines and photocopiers, grain silos igniting instantly, and paper being attracted to a charging scale. While Gauss Law, also known as Gauss&#8217; flux theorem of Gauss&#8217; theorem, is the law that describes the relationship between electric charge distribution and the consequent electric field. Some key topics covered under this unit are electric charges, electric flux, electric fields, application of Gauss law.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Gauss&#8217;s law and electrostatics potential is a significant unit in the NEET 2022 Physics Syllabus, with approximately 3-4 questions anticipated yearly.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">According to last year&#8217;s study, Gauss&#8217;s law and electrostatics potential questions in the NEET 2022 Physics examination will be moderately challenging. Thus, students must refer to these revision notes to help them score well in this section.\u00a0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Check the entire <\/span><a href=\"https:\/\/www.aakash.ac.in\/neet-physics-syllabus\"><span style=\"font-weight: 400;\">NEET Physics Syllabus 2022<\/span><\/a><span style=\"font-weight: 400;\">\u00a0here!<\/span><\/p>\n<table>\n<tbody>\n<tr>\n<td><b>Table of Contents<\/b><\/td>\n<\/tr>\n<tr>\n<td><span style=\"font-weight: 400;\">NEET 2022 Physics: Gauss&#8217;s Law and Electrostatic Potential Notes<\/span><\/td>\n<\/tr>\n<tr>\n<td><span style=\"font-weight: 400;\">Conclusion<\/span><\/td>\n<\/tr>\n<tr>\n<td><span style=\"font-weight: 400;\">FAQs<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h4><b>NEET 2022 Physics: Gauss&#8217;s Law and Electrostatic Potential Notes<\/b><\/h4>\n<h4><b><a href=\"https:\/\/www.aakash.ac.in\/important-concepts\/physics\/gauss-law\">What is Gauss&#8217;s Law<\/a>?<\/b><\/h4>\n<p><span style=\"font-weight: 400;\">Gauss law defines the proportional relationship between the total quantity of electric flux travelling through any closed path to the encased electric charge. Gauss Law is mostly used to determine the electrostatic potential due to continuous symmetries, such as:<\/span><\/p>\n<ul>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Uniformly charged straight wire<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Uniformly charged thin spherical shell\u00a0<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Uniformly charged infinite plate sheet<\/span><\/li>\n<\/ul>\n<p><span style=\"font-weight: 400;\">Gauss Law, often known as Gauss&#8217; theorem or Gauss&#8217; flux theorem, is the law that explains the relationship between electric charge distribution and the consequent electric potential. The surface&#8217;s electrical field is computed using Coulomb&#8217;s law, but the Gauss law is necessary to determine the dispersion of the electric current on a closed path. It describes the static electricity confined or existing in the sealed closed path.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Gauss law refers to the total flux available within a closed surface being 1\/0 times the total electric charge contained by the closed surface.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">For instance, a point charge q is placed within a cube with edge&#8217; a&#8217;. According to the gauss theorem, the flux across each surface of the cube will be &#8220;q\/6\u03c0\u03b50&#8221;.<\/span><\/p>\n<p>Also See: <a href=\"https:\/\/www.aakash.ac.in\/blog\/neet-physics-important-notes-on-electric-charges-and-fields\/\">Electric Charges and Fields Notes<\/a><\/p>\n<h4><b>Application of Gauss&#8217;s Law<\/b><\/h4>\n<p><span style=\"font-weight: 400;\">Gauss&#8217;s law can be utilised effectively with anomalous symmetry to solve complex electrostatic issues, such as spherical, cylindrical, or planar symmetry. It also aids in the computation of the electrical field, which is highly complicated and requires difficult integration.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">To evaluate the electrical field directly, students can use Gauss&#8217;s theorem.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Gauss law has the following major applications:<\/span><\/p>\n<ul>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Electric field because of uniformly charged infinite straight wire.