This becomes really a very conceptual doubt . I have attached the figure below which explains my question best.Difference in Flux The dimension of the finite square sheet (yes , it's a square) is l. The Gaussian surface taken in finite sheet case is a Cuboidal box grazing the sides of the sheet and extending to a height h (i.e dimensions of the Cuboidal Gaussian surface are l,l,h)

My questions are 1)Why has E been taken out of the Integral in the infinite sheet case ? 2) Is the Gaussian surface taken in the finite sheet case correct ? 3)Why will the Flux due to finite charged sheet not be pa^2/(2epsilon) (p is the charge density ? 4)Where is the difference in Flux through finite and infinite sheets coming from? 5) Finally what would be the E due to finite sheet at height h?


In the infinite sheet, you have a laterally constant vertical electric field. Therefore you can take it out of the Gauss integral. In the case of the finite charge sheet, the vertical electric fields in the top and bottom Gauss surfaces are not constant, they vary laterally. Furthermore, you have to take the sidewall contributions of the Gauss integral into account because in the finite sheet case you also have lateral electric fields there. The calculation of E at hight h in the finite case is not trivial. In general, you will probably need a numerical solution of the electrostatic potential. If you could convert the problem to a 2-dimensional potential problem (e.g. using rotational symmetry for a circular charge sheet) you could try to solve it analytically by conformal mapping.

  • $\begingroup$ Thanks. I have marked you answer right. I was thinking the same , just wanted to confirm. Just one thing . In the finite sheet case , why would we need to take the Flux through side walls ?Isn't the field only vertical everywhere ? $\endgroup$ – Shashaank Sep 6 '16 at 20:44
  • $\begingroup$ In the finite case, the field is not vertical everywhere. There are lateral field components at the side walls. You have to include their flux in the Gauss surface integral. Therefore, in the finite case, Gauss law is not useful for finding the electric field. $\endgroup$ – freecharly Sep 6 '16 at 22:10

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