1. No.
2. Field due to A uniformly charged infinite plane sheet:
Consider an infinite thin plane sheet of change of density σ.
To find electric field at a point P (at a distance ‘r’ from sheet), imagine a Gaussian surface in the form of cylinder having area of cross section ‘ds’.
According to Gauss’s law we can write,
But electric field passes only through end surfaces ,so we get ∫ds = 2ds
i.e. E 2ds = \(\frac{\sigma ds}{\varepsilon_0}\)
E = \(\frac{\sigma ds}{2ds\varepsilon_0},\) \(E =\frac{\sigma}{2\varepsilon_0}\)
E is directed away from the charged sheet, if σ is positive and directed towards the sheet if σ is negative.