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Electric field because of a uniformly charged thin spherical shell.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Electric field because of uniformly charged infinite plate sheets.<\/span><\/li>\n<\/ul>\n<p><strong>Also Read:<\/strong><\/p>\n<p><a href=\"https:\/\/www.aakash.ac.in\/blog\/how-to-prepare-for-neet-physics-exam\/\">NEET Physics Preparation: How to prepare for NEET Physics 2022 Exam?<\/a><\/p>\n<p><a href=\"https:\/\/www.aakash.ac.in\/blog\/how-to-prepare-for-neet-physics\/\">How to Prepare for NEET Physics<\/a><\/p>\n<p><a href=\"https:\/\/www.aakash.ac.in\/blog\/10-tips-to-solve-neet-physics-problems-easily\/\">10 Tips to Solve NEET 2022 Physics Problems Easily<\/a><\/p>\n<h4><b>Electric Field due to Infinitely Charged Straight Wire<\/b><\/h4>\n<p><span style=\"font-weight: 400;\">Imagine an infinitely long wire with a constant load density of &#8220;\u03bb&#8221; with length &#8220;l&#8221;. Assume the Gaussian cylindrical field is related to wire symmetry when calculating the electric fields.\u00a0<\/span><\/p>\n<ul>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">The electric flow throughout the end of the cylindrical surface will be zero since the electric fields and the area vector are perpendicular to one another.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">The only electric flux that flows would be through the curved Gaussian surface.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">The magnitude of the electric field will be constant because it is perpendicular to each point of the curved surface.\u00a0<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">The curved cylindrical surface will have a surface area of 2\u03c0rl. The electric flow along the curve is E \u00d7 2\u03c0rl.<\/span><\/li>\n<\/ul>\n<p><span style=\"font-weight: 400;\">And as per Gauss&#8217;s law,\u00a0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u03a6 = q \/ \u03b50<\/span><\/p>\n<p><span style=\"font-weight: 400;\">E \u00d7 2\u03c0rl = \u03bbl \/ \u03b50<\/span><\/p>\n<p><span style=\"font-weight: 400;\">E = \u03bb \/ 2\u03c0\u03b50r<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Vector can define the above relationship as<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u2192E = \u03bb \/ 2\u03c0\u03b50r n<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Where &#8220;n&#8221; denotes a circular unit vector pointing toward the applied field E.<\/span><\/p>\n<p><b>Point 1: <\/b><span style=\"font-weight: 400;\">If the linear charge density is positive, the electric field will be circumferentially outward; if the linear charge density is negative, the electric field will be circumferentially inward.<\/span><\/p>\n<p><b>Point 2: <\/b><span style=\"font-weight: 400;\">Gaussian surface only considers the enclosed charge within it.Point\u00a0<\/span><\/p>\n<p><b>Point 3:<\/b><span style=\"font-weight: 400;\"> It is critical to assume that the wire is infinitely long because otherwise, the electric current would not be perpendicular to the curved cylindrical Gaussian surface and would be at an angle with it. Check all the <\/span><a href=\"https:\/\/www.aakash.ac.in\/important-concepts\/physics\"><span style=\"font-weight: 400;\">Important Physics Concepts <\/span><\/a><span style=\"font-weight: 400;\">\u00a0for free here!<\/span><\/p>\n<h4><b>Electric Field due to Infinite Plate Sheet<\/b><\/h4>\n<p><span style=\"font-weight: 400;\">Consider an infinite plane sheet with a surface charge density \u03c3 and a cross-sectional area of A. The view of the infinite plane sheet is illustrated below:<\/span><\/p>\n<p><strong>Infinite Charge Sheet<\/strong><\/p>\n<p><span style=\"font-weight: 400;\">The electric field created by the infinite charge sheet would be perpendicular to the sheet&#8217;s plane. Imagine a cylindrical Gaussian surface with an axis normal to the sheet&#8217;s plane. Students can use Gauss&#8217;s law to calculate the electric field E.