Question

Use Gauss’ law to derive this (23-13) equation. Please show steps.

Gauss' Law: Planar Symmetry Sheet 7 shows a portion of a thin, infinite, nonconducting s ve) surface charge density ơ.A sheet of thi one side, can serve as a simple model. Le n front of the sheet. ing heet with a uni- plastic wrap, uniformly t us find the electric field Gaussian surface is a closed cylinder with end caps of are ierce the sheet perpendicularly as shown. From symmetry, E must lar to the sheet and hence to the end caps. Furthermore, since the tive, E is directed away from the sheet, and thus the electric field e two Gaussian end caps in an outward direction. Because the field erce the curved surface, there is no flux through this portion of the ce. Thus E . dA is simply E dA; then Gauss' law, a A, charge enclosed by the Gaussian surface. This gives E 2ε (sheet of charge). (23-13) idering an infinite sheet with uniform charge density, this result t at a finite distance from the sheet. Equation 23-13 agrees with e found by integration of electric field components. tive view portion of a sheet, uni- side to sur- A closed rface passes Gaussian tsurface ttt There is flux on

Answer 1

We know $\phi=\oint E.dA=\dfrac{q_{enclosed}}{\epsilon _0}$

For an infinite sheet of charge having charge density we will consider a gaussian cylindrical surface.

For an infinite sheet of charge the field will be perpendicular to the surface.

Thus the curved part of cylinder don not contribute to flux and only the ends of cylinder will contribute to flux.

Let the area of one end surface be A.

Now, Qenlcosed= total charge inside gausssian surface=charge density*Area=$\sigma A$

Thus,

$\phi=\oint E.dA=\dfrac{q_{enclosed}}{\epsilon _0}$

$\phi=E\oint dA=\dfrac{\sigma A}{\epsilon _0}$

Total area=A+A=2A (area of both ends)

$\phi=E(2A)=\dfrac{\sigma A}{\epsilon _0}$

$E=\dfrac{\sigma A}{2A\epsilon _0}=\dfrac{\sigma }{2\epsilon _0}$