Consider a thin, flat, infinite, positively charged conducting sheet, with a uniform surface charge density σ. let the permittivity of the surrounding medium be ε. The charge density has a planar symmetry, i.e., it appears the same from all points on a plane parallel to the sheet. Then, by symmetry, the electric field intensity \(\vec E\) (1) is perpendicular to the sheet (in this case, outwards) at all points outside the sheet, and (2) has the same magnitude E at any given distance on either side of the sheet.
To find E at a point P outside the sheet, imagine a Gaussian surface in the form of a small closed cylinder. Its axis is perpendicular to the sheet, with the point P on one end face, shown in below figure. The cylinder encloses a small area dS of the sheet. So, the charge enclosed by the cylinder
= σdS …………. (1)
The flux through one end = EdS
∴ The flux through the Gaussian surface,
Φ = 2EdS …………… (2)
By Gauss’s theorem,
εΦ = net charge enclosed
∴ ε(2EdS) = σdS …………. (3)
∴ E = \(\cfrac{σ}{2ε}\) = \(\cfrac{σ}{2kε_0}\)…………. (4)
where ε0 is the permittivity of free space and k = ε/ε0 is the relative permittivity (dielectric constant) of the surrounding medium.
Electric feild intensity outside an infinite, positively xharged conducting sheet
Equation (4) shows that the magnitude of the electric field intensity outside the charged sheet is uniform and independent of the distance from the sheet
[Notes : (1) The electric field lines are straight, parallel to each other, and perpendicular to the sheet. (2) If the sheet is negatively charged,\(\vec E\) on either side of the sheet is directed towards the sheet.]