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Gauss&#8217;s law states that:<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u03d5 = q \/ \u03f50<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Charge q will be \u03c3A as a consequence of continuous charge distribution. The gaussian surface only looks at electric flux from the two ends when discussing net electric flux. The curved surface area and an electric field are regular, resulting in zero electric flux. As a result, the net electric flux would be<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u03a6 = EA \u2013 (\u2013 EA)<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u03a6 = 2EA<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Which can be written as,<\/span><\/p>\n<p><span style=\"font-weight: 400;\">2EA = \u03c3A \/ \u03f50<\/span><\/p>\n<p><span style=\"font-weight: 400;\">The term A cancels out, implying that the electric field generated by an infinite plane sheet is reflective of cross-section area A and equals<\/span><\/p>\n<p><span style=\"font-weight: 400;\">E = \u03c32 \/ \u03f50<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Vector can write the above equation as<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u2192 E = \u03bb \/ 2\u03c0\u03f50r n<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Where,<\/span><\/p>\n<p><span style=\"font-weight: 400;\">n = direction of the current field perpendicular to and away from the infinite sheet<\/span><\/p>\n<p><b>Point 1:<\/b><span style=\"font-weight: 400;\"> If the surface charge density is positive, the electric field is directed away from the infinite sheet; if the surface charge density is negative, the electric field is directed towards the infinite sheet.<\/span><\/p>\n<p><b>Point 2:<\/b><span style=\"font-weight: 400;\"> The infinite sheet&#8217;s electric field is reflective of its position.<\/span><\/p>\n<h4><b>Electric Field due to thin Circular Shell<\/b><\/h4>\n<p><span style=\"font-weight: 400;\">Imagine the radius &#8220;R&#8221; and the thin circular shell of the density of the surface electric charge. The shell exhibits circular symmetry, as may be shown from the observation. The electric field caused by the spherical shell can be also evaluated in two ways:<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Electric Field Outside the Spherical Shell<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Electric Field Inside the Spherical Shell<\/span><\/p>\n<ul>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Electric Field Outside the Spherical Shell<\/span><\/li>\n<\/ul>\n<p><span style=\"font-weight: 400;\">To determine an electric field outside the spherical shell, consider a point P outside the shell at a distance r from the shell&#8217;s centre. Gaussian spherical objects will use symmetry to create a surface with radius r and centre O.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">As all points are equally spaced &#8220;r&#8221; from the centre of the sphere, the Gaussian surface will move through P and insight a consistent electric field. Then,<\/span><\/p>\n<p><span style=\"font-weight: 400;\">As per gauss&#8217;s law,<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u03d5 = q \/ \u03f50<\/span><\/p>\n<p><span style=\"font-weight: 400;\">The enclosed charge q within the Gaussian surface is \u03c3 \u00d7 4 \u03c0R<\/span><b>2<\/b><span style=\"font-weight: 400;\">. The total electric flux perpendicular to the Gaussian surface would be<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u03a6 = E \u00d7 4 \u03c0r2<\/span><\/p>\n<p><span style=\"font-weight: 400;\">According to Gauss&#8217;s law, it can also be written as,<\/span><\/p>\n<p><span style=\"font-weight: 400;\">E \u00d7 4\u03c0r<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\"> = \u03c3 \u00d7 4\u03c0 R<\/span><span style=\"font-weight: 400;\">3<\/span><span style=\"font-weight: 400;\"> \/ \u03f50<\/span><\/p>\n<p><span style=\"font-weight: 400;\">E = \u03c3R<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\"> \/ \u03f50r<\/span><span style=\"font-weight: 400;\">2<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Using the value of surface charge density \u03c3 as q \/4\u03c0R<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">, students can also write the electric field as<\/span><\/p>\n<p><span style=\"font-weight: 400;\">In vector form, the electric field can be represented as<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u2192 E = kq \/ r<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\"> \u0155<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Where,\u00a0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u0155 = radius vector representing the electric field&#8217;s direction<\/span><\/p>\n<p><b>Point:<\/b><span style=\"font-weight: 400;\"> If the surface charge density is negative, the electric field will be directed circumferentially inward.