Question

State Gauss’s law in electrostatic. Using this law derive an expression for the electric field due to a uniformly charged infinite plane sheet.

Answer 1

Electric field due to a uniformly charged infinite plane sheet : Suppose a thin non-conducting infinite sheet of uniform surface, charge density σ.

Electric field intensity on either side of the sheet must be perpendicular to the plane of sheet having same magnitude at all points equidistant from sheet.

Let P be any point at a distance r from the sheet.

Let the small area element \(\vec E = d s \hat n\)

\(\vec E\) and \(\hat n\) are perpendicular, on the surface of imagined cylinder, so electric flux is zero.

\(\vec E\) and \(\hat n\) are parallel on the two cylindrical edges P and Q, which contributes electric flux.

∴ Electric flux over the edges P and Q of the cylinder is

\(2 \phi = \frac q {\in _0}\)

⇒ \(2 \oint\vec E.\vec {ds} = \frac q{\in _0}\)      \(\left[\because \phi = \oint \vec E. dec (ds)\right]\)

\(2 \oint \vec E.\vec {ds} = \frac q{\in_0}\)

\(2 \oint E {ds} = \frac q{\in_0}\)      \([\because \vec E \perp \vec {ds}]\)

\(2E\pi r^2 = \frac q{\in_0}\) 

⇒ \(\frac q{2\pi \in_0r^2}\)

∴ The charge density σ = \(\frac q S\)

⇒ q = πr²σ [where S-area of circle]

\(E = \frac{\pi r^2 \sigma}{2\pi \in_0r^2}\)

\(E = \frac{ \sigma}{2\in_0}\)

∴ vecotrically \(\vec E = \frac \sigma {2 \in_0} \hat n\)

where \(\hat n\) is a unit vector normal to the plane and going away from it.

When σ > 0, E is directed away from both sides. Hence electric field intensity is independent of r.

When σ > 0, E is directed away from both sides. Hence electric field intensity is independent of r.

Note: For conducting sheet, the surface charge density on both the surface of sheet will be same.

\(\therefore E = \frac \sigma { \in_0} \)

Answer 2

Gauss Theorem: The net outward electric flux through a closed surface is equal to 1/ε₀ times the net charge enclosed within the surface i.e., 

Let electric charge be uniformly distributed over the surface of a thin, non-conducting infinite sheet. Let the surface charge density (i.e., charge per unit surface area) be s. We have to calculate the electric field strength at any point distance r from the sheet of charge.

To calculate the electric field strength near the sheet, we now consider a cylindrical Gaussian surface bounded by two plane faces A and B lying on the opposite sides and parallel to the charged sheet and the cylindrical surface perpendicular to the sheet (fig). By symmetry the electric field strength at every point on the flat surface is the same and its direction is normal outwards at the points on the two plane surfaces and parallel to the curved surface. Total electric flux

As  σ is charge per unit area of sheet and a is the intersecting area, the charge enclosed by Gaussian surface = σa

According to Gauss’s theorem,

Thus electric field strength due to an infinite flat sheet of charge is independent of the distance of the point and is directed normally away from the charge. If the surface charge density σ is negative the electric field is directed towards the surface charge.