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  • two sheets one of them conductor and second insulator have same charge density (sigma) difference between their electric field.

two sheets one of them conductor and second insulator have same charge density (sigma) difference between their electric field. Pls explain sir it is most confusion creating question for me

Answers (1)

@Ankush

Non-Conductor : To calculate the electric feld using Gauss’s Law, we assume a Gaussian surface in the form of a right cylinder with cross sectional area A and height 2r, chosen to cut through the plane perpendicularly ! Because the electric feld is perpendicular to the plane everywhere, the electric feld will be parallel to the walls of the cylinder and perpendicular to the ends of the cylinder ! Using Gauss’ Law we get

flux\ \phi =2EA=\frac{\sigma A}{\epsilon _0}\\E=\frac{\sigma }{2\epsilon _0}

Assume that we have a thin, infnite conductor (metal plate) with positive charge ! Te charge density in this case is also the charge per unit area, σ, but it’s on both surfaces, there is equal surface charge on both sides ! From symmetry, we can see that the electric feld will be perpendicular to the surface of the sheet.

To calculate the electric feld using Gauss’s Law, we assume a Gaussian surface in the form of a right cylinder with cross sectional area A and height r, chosen to cut through one side of the plane perpendicularly . Te feld inside the conductor is zero so the end inside the conductor does not contribute to the integral ! Because the electric feld is perpendicular to the plane everywhere, the electric feld will be parallel to the walls of the cylinder and perpendicular to the end of the cylinder outside the conductor . Using Gauss’s Law we get the electric feld from a charged fat conductor.

flux\ \phi =EA=\frac{\sigma A}{\epsilon _0}\\E=\frac{\sigma }{\epsilon _0}

???????

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Safeer PP

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