<\/span><\/p>\n<ul>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Electric Field Inside the Spherical Shell<\/span><\/li>\n<\/ul>\n<p><span style=\"font-weight: 400;\">Consider point P inside the shell to calculate the electric field within the spherical shell. The spherical Gaussian surface that crosses through P is symmetrically focused at O and has a radius of r. Now, according to Gauss&#8217;s law,<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u03d5 = q \/ \u03f5<\/span><span style=\"font-weight: 400;\">0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">The net electric flux would be E \u00d7 4\u03c0r<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">. However, because surface charge density is spread outside the surface, the enclosed charge q will be 0, implying that there is no charge or current within the spherical shell. Then, according to Gauss&#8217;s Law,<\/span><\/p>\n<p><span style=\"font-weight: 400;\">E = 4<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u03c0r<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\"> = 0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">E = 0<\/span><\/p>\n<p><b>Point:<\/b><span style=\"font-weight: 400;\"> Because there is no confined charge, there is no electric current within the spherical shell.<\/span><\/p>\n<p><strong>Check our visual story on NEET 2022 Physics Syllabus &amp; Important Topics<\/strong><\/p>\n<blockquote class=\"wp-embedded-content\" data-secret=\"6oK1zIgCmr\"><p><a href=\"https:\/\/www.aakash.ac.in\/blog\/web-stories\/neet-physics-syllabus-2022-important-chapters-and-topics\/\">NEET 2022 Physics Syllabus &#038; Important Topics<\/a><\/p><\/blockquote>\n<p><iframe loading=\"lazy\" class=\"wp-embedded-content\" sandbox=\"allow-scripts\" security=\"restricted\" style=\"position: absolute; clip: rect(1px, 1px, 1px, 1px);\" title=\"&#8220;NEET 2022 Physics Syllabus &#038; Important Topics&#8221; &#8212; Aakash Blog\" src=\"https:\/\/www.aakash.ac.in\/blog\/web-stories\/neet-physics-syllabus-2022-important-chapters-and-topics\/embed\/#?secret=u90K1NrWZ7#?secret=6oK1zIgCmr\" data-secret=\"6oK1zIgCmr\" width=\"360\" height=\"600\" frameborder=\"0\" marginwidth=\"0\" marginheight=\"0\" scrolling=\"no\"><\/iframe><\/p>\n<h4><b>Gauss&#8217;s Law Formula<\/b><\/h4>\n<p><span style=\"font-weight: 400;\">According to Gauss&#8217;s theorem, the total charge concealed in a closed path is directly proportional to the total flux concealed by the surface. As a result, if \u03a6 is the total flux and \u03b50 is the electrically constant, the total electric charge Q concealed by the surface is as follows:<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Q = \u03a6\u03b50<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Students can write the formula for Gauss&#8217;s law as follows,<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u03a6 = Q \/ \u03f5<\/span><span style=\"font-weight: 400;\">o<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Where,<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Q = total charge within the mentioned surface,<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u03b5<\/span><span style=\"font-weight: 400;\">0<\/span><span style=\"font-weight: 400;\"> = the electric constant.<\/span><\/p>\n<h4><b>Gauss&#8217;s Theorem<\/b><\/h4>\n<p><span style=\"font-weight: 400;\">According to Gauss&#8217;s theorem, the net flux through a closed path is proportional to the net charge in the volume sealed by the closed path.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u03a6 = \u2192 E.d \u2192 A = qnet\/\u03b5<\/span><span style=\"font-weight: 400;\">0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">In layman&#8217;s terms, Gauss&#8217;s theorem connects the &#8216;flow&#8217; of electric potential lines (flux) to charges on the closed path. The net electric flux is zero if no charges are closed on the path.