際際滷

際際滷Share a Scribd company logo

EMTL Class10.pptx

Download as PPTX, PDF
G
Gnanasambanthan Rajendran

This document discusses Gauss's law and its applications in solving complex electrostatic problems involving unique symmetries like cylindrical, spherical, or planar symmetry. It provides 4 examples of using Gauss's law to calculate electric field intensity due to (1) a point charge, (2) an infinite line charge, (3) an infinite plane sheet, and (4) a charged sphere. It also briefly defines dielectrics as materials where charges are bound and unable to migrate in response to an applied electric field, instead causing polarization.

1 of 10
Download to read offline
Electromagnetic waves & Transmission lines By M. Gnanapriya, Head Of the Department of ECE Gokula Krishna College of Engineering, Sullurpet Date: 02.06.2021 Gausss law& its applications convection & conduction current 22-09-2022 1
22-09-2022 2
Gauss law can be used to solve complex electrostatic problems involving unique symmetry like cylindrical , spherical or planer symmetry. 1. Electric field due to point charge Consider a point charge located at origin. We choose a spherical Gaussian surface of radius r where the electric field intensity is going to be measured. According to Gauss law r 22-09-2022 3
2. Electric field due to an infinite line charge Consider an infinite line of uniform charge density l c/m Lies along z axis. To calculate electric field intensity we assume a cylindrical Gaussian surface of length l to satisfy symmetry condition. The electric field E is the radial in the direction of flux. So the flux through the end of cylindrical surface is zero. Ie. E.ds2=E.ds3=0 (COS90=0) According to Gauss law l 22-09-2022 4
3. Electric field due to an infinite Plane sheet Consider an infinite sheet uniform charge density s c/m 2. To calculate electric field intensity we assume a cylindrical Gaussian surface whose axis is normal to the plane of sheet. We shall consider the electric flus only from tow ends of the cylinder. Because curved space areal & E are normal to each other so the flux it produces is zero. According to Gauss law 22-09-2022 5
4. Electric field intensity outside the sphere charge Consider a sphere of radius R with uniform charge density v c/m3. To calculate electric field intensity we assume a spherical Gaussian surface of radius r which experiences uniform electric field E because of sphere charge. According to Gauss law 22-09-2022 6
22-09-2022 7
22-09-2022 8
22-09-2022 9
22-09-2022 10 DIELECTRIC In a dielectric or insulator charges are bound and they are not free to migrate. The conductivity of an insulator ideally zero. If a field is applied to an insulator there may not be any migration of charge but it can produce polarization of the dielectric displacement of electrons with respect to equilibrium position. Example: wood, glass, paraffins (plastics) are good insulators

More Related Content

EMTL Class10.pptx

  • 1. Electromagnetic waves & Transmission lines By M. Gnanapriya, Head Of the Department of ECE Gokula Krishna College of Engineering, Sullurpet Date: 02.06.2021 Gausss law& its applications convection & conduction current 22-09-2022 1
  • 3. Gauss law can be used to solve complex electrostatic problems involving unique symmetry like cylindrical , spherical or planer symmetry. 1. Electric field due to point charge Consider a point charge located at origin. We choose a spherical Gaussian surface of radius r where the electric field intensity is going to be measured. According to Gauss law r 22-09-2022 3
  • 4. 2. Electric field due to an infinite line charge Consider an infinite line of uniform charge density l c/m Lies along z axis. To calculate electric field intensity we assume a cylindrical Gaussian surface of length l to satisfy symmetry condition. The electric field E is the radial in the direction of flux. So the flux through the end of cylindrical surface is zero. Ie. E.ds2=E.ds3=0 (COS90=0) According to Gauss law l 22-09-2022 4
  • 5. 3. Electric field due to an infinite Plane sheet Consider an infinite sheet uniform charge density s c/m 2. To calculate electric field intensity we assume a cylindrical Gaussian surface whose axis is normal to the plane of sheet. We shall consider the electric flus only from tow ends of the cylinder. Because curved space areal & E are normal to each other so the flux it produces is zero. According to Gauss law 22-09-2022 5
  • 6. 4. Electric field intensity outside the sphere charge Consider a sphere of radius R with uniform charge density v c/m3. To calculate electric field intensity we assume a spherical Gaussian surface of radius r which experiences uniform electric field E because of sphere charge. According to Gauss law 22-09-2022 6
  • 10. 22-09-2022 10 DIELECTRIC In a dielectric or insulator charges are bound and they are not free to migrate. The conductivity of an insulator ideally zero. If a field is applied to an insulator there may not be any migration of charge but it can produce polarization of the dielectric displacement of electrons with respect to equilibrium position. Example: wood, glass, paraffins (plastics) are good insulators