Conducting and non-conducting infinite sheets of charge have same $E$? ($E$= electric field intensity)? - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-07-22T23:34:57Z https://physics.stackexchange.com/feeds/question/276821 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/276821 1 Conducting and non-conducting infinite sheets of charge have same $E$? ($E$= electric field intensity)? Absaed https://physics.stackexchange.com/users/117145 2016-08-28T07:15:13Z 2018-12-10T10:54:40Z <p>From my textbook the description is given below to the image.</p> <p><img src="https://i.stack.imgur.com/CAd7E.jpg" alt=""></p> https://physics.stackexchange.com/questions/276821/-/278647#278647 1 Answer by freecharly for Conducting and non-conducting infinite sheets of charge have same $E$? ($E$= electric field intensity)? freecharly https://physics.stackexchange.com/users/129209 2016-09-06T23:10:54Z 2016-09-08T23:15:34Z <p>For Gauss' Law, which is derived from Poisson's equation, only the charge density matters irrespective where it resides. Therefore, the electric fields are equal for the conducting or non-conducting infinite sheets with equal total areal charge densities 𝜎. It should be noted that in the case of the conducting infinite sheet, equal sheet charge densities exist at both conductor/vacuum interfaces whose sum represent the total areal charge density 𝜎 of the conductor sheet.</p> https://physics.stackexchange.com/questions/276821/-/278649#278649 0 Answer by Sean E. Lake for Conducting and non-conducting infinite sheets of charge have same $E$? ($E$= electric field intensity)? Sean E. Lake https://physics.stackexchange.com/users/47360 2016-09-06T23:17:43Z 2016-09-06T23:17:43Z <p>The answer is that the change in the component of the electric field that is perpendicular to the sheet of charge is: $$\hat{n}\cdot \Delta \mathbf{E} = \frac{\sigma}{\epsilon_0}.$$ Specifically, $\hat{n}$ is the unit vector perpendicular to the sheet of surface charge. This is true no matter where the sheet is as long as it is smooth enough to be flat if you zoom in enough. In the case of the conductor you have a condition that the electric field inside of it is $\mathbf{0}$, so the electric field outside will be $\sigma / \epsilon_0$. With an isolated infinite sheet that is the only source of electric fields in the universe you have a symmetry that the field on either side has to have the same magnitude, even if the directions are opposite, giving $|\mathbf{E}| = \frac{\sigma}{2\epsilon_0}$.</p> https://physics.stackexchange.com/questions/276821/-/446334#446334 0 Answer by GOPAL CHANDRA SAHA for Conducting and non-conducting infinite sheets of charge have same $E$? ($E$= electric field intensity)? GOPAL CHANDRA SAHA https://physics.stackexchange.com/users/215633 2018-12-10T10:54:40Z 2018-12-10T10:54:40Z <p>ACTUALLY THE ELECTRIC FIELD INTENSITY REMAINS SAME I) IN CASE OF NON CONDUCTING SHEET THE CHARGE LIES ON ONE SIDE OF THE SHEET SO CHARGE DENSITY IS CHARGE DIVIDED BY SURFACE AREA( ONE SIDE OF THE SHEET) ELECTRIC FIELD INTENSITY = CHARGE DENSITY/2*EPSILON II) IN CASE OF CONDUCTING SHEET THE CHARGE LIES ON BOTH SIDE OF THE SHEET SO CHARGE DENSITY IS CHARGE DIVIDED BY SURFACE AREA OF BOTH SIDE OF THE SHEET. SO CHARGE DENSITY GETS HALVED. NOW ELECTRIC FIELD INTENSITY = CHARGE DENSITY/EPSILON , AS CHARGE DENSITY GETS HALVED SO AGAIN ELECTRIC FIELD INTENSITY REMAINS SAME.</p>