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">It indicates that the number of electric potential accessing the surface equals the number of field lines exiting it.<\/span><\/p>\n<h4><b>Important Benefits of Gauss&#8217;s Theorem<\/b><\/h4>\n<p><span style=\"font-weight: 400;\">The electrical flux from any enclosed surface is caused by the electrical fields&#8217; sinks (negative charges) and sources (positive charges). Any electrostatics outside the surface does not affect the flow of electricity. Furthermore, only electrostatic interactions can act as electrical field sources or sinks. For instance, adjusting magnetic fields cannot act as electrical field sources or sinks.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Because the area on the left encloses a net charge, its net flux is non-zero. The net flux for the area on the right is zero because it contains no charge.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Students should note that Gauss&#8217;s law is simply a reaffirmation of Coulomb&#8217;s law. Students can easily obtain Coulomb&#8217;s law by applying the Gauss theorem to a charged object encircled by a sphere.<\/span><\/p>\n<h4><b>Conclusion<\/b><\/h4>\n<p><span style=\"font-weight: 400;\">Preparing for the NEET 2022 Physics Gauss&#8217; law and potential electrostatics exam is an important step toward understanding the potential of keeping energy. Students will learn the responses to questions which may have been presented in the examinations by using these revision notes. Besides being an important topic in the NEET 2022 Physics syllabus, questions on Gauss&#8217; law and electrostatic potential carry significant points. Reading these notes will provide students with further insights on this topic which will ultimately help them in improving their scores!<\/span><\/p>\n<blockquote class=\"wp-embedded-content\" data-secret=\"CN5yU2zEBX\"><p><a href=\"https:\/\/www.aakash.ac.in\/blog\/how-many-laws-of-physics-are-there\/\">How Many Laws of Physics are there?<\/a><\/p><\/blockquote>\n<p><iframe loading=\"lazy\" class=\"wp-embedded-content\" sandbox=\"allow-scripts\" security=\"restricted\" style=\"position: absolute; clip: rect(1px, 1px, 1px, 1px);\" title=\"&#8220;How Many Laws of Physics are there?&#8221; &#8212; Aakash Blog\" src=\"https:\/\/www.aakash.ac.in\/blog\/how-many-laws-of-physics-are-there\/embed\/#?secret=cz9Cxlh2kh#?secret=CN5yU2zEBX\" data-secret=\"CN5yU2zEBX\" width=\"500\" height=\"282\" frameborder=\"0\" marginwidth=\"0\" marginheight=\"0\" scrolling=\"no\"><\/iframe><\/p>\n<h4><b>FAQs<\/b><\/h4>\n<p><b>1. What is Coulomb&#8217;s law?<\/b><\/p>\n<p>The force between two charges is directly proportional to the product of their magnitudes and indirectly proportional to the square of their distance.<\/p>\n<p><span style=\"font-weight: 400;\">It is mathematically represented as<\/span><\/p>\n<p><span style=\"font-weight: 400;\">F \u221d | q1 | | q2 |<\/span><\/p>\n<p><span style=\"font-weight: 400;\">F \u221d 1 \/ r2<\/span><\/p>\n<p><span style=\"font-weight: 400;\">F = k | q1 | | q2 | \/ r2<\/span><\/p>\n<p><span style=\"font-weight: 400;\">where the proportionality constant is k = 1 \/ 4\u03c0\u03f50 = 9 \u0425 109 Nm<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">C<\/span><span style=\"font-weight: 400;\">2<\/span><\/p>\n<p><span style=\"font-weight: 400;\">The permissibility of free space is \u03f50 = 8.85 \u0425 10<\/span><span style=\"font-weight: 400;\">-12<\/span><span style=\"font-weight: 400;\">C<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">N<\/span><span style=\"font-weight: 400;\">-1<\/span><span style=\"font-weight: 400;\">m<\/span><span style=\"font-weight: 400;\">-2<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Coulomb&#8217;s law in vector form is \u2192F12 = k | q1 | | q2 | \/ r<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\"> r<\/span><\/p>\n<p><b>2. What is Electric Potential?<\/b><\/p>\n<p><span style=\"font-weight: 400;\">The electric or electrostatic potential is the volume of work done to move a unit positive test current from infinity to any place along any path. It is represented by the letter V.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">V = W \/ q0 = work done in bringing unit positive test charge from infinity to some point\/unit positive test charge.\u00a0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Electric potential is a vector value with volt as the SI unit.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">The electric potential related to a point charge q at separation r is equal to V = 1 \/ 4\u03c0\u03f5 X q \/ r<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Students will answer electric potential due to several charges q1, q2,&#8230; qn.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">V= V1+V2..+Vn<\/span><\/p>\n<p><span style=\"font-weight: 400;\">= 1 \/ 4\u03c0\u03f5 X ( q1\/r1 + q2\/r2 + \u2026 + qn\/rn)\u00a0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">= 1 \/ 4\u03c0\u03f5 \u2211 qi \/ ri<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Electric Field and Potential Relationship<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Students can describe the relation between electric field and potential as<\/span><\/p>\n<p><span style=\"font-weight: 400;\">E = -dV \/ dr<\/span><\/p>\n<p><span style=\"font-weight: 400;\">where,\u00a0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">E represents the electric field.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">V represents the potential difference.<\/span><\/p>\n<p><b>3. Define conductors, semiconductors or insulators.<\/b><\/p>\n<p><span style=\"font-weight: 400;\">A conductor is a body through which an electric charge can easily pass. Metals are good examples of conductors.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Insulators, often known as dielectrics, are materials that do not conduct electricity. Rubber, glass, plastics, wool, and other materials are examples of insulators.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Semiconductors are the bodies that exist between conductors and insulators. Examples include silicon, germanium, and others.<\/span><\/p>\n<p><b>4. Define electric lines of force?<\/b><\/p>\n<p>An electric line of force is a straight or curved path through which a positive unit charge is compelled to go when free to do that in an electric field. The velocity of a unit positive charge determines the direction of the line of force.<\/p>\n<p><span style=\"font-weight: 400;\">Properties:-<\/span><\/p>\n<p><span style=\"font-weight: 400;\">The force lines diverge from a positive charge conductor toward a negative charge conductor.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">A forced line begins with a positive charge and finishes with a negative charge. It indicates that the line of force begins at a greater potential and ends at a lower potential.<\/span><\/p>\n<p><b>5. What are the important pointers of Gauss&#8217;s law?<\/b><\/p>\n<p><span style=\"font-weight: 400;\">The primary application of Gauss law is to determine the electric field produced by:<\/span><\/p>\n<ul>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">A uniformly charged infinite straight wire.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">An endless plate sheet with a uniform charge.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">A thin uniformly charged spherical shell.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Gauss&#8217;s law explains the electrical charge confined in the enclosed or electrical charge in the enclosed closed surface.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Gauss Law states that the total flux associated with a closed path is 1\/0 times the charge encompassed by the closed surface.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">The Gauss law formula is written as \u03a6 = Q\/\u03b50. Q is the total charge on the given surface, and 0 is the electric constant.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Students can utilise Gauss&#8217; law to address complex electrostatic issues with unique symmetry such as spherical, cylindrical, or planar.<\/span><\/li>\n<\/ul>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Electrostatics is the analysis of static or slow-moving electrically charged particles. Electrostatics can be seen in the following phenomena: laser machines and photocopiers, grain silos igniting instantly, and paper being attracted to a charging scale. While Gauss Law, also known as Gauss&#8217; flux theorem of Gauss&#8217; theorem, is the law that describes the relationship between [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":217586,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[14],"tags":[],"class_list":["post-217473","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-neet-coaching"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.0 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Gauss&#039;s law and Electrostatic Potential Notes: NEET Physics Chapter 11<\/title>\n<meta name=\"description\" content=\"Gauss&#039;s law and Electrostatic Potential Notes: This article will talk about NEET Physics chapter Gauss&#039;s law and Electrostatic Potential &amp; its concepts and more on aakash.ac.in\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.aakash.ac.in\/blog\/gausss-law-and-electrostatic-potential-neet-physics-chapter-11-notes\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Gauss&#039;s law and Electrostatic Potential Notes: NEET Physics Chapter 11\" \/>\n<meta property=\"og:description\" content=\"Gauss&#039;s law and Electrostatic Potential Notes: This article will talk about NEET Physics chapter Gauss&#039;s law and Electrostatic Potential &amp; its concepts and more on aakash.ac.in\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.aakash.ac.in\/blog\/gausss-law-and-electrostatic-potential-neet-physics-chapter-11-notes\/\" \/>\n<meta property=\"og:site_name\" content=\"Aakash Blog\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/aakasheducation\" \/>\n<meta property=\"article:published_time\" content=\"2022-07-12T10:47:03+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2023-05-03T06:28:34+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/blogcdn.aakash.ac.in\/wordpress_media\/2022\/07\/Blog-Image-28.jpg\" \/>\n\t<meta property=\"og:image:width\" content=\"1200\" \/>\n\t<meta property=\"og:image:height\" content=\"900\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/jpeg\" \/>\n<meta name=\"author\" content=\"Team @Aakash\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@aksblog\" \/>\n<meta name=\"twitter:site\" content=\"@AESL_Official\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Team @Aakash\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"11 minutes\" \/>\n<!-- \/ Yoast SEO plugin. -->","yoast_head_json":{"title":"Gauss's law and Electrostatic Potential Notes: NEET Physics Chapter 11","description":"Gauss's law and Electrostatic Potential Notes: This article will talk about NEET Physics chapter Gauss's law and Electrostatic Potential & its concepts and more on aakash.ac.in","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/www.aakash.ac.in\/blog\/gausss-law-and-electrostatic-potential-neet-physics-chapter-11-notes\/","og_locale":"en_US","og_type":"article","og_title":"Gauss's law and Electrostatic Potential Notes: NEET Physics Chapter 11","og_description":"Gauss's law and Electrostatic Potential Notes: This article will talk about NEET Physics chapter Gauss's law and Electrostatic Potential & its concepts and more on aakash.ac.in","og_url":"https:\/\/www.aakash.ac.in\/blog\/gausss-law-and-electrostatic-potential-neet-physics-chapter-11-notes\/","og_site_name":"Aakash Blog","article_publisher":"https:\/\/www.facebook.com\/aakasheducation","article_published_time":"2022-07-12T10:47:03+00:00","article_modified_time":"2023-05-03T06:28:34+00:00","og_image":[{"width":1200,"height":900,"url":"https:\/\/blogcdn.aakash.ac.in\/wordpress_media\/2022\/07\/Blog-Image-28.jpg","type":"image\/jpeg"}],"author":"Team @Aakash","twitter_card":"summary_large_image","twitter_creator":"@aksblog","twitter_site":"@AESL_Official","twitter_misc":{"Written by":"Team @Aakash","Est. reading time":"11 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/www.aakash.ac.in\/blog\/gausss-law-and-electrostatic-potential-neet-physics-chapter-11-notes\/","url":"https:\/\/www.aakash.ac.in\/blog\/gausss-law-and-electrostatic-potential-neet-physics-chapter-11-notes\/","name":"Gauss's law and Electrostatic Potential Notes: NEET Physics Chapter 11","isPartOf":{"@id":"https:\/\/www.aakash.ac.in\/blog\/#website"},"primaryImageOfPage":{"@id":"https:\/\/www.aakash.ac.in\/blog\/gausss-law-and-electrostatic-potential-neet-physics-chapter-11-notes\/#primaryimage"},"image":{"@id":"https:\/\/www.aakash.ac.in\/blog\/gausss-law-and-electrostatic-potential-neet-physics-chapter-11-notes\/#primaryimage"},"thumbnailUrl":"https:\/\/blogcdn.aakash.ac.in\/wordpress_media\/2022\/07\/Blog-Image-28.jpg","datePublished":"2022-07-12T10:47:03+00:00","dateModified":"2023-05-03T06:28:34+00:00","author":{"@id":"https:\/\/www.aakash.ac.in\/blog\/#\/schema\/person\/cf47f7cea4939aa1d6f7066a7d62eff9"},"description":"Gauss's law and Electrostatic Potential Notes: This article will talk about NEET Physics chapter Gauss's law and Electrostatic Potential & its concepts and more on aakash.ac.in","breadcrumb":{"@id":"https:\/\/www.aakash.ac.in\/blog\/gausss-law-and-electrostatic-potential-neet-physics-chapter-11-notes\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/www.aakash.ac.in\/blog\/gausss-law-and-electrostatic-potential-neet-physics-chapter-11-notes\/"]}]},{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/www.aakash.ac.in\/blog\/gausss-law-and-electrostatic-potential-neet-physics-chapter-11-notes\/#primaryimage","url":"https:\/\/blogcdn.aakash.ac.in\/wordpress_media\/2022\/07\/Blog-Image-28.jpg","contentUrl":"https:\/\/blogcdn.aakash.ac.in\/wordpress_media\/2022\/07\/Blog-Image-28.jpg","width":1200,"height":900,"caption":"Gauss's law and Electrostatic Potential: NEET Physics Chapter 11 Notes"},{"@type":"BreadcrumbList","@id":"https:\/\/www.aakash.ac.in\/blog\/gausss-law-and-electrostatic-potential-neet-physics-chapter-11-notes\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/www.aakash.ac.in\/blog\/"},{"@type":"ListItem","position":2,"name":"NEET Coaching","item":"https:\/\/www.aakash.ac.in\/blog\/category\/neet-coaching\/"},{"@type":"ListItem","position":3,"name":"Gauss&#8217;s law and Electrostatic Potential: NEET Physics Chapter 11 Notes"}]},{"@type":"WebSite","@id":"https:\/\/www.aakash.ac.in\/blog\/#website","url":"https:\/\/www.aakash.ac.in\/blog\/","name":"Aakash Blog","description":"Medical, IIT-JEE &amp; Foundations","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/www.aakash.ac.in\/blog\/?s={search_term_string}"},"query-input":{"@type":"PropertyValueSpecification","valueRequired":true,"valueName":"search_term_string"}}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/www.aakash.ac.in\/blog\/#\/schema\/person\/cf47f7cea4939aa1d6f7066a7d62eff9","name":"Team @Aakash","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/www.aakash.ac.in\/blog\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/8e3389fb55661953db09dd08e8c113cb006c80494aef17a284c4a0a00976005f?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/8e3389fb55661953db09dd08e8c113cb006c80494aef17a284c4a0a00976005f?s=96&d=mm&r=g","caption":"Team @Aakash"},"sameAs":["https:\/\/x.com\/aksblog"],"url":"https:\/\/www.aakash.ac.in\/blog\/author\/aksblog\/"}]}},"_links":{"self":[{"href":"https:\/\/www.aakash.ac.in\/blog\/wp-json\/wp\/v2\/posts\/217473","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.aakash.ac.in\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.aakash.ac.in\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.aakash.ac.in\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.aakash.ac.in\/blog\/wp-json\/wp\/v2\/comments?post=217473"}],"version-history":[{"count":5,"href":"https:\/\/www.aakash.ac.in\/blog\/wp-json\/wp\/v2\/posts\/217473\/revisions"}],"predecessor-version":[{"id":269305,"href":"https:\/\/www.aakash.ac.in\/blog\/wp-json\/wp\/v2\/posts\/217473\/revisions\/269305"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.aakash.ac.in\/blog\/wp-json\/wp\/v2\/media\/217586"}],"wp:attachment":[{"href":"https:\/\/www.aakash.ac.in\/blog\/wp-json\/wp\/v2\/media?parent=217473"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.aakash.ac.in\/blog\/wp-json\/wp\/v2\/categories?post=217473"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.aakash.ac.in\/blog\/wp-json\/wp\/v2\/tags?post=217473